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Earlier I posted the handwritten notes on how to calculate eigenvalues and eigenfunctions of a boundary eigenvalue problem with the boundary conditions of the "third kind", also called Robin boundary conditions. This was in the context of solving the string equation with the boundary conditions and the initial conditions as follows:
Let $L \gt 0$, \(c\gt 0\), and $h \in \mathbb{R}.$ Let $f:[0,L]\to\mathbb{R}$ and $g:[0,L]\to\mathbb{R}$ be continuous functions whose derivatives are piecewise continuous. We consider the vibrating string equation with a Robin boundary condition at $0$ and the Diriclet boundary condition at $L$: \begin{alignat*}{2} &\text{PDE:} \qquad & & \frac{\partial u^2}{\partial t^2}(x,t) = c^2 \frac{\partial u^2}{\partial x^2}(x,t) \quad \text{where} \quad x \in [0,\pi] \quad \text{and} \quad t \in [0,+\infty) \\ &\text{BCs:} \qquad & & u(0,t) - h \frac{\partial u}{\partial x}(0,t) = 0 \quad \text{and} \quad u(L,t) = 0 \quad \text{for all} \quad t \in [0,+\infty), \\ &\text{ICs:} \qquad & & u(x,0) = f(x) \quad \text{and} \quad \frac{\partial u}{\partial t}(x,0) = g(x) \quad \text{for all} \quad x \in [0,L]. \end{alignat*}
In the earlier notes I stopped after I calculated the natural modes of vibration of the above string. I focused on a string with only positive eigenvalues.
Since in Problem 4 on Assignment 3 I ask you to provide a complete solution of a problem which is a variation on the problem stated above, prompted by students questions in Discussions on Canvas and email, I continued the notes posted above. The previously posted notes finished with finding the natural modes of vibrations of the string described by the PDE and BCs stated above. (I focussed on the problem with positive eigenvalues only.) In the addition to the notes, I present how to find the solution satisfying the initial conditions. Here are the complete notes.
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The earlier posted notebook Super_glued_ends_solution_v12.nb contains all the details of a Mathematica implementation of finding a solution of the vibrating string equation with several different Robin boundary conditions. This Mathematica file is important for the Mathematica part of Problem 4 on Assignment 3. A pdf printout of this notebook is Super_glued_ends_solution_v12.pdf. You can download this Mathematica notebook here Super_glued_ends_solution_v12.nb
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We finished the problem that we started yesterday. We continued the Mathematica notebook started yesterday. I did not change the name of the notebook, I just continued writing new content in it. A pdf printout of this notebook is 20231204_super_glue.pdf. You can download this updated Mathematica notebook here 20231204_super_glue.nb
For the statement of the problem see the post yesterday. Also, in the post yesterday you can find a link to a more comprehensive Mathematica notebook in which I pay attention to all the details of the graphical presentation of vibrating strings.
- There was a question in Discussions on Canvas today related to Problem 4 on Assignment 3. I try hard to make my writing precise and unambiguous. However, that is not easy; pictures, and animations can help. In the items below I present how solutions of Problem 3, Problem 4, and Problem 6 on Assignment 3 should look when animated in Mathematica. Ideally you should be able to reproduce these animations on your own. If you do not have time to do the Mathematica work, look at your solutions, and try to see in your formulas how vibrations would occur. I am hoping that looking at the animations below and your formulas for the solutions you can see how animations occur.
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Below I present two animations which are the final result of solving Problem 3 on Assignment 3. The first string is set in motion by the initial displacement which can be seen in the first scene of the animation. The second string is set in motion by the initial velocity which equal to the initial displacement of the first string.
Place the cursor over the image to start vibrations.
Place the cursor over the image to start vibrations.
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Below I present an animation which is the final result of solving Problem 4 on Assignment 3. A string with two rigid ends is set in motion by the given initial displacement which can be seen in the first scene of the animation.
Place the cursor over the image to start vibrations.
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Below I present two animations which are the final result of solving Problem 6 on Assignment 3. The first string is set in motion by the initial displacement which can be seen in the first scene of the animation. The second string is set in motion by the given initial velocity \(\sin\bigl(2\pi x/5\bigr)\).
Place the cursor over the image to start vibrations.
Place the cursor over the image to start vibrations.
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Here I give a hint for Problem 6 on Assignment 3. In part (a) I ask you to establish a mathematical model for the vibrations of a string with the rigid part over the interval \([1,2]\). The string is placed over the interval \([0,5]\). The value of \(x\) in a mathematical model for the vibrations of this string should be restricted to the union \begin{equation*} [0,1] \cup [2,5]. \end{equation*} The fact that the string is rigid over \([1,2]\) should be reflected by boundary conditions involving the values of the function \(u(x,t)\) and its derivative at the points \(x=1\) and \(x=2\). Stating all the conditions carefully, you end up with four boundary conditions: one at \(x=0\), two involving \(x=1\) and \(x=2\), and one at \(x=5\).
Dealing with four boundary conditions is not easy, so to find eigenvalues and eigenfunctions I suggest that you start with the "general" solution on \([0,1] \) of the form \begin{equation*} C_1 \sin(\mu x) \end{equation*} which automatically satisfies the boundary condition at \(x=0\) and with the "general" solution on \([2,5] \) of the form \begin{equation*} C_2 \sin\bigl(\mu (5-x)\bigr) \end{equation*} which automatically satisfies the boundary condition at \(x=5\). In this setting we are left with only two remaining boundary conditions that we can use to set up a linear system for the unknowns \(C_1\) and \(C_2\) which will lead towards an equation which will determine \(\mu\)-s.
What I presented in the preceding paragraph applies only to positive eigenvalues. You need to make necessary adjustments for the negative eigenvalues and zero.
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Let \(L \gt 0\), \(c\gt 0\), and $h \in \mathbb{R}.$ Let $f:[0,L]\to\mathbb{R}$ and $g:[0,L]\to\mathbb{R}$ be continuous functions whose derivatives are piecewise continuous. Today we considered the vibrating string equation with a Robin boundary condition at $L$ and the Diriclet boundary condition at $0$:
\begin{alignat*}{2} &\text{PDE:} \qquad & & \frac{\partial u^2}{\partial t^2}(x,t) = c^2 \frac{\partial u^2}{\partial x^2}(x,t) \quad \text{where} \quad x \in [0,\pi] \quad \text{and} \quad t \in [0,+\infty) \\ &\text{BCs:} \qquad & & u(0,t) = 0 \quad \text{and} \quad u(L,t) + h \frac{\partial u}{\partial x}(L,t) = 0 \quad \text{for all} \quad t \in [0,+\infty), \\ &\text{ICs:} \qquad & & u(x,0) = f(x) \quad \text{and} \quad \frac{\partial u}{\partial t}(x,0) = g(x) \quad \text{for all} \quad x \in [0,L]. \end{alignat*}
The emphasis today was on a review of the method and the demonstration of implementation of the method in Mathematica.
- In class I created a small Mathematica notebook with the basics of the implementation in Mathematica. A pdf printout of this notebook is 20231204_super_glue.pdf. You can download this Mathematica notebook here 20231204_super_glue.nb
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The notebook Super_glued_ends_solution_v12.nb contains all the details of a Mathematica implementation of finding a solution of the vibrating string equation with several different Robin boundary conditions. This Mathematica file is important for the Mathematica part of Problem 4 on Assignment 3. A pdf printout of this notebook is Super_glued_ends_solution_v12.pdf. You can download this Mathematica notebook here Super_glued_ends_solution_v12.nb
- The notebook Super_glued_ends_solution_v12.nb contains a Mathematica implementation of finding a solution of the vibrating string equation with a Robin boundary condition. This Mathematica file is essential for the Mathematica part of Problem 4 on Assignment 3. A pdf printout of this notebook is Super_glued_ends_solution_v12.pdf. You can download the Mathematica notebook here Super_glued_ends_solution_v12.nb
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Let $L \gt 0$, \(c\gt 0\), and $h \in \mathbb{R}.$ Let $f:[0,L]\to\mathbb{R}$ and $g:[0,L]\to\mathbb{R}$ be continuous functions whose derivatives are piecewise continuous. We consider the vibrating string equation with a Robin boundary condition at $0$ and the Diriclet boundary condition at $L$:
\begin{alignat*}{2}
&\text{PDE:} \qquad & & \frac{\partial u^2}{\partial t^2}(x,t) = c^2 \frac{\partial u^2}{\partial x^2}(x,t)
\quad \text{where} \quad x \in [0,\pi] \quad \text{and} \quad t \in [0,+\infty) \\
&\text{BCs:} \qquad & & u(0,t) - h \frac{\partial u}{\partial x}(0,t) = 0 \quad \text{and} \quad u(L,t) = 0
\quad \text{for all} \quad t \in [0,+\infty), \\
&\text{ICs:} \qquad & & u(x,0) = f(x) \quad \text{and} \quad \frac{\partial u}{\partial t}(x,0) = g(x)
\quad \text{for all} \quad x \in [0,L].
\end{alignat*}
On Thursday and Friday we analyzed "natural modes of vibration" of this string for all possible cases of $L \gt 0$ and $h \in \mathbb{R}.$ There are three distinct cases:
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Case 1. $-L \lt h \lt 0$ In this case, there is one negative eigenvalue and countably many positive eigenvalues. For a general choice of the initial shape $f(x)$ and the initial velocity $g(x)$ the string will break in this case. See the the animation below.
Place the cursor over the image to start vibrations.
However, if we set the initial velocity $g(x)= 0$ and we make a special choice of the initial shape $f(x)$ that is orthogonal to the eigenfunction corresponding to the negative eigenvalue than the string does not break. See the animation below.
Place the cursor over the image to start vibrations.
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Case 2. $h = -L$ In this case, $0$ is an eigenvalue and there are countably many positive eigenvalues. For a general choice of the initial shape $f(x)$ and no initial velocity, that is $g(x) = 0,$ the string will not break. This is illustrated in the animation below with $L = \pi$ and $h=-\pi$.
Place the cursor over the image to start vibrations.
For a general choice of the initial velocity $g(x)$ the string breaks. We did not illustrate this case.
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Case 3. $h \geq 0$ or $h \leq -L$ In this case, there are no negative eigenvalues, $0$ is not an eigenvalue and there are countably many positive eigenvalues. See the animation below with $L =\pi$ and $h=-4$.
Place the cursor over the image to start vibrations.
In the animation below we used $L =\pi$ and $h=1$.
Place the cursor over the image to start vibrations.
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- These are the complete notes of what we did on Thursday and Friday. The relevant part from the textbook is Section 5.8.
- Today we worked on solving boundary eigenvalue problems with the boundary conditions of the "third kind" or Robin boundary conditions. I wrote the notes presenting this method. The relevant part from the textbook is Section 5.8.
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In the animation below the parameter $h=1$. The boundary condition at the endpoint $0$ is
\[
u(0,t) - 1 \frac{\partial u}{\partial x}(0,t) = 0.
\]
This is a Robin boundary condition.
Place the cursor over the image to start vibrations.
Notice that the red part of the string is rigid, while the orange part is governed by the vibrating string equation.
- Today we presented a derivation of the vibrating string equation with a variety of associated boundary conditions. This is the content of Sections 4.2 and Sections 4.3 in the book.
- Next we will present a method of solving boundary eigenvalue problems with the boundary conditions of the "third kind" or Robin boundary conditions (the boundary conditions discussed today). I wrote the notes presenting this method. The relevant part from the textbook is Section 5.8.
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Today we introduced the boundary conditions of the "third kind" or Robin boundary conditions for the vibrating string equation. In my interpretation, these boundary conditions describe a string with a rigid end as in the animation below. In the animation below the parameter $h=1$. The boundary condition at the endpoint $0$ is
\[
u(0,t) - 1 \frac{\partial u}{\partial x}(0,t) = 0.
\]
This is a Robin boundary condition.
Place the cursor over the image to start vibrations.
Notice that the red part of the string is rigid, while the orange part is governed by the vibrating string equation.
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A fundamental aspect of our course is the study of Fourier series of functions. We set a positive real number \( L > 0 \) and focus on real-valued, piecewise continuous functions defined on the interval \( [-L, L] \). We have explored the computation of a Fourier series for such functions and examined the criteria for the convergence of these series. Typically, a Fourier series of a function is an infinite series of trigonometric functions. Specifically, a Fourier series is an infinite series of functions whose partial sums are linear combinations of the constant function \(1\) and the trigonometric functions: \[ 1, \quad \cos\left(\frac{k \pi}{L} x\right), \quad \sin\left(\frac{k \pi}{L} x\right), \quad \text{where} \quad k \in \mathbb{N}. \]
As always on this website \(\mathbb{N}\) denotes the set of positive integers. To denote the set of nonnegative integers we will use the union \(\{0\}\cup\mathbb{N}.\)
- Since we work with trigonometric functions a lot, it is helpful to review beautiful world of trigonometric identities. The Wikipedia page List_of_trigonometric_identities is a good place for such a review.
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One particular kind of trigonometric identities are Power reduction formulas. Below I list several such formulas.
In the context of Fourier series, the power reduction formulas are important since they are in fact finite Fourier series (here \(L=\pi\)) for the powers of trigonometric functions.
\begin{align*} (\sin x)^2 & = \frac{1}{2}-\frac{1}{2} \cos (2 x) \\ (\sin x)^3 & = \frac{3}{4}\sin(x) - \frac{1}{4} \sin(3 x)\\ (\sin x)^4 & = \frac{3}{8} -\frac{1}{2} \cos (2 x)+\frac{1}{8} \cos (4 x)\\ (\sin x)^5 & = \frac{5}{8}\sin (x) - \frac{5}{16} \sin (3 x)+\frac{1}{16} \sin (5 x)\\ (\sin x)^6 & = \frac{5}{16} - \frac{15}{32} \cos (2 x)+\frac{3}{16} \cos (4x)-\frac{1}{32} \cos (6x) \end{align*}
\begin{align*} (\cos x)^2 & = \frac{1}{2}+\frac{1}{2} \cos (2 x) \\ (\cos x)^3 & = \frac{3}{4}\cos (x)+\frac{1}{4} \cos (3 x)\\ (\cos x)^4 & = \frac{3}{8} + \frac{1}{2} \cos (2 x) + \frac{1}{8} \cos(4 x)\\ (\cos x)^5 & = \frac{5}{8}\cos (x)+\frac{5}{16} \cos (3 x)+\frac{1}{16} \cos (5 x)\\ (\cos x)^6 & = \frac{5}{16} + \frac{15}{32} \cos (2 x) + \frac{3}{16} \cos (4x) + \frac{1}{32} \cos (6x) \end{align*}
\begin{align*} (\sin x)(\cos x) & = \frac{1}{2} \sin (2 x) \\ (\sin x)(\cos x)^2 & = \frac{1}{4} \sin (x) + \frac{1}{4} \sin (3 x) \\ (\sin x)(\cos x)^3 & = \frac{1}{4} \sin (2 x) + \frac{1}{8} \sin (4 x) \\ (\sin x)(\cos x)^4 & = \frac{1}{8} \sin (x) + \frac{3}{16} \sin (3 x) + \frac{1}{16} \sin (5 x) \\ (\sin x)(\cos x)^5 & = \frac{5}{32} \sin (2x) + \frac{1}{8} \sin (4 x) + \frac{1}{32} \sin (6 x) \\ (\sin x)^3 & = \frac{3}{4} \sin (x) - \frac{1}{4} \sin (3 x) \\ (\sin x)^3(\cos x) & = \frac{1}{4} \sin (2 x) - \frac{1}{8} \sin (4 x) \\ (\sin x)^3 (\cos x)^2 & = \frac{1}{8} \sin (x) + \frac{1}{16} \sin (3 x) - \frac{1}{16} \sin (5 x) \\ (\sin x)^3(\cos x)^3 & = \frac{3}{32} \sin (2 x) - \frac{1}{32} \sin (6 x) \\ (\sin x)^3(\cos x)^4 & = \frac{3}{64} \sin (x) + \frac{3}{64} \sin (3 x) - \frac{1}{64} \sin (5 x) - \frac{1}{64} \sin (7 x) \\ (\sin x)^3(\cos x)^5 & = \frac{3}{64} \sin (2 x) + \frac{1}{64} \sin (4 x) - \frac{1}{64} \sin (6 x) - \frac{1}{128} \sin (8 x) \\ \end{align*}
In the context of this class, each of the above formulas gives a (finite) Fourier series of the function on the left-hand side of the equality sign.
Also, the above formulas, though not immediately apparent, contain numerous integrals in a disguised form. For example, \[ \int_{-\pi} ^{\pi} (\sin x)^3(\cos x)^4 \sin (3 x) dx = \frac{3 \pi}{64}, \] but also \[ \int_{-\pi} ^{\pi} (\sin x)^3(\cos x)^4 \sin (4 x) dx = 0. \] Do you see why?
- The preceding item is a hint for Problem 3 on Assignment 3. Notice that the solutions of Problem 3 are exact solutions given as finite sums, not infinite series.
- There are general Power reduction formulas. For example, for all \(m\in\mathbb{N}\) we have \begin{align*} (\cos x)^{2m} & = \frac{1}{2^{2m}} \binom{2m}{m} + \frac{1}{2^{2m-1}} \sum_{k=1}^{m} \binom{2m}{m-k} \cos(2k x) \\ (\cos x)^{2m+1} & =\frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m+1}{m-k} \cos\bigl((2k+1)x\bigr) \\ (\sin x)^{2m} & = \frac{1}{2^{2 m}} \binom{2 m}{m} + \frac{1}{2^{2 m-1}} \sum_{k=1}^m (-1)^k \binom{2 m}{m-k} \cos\bigl(2 k x \bigr), \\ (\sin x)^{2m+1} & = \frac{1}{2^{2m}} \sum_{k=0}^{m} (-1)^{k} \binom{2m+1}{m-k} \sin{\big((2k+1)x\big)} \end{align*}
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In the above formulas, for \( n, k \in \{0\}\cup\mathbb{N}\) with \(k \leq n \), the symbol \( \binom{n}{k} \) (read as "n choose k") denotes the Binomial coefficient. The definition is:
\[ \binom{n}{k} = \frac{n!}{k! \, (n-k)!}, \]
where for \( m \in \mathbb{N} \), \( m! \) (read as "m factorial") is the product of all positive integers up to \( m \). By convention \( 0! = 1 \).
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Recursive definitions of functions defined on nonnegative integers are at the cornerstone of mathematics. Therefore, it is appropriate here to point out to the recursive definitions of the factorial and the binomial coefficients.
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The recursive definition of the factorial is as follows:
The Base Case: \(0!=1\)
The Recursive Step: For all \(m\in\mathbb{N}\) we set \(m! = \bigl( (m-1)! \bigr) \mkern 2px m\).
For more details, see Factorial.
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The recursive definition of the binomial coefficients is as follows:
The Base Case: \begin{equation*} \text{For all} \ \ n \in \{0\}\cup\mathbb{N} \quad \text{we set} \quad \binom{n}{0} = 1 \quad \text{and} \quad \binom{n}{n} = 1. \end{equation*} The Recursive Step: \begin{equation*} \text{For all} \ \ n \in \mathbb{N} \ \ \text{and} \ \ k \in \{1,\ldots,n\} \quad \text{we set} \quad \binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}. \end{equation*} At each line below, the recursive step with specific values for \(n\) and \(k\) and the previously evaluated values (that is why it is called a recursion, see the next item below) for the binomial coefficients yields:
\begin{alignat*}{2} &\text{For } n=2, \ k=1 \qquad &&\binom{2}{1} = \binom{1}{0} + \binom{1}{1} = 1 + 1 = 2, \\ &\text{For } n=3, \ k=1 &&\binom{3}{1} = \binom{2}{0} + \binom{2}{1} = 1 + 2 = 3, \\ &\text{For } n=3, \ k=2 &&\binom{3}{2} = \binom{2}{1} + \binom{2}{2} = 2 + 1 = 3, \\ &\text{For } n=4, \ k=1 &&\binom{4}{1} = \binom{3}{0} + \binom{3}{1} = 1 + 3 = 4, \\ &\text{For } n=4, \ k=2 &&\binom{4}{2} = \binom{3}{1} + \binom{3}{2} = 3 + 3 = 6, \\ &\text{For } n=4, \ k=3 &&\binom{4}{3} = \binom{3}{2} + \binom{3}{3} = 3 + 1 = 4, \\ &\text{For } n=5, \ k=1 &&\binom{5}{1} = \binom{4}{0} + \binom{4}{1} = 1 + 4 = 5, \\ &\text{For } n=5, \ k=2 &&\binom{5}{2} = \binom{4}{1} + \binom{4}{2} = 4 + 6 = 10, \\ &\text{For } n=5, \ k=3 &&\binom{5}{3} = \binom{4}{2} + \binom{4}{3} = 6 + 4 = 10, \\ &\text{For } n=5, \ k=4 &&\binom{5}{4} = \binom{4}{3} + \binom{4}{4} = 4 + 1 = 5, \\ & & & \quad \quad \mkern 12px \vdots \end{alignat*}
For more details about this recursion, see Pascal's triangle.
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The recursive definition of the factorial is as follows:
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In our exploration of power reduction formulas, we naturally encountered binomial coefficients. This prompted a review of the definition of the binomial coefficients, for which I sought assistance from ChatGPT. I was somewhat surprised that in ChatGPT's basic writing about binomial coefficients, ChatGPT included the recursive definition of the binomial coefficients. That made an impression on me. I realized that I had not been highlighting the fundamental role of recursive definitions in mathematics as much as I should.
Here are two additional questions that I posed to ChatGPT, further exploring these concepts:
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Above is a snippet of my interaction with ChatGPT. That is an adequate way to share my interaction on the website. A drawback of sharing a snippet is that one cannot copy-paste from a PNG file.
Next I asked ChatGPT to organize my question and its response in a form that I can post on this website. I was simply stunned with the clarity of ChatGPT's response. You can see the result below.
The only problem I encountered was when I asked ChatGPT to design the icons for the user CU and OpenAI, which you can see in the snippet. I asked ChatGPT to create those yellow and purple icons as close as possible. However, ChatGPT refused based on the copyright concerns. The icons that I show below are created by ChatGPT, the closest to the original ones that I could get.
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You:
What is a recursion? What does it mean to "define a function recursively"? Assume that the audience is familiar with the concepts of integers and functions.
ChatGPT:
Recursion is a concept in mathematics and computer science where a function is defined in terms of itself. It involves two key components: the base case and the recursive step.
1. Base Case: This is a simple instance of the problem which can be solved directly, without further recursion. It acts as a stopping point for the recursion. For example, in the factorial function, the base case is \(0! = 1\).
2. Recursive Step: This is where the function calls itself with a simpler or smaller argument. Each recursive call brings the problem closer to the base case. For the factorial function \(n!\), the recursive step is \(n! = n \times (n-1)!\), where \(n > 0\).
To "define a function recursively" means to express the function in terms of simpler instances of itself, leading eventually to the base case. This approach is powerful because it allows for elegant solutions to complex problems by breaking them down into simpler sub-problems.
In the context of the binomial coefficient \(\binom{n}{k}\), a recursive definition involves expressing \(\binom{n}{k}\) in terms of other binomial coefficients with smaller values of \(n\) and/or \(k\), along with base cases like \(\binom{n}{0} = 1\) and \(\binom{n}{n} = 1\).
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I have enjoyed the interaction between my mind and computer since I started using computers in the early 1990s. Only recently, I encountered a beautiful visualization of the relationship between the mind and computer expressed by Steve Jobs: In an old interview, Steve Jobs explained the usefulness of computers with a beautiful image. He said: "For me, a computer has always been a bicycle for the mind." That is what ChatGPT is. And isn't that wonderful!
See the Bicycle for the Mind video. Or, a longer
version.
- Today we discussed the vibrating string equation. We considered a string which is fixed at both endpoints. This is in Section 4.4 in the book.
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Below I present the first six natural modes of vibration of a string with fixed endpoints. For each picture hover the cursor over the image to start vibrations.
the first harmonic or fundamental
the second harmonic
the third harmonic
the fourth harmonic
the fifth harmonic
the sixth harmonic
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Below is a string with fixed endpoints set in motion by initial displacement only, no initial velocity.
Place the cursor over the image to start vibrations.
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The above animations are created in the Mathematica notebook linked below
- Natural_modes_of_vibration.nb Here is the pdf printout of the same notebook.
- This is in Sections 4.4 in the book.
- I created this small Mathematica notebook in which I explain how to create animations and export them as common animated files. Here is the pdf printout of the same notebook.
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Below is an animated gif that I produced
Below is an animated png that I produced
There is not much difference in quality. For more complicated pictures I think that png format is better.
- Unfortunately I did not find a simple way to include animated files in pdf documents. As you can see from my website, it is easy to include animated gifs and animated pngs in html files. These files show nicely in browsers.
- Recall the theorems from Thursday in which deduced the Fourier series for the derivative $f'(x)$ and the indefinite integral $F(x) = \int_{0}^x f(\xi) d\xi$ based on the Fourier series of $f(x)$.
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Differentiation Theorem. Let $L \gt 0$ and let $f:[-L,L] \to \mathbb R$ be a function. Assume that
- $f$ is continuous on the closed interval $[-L,L],$
- $f$ is piecewise smooth function on $[-L,L],$
- $f'$ is piecewise smooth function on $[-L,L],$
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Integration Theorem. Let $L \gt 0$ and let $f:[-L,L] \to \mathbb R$ be a function. Assume that
- $f$ is piecewise smooth function on $[-L,L],$
- $\displaystyle a_0 = \frac{1}{2L} \int_{-L}^{L} f(\xi) d\xi = 0,$
- $\displaystyle F(x) = \int_{0}^{x} f(\xi) d\xi.$
- On Friday we deduced several Fourier series without performing any integration. We just used algebra suggested by the preceding two theorems. We can apply this method whenever the derivative of a function is particularly simple and we know its Fourier series or there is a repetition in the derivative and integral. The later is the case for $\exp(x)$, $\cosh(x)$, $\sinh(x)$, $\cos(x)$, $\sin(x)$.
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We can apply the first differentiation theorem to calculate the Fourier coefficients of the Fourier series of the function $f(x) = \exp(x)$ say on the interval $[-1,1]$. The constant Fourier coefficient is \[ a_0 = \frac{1}{2} \int_{-1}{1} e^x dx = \frac{1}{2} \biggl. e^x \biggr|_{-1}^{1} = \frac{e - e^{-1}}{2} = \sinh(1). \] Since $\exp(x)$ is its own derivative we have the following equalities for the Fourier coefficients: \begin{equation*} a_k = 2 (-1)^{k} \sinh(1) + k \mkern2px\pi \, b_k, \quad b_k = - k\mkern2px\pi \, a_k \quad \text{for all} \quad k\in\mathbb{N}. \end{equation*} Therefore, substituting the expression for $b_k$ into the first equation and solving for $a_k$ we get \[ a_k = 2 \sinh(1) \frac{(-1)^k }{1+(k \pi)^2}, \quad b_k = - 2 \pi \sinh(1)\frac{(-1)^k k }{1+(k \pi)^2} \quad \text{for all} \quad k\in\mathbb{N}. \]
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In the next example we apply the differentiation theorem to calculate the coefficients of the Fourier series of the function $f(x) = x$ say on the interval $[-1,1]$. We immediately observe that the function $f$ is odd. Therefore all the coefficients $a_k$ with $k \in \{0\}\cup \mathbb{N}$ are equal to $0$. Therefore we need to deal only with the coefficients $b_k$ with $k\in \mathbb{N}$.
Next we observe that the Fourier series of the derivative $f'(x)$ is \[ f'(x) = 1. \] Thus, the only Fourier coefficient of the derivative is $1$ and all other coefficients are $0$. Substituting this in our formula for the Fourier coefficients of the derivative, using the fact that $a_k = 0$ for all $k \in \{0\}\cup \mathbb{N}$, we get the identities \begin{align*} \frac{1}{2L} \bigl( f(L) - f(-L) \bigr) & = 1 \\ \frac{1}{L} \left( (-1)^{k} \bigl( f(L) - f(-L) \bigr) + k\pi \, b_k \right) & = 0. \end{align*} Substituting $L=1$ and $f(x) = x$ we get \begin{align*} \frac{1}{2} \bigl( 1 - (-1) \bigr) & = 1 \\ (-1)^{k} \bigl( 1 - (-1) \bigr) + k\pi \, b_k & = 0. \end{align*} The first equality holds no information, and the second one yields \[ b_k = -\frac{2}{\pi} \frac{(-1)^{k}}{k} \quad \text{for all} \quad k \in \mathbb{N}. \] Thus the Fourier series for the Fourier periodic extension of the function $f(x) = x$ with $x \in (-1,1)$ is \[ \tilde{f}_{\!\!\rm Fourier}(x) = -\frac{2}{\pi} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} \sin\bigl(k\mkern2px \pi \mkern2px x\bigr) \]
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In the next example we apply both the differentiation theorem and the integration theorem to calculate the coefficients of the Fourier series of the function $f(x) = \sin(x)$ say on the interval $[-\pi/2,\pi/2]$. We immediately observe that the function $f$ is odd. Therefore all the Fourier coefficients $a_k$ with $k \in \{0\}\cup \mathbb{N}$ are equal to $0$. Therefore we need to deal only with the Fourier coefficients $b_k$ with $k\in \mathbb{N}$.
We use the fact that we know that \[ f'(x) = \cos(x) \quad \text{and} \quad F(x) = 1 - \cos(x) \quad \text{for all} \quad x \in x\in (-\pi/2,\pi/2). \] Consequently we have the following relationship between the Fourier coefficients of the Fourier series for the derivative and the integral: \begin{align*} \frac{1}{2L} \bigl( f(L) - f(-L) \bigr) & = 1 - \frac{L}{\pi}\sum_{k=1}^{\infty} \frac{b_k}{k} \\ \frac{1}{L} \left( (-1)^{k} \bigl( f(L) - f(-L) \bigr) + k\pi \, b_k \right) & = -\left(- \frac{L}{\pi} \frac{b_k}{k}\right). \end{align*} Substituting $L=\pi/2$ and $f(x) = \sin(x)$ we get \begin{align*} \frac{1}{\pi} \bigl(1 - (-1) \bigr) & = 1 - \frac{1}{2}\sum_{k=1}^{\infty} \frac{b_k}{k} \\ \frac{2}{\pi} \left( (-1)^{k} \bigl( 1 - (-1) \bigr) + k \mkern2px \pi \, b_k \right) & = \frac{1}{2} \frac{b_k}{k}. \end{align*} That is \begin{align*} \frac{2}{\pi} & = 1 - \frac{1}{2}\sum_{k=1}^{\infty} \frac{b_k}{k} \\ \frac{4}{\pi} (-1)^{k} + 2 k \mkern2px \, b_k & = \frac{1}{2} \frac{b_k}{k}. \end{align*} Solving for $b_k$ from the second equation yields \[ b_k = - \frac{4}{\pi} (-1)^{k} \frac{1}{ 2 k - \frac{1}{2k} } = \frac{8}{\pi} \frac{(-1)^{k} k}{1 - 4 k^2} \quad \text{for all} \quad k \in \mathbb{N}. \]
Thus the Fourier series for the Fourier periodic extension of the function $f(x) = \sin(x)$ with $x \in (-\pi/2,\pi/2)$ is \[ \tilde{f}_{\!\!\rm Fourier}(x) = \frac{8}{\pi} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{1-4 k^2} \sin\bigl(2 k \mkern2px x\bigr). \]
- We can confirm the calculations in the preceding three items by plotting the Fourier periodic extension of all three functions $f(x)$ on the corresponding intervals $[-L,L]$ and its approximation by its Fourier series in a small Mathematica notebook. A pdf printout of this notebook is 20221107_PerExt.pdf
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We first prove a Differentiation Theorem for Fourier Series.
Differentiation term-by-term Theorem. Let $L \gt 0$ and let $f:[-L,L] \to \mathbb R$ be a function. Assume that
- $f$ is continuous on the closed interval $[-L,L],$
- $f$ is piecewise smooth function on $[-L,L],$
- $f'$ is piecewise smooth function on $[-L,L],$
Proof Since $f'$ is piecewise smooth its Fourier series converges pointwise to the Fourier periodic extension of $f'.$ Denote the coefficients of this Fourier series by $\alpha_0,$ $\alpha_k$ and $\beta_k.$ Let us calculate $\alpha_0.$ Since $f$ is continuous and $f'$ is piecewise continuous the Fundamental Theorem of Calculus applies and it yields \[ \alpha_0 = \frac{1}{2L} \int_{-L}^L f'(x) dx = \frac{1}{2L} \bigl( f(L) - f(-L) \bigr). \] Let us calculate $A_k$. Since $f(x)\cos\bigl(k\pi x/L\bigr)$ is continuous and $f'$ is piecewise continuous the Integration by Parts Theorem applies and it yields \begin{align*} \alpha_k & = \frac{1}{L} \int_{-L}^L f'(x) \cos\left(\frac{k \pi}{L} x\right) dx \\ & = \frac{1}{L} \left( f(L) \cos\bigl(k\pi\bigr) - f(-L) \cos\bigl(-k\pi\bigr) + \frac{k \pi}{L} \int_{-L}^L f(x) \sin\left(\frac{k \pi}{L} x\right) dx \right) \\ & = \frac{1}{L} \left( (-1)^k \bigl( f(L) - f(-L) \bigr) + \frac{k \pi}{L} \int_{-L}^L f(x) \sin\left(\frac{k \pi}{L} x\right) dx \right) \\ & = \frac{1}{L} \left( (-1)^k \bigl( f(L) - f(-L) \bigr) + k \pi \, b_k \right) \\ \end{align*} Let us calculate $B_k$. Since $f(x)\sin\bigl(k\pi x/L\bigr)$ is continuous and $f'$ is piecewise continuous the Integration by Parts Theorem applies and it yields \begin{align*} \beta_k & = \frac{1}{L} \int_{-L}^L f'(x) \sin\left(\frac{k \pi}{L} x\right) dx \\ & = \frac{1}{L} \left( f(L) \sin\bigl(k\pi\bigr) - f(-L) \sin\bigl(-k\pi\bigr) - \frac{k \pi}{L} \int_{-L}^L f(x) \cos\left(\frac{k \pi}{L} x\right) dx \right) \\ & = \frac{1}{L} \left( - k \pi \, a_k \right) \end{align*} -
The book proves the following Differentiation Theorem for Fourier sine series.
- $f$ is continuous on the closed interval $[0,L],$
- $f$ is piecewise smooth function on $[0,L],$
- $f'$ is piecewise smooth function on $[0,L].$
-
Here we prove Integration term-by-term Theorem for Fourier series.
Integration term-by-term Theorem. Let $L \gt 0$ and let $f:[-L,L] \to \mathbb R$ be a function. Assume that
- $f$ is piecewise smooth function on $[-L,L],$
- $\displaystyle a_0 = \frac{1}{2L} \int_{-L}^{L} f(\xi) d\xi = 0,$
- $\displaystyle F(x) = \int_{0}^{x} f(\xi) d\xi.$
Proof of Integration term-by-term Theorem Since we assume \[ 0 = \int_{-L}^{L} f(\xi) d\xi = -\int_{0}^{-L} f(\xi) d\xi + \int_{0}^{L} f(\xi) d\xi = -F(-L) + F(L), \] the periodic extension $\widetilde F$ of $F$ is continuous and piecewise smooth. Therefore its Fourier series converges uniformly to $\widetilde F$. Let us now calculate the Fourier series of $F.$ Denote the coefficients of this Fourier series by $A_0,$ $A_k$ and $B_k.$ It is tricky to calculate $A_0,$ so we postpone it for the end of the proof. However, in each practical situation this will be the easiest one to calculate since it is the average value of $F$ on $[-L,L].$ The difficulty here arises from the fact that the average value of $F$ on $[-L,L]$ cannot be expressed directly in terms of $f.$
Let us calculate $A_k$. Since $F(x)\cos\bigl(k\pi x/L\bigr)$ is continuous and $f(x)$ is piecewise continuous the Integration by Parts Theorem applies and it yields \begin{align*} A_k & = \frac{1}{L} \int_{-L}^L F(\xi) \cos\left(\frac{k \pi}{L} \xi \right) dx \\ & = \frac{1}{L} \left( \frac{L}{k\pi} F(L) \sin\bigl(k\pi\bigr) - \frac{L}{k\pi} F(-L) \sin\bigl(-k\pi\bigr) - \frac{L}{k\pi} \int_{-L}^L f(\xi) \sin\left(\frac{k \pi}{L} \xi\right) d\xi \right) \\ & = - \frac{L}{k \pi} b_k \end{align*} Let us calculate $B_k$. Since $F(x)\sin\bigl(k\pi x/L\bigr)$ is continuous and $f$ is piecewise smooth the Integration by Parts Theorem applies and it yields \begin{align*} B_k & = \frac{1}{L} \int_{-L}^L F(\xi) \sin\left(\frac{k \pi}{L} \xi \right) d\xi \\ & = \frac{1}{L} \left( - \frac{L}{k\pi} F(L) \cos\bigl(k\pi\bigr) + \frac{L}{k\pi} F(-L) \cos\bigl(k\pi\bigr) + \frac{L}{k\pi} \int_{-L}^L f(\xi) \cos\left(\frac{k \pi}{L} \xi\right) d\xi \right) \\ & = \frac{L}{k \pi} a_k \end{align*} Hence, for all $x\in \mathbb{R}$ we have \[ {\widetilde F}(x) = A_0 - \frac{L}{\pi} \sum_{k=1}^{\infty} \frac{b_k}{k} \cos\left(\frac{k\pi}{L} x \right) + \frac{L}{\pi} \sum_{k=1}^{\infty} \frac{a_k}{k} \sin\left(\frac{k\pi}{L} x \right). \] In particular the preceding equality holds for $x=0.$ Since $F(0) = 0$ we have \[ 0 = A_0 - \frac{L}{\pi} \sum_{k=1}^{\infty} \frac{b_k}{k}. \] This completes the proof. - The preceding theorem states that if $a_0 = 0$ in a Fourier series of a piecewise smooth function $f,$ then the Fourier series of $F$ can be obtained by term-by-term integration of the Fourier series of $f.$ This claim is the consequence of the fact that integrating $k$-th terms we obtain \[ \int_{0}^{x} a_k \cos\left(\frac{k\pi}{L} \xi \right) d\xi = \frac{La_k}{k\pi} \sin\left(\frac{k\pi}{L} x \right) \] and \[ \int_{0}^{x} b_k \sin\left(\frac{k\pi}{L} \xi \right) d\xi = -\frac{Lb_k}{k\pi} \cos\left(\frac{k\pi}{L} x \right) + \frac{Lb_n}{k\pi}, \] which are exactly the terms of the Fourier series obtained in the preceding Theorem.
-
We can eliminate the condition that $a_0 = 0$ in the preceding theorem in the following way:
Integration term-by-term Theorem. Let $L \gt 0$ and let $f:[-L,L] \to \mathbb R$ be a function. Assume that
- $f$ is piecewise smooth function on $[-L,L],$
- $\displaystyle a_0 = \frac{1}{2L} \int_{-L}^{L} f(\xi) d\xi,$
- $\displaystyle F(x) = \int_{0}^{x} f(\xi) d\xi.$
-
Applying the above theorems one can calculate the coefficients of the Fourier series for functions whose derivatives and integrals replicate itself. For example for function $e^x$ or $\cosh(x),$ or a periodic extension of $\sin(x)$ restricted to $(-\pi/2, \pi/2)$, of and similar.
We can apply the first differentiation theorem to calculate the coefficients of the Fourier series of the function $\exp(x)$ say on the interval $[-1,1]$. The constant coefficient is \[ a_0 = \sinh(1). \] Since $\exp(x)$ is its own derivative we have the following equalities: \begin{equation*} a_k = 2 (-1)^{k} \sinh(1) + k\pi \, b_k, \quad b_k = - k\pi \, a_k \quad \text{for all} \quad k\in\mathbb{N}. \end{equation*} Therefore, substituting the expression for $b_k$ into the first equation and solving for $a_k$ we get \[ a_k = \frac{2 (-1)^k \sinh(1)}{1+(k \pi)^2}, \quad b_k = -\frac{2 (-1)^k k \pi \sinh(1) }{1+(k \pi)^2} \quad \text{for all} \quad k\in\mathbb{N}. \] We can confirm this by plotting the Fourier periodic extension of $\exp(x)$ on [-1,1]$ and its approximation by its Fourier series in a small Mathematica notebook.
- I post three Mathematica notebooks and their pdf printouts. A prudent practice in working with Mathematica notebooks is to before saving a notebook. is an item in the menu list . After you it is a good idea to (this will evaluate the entire notebook), in the menu list . If your notebook does not evaluate properly, it is a good idea to fix errors immediately, rather than leaving it for some later time. I recommend to before saving a notebook, since this will make the saved notebook much smaller.
- I recommend that you enable autosaving in your Mathematica notebooks. Please see this site for the instructions how to change this setting.
- The notebook Coloring_Functions_v12.nb contains a method that I used to create the animation of the diffusion of dye in a narrow glass container. A pdf printout of this notebook is Coloring_Functions_v12.pdf
- The notebook Heat_eq_3BCs_v12.nb contains a Mathematica implementation of finding solutions of the heat equation under three different boundary conditions. In this notebook for each boundary conditions and initial conditions I write a separate solution. I think that this method is safer than the method in the notebook in the preceding item. A pdf printout of this notebook is Heat_eq_3BCs_v12.pdf
- You can download these two Mathematica notebooks at the following links:
- To get started with Mathematica read my Mathematica page. On this page, there are links to several videos that might be helpful to efficiently get started with Mathematica. Mathematica is available in the computer labs in BH 215 and BH 209.
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Today we discussed Fourier series of a piecewise smooth function. The relevant section is
Section 3.2: the assigned problems are 3.2.1, 3.2.2, 3.2.3, 3.2.4.
- Definition. Let $L \gt 0$ and let $f:[-L, L] \to \mathbb R$ be a piecewise continuous function. The series \[ a_0 + \sum_{k=1}^{+\infty} \biggl( a_k \cos\Bigl(\!\tfrac{k \pi}{L} x\!\Bigr) + b_k \sin\Bigl(\!\tfrac{k \pi}{L} x\!\Bigr) \biggr) \] where \[ a_0 = \frac{1}{2L} \int_{-L}^L f(\xi) d\xi \] and, for $k \in {\mathbb N}$, \begin{align*} a_k &= \frac{1}{L}\int_{-L}^L f(\xi) \cos\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \\ b_k &= \frac{1}{L}\int_{-L}^L f(\xi) \sin\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \end{align*} is called the Fourier series of $f$. Definition. For $n \in {\mathbb N}$, the $n$-th partial sum of the Fourier series of $f$ is \[ (S_n f)(x) = a_0 + \sum_{k=1}^{n} \biggl( a_k \cos\Bigl(\!\tfrac{k \pi}{L} x\!\Bigr) + b_k \sin\Bigl(\!\tfrac{k \pi}{L} x\!\Bigr) \biggr) \]
- Pointwise Convergence Theorem. If $f$ is piecewise smooth on $[-L,L]$, then for every $x \in {\mathbb R}$ we have \[ \lim_{n \to +\infty} (S_n f)(x) = \tilde{f}_{\!\!\rm Fourier}(x). \] That is, the Fourier series of $f$ converges pointwise to the Fourier periodic extension of $f$.
- Remark. Notice that if the periodic extension of $f$ is a continuous function, then the Fourier periodic extension of $f$ coincides with the periodic extension of $f$. In other words, if $\tilde{f}$ is a continuous function, then $\tilde{f}_{\!\!\rm Fourier} = \tilde{f}$.
- Uniform Convergence Theorem. If $f$ is piecewise smooth and the periodic extension of $f$ is continuous, then the sequence of functions $\bigl\{S_n f\bigr\}_{n=1}^{+\infty}$ converges uniformly on ${\mathbb R}$ to $\tilde{f}$. This means that for every $\varepsilon > 0$ there exists $N_\varepsilon$ such that for all $n \gt N_\varepsilon$ and for all $x \in {\mathbb R}$ we have \[ \bigl|(S_n f)(x) - \tilde{f}(x) \bigr| < \varepsilon. \]
-
Today we discussed Fourier series of a piecewise smooth function. The relevant sections are Section 3.1 and Section 3.2: the assigned problems are 3.2.1, 3.2.2, 3.2.3, 3.2.4.
- Definition. Let $L \gt 0$ and let $f:[-L, L] \to \mathbb R$ be a piecewise continuous function. The series \[ a_0 + \sum_{k=1}^{+\infty} \biggl( a_k \cos\Bigl(\!\tfrac{k \pi}{L} x\!\Bigr) + b_k \sin\Bigl(\!\tfrac{k \pi}{L} x\!\Bigr) \biggr) \] where \[ a_0 = \frac{1}{2L} \int_{-L}^L f(\xi) d\xi \] and, for $k \in {\mathbb N}$, \begin{align*} a_k &= \frac{1}{L}\int_{-L}^L f(\xi) \cos\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \\ b_k &= \frac{1}{L}\int_{-L}^L f(\xi) \sin\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \end{align*} is called the Fourier series of $f$.
-
To understand the convergence of the Fourier series of a function $f:[-L,L] \to \mathbb{R}$ we first need to define the concept of a piecewise smooth function and its periodic extension. This is studied in Chapter 3 in the textbook. However, the book does not have formal definitions of piecewise smooth function and its periodic extension. Therefore, I give few formal definitions here. For more see my webpage about
Fourier periodic extension.
-
Definition.
Let $a$ and $b$ be real numbers such that $a \lt b$. A function $f:[a,b] \to \mathbb R$ is said to be piecewise continuous on $[a,b]$ if the following conditions are satisfied:
- there exists a finite set $\{x_1,\ldots,x_n\} \subset (a,b)$ such that $x_1 \lt \cdots \lt x_n$ and $f$ is continuous on each interval \[ (a,x_1), \quad (x_k,x_{k+1}), \ \ k \in \{1,\ldots,n-1\}, \quad (x_n,b); \]
- all the following one-sided limits exist \[ \lim_{x\downarrow a} f(x), \quad \lim_{x\uparrow x_k} f(x), \quad \lim_{x\downarrow x_k} f(x), \ k \in \{1,\ldots,n\}, \quad \lim_{x\uparrow b} f(x). \]
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Definition. Let $a$ and $b$ be real numbers such that $a \lt b$. A function $f:[a,b] \to \mathbb R$ is said to be piecewise smooth on $[a,b]$ if the following conditions are satisfied:
- there exists a finite set $\{x_1,\ldots,x_n\} \subset (a,b)$ such that $x_1 \lt \cdots \lt x_n$ and $f$ is continuous and it has a continuous derivative $f'$ on each interval \[ (a,x_1), \quad (x_k,x_{k+1}), \ k \in \{1,\ldots,n-1\}, \quad (x_n,b); \]
- all the following one-sided limits exist \[ \lim_{x\downarrow a} f(x), \quad \lim_{x\uparrow x_k} f(x), \quad \lim_{x\downarrow x_k} f(x), \ k \in \{1,\ldots,n\}, \quad \lim_{x\uparrow b} f(x); \]
- all the following one-sided limits exist \[ \lim_{x\downarrow a} f'(x), \quad \lim_{x\uparrow x_k} f'(x), \quad \lim_{x\downarrow x_k} f'(x), \ k \in \{1,\cdots,n\}, \quad \lim_{x\uparrow b} f'(x). \]
- Definition. Let $a$ and $b$ be real numbers such that $a \lt b$ and let $f:(a,b] \to \mathbb R$ be a function. The function $\tilde{f}: {\mathbb R} \to {\mathbb R}$ defined by \[ \tilde{f}(x) = f\left(\!x-\left\lceil\!\dfrac{x-b}{b-a}\! \right\rceil (b-a)\!\right), \quad x \in {\mathbb R}. \] is called the periodic extension of $f$.
- Definition. Let $a$ and $b$ be real numbers such that $a \lt b$ and let $f:[a,b) \to \mathbb R$ be a function. The function $\tilde{f}: {\mathbb R} \to {\mathbb R}$ defined by \[ \tilde{f}(x) = f\left(\!x- \left\lfloor\!\dfrac{x-a}{b-a}\! \right\rfloor (b-a)\!\right), \quad x \in {\mathbb R}. \] is called the periodic extension of $f.$
- Explanation. The reasoning for the above definition of the periodic extension is as follows. For every $x\in\mathbb R$ there exists unique $k \in \mathbb Z$ such that \[ a + k (b-a) \leq x \lt a + (k+1)(b-a). \] The last expression is equivalent to \[ k \leq \frac{x - a}{b-a} \lt k+1. \] Therefore \[ k = \left\lfloor \frac{x - a}{b-a} \right\rfloor. \] Consequently \[ x - \left\lfloor\!\dfrac{x-a}{b-a} \right\rfloor(b-a) \in [a,b). \]
-
Definition.
Let $a$ and $b$ be real numbers such that $a \lt b$. A function $f:[a,b] \to \mathbb R$ is said to be piecewise continuous on $[a,b]$ if the following conditions are satisfied:
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I do understand that the last two definitions might look somewhat weird. The only reason for that is that the ceiling function and the floor function are almost completely absent from our curriculum. That is a fault of our curriculum.
The floor function is defined as follows: For $x \in \mathbb{R}$ we set \[ \lfloor x \rfloor = \max \bigl\{ k \in \mathbb{Z} : k \leq x \bigr\}. \] In words: The floor function of a real number $x$ is defined as the largest integer less than or equal to $x$. This means it rounds $x$ down to the nearest integer. For example, $\lfloor \pi \rfloor = 3$, $\lfloor -e \rfloor = -3$.
The ceiling function is defined as follows: For $x \in \mathbb{R}$ we set \[ \lceil x \rceil = \min \bigl\{ k \in \mathbb{Z} : x \leq k \bigr\}. \] In words: The ceiling function of a real number $x$ is the smallest integer greater than or equal to $x$. This means it rounds $x$ up to the nearest integer. For example, $\lceil \pi \rceil = 4$, $\lceil -e \rceil = -2$.
The book gives a descriptive definition in English of the concept of a periodic extension. The above formula involving the ceiling and floor function is the only way that I was able to translate the definition from English into Mathish. The figures below illustrate the definition with some simple functions $f$. Here is the Mathematica notebook which I used to produce these figures.
In the figure below the function $f$ is the restriction of the function $x \mapsto x$ (in blue) to the interval $[1,4)$. The red function is the periodic extension.
In the figure below the function $f$ is the restriction of the function $x \mapsto x^2-2$ (in blue) to the interval $[-2,2)$. The red function is the periodic extension.
In the figure below the function $f$ is the restriction of the function $x \mapsto \cos(x)$ (in blue) to the interval $[0,\pi)$. The red function is the periodic extension.
- Read Section 3.1 and Section 3.2 in the book. This is a small notebook that I wrote in class today.
- I promised to post pictures illustrating a "geometric" proof that \[ \int_{-\pi}^{\pi} \bigl(\cos (k\mkern 1px x) \bigr)^2 dx = \pi \qquad \text{whenever} \qquad k \in \mathbb{N}. \] Similar pictures can be created on any interval $[-L, L]$ with $L \gt 0$. Those pictures would prove \[ \int_{-L}^{L} \biggl(\cos \Bigl( \frac{k\mkern 1px \pi}{L} x \Bigr) \biggr)^2 dx = L \qquad \text{whenever} \qquad k \in \mathbb{N}. \]
-
Below are pictures and animations that first justify the fact
\[
\int_{0i}^{\pi/2} \bigl(\cos x \bigr)^2 dx = \frac{\pi}{4}
\]
In fact, there are no integrals in these
visual and colorful proofs. These questions could be explored in any trigonometry class. For example, one could print colored pictures from below and ask students to use scissors to perform a proof. That is well known method of proofs: Scissors Method. -
A trigonometry question: Given that the equation of the red curve in the picture below is
\[
y = (\cos x)^2,
\]
calculate the orange area?
-
The following picture could be regarded as a hint. Or, you can print out the picture below hand it out to trigonometry students, and ask them to use scissors to do the calculation. Colors help.
-
Another task for trigonometry students:
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Animations, always helpful, hopefully.
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Try larger domains.
- Today, we finished exploration of solving Laplace's equation on a rectangle. I wrote a webpage Laplace's equation on a Rectangle with my approach to this problem. That is Section 2.5.1 in the textbook.
-
Recall that in Problem 5 on Assignment 1 we solved the following problem:
Problem. Let $K$ and $L$ be positive real numbers. Find the real function $u$ defined on the rectangle \[ [0, K] \times [0, L] = \bigl\{(x,y) \in \mathbb{R}^2 : 0 \leq x \leq K \ \ \text{and} \ \ 0 \leq y \leq L \bigr\} \] such that $u$ satisfies Laplace's partial differential equation: \begin{equation*} \frac{\partial^2 u}{\partial x^2}(x,y) + \frac{\partial^2 u}{\partial y^2}(x,y) = 0 \end{equation*} and the boundary conditions \begin{alignat*}{2} u(x,0) & = \frac{b-a}{K} x + a, & \qquad u(x,L) & =\frac{c-d}{K} x + d \quad \text{for all} \quad x \in [0, K], \\ u(0,y) & =\frac{d-a}{L} y + a, & \qquad u(K,y) & = \frac{c-b}{L} y + b \quad \text{for all} \quad y \in [0,L], \\ \end{alignat*} where $a, b, c$ and $d$ are arbitrary real numbers.
Notice that the above boundary conditions are chosen such that \[ u(0,0) = a, \quad u(K,0) = b, \quad u(K,L) = c, \quad u(0,L) = d. \]
Solution. The solution obtained in Problem 5 on Assignment 1 was \[ u(x,y) = a + \frac{b-a}{K} x + \frac{d-a}{L} y + \frac{a-b+c-d}{K L}xy. \]
Thus, we know how to solve Laplace's equation on a rectangle whenever the boundary conditions are line segments.
- Example. Let $K = 3$ and $L = 2$. Find the real function $u$ defined on the rectangle \[ [0, 3] \times [0,2] = \bigl\{(x,y) \in \mathbb{R}^2 : 0 \leq x \leq 3 \ \ \text{and} \ \ 0 \leq y \leq 2 \bigr\} \] such that $u$ satisfies Laplace's partial differential equation: \begin{equation*} \frac{\partial^2 u}{\partial x^2}(x,y) + \frac{\partial^2 u}{\partial y^2}(x,y) = 0 \end{equation*} and the boundary conditions \begin{alignat*}{2} u(x,0) & = \frac{4}{3} x + \frac{1}{2}, & \qquad u(x,2) & = -\frac{1}{3} x + \frac{7}{2} \quad \text{for all} \quad x \in [0, 3], \\ u(0,y) & =\frac{3}{2} y + \frac{1}{2}, & \qquad u(3,y) & = -y + \frac{9}{2} \quad \text{for all} \quad y \in [0,L], \\ \end{alignat*} Solution. To determine the solution we calculate \[ a=u(0,0) = \frac{1}{2}, \quad b=u(3,0) = \frac{9}{2}, \quad c=u(3,2) = \frac{5}{2}, \quad d=u(0,2) = \frac{7}{2}. \] Hence, the solution is \[ u(x,y) = \frac{1}{2} +\frac{4}{3} x + \frac{3}{2} y - \frac{5}{6}x y. \] And, as it should always be, illustrate with images created in Mathematica:
- Today, we began exploration of solving Laplace's equation on a rectangle. I wrote a webpage Laplace's equation on a Rectangle with my approach to this problem. That is Section 2.5.1 in the textbook.
- The notebook Equilibrium_temperature_2D_v12.nb contains a Mathematica implementation of finding the solution of the Laplace equation on a rectangle. A pdf printout of this notebook is Equilibrium_temperature_2D_v12.pdf. You can download the notebook here Equilibrium_temperature_2D_v12.nb
- In the context of separation of variables in Laplace's equation and beyond, we often encounter trigonometric and hyperbolic functions. Below, I will provide a brief overview of relevant concepts.
-
Let $P$ and $Q$ be continuous real valued functions defined on $\mathbb{R}$. Let $Y_1 : \mathbb{R} \rightarrow \mathbb{R}$ and $Y_2: \mathbb{R} \rightarrow \mathbb{R}$ be linearly independent solutions of the homogeneous linear equation (HLE) \begin{equation*} Y^{\prime\prime}(x)+P(x) Y^{\prime}(x)+Q(x)Y(x)= 0, \ x \in \mathbb{R}. \end{equation*} Then all solutions of the HLE are given by the formula \[ Y(x) = c_1 Y_1(x) + c_2 Y_2(x), \ \ x \in \mathbb{R}, \] where $c_1$ and $c_2$ are arbitrary constants.
The solution \[ Y(x) = c_1 Y_1(x) + c_2 Y_2(x), \ \ x \in \mathbb{R}, \] is called the general solution of the HLE.
A pair of linearly independent solutions of the HLE is called a fundamental set of solutions of the HLE.
Below, we always have $x \in \mathbb{R}$.
-
Let us take a closer look at two fundamental trigonometric functions
\[
x \mapsto \cos(x) \quad \text{and} \quad x \mapsto \sin(x).
\]
- Their derivatives are \[ \bigl( \cos(x) \bigr)' = - \sin(x) \quad \text{and} \quad \bigl( \sin(x) \bigr)' = \cos(x), \] and their second derivatives are \[ \bigl( \cos(x) \bigr)'' = - \cos(x) \quad \text{and} \quad \bigl( \sin(x) \bigr)'' = -\sin(x). \]
-
Therefore, $\cos(x)$ and $\sin(x)$ satisfy the following linear homogeneous second-order differential equation with constant coefficients \[ Y''(x) + Y(x) = 0. \] Since the preceding differential equation is linear homogeneous equation, any linear combination of $\cos(x)$ and $\sin(x)$ is also a solution. In fact, all solutions of the preceding differential equation are given by the following expression \[ c_1 \cos(x) + c_2 \sin(x), \quad \text{where} \quad c_1, c_2 \in \mathbb{R}. \] The preceding expression is called the general solution of $Y''(x) + Y(x) = 0.$
-
There is something special about $\cos(x)$ and $\sin(x)$ that I must mention here. Look at the values of $\cos(x)$ and $\sin(x)$ and its derivatives at $0$: \[ \begin{array}{c} \Big.\bigl(\cos(x)\bigr)\Big|_{x=0} = 1 \\ \Big.\bigl(\cos(x)\bigr)'\Big|_{x=0} = 0 \end{array} \qquad \text{and} \qquad \begin{array}{c} \Big.\bigl(\sin(x)\bigr)\Big|_{x=0} = 0 \\ \Big.\bigl(\sin(x)\bigr)'\Big|_{x=0} = 1 \end{array} \]
- There are many interesting functions that hide in the general solution. For example, for arbitrary $a, b \in \mathbb{R}$ the functions \[ x \mapsto b \cos(x - a), \quad x \mapsto b \sin(x-a), \quad x \mapsto b \cos(a - x), \quad x \mapsto b \sin(a-x). \] solve the differential equation $Y''(x) + Y(x) = 0.$ To convince yourself that this is true, just find the second derivative by using the chain rule: \begin{multline*} \bigl( b \cos(x - a)\bigr)' = -b \sin(x - a) , \quad \bigl( b \sin(x-a)\bigr)' = b \cos(x-a), \\ \bigl( b \cos(a - x) \bigr)' = b \sin(a - x), \quad \bigl( b \sin(a-x)\bigr)' = - b \cos(a-x), \end{multline*} and \begin{multline*} \bigl( b \cos(x - a)\bigr)'' = -b \cos(x - a) , \quad \bigl( b \sin(x-a)\bigr)'' = -b \sin(x-a), \\ \bigl( b \cos(a - x) \bigr)'' = -b \cos(a - x), \quad \bigl( b \sin(a-x)\bigr)'' = - b \sin(a-x). \end{multline*}
-
In this item we introduce two fundamental hyperbolic functions. We start with the linear homogeneous second-order differential equation with constant coefficients
\[
Y''(x) - Y(x) = 0.
\]
Notice that this equation is just a slight variation on the differential equation $Y''(x) + Y(x) = 0$ which leads to the famous trigonometric functions $\cos(x)$ and $\sin(x)$.
- What is the general solution of \[ Y''(x) - Y(x) = 0? \] Consider the functions \[ x \mapsto e^x \quad \text{and} \quad x \mapsto e^{-x}, \] and their derivatives \[ \bigl(e^x\bigr)' = e^x \quad \text{and} \quad \bigl(e^{-x}\bigr)' = - e^{-x}, \] and \[ \bigl(e^x\bigr)'' = e^x \quad \text{and} \quad \bigl(e^{-x}\bigr)'' = e^{-x}. \] Therefore, the general solution of $Y''(x) - Y(x) = 0$ is \[ c_1 e^{x} + c_2 e^{-x}, \quad c_1, c_2 \in \mathbb{R}. \]
- There are two solutions in the general solution given in the preceding item that stand out. Those are \[ x \mapsto \frac{1}{2} e^x + \frac{1}{2} e^{-x} = \cosh(x) \quad \text{and} \quad x \mapsto \frac{1}{2} e^x - \frac{1}{2} e^{-x} = \sinh(x). \] The function $\cosh$ is called the hyperbolic cosine and the function $\cosh$ is called the hyperbolic sine. You can learn more about them at the Wikipedia page Hyperbolic Functions.
- To familiarize ourselves with these functions, we take their derivatives \[ \bigl(\cosh(x)\bigr)' = \sinh(x) \quad \text{and} \quad \bigl( \sinh(x) \bigr)' = \cosh(x), \] and \[ \bigl(\cosh(x)\bigr)'' = \cosh(x) \quad \text{and} \quad \bigl( \sinh(x) \bigr)'' = \sinh(x), \] So, they clearly solve $Y''(x) - Y(x)=0$. But, more importantly, they have special properties. Look at the values of $\cosh(x)$ and $\sinh(x)$ and its derivatives at $0$: \[ \begin{array}{c} \Big.\bigl(\cosh(x)\bigr)\Big|_{x=0} = 1 \\ \Big.\bigl(\cosh(x)\bigr)'\Big|_{x=0} = 0 \end{array} \qquad \text{and} \qquad \begin{array}{c} \Big.\bigl(\sinh(x)\bigr)\Big|_{x=0} = 0 \\ \Big.\bigl(\sinh(x)\bigr)'\Big|_{x=0} = 1 \end{array} \] Very convenient!
- The hyperbolic trigonometric functions $\cosh(x)$ and $\sinh(x)$ give us another way to write the general solution of $Y''(x) - Y(x) = 0$: \[ c_1 \cosh(x) + c_2 \sinh(x), \quad \text{where} \quad c_1, c_2 \in \mathbb{R}. \]
- There are many interesting functions that hide in the general solution given in the preceding item. For example, for arbitrary $a, b \in \mathbb{R}$ the functions \[ x \mapsto b \cosh(x - a), \quad x \mapsto b \sinh(x-a), \quad x \mapsto b \cosh(a - x), \quad x \mapsto b \sinh(a-x). \] solve the differential equation $Y''(x) - Y(x) = 0.$ To convince yourself that this is true, just find the second derivative by using the chain rule: \begin{multline*} \bigl( b \cosh(x - a)\bigr)' = b \sin(x - a) , \quad \bigl( b \sinh(x-a)\bigr)' = b \cosh(x-a), \\ \bigl( b \cosh(a - x) \bigr)' = - b \sinh(a - x), \quad \bigl( b \sinh(a-x)\bigr)' = - b \cosh(a-x), \end{multline*} and \begin{multline*} \bigl( b \cosh(x - a)\bigr)'' = b \cosh(x - a) , \quad \bigl( b \sinh(x-a)\bigr)'' = b \sinh(x-a), \\ \bigl( b \cosh(a - x) \bigr)'' = b \cosh(a - x), \quad \bigl( b \sinh(a-x)\bigr)'' = b \sinh(a-x). \end{multline*}
-
I did not expect this to be that long. But, I find it to be a nice story. The story continues by considering the trigonometric functions with different frequencies. For $\mu \gt 0$ consider
\[
x \mapsto \cos(\mu x) \quad \text{and} \quad x \mapsto \sin(\mu x).
\]
- The derivatives of the trigonometric functions given above are \[ \bigl( \cos(\mu x) \bigr)' = - \mu \sin(\mu x) \quad \text{and} \quad \bigl( \sin(\mu x) \bigr)' = \mu \cos(\mu x), \] and their second derivatives are \[ \bigl( \cos(\mu x) \bigr)'' = - \mu^2 \cos(\mu x) \quad \text{and} \quad \bigl( \sin(\mu x) \bigr)'' = -\mu^2 \sin(\mu x). \]
-
Therefore, $\cos(\mu x)$ and $\sin(\mu x)$ satisfy the following linear homogeneous second-order differential equation with constant coefficients \[ Y''(x) + \mu^2 Y(x) = 0. \] This equation can be written as an "eigenvalue equation" \[ -Y''(x) = \mu^2 Y(x) \] Since the preceding differential equation is linear homogeneous equation, any linear combination of $\cos(\mu x)$ and $\sin(\mu x)$ is also a solution. In fact, all solutions of the preceding differential equation are given by the following expression \[ c_1 \cos(\mu x) + c_2 \sin(\mu x), \quad \text{where} \quad c_1, c_2 \in \mathbb{R}. \] The preceding expression is called the general solution of $-Y''(x) = \mu^2 Y(x).$
- There are many interesting functions that hide in the general solution given in the preceding item. For example, for arbitrary $a, b \in \mathbb{R}$ the functions \[ x \mapsto b \cos\bigl(\mu (x - a) \bigr), \quad x \mapsto b \sin\bigl(\mu (x - a) \bigr), \quad x \mapsto b \cos\bigl(\mu (a - x) \bigr), \quad x \mapsto b \sin\bigl(\mu (a - x) \bigr) \] solve the differential equation $-Y''(x) = \mu^2 Y(x).$ To convince yourself that this is true, just find the second derivative by using the chain rule: \begin{multline*} \bigl( b \cos\bigl(\mu (x - a) \bigr)\bigr)' = - \mu b \sin\bigl(\mu (x - a) \bigr) , \quad \bigl( b \sin\bigl(\mu (x - a) \bigr)\bigr)' = \mu b \cos\bigl(\mu (x - a) \bigr), \\ \bigl( b \cos\bigl(\mu (a - x) \bigr) \bigr)' = \mu b \sin\bigl(\mu (a - x) \bigr), \quad \bigl( b \sin\bigl(\mu (a - x) \bigr)\bigr)' = - \mu b \cos\bigl(\mu (a - x) \bigr), \end{multline*} and \begin{multline*} \bigl( b \cos\bigl(\mu (x - a) \bigr)\bigr)'' = - \mu^2 b \cos\bigl(\mu (x - a) \bigr) , \quad \bigl( b \sin(x-a)\bigr)'' = -\mu^2 b \sin\bigl(\mu (x - a) \bigr), \\ \bigl( b \cos\bigl(\mu (a - x) \bigr) \bigr)'' = -\mu^2 b \cos\bigl(\mu (a - x) \bigr), \quad \bigl( b \sin\bigl(\mu (a - x) \bigr)\bigr)'' = -\mu^2 b \sin\bigl(\mu (a - x) \bigr). \end{multline*}
-
The story continues by considering the hyperbolic functions with scaled independent variable. For $\mu \gt 0$ consider
\[
x \mapsto \cosh(\mu x) \quad \text{and} \quad x \mapsto \sinh(\mu x).
\]
- The derivatives of the hyperbolic functions given above are \[ \bigl( \cosh(\mu x) \bigr)' = \mu \sinh(\mu x) \quad \text{and} \quad \bigl( \sinh(\mu x) \bigr)' = \mu \cosh(\mu x), \] and their second derivatives are \[ \bigl( \cosh(\mu x) \bigr)'' = \mu^2 \cosh(\mu x) \quad \text{and} \quad \bigl( \sinh(\mu x) \bigr)'' = \mu^2 \sinh(\mu x). \]
-
Therefore, $\cosh(\mu x)$ and $\sinh(\mu x)$ satisfy the following linear homogeneous second-order differential equation with constant coefficients \[ Y''(x) - \mu^2 Y(x) = 0. \] This equation can be written as an "eigenvalue equation" \[ Y''(x) = \mu^2 Y(x) \] Since the preceding differential equation is linear homogeneous equation, any linear combination of $\cosh(\mu x)$ and $\sinh(\mu x)$ is also a solution. In fact, all solutions of the preceding differential equation are given by the following expression \[ c_1 \cosh(\mu x) + c_2 \sinh(\mu x), \quad \text{where} \quad c_1, c_2 \in \mathbb{R}. \] The preceding expression is called the general solution of $Y''(x) = \mu^2 Y(x).$
- There are many interesting functions that hide in the general solution given in the preceding item. For example, for arbitrary $a, b \in \mathbb{R}$ the functions \[ x \mapsto b \cosh\bigl(\mu (x - a) \bigr), \quad x \mapsto b \sinh\bigl(\mu (x - a) \bigr), \quad x \mapsto b \cosh\bigl(\mu (a - x) \bigr), \quad x \mapsto b \sinh\bigl(\mu (a - x) \bigr) \] solve the differential equation $Y''(x) = \mu^2 Y(x).$ To convince yourself that this is true, just find the second derivative by using the chain rule: \begin{multline*} \bigl( b \cosh\bigl(\mu (x - a) \bigr)\bigr)' = \mu b \sinh\bigl(\mu (x - a) \bigr) , \quad \bigl( b \sinh\bigl(\mu (x - a) \bigr)\bigr)' = \mu b \cosh\bigl(\mu (x - a) \bigr), \\ \bigl( b \cosh\bigl(\mu (a - x) \bigr) \bigr)' = -\mu b \sinh\bigl(\mu (a - x) \bigr), \quad \bigl( b \sinh\bigl(\mu (a - x) \bigr)\bigr)' = - \mu b \cosh\bigl(\mu (a - x) \bigr), \end{multline*} and \begin{multline*} \bigl( b \cosh\bigl(\mu (x - a) \bigr)\bigr)'' = \mu^2 b \cosh\bigl(\mu (x - a) \bigr) , \quad \bigl( b \sinh(x-a)\bigr)'' = -\mu^2 b \sinh\bigl(\mu (x - a) \bigr), \\ \bigl( b \cosh\bigl(\mu (a - x) \bigr) \bigr)'' = \mu^2 b \cosh\bigl(\mu (a - x) \bigr), \quad \bigl( b \sinh\bigl(\mu (a - x) \bigr)\bigr)'' = \mu^2 b \sinh\bigl(\mu (a - x) \bigr). \end{multline*}
- Today we discussed the details of what I wrote up yesterday. It is Subsection 2.5.2, Laplace's Equation Inside a Circular Disk in the textbook.
- Today we talked about Laplace's Equation Inside a Circular Disk, Subsection 2.5.2. The method that we presented in a disk with a radius equal to a given $a$, where $a$ is a fixed positive number applies to geometric shapes which are "friendly" to polar coordinates. Below, for simplicity, I use $a = 1$.
- Recall that the Laplacian in polar coordinates is \[ \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial w}{\partial r}(r,\theta) + \frac{\partial^2 w}{\partial r^2}(r,\theta) = \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial w}{\partial r}(r,\theta) \right) \]
- We applied the Method of Separation of Variables applied to the homogeneous Laplace's in a unit disk \[ \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial w}{\partial r}(r,\theta) \right) = 0, \quad r \in [0,+\infty), \quad \theta \in [-\pi, \pi], \] subject to the Boundary Condition: \[ w(1,\theta) = f(\theta), \qquad \theta \in [-\pi, \pi]. \]
- The separation of variables, as before is, \[ w(r,\theta) = A(r) B(\theta). \] Substitution of $w(r,\theta) = A(r) B(\theta)$ into Laplace's equation in polar coordinates yields \[ \frac{1}{r^2} A(r) B''(\theta) + \frac{1}{r} \bigl(r A'(r) \bigr)' B(\theta) = 0. \] Next, multiply the preceding equality with $\displaystyle \frac{r^2}{A(r) B(\theta)}$ to get \[ \frac{B''(\theta)}{B(\theta)} + \frac{r \bigl(r A'(r) \bigr)'}{A(r)} = 0. \] Finally, the variables are separated, with the separation constant $\lambda$: \[ \frac{r \bigl(r A'(r) \bigr)'}{A(r)} = - \frac{B''(\theta)}{B(\theta)} = \lambda. \]
- Physical consideration lead to the boundary conditions for the function $B(\theta)$: \[ -B''(\theta) = \lambda B(\theta), \quad B(-\pi) = B(\pi), \quad B'(-\pi) = B'(\pi). \] This is a Boundary-Eigenvalue Problem. The eigenvalues are $0, 1^2, 2^2, 3^2, \ldots$, that is the eigenvalues are the squares of the nonnegative integers. The corresponding eigenfunctions are as follows: \begin{align*} \text{eigenvalue} \ \ 0, & \qquad \text{eigenfunction} \ \ 1 \ \ \text{(the constant function)} \\ \text{eigenvalue} \ \ k^2, & \qquad \text{eigenfunctions} \ \ \cos(k\mkern 1pt \theta), \ \ \sin(k\mkern 1pt \theta), \quad k \in\mathbb{N}. \\[-7pt] & \mkern 4cm \text{(two linearly independent eigenfunctions)} \end{align*} As always, $\mathbb{N}$ denotes the set of all positive integers.
- The second order differential equation for the function $A(r)$ is \[ r^2 A''(r) + r A'(r) - k^2 A(r) = 0, \quad k \in \{0\} \cup \mathbb{N}. \] For $k = 0$ the general solution of this equation is \[ A_0(r) = c_1 \mkern 1pt 1 + c_2 \ln(r). \] For $k \in \mathbb{N}$ the general solution of this equation is \[ A_k(r) = c_1 \mkern 1pt r^{k} + c_2 \mkern 1pt r^{-k}. \] Now recall that $A(r)$ is a part of the temperature function $w(r,\theta) = A(r) B(\theta)$. Therefore, $A(r)$ must be defined at the origin, that is at $r=0$. Consequently, we must reject the functions $\ln(r)$, $r^{-k}$ as possible solutions. In conclusion, for $k=0$, the only solution is $1$ the constant function, and for $k\in\mathbb{N}$ the only solution is $r^k$.
- At the end of this part we have the following double sequence of solutions of Laplace's equation: \[ 1, \qquad r^k \cos (k \mkern 1pt \theta ), \quad r^k \sin (k \mkern 1pt \theta ), \quad \text{where} \quad k \in \mathbb{N}. \]
- It is always fun to verify that the above solutions satisfy Laplace's equation: \[ \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + r \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r} (r,\theta) \right) = 0, \quad r \in [0,+\infty), \quad \theta \in [-\pi, \pi], \] For example $r^k \sin (k \mkern 1pt \theta )$: \begin{align*} \frac{\partial^2}{\partial \theta^2}\bigl(r^k \sin (k \mkern 1pt \theta )\bigr) + r \frac{\partial}{\partial r} \left( r \frac{\partial}{\partial r} \bigl(r^k \sin (k \mkern 1pt\theta )\bigr) \right) & = - k^2 r^k \sin (k \mkern 1pt \theta ) + r \frac{\partial}{\partial r}\bigl( r k r^{k-1} \sin (k \mkern 1pt \theta ) \bigr) \\ & = - k^2 r^k \sin (k \mkern 1pt \theta ) + r \frac{\partial}{\partial r}\bigl( k r^{k} \sin (k \mkern 1pt \theta ) \bigr) \\ & = - k^2 r^k \sin (k \mkern 1pt \theta ) + r k^2 r^{k-1} \sin (k \mkern 1pt \theta ) \\ & = 0. \end{align*}
- To satisfy the boundary condition \[ w(1,\theta) = f(\theta) \quad \theta \in [-\pi, \pi], \] we look for the solution in the form of a Fourier series \[ w(r,\theta) = a_0 + \sum_{k=1}^{n} a_k r^k \cos(k \mkern 1pt\theta) + \sum_{k=1}^{n} b_k r^k \sin(k \mkern 1pt \theta), \] where $a_0$, $a_k$, and $b_k$ are real numbers to be determined from the boundary condition: \[ f(\theta) = w(1,\theta) = a_0 + \sum_{k=1}^{n} a_k \cos(k \mkern 1pt\theta) + \sum_{k=1}^{n} b_k \sin(k \mkern 1pt \theta), \quad \theta \in [-\pi, \pi]. \] Next we use the fact that the functions \[ 1, \qquad \cos (k \mkern 1pt \theta ), \quad \sin (k \mkern 1pt \theta ), \quad \text{where} \quad k \in \mathbb{N}, \] are mutually orthogonal relative to the inner product \[ \bigl\langle u(\theta), v(\theta) \bigr\rangle = \int_{-\pi}^{\pi} u(\theta) v(\theta) d \theta \] and \[ \int_{-\pi}^{\pi} 1 d \theta = 2\pi, \quad \int_{-\pi}^{\pi} \bigl(\cos(k \mkern 1pt\theta)\bigr)^2 d \theta = \pi, \quad \int_{-\pi}^{\pi} \bigl(\sin(k \mkern 1pt \theta)\bigr)^2 d \theta = \pi, \] to calculate \begin{align*} a_0 & = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\theta) d \theta, \\ a_k & = \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \cos(k \mkern 1pt\theta) d \theta, \quad k \in \mathbb{N}, \\ b_k & = \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \sin(k \mkern 1pt\theta) d \theta, \quad k \in \mathbb{N}. \end{align*} With these coefficients in the above Fourier series we have the solution of the boundary value problem on the unit disk.
- This method is particularly rewarding when the boundary function $f(\theta)$ is a finite linear combination of the functions \[ 1, \qquad \cos (k \mkern 1pt \theta ), \quad \sin (k \mkern 1pt \theta ), \quad \text{where} \quad k \in \mathbb{N}. \] I listed several such functions on Wednesday, November 1. Here are several more \begin{align*} (\cos x)^2 & = \frac{1}{2}+\frac{1}{2} \cos (2 x) \\ (\cos x)^4 & = \frac{3}{8} +\frac{1}{2} \cos (2 x)+\frac{1}{8} \cos (4 x)\\ (\cos x)^6 & = \frac{5}{16} + \frac{15}{32} \cos (2 x)+\frac{3}{16} \cos (4x)+\frac{1}{32} \cos (6x) \end{align*} These are trigonometric identities that can be deduced by the repeated application of the double-angle identities and angle sum and difference identities, or they can be found at Wikipedia page Power-reduction formulae.
- Let \[ f(\theta) = 2 \bigl(\cos(\theta/2) \bigr)^6, \quad \theta \in [-\pi, \pi], \] in the boundary condition for Laplace's equation on a disk. Based on the last identity in the preceding item we have the following identity for the given boundary condition: \[ 2 \bigl(\cos(\theta/2) \bigr)^6 = \frac{5}{8}+\frac{15}{16} \cos (\theta ) +\frac{3}{8}\cos (2 \theta ) + \frac{1}{16} \cos (3 \theta ). \] From this formula we can read out the coefficients \[ a_0 = \frac{5}{8}, \quad a_1= \frac{15}{16}, \quad a_2 = \frac{3}{8}, \quad a_3 = \frac{1}{16}, \] and all other coefficients are zero. Surprisingly, from these trigonometric identities we can also deduce the following integrals \begin{align*} \frac{5}{8} & = \frac{1}{2\pi} \int_{-\pi}^{\pi} 2 \bigl(\cos(\theta/2) \bigr)^6 d \theta \\ \frac{15}{16} & = \frac{1}{\pi} \int_{-\pi}^{\pi} 2 \bigl(\cos(\theta/2) \bigr)^6 \cos(\theta) d \theta \\ \frac{3}{8} & = \frac{1}{\pi} \int_{-\pi}^{\pi} 2 \bigl(\cos(\theta/2) \bigr)^6 \cos(2 \theta) d \theta \\ \frac{1}{16} & = \frac{1}{\pi} \int_{-\pi}^{\pi} 2 \bigl(\cos(\theta/2) \bigr)^6 \cos(3 \theta) d \theta \end{align*}
-
Hence, the exact solution of the boundary value problem on the unit disk with this specific boundary condition is
\[
w(r,\theta) = \frac{5}{8}+\frac{15}{16} r \cos (\theta )
+\frac{3}{8} r^2 \cos (2 \theta ) + \frac{1}{16} r^3 \cos (3 \theta ).
\]
The Mathematica code for the right picture above (it uses some definitions in the preceding command):MyOptionsD = Sequence[ImageResolution -> 600, Axes -> True, AxesLabel -> {None, None, None}, AxesOrigin -> {Automatic, Automatic, Automatic}, Boxed -> True, DisplayFunction -> Identity, FaceGrids -> None, FaceGridsStyle -> Automatic, ImageSize -> 500, Lighting -> {{"Ambient", White}}, BoundaryStyle -> None, Method -> {"DefaultGraphicsInteraction" -> {"Version" -> 1.2`, "TrackMousePosition" -> {True, False}, "Effects" -> {"Highlight" -> {"ratio" -> 2}, "HighlightPoint" -> {"ratio" -> 2}, "Droplines" -> {"freeformCursorMode" -> True, "placement" -> {"x" -> "All", "y" -> "None"}}}}, "RotationControl" -> "Globe"}, PlotRangePadding -> {Scaled[0.02`], Scaled[0.02`], Scaled[0.02`]}, Ticks -> {Automatic, Automatic, Automatic}, ViewPoint -> {-1.0311155006393165`, 2.874130096265292`, 1.4581416303238175`}, ViewVertical -> {0.15265090977010184`, -0.40303458278204485`, 0.9023640201316004`}]; fbth[\[Theta]_] = 2 Cos[\[Theta]/2]^6; gtfbth = ParametricPlot3D[{Cos[\[Theta]], Sin[\[Theta]], fbth[\[Theta]]}, {\[Theta], -Pi, Pi}, PlotStyle -> {RGBColor[0, 0.6, 0], Thickness[0.01]}, PlotPoints -> {100}]; disks = Graphics3D[ { {FaceForm[RGBColor[0.75, 0.75, 0.75]], Opacity[0.25], Polygon[{Cos[#], Sin[#], 0} & /@ Range[-Pi, Pi, Pi/64]]}, {GrayLevel[0.5], Thickness[0.003], Line[{Cos[#], Sin[#], 0} & /@ Range[-Pi, Pi, Pi/64]]} }, Lighting -> {{"Ambient", White}} ]; EqSolDiskBC = Show[{gtfbth, disks}, PlotRange -> {{-1.05, 1.05}, {-1.05, 1.05}, {-.05, 2.05}}, BoxRatios -> {1, 1, 1}, MyOptionsD]
Clear[wwSd]; wwSd[r_, \[Theta]_] := 5/8 + 15 /16 r Cos[\[Theta]] + 3/8 r^2 Cos[2 \[Theta]] + 1/16 r^3 Cos[3 \[Theta]]; soluD = ParametricPlot3D[{r Cos[\[Theta]], r Sin[\[Theta]], wwSd[r, \[Theta]]}, {r, 0, 1}, {\[Theta], -Pi, Pi}, PlotStyle -> {Opacity[0.7]}, PlotPoints -> {30, 100}, Mesh -> {10, 50}, MeshStyle -> {{Thickness[0.002], RGBColor[0, 0.5, 0.5], Opacity[0.5]}, {Thickness[0.002], RGBColor[0, 0.5, 0.5], Opacity[0.5]}}, BoundaryStyle -> Directive[{Thickness[0.002], RGBColor[0, 0.5, 0.5], Opacity[0.5]}]]; EqSolDisk = Show[{soluD, gtfbth, disks}, PlotRange -> {{-1.05, 1.05}, {-1.05, 1.05}, {-.05, 2.05}}, BoxRatios -> {1, 1, 1}, MyOptionsD]
- The best first illustration of the solution of a specific heat equation with the Dirichlet Boundary Conditions and a specific Initial Condition is as follows: \[ \frac{\partial u}{\partial t}(x,t) = \frac{\partial^2 u}{\partial x^2}(x,t) \qquad x \in [0,\pi], \quad t \geq 0, \] subject to the Boundary Conditions: \[ u(0,t) = 0 \quad \text{and} \quad u(\pi,t) = 0 \quad \text{for all} \quad t \geq 0, \] and the Initial Condition \[ u(x,0) = \bigl(\sin x \bigr)^5, \quad 0 \leq x \leq \pi. \]
-
Solution.
- For this PDE with BCs, the sequence of special solutions is \[ u_k(x,t) = \exp\left(\!- k^2t \right) \sin\left(k \mkern 1mu x\right) \quad \text{where} \quad k \in \mathbb{N}. \] Or, we can create more solutions by writing linear combinations of the special solutions: \[ u(x,t) = \sum_{k = 1}^{n} c_k e^{- k^2t} \sin\left(k \mkern 1mu x\right). \]
- To satisfy the Initial Condition we need to find the coefficients $c_k$ such that \[ \bigl(\sin x \bigr)^5 = u(x,0) = \sum_{k = 1}^{n} c_k \sin\left(k \mkern 1mu x\right) \] That is we need $c_1, c_2, \ldots$ such that \[ \bigl(\sin x \bigr)^5 =c_1 \sin\left(x\right) + c_2 \sin\left(2 \mkern 1mu x\right) + c_3 \sin\left(3 \mkern 1mu x\right) + c_5 \sin\left(5 \mkern 1mu x\right) + \cdots \]
- In class we presented a universal way of calculating such coefficients $c_k$: \[ c_k = \frac{2}{\pi} \int_0^\pi \bigl(\sin x \bigr)^5 \mkern 2mu \sin\left(k \mkern 1mu x\right) dx. \]
- However, for this special function $\bigl(\sin x \bigr)^5$ one can find a trigonometric formula among many presented at the Wikipedia page Trigonometric Identities, specifically Power-reduction formulae, \[ \bigl(\sin x \bigr)^5 = \frac{5}{8} \sin\left(x\right) - \frac{5}{16} \sin\left(3 \mkern 1mu x\right) + \frac{1}{16} \sin\left(5 \mkern 1mu x\right). \] The preceding formula is an identity. Thus \[ c_1 = \frac{5}{8}, \ c_2 =0, \ c_3=- \frac{5}{16}, \ c_4 = 0, \ c_5 = \frac{1}{16}, \ c_k = 0 \ \ \text{for all} \ \ k \geq 6. \]
- Thus, the solution of the given problem is \[ u(x,t) = \frac{5}{8} e^{-t} \sin\left(x\right) - \frac{5}{16} e^{-9\mkern 1mu t} \sin\left(3 \mkern 1mu x\right) + \frac{1}{16} e^{-25\mkern 1mu t} \sin\left(5 \mkern 1mu x\right). \]
- From the formula that we found on Wikipedia, and from the general theory that we developed in class, we can deduce without any additional calculations that \begin{align*} \frac{2}{\pi} \int_0^\pi \bigl(\sin x \bigr)^5 \mkern 2mu \sin\left(x\right) dx & = \frac{5}{8}, \\ \frac{2}{\pi} \int_0^\pi \bigl(\sin x \bigr)^5 \mkern 2mu \sin\left(2 \mkern 1mu x\right) dx & = 0, \\ \frac{2}{\pi} \int_0^\pi \bigl(\sin x \bigr)^5 \mkern 2mu \sin\left(3 \mkern 1mu x\right) dx & = - \frac{5}{16}, \\ \frac{2}{\pi} \int_0^\pi \bigl(\sin x \bigr)^5 \mkern 2mu \sin\left(4 \mkern 1mu x\right) dx & = 0, \\ \frac{2}{\pi} \int_0^\pi \bigl(\sin x \bigr)^5 \mkern 2mu \sin\left(5 \mkern 1mu x\right) dx & = \frac{1}{16}, \\ \frac{2}{\pi} \int_0^\pi \bigl(\sin x \bigr)^5 \mkern 2mu \sin\left(k \mkern 1mu x\right) dx & = 0, \ \ k \geq 6. \\ \end{align*}
- In class I created a small Mathematica notebook 20231031.nb in which I illustrated this solution.
- I also created 20231031.pdf, for you to enjoy this file without access to Mathematica.
- We continued with the discussion of Fourier's Method of Separation of Variables. The relevant sections of the textbook is 2.3.1, 2.3.2, 2.3.3, 2.3.4.
- We considered the homogeneous heat equation in a one-dimensional rod of length $L$ subject to the Dirichlet boundary conditions: \[ \frac{\partial u}{\partial t}(x,t) = \kappa \frac{\partial^2 u}{\partial x^2}(x,t) \qquad x \in [0,L], \quad t \geq 0, \] subject to the Boundary Conditions: \[ u(0,t) = 0 \quad \text{and} \quad u(L,t) = 0 \quad \text{for all} \quad t \geq 0. \] Notice that in the heat equation $\frac{\partial u}{\partial t} = \kappa \frac{\partial^2 u}{\partial x^2}$ I write $\kappa$, the Greek lower case letter kappa, instead of $k$ in the book. This constant is called thermal diffusivity.
- The brilliant idea to find solutions of the problem stated in the preceding item is to look for a solution in which the independent variables are separated in two functions $A$ and $B$: \[ u(x,t) = A(x) \mkern 1mu B(t). \] The brilliancy of the above idea is that one unpleasant unknown function $u(x,t)$ of two independent variables is replaced by two unknown functions $A(x)$ and $B(t)$, but each one of which is a function of a single independent variable. There is no guarantee that this idea will work. But, it is brilliant and amazing since it works, as we will see later.
-
The solutions for the functions $A(x)$ and $B(t)$ are : \begin{equation*} B(t) = \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right), \qquad A(x) = \sin\left(\frac{m \pi}{L} x\right), \end{equation*} where $\kappa$ and $L$ are given positive real numbers and $m$ is any positive integer.
Thus, we have obtained the sequence of solutions for $u_m(x,t)$ given by \[ u_m(x,t) = \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right) \sin\left(\frac{m \pi}{L} x\right) \quad \text{where} \quad m \in \mathbb{N}. \] We use $\mathbb{N}$ to denote the set of all positive integers. A good exercise is to verify that the preceding formula for $u(x,t)$ indeed solves the heat equation with the Dirichlet boundary conditions, the problem stated above.
- In the next two items I list solutions of the heat equation with different boundary conditions.
- The Method of Separation of Variables applied to the homogeneous heat equation in a one-dimensional rod of length $L$ subject to the Neumann boundary conditions: \[ \frac{\partial u}{\partial t}(x,t) = \kappa \frac{\partial^2 u}{\partial x^2}(x,t) \qquad x \in [0,L], \quad t \geq 0, \] subject to the Boundary Conditions: \[ \frac{\partial u}{\partial x}(0,t) = 0 \quad \text{and} \quad \frac{\partial u}{\partial x}(L,t) = 0 \quad \text{for all} \quad t \geq 0, \] leads to the important sequence of special solutions \[ 1, \qquad \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right) \cos\left(\frac{m \pi}{L} x\right) \quad \text{where} \quad m \in \mathbb{N}. \]
- The Method of Separation of Variables applied to the homogeneous heat equation in a one-dimensional ring of length $2L$ subject to the periodic boundary conditions: \[ \frac{\partial u}{\partial t}(x,t) = \kappa \frac{\partial^2 u}{\partial x^2}(x,t) \qquad x \in [-L,L], \quad t \geq 0, \] subject to the Boundary Conditions: \[ u(-L,t) = u(L,t) \quad \text{and} \quad \frac{\partial u}{\partial x}(-L,t) = \frac{\partial u}{\partial x}(L,t) \quad \text{for all} \quad t \geq 0, \] leads to the important sequence of special solutions \[ 1, \qquad \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right) \cos\left(\frac{m \pi}{L} x\right), \qquad \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right) \sin\left(\frac{m \pi}{L} x\right) \quad \text{where} \quad m \in \mathbb{N}. \]
-
For homework do the following two problems:
- Consider a one-dimensional rod placed over the interval $[0,2]$ without heat sources. The rod is made of a uniform material whose physical constants are all equal to $1.$ This rod is subject to the following Boundary Conditions \[ u(0,t) = 0 \quad \text{and} \quad u(2,t) = 0 \quad \text{for all} \quad t \geq 0 \] and the Initial Condition \[ u(x,0) = \sin(\pi x) \quad \text{for all} \quad x \in [0,2]. \] Find the temperature of this rod at all times $t \geq 0.$
- Consider a one-dimensional rod placed over the interval $[0,2]$ without heat sources. The rod is made of a uniform material whose physical constants are all equal to $1.$ This rod is subject to the following Boundary Conditions \[ u(0,t) = 1 \quad \text{and} \quad u(2,t) = 3 \quad \text{for all} \quad t \geq 0 \] and the Initial Condition \[ u(x,0) = 1+x + \sin(\pi x) \quad \text{for all} \quad x \in [0,2]. \] Find the temperature of this rod at all times $t \geq 0.$
-
In the rest of the post today, I wanted to celebrate the function that is a solution of Problem 2 on Assignment 1
\[
u(x,y) = \frac{\sin(x y)}{y}.
\]
It is not surprising to make a claim that the domain of this function is the set
\[
\bigl\{ (x,y) \in \mathbb{R}^2 \, : \, y \neq 0\bigr\}.
\]
However, the plot of a graph of this function indicates that this function is defined on the entire plane $\mathbb{R}^2$.
A graph might be even nicer without a Mesh.Plot3D[x Sinc[y x], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, PlotStyle -> {Opacity[0.85]}, Exclusions -> None, PlotPoints -> {151, 151}, Mesh -> {55, 55}, MeshStyle -> {{GrayLevel[0.35], Thickness[0.001]}, {GrayLevel[0.35], Thickness[0.001]}}, ClippingStyle -> False, PlotRange -> {-2 Pi, 2 Pi}, ImageSize -> 800, Axes -> False, Boxed -> True, BoxRatios -> {1, 1, 1}, ViewAngle -> Automatic, ViewCenter -> {0.5`, 0.5`, 0.5`}, ViewMatrix -> Automatic, ViewPoint -> {2.5075250998820904`, -1.6385753275514336`, \ -1.5739722899089035`}, ViewProjection -> Automatic, ViewRange -> All, ViewVector -> Automatic, ViewVertical -> {-0.34286271463298007`, 0.30614344718399633`, -0.8880998528661083`}]
Plot3D[x Sinc[y x], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, PlotStyle -> {Opacity[1]}, Exclusions -> None, PlotPoints -> {151, 151}, Mesh -> {0, 0}, ClippingStyle -> False, PlotRange -> {-2 Pi, 2 Pi}, ImageSize -> 800, Axes -> False, Boxed -> False, BoxRatios -> {1, 1, 1}, ViewAngle -> Automatic, ViewCenter -> {0.5`, 0.5`, 0.5`}, ViewMatrix -> Automatic, ViewPoint -> {2.5075250998820904`, -1.6385753275514336`, \ -1.5739722899089035`}, ViewProjection -> Automatic, ViewRange -> All, ViewVector -> Automatic, ViewVertical -> {-0.34286271463298007`, 0.30614344718399633`, -0.8880998528661083`}]
-
The explanation for the fact that the function behaves well along the $x$-axis is the existence of the following limit
\[
\lim_{y\to 0} \frac{\sin(xy)}{y} = x \lim_{y\to 0} \frac{\sin(xy)}{xy} = x \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = x \quad \text{for all} \quad x \in \mathbb{R}.
\]
The above limit is a consequence of the most important trigonometric limit
\[
\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1.
\]
The preceding limit is a motivation for the definition of the following special function
\[
\operatorname{sinc}(x) = \begin{cases}
\dfrac{\sin(x)}{x} & \quad \text{if} \quad x \in \mathbb{R}\setminus\{0\}, \\
1 & \quad \text{if} \quad x = 0.
\end{cases}
\]
The function $\operatorname{sinc}(x)$ is usually not studied in undergraduate math classes. However, it appears in many engineering applications. Look it up on Wikipedia. It is important to emphasise that the function $\operatorname{sinc}(x)$ is smooth on $\mathbb{R}$, that is, it has continuous derivatives of all orders. Below is a graph of $\operatorname{sinc}(x)$.
The Mathematica code for the above picture:
Plot[Sinc[x], {x, -3 Pi - 3.4, 3 Pi + 3.4}, PlotStyle -> {{Red, Thickness[0.004]}}, PlotRange -> {{-3 Pi - 1.4, 3 Pi + 1.4}, {-0.7, 1.4}}, Ticks -> {Range[-5 Pi, 5 Pi, Pi/2], Range[-2, 2, 0.5]}, AspectRatio -> Automatic, Background -> Black, ImageSize -> 800]
The lime curves in the picture above are sine curves with different amplitudes and frequencies: in the formula $x\operatorname{sinc}(xy)$, we keep $y$ fixed and $x$ varies. The exception is $y=0$. In this case, the lime curve is a straight line. In some sense, this picture indicates that the lime straight line is a sine that forgot to oscillate.
The Mathematica code for the above picture:Plot3D[x Sinc[y x], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, PlotStyle -> {White, Opacity[0.25]}, Exclusions -> None, PlotPoints -> {151, 151}, Mesh -> {0, 75}, MeshStyle -> {{RGBColor[0.196, 0.804, 0.196], Thickness[0.0015]}}, ClippingStyle -> False, PlotRange -> {-2 Pi, 2 Pi}, ImageSize -> 800, Axes -> False, Boxed -> False, BoxRatios -> {1, 1, 1}, Background -> Black, ViewAngle -> Automatic, ViewCenter -> {0.5`, 0.5`, 0.5`}, ViewMatrix -> Automatic, ViewPoint -> {2.5075250998820904`, -1.6385753275514336`, \ -1.5739722899089035`}, ViewProjection -> Automatic, ViewRange -> All, ViewVector -> Automatic, ViewVertical -> {-0.34286271463298007`, 0.30614344718399633`, -0.8880998528661083`}]
The red curves in the picture above are graphs of sinc functions with different amplitudes and frequencies: in the formula $x\operatorname{sinc}(xy)$, we keep $x$ fixed and $y$ varies.
The Mathematica code for the above picture:Plot3D[x Sinc[y x], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, PlotStyle -> {White, Opacity[0.25]}, Exclusions -> None, PlotPoints -> {151, 151}, Mesh -> {75, 0}, MeshStyle -> {{RGBColor[{1, 0, 0}], Thickness[0.0015]}}, ClippingStyle -> False, PlotRange -> {-2 Pi, 2 Pi}, ImageSize -> 800, Axes -> False, Boxed -> False, BoxRatios -> {1, 1, 1}, Background -> Black, ViewAngle -> Automatic, ViewCenter -> {0.5`, 0.5`, 0.5`}, ViewMatrix -> Automatic, ViewPoint -> {2.5075250998820904`, -1.6385753275514336`, \ -1.5739722899089035`}, ViewProjection -> Automatic, ViewRange -> All, ViewVector -> Automatic, ViewVertical -> {-0.34286271463298007`, 0.30614344718399633`, -0.8880998528661083`}]
In the picture above, I show both sines and sincs together, lime and red together.
The Mathematica code for the above picture:Plot3D[x Sinc[y x], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, PlotStyle -> {White, Opacity[0.25]}, Exclusions -> None, PlotPoints -> {151, 151}, Mesh -> {75, 75}, MeshStyle -> {{RGBColor[{1, 0, 0}], Thickness[0.002]}, {RGBColor[0.196, 0.804, 0.196], Thickness[0.002]}}, ClippingStyle -> False, PlotRange -> {-2 Pi, 2 Pi}, ImageSize -> 800, Axes -> False, Boxed -> False, BoxRatios -> {1, 1, 1}, Background -> Black, ViewAngle -> Automatic, ViewCenter -> {0.5`, 0.5`, 0.5`}, ViewMatrix -> Automatic, ViewPoint -> {2.5075250998820904`, -1.6385753275514336`, \ -1.5739722899089035`}, ViewProjection -> Automatic, ViewRange -> All, ViewVector -> Automatic, ViewVertical -> {-0.34286271463298007`, 0.30614344718399633`, -0.8880998528661083`}]
- I wrote a summary of what we will need about Homogeneous Second Order Linear Differential Equations. The functions $\cosh x$ and $\sinh x$ appear in this summary in Problem 7 and in Remark 9.
- Today we discussed Fourier's Method of Separation of Variables. The relevant sections of the textbook is 2.3.1, 2.3.2, 2.3.3, 2.3.4.
- We considered the homogeneous heat equation in a one-dimensional rod of length $L$ subject to the Dirichlet boundary conditions: \[ \frac{\partial u}{\partial t}(x,t) = \kappa \frac{\partial^2 u}{\partial x^2}(x,t) \qquad x \in [0,L], \quad t \geq 0, \] subject to the Boundary Conditions: \[ u(0,t) = 0 \quad \text{and} \quad u(L,t) = 0 \quad \text{for all} \quad t \geq 0. \] Notice that in the heat equation $\frac{\partial u}{\partial t} = \kappa \frac{\partial^2 u}{\partial x^2}$ I write $\kappa$, the Greek lower case letter kappa, instead of $k$ in the book. This constant is called thermal diffusivity. It is a complicated constant given by \[ \kappa = \frac{K_0}{c \, \rho}, \] where $K_0$ is the thermal conductivity constant from Fourier's Law, $c$ is the specific heat and $\rho$ is the mass density of the material. I do not like $k$ since $k$ is commonly used as a counting index in series, and we will be using infinite series all the time in the rest of the quarter.
- The brilliant idea to find solutions of the problem stated in the preceding item is to look for a solution in which the independent variables are separated in two functions $A$ and $B$: \[ u(x,t) = A(x) \mkern 1mu B(t). \] The brilliancy of the above idea is that one unpleasant unknown function $u(x,t)$ of two independent variables is replaced by two unknown functions $A(x)$ and $B(t)$, but each one of which is a function of a single independent variable. There is no guarantee that this idea will work. But, it is brilliant and amazing since it works, as we will see later.
-
Although we did not finish the process until the end, I will state the final solutions for $A(x)$ and $B(t)$: \begin{align*} B(t) & = \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right), \\ A(x) & = \sin\left(\frac{m \pi}{L} x\right), \end{align*} where $\kappa$ and $L$ are given positive real numbers and $m$ is any positive integer.
Thus, we have obtained the sequence of solutions for $u(x,t)$ given by \[ u(x,t) = \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right) \sin\left(\frac{m \pi}{L} x\right) \quad \text{where} \quad m \in \mathbb{N}. \] We use $\mathbb{N}$ to denote the set of all positive integers. A good exercise is to verify that the preceding formula for $u(x,t)$ indeed solves the heat equation with the Dirichlet boundary conditions, the problem stated above.
- To help you with Problem 3 on Assignment 1, today in class i created a problem with a similar setting but, instead of a rod it involves an annulus. So, it requires us to work with polar coordinates.
-
Problem 1.
Consider the heat equation for a uniform annulus between the two concentric circles, both circles are centered at the origin. The inner circle has the radius $1$ and the outer circle has the radius $b$ with $b \gt 1$. All other relevant physical constants are equal to $1$. We assume that the temperature $w(r,t)$ is circularly symmetric (that is the temperature is constant along the circles centered at the origin). The heat equation in this case is
\begin{equation*}
\frac{\partial w}{\partial t}(r,t) = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,t) \right) + r \quad \text{subject to} \quad \frac{\partial w}{\partial r}(1,t)=5, \quad\frac{\partial w}{\partial r}(b,t)=-4.
\end{equation*}
Notice that there are internal sources $Q(r) = r$ of thermal energy present in the annulus.
- For what value of $b \gt 1$ the given partial differential equation with the boundary conditions has an equilibrium temperature distribution?
- What is the heat energy generated per second in the annulus by the internal sources of heat energy? Use this result and the boundary conditions to support your answer in i.
- Suppose that the initial temperature distribution is $w(r,0) = f(r)$. Find the total thermal energy in the rod as a function of time. Use the answer obtained here to support your answer in i.
- Solution. Part i. An equilibrium solution of the given partial differential equation must satisfy \[ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,t) \right) + r = 0. \] Solving this equation using the methods of integral calculus we obtain \[ w(r) = -\frac{1}{9} r^3 + c_1 \ln(r) + c_2. \] Next we need to determine the constants $c_1$ and $c_2$ such that \[ \frac{\partial w}{\partial r}(1)=5, \quad\frac{\partial w}{\partial r}(b)=-4. \] Thus we need to solve \[ -\frac{1}{3} 1^2 + c_1 \frac{1}{1} = 5, \quad -\frac{1}{3} b^2 + c_1 \frac{1}{b} = -4. \] Thus, $c_1 = 16/3,$ and $b$ must satisfy \[ -\frac{1}{3} b^2 + \frac{16}{3b} = -4, \] that is \[ b^3 - 12 b -16 = 0. \] Since \[ b^3 - 12 b -16 =(b-4)(b+2)(b+2), \] we deduce that for $b=4$, the given problem has an equilibrium solution.
-
Part ii. The internal sources in the annulus are given by $Q(r) = r$ where $1 \leq r \leq b$. The units of this function are $\displaystyle \frac{\operatorname{cal}}{\operatorname{cm}^3 \operatorname{sec}}.$ To get the total heat energy generated we need to integrate the function $Q(r)$ over the annulus. Since the height of the annulus is $1$, the total heat energy generated in the annulus is \[ \int_{0}^{2\pi} \int_1^b r \mkern 1mu r \mkern 1mu dr \mkern 1mu d\theta = \frac{2}{3}\bigl(b^3 - 1\bigr)\mkern 1mu \pi. \]
The given boundary conditions tell us about the heat energy that leaves the annulus. Based on Fourier's Law the flux along the inner circle is $-5$ in units $\frac{\operatorname{cal}}{\operatorname{cm}^2 \operatorname{sec}}.$ To get the total heat that leaves the annulus along the inner circle we multiply $5$ with the surface area $2\pi$, getting $10\pi$. The flux along the outer circle is $4$ in units $\frac{\operatorname{cal}}{\operatorname{cm}^2 \operatorname{sec}}.$ To get the total heat that leaves the annulus along the outer circle we multiply $4$ with the surface area $2b\pi$, getting $8 b \pi$. Thus the total amount of the heat energy that leaves the annulus is $10\pi+8 b \pi$.
For the annulus to have the equilibrium temperature we must have \[ \frac{2}{3}\bigl(b^3 - 1\bigr)\mkern 1mu \pi = 10\pi+8 b \pi. \] The last equation simplifies to \[ b^3 - 12 b - 16 = 0, \] the same equation we obtained in i.
- Part iii. This part is a little tricky. It follows the reasoning presented in the derivation of the heat equation. Denote by $E(t)$ the amount of the heat energy in the annulus at time $t$. By the definition of the heat energy it depends on the temperature, the mass density and the specific heat. Since in this problem the mass density and the specific heat are bothe equal to $1$, the heat energy of the annulus is \[ E(t) = \int_0^{2\pi} \int_0^b w(r,t)\mkern 1mu r \mkern 1mu dr d\theta = 2\pi \int_0^b w(r,t)\mkern 1mu r \mkern 1mu dr, \] with the initial condition \[ E(0) = \int_0^{2\pi} \int_0^b f(r)\mkern 1mu r \mkern 1mu dr d\theta = 2 \pi \int_0^b f(r)\mkern 1mu r \mkern 1mu dr. \] The above formula for $E(t)$ is not very useful, but, we can differentiate with respect to $t$ and get \[ E'(t) = \frac{d}{dt} 2\pi \int_0^b w(r,t)\mkern 1mu r \mkern 1mu dr = 2\pi \int_0^b \frac{\partial w}{\partial t}(r,t)\mkern 1mu r \mkern 1mu dr \] Now we recall the given equation \begin{equation*} \frac{\partial w}{\partial t}(r,t) = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,t) \right) + r, \end{equation*} and substitute in the integral expression for $E'(t)$: \begin{align*} E'(t) & = 2\pi \int_0^b \frac{\partial w}{\partial t}(r,t)\mkern 1mu r \mkern 1mu dr \\ & = 2\pi \int_0^b \left( \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,t) \right) + r \right)\mkern 1mu r \mkern 1mu dr \\ & = 2\pi \int_0^b \left( \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,t) \right) + r^2 \right)\mkern 1mu dr \\ & = 2\pi \int_0^b \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,t) \right)\mkern 1mu dr + 2\pi \int_0^b r^2 \mkern 1mu dr \\ & = 2\pi \left. \left( r \frac{\partial w}{\partial r}(r,t) \right) \right|_{0}^b + \left. \frac{2\pi}{3} r^3 \right|_{0}^b \\ & = 2\pi \left( b \frac{\partial w}{\partial r}(b,t) - 1 \frac{\partial w}{\partial r}(1,t) \right) + \frac{2\pi}{3}\bigl(b^3 - 1\bigr) \\ \text{here we use the given BCs} & = 2\pi \bigl( b (-4) - 5 \bigr) + \frac{2\pi}{3}\bigl(b^3 - 1\bigr) \\ & = 2\pi \left( -4 b - 5 + \frac{1}{3} \bigl(b^3 - 1\bigr) \right) \\ & = \frac{2\pi}{3} \bigl( b^3 -12 b - 16\bigr) \end{align*} Thus \[ E'(t) = \frac{2\pi}{3} \bigl( b^3 -12 b - 16\bigr), \] Hence \[ E(t) = \frac{2\pi}{3} \bigl( b^3 -12 b - 16\bigr) t + E(0) = \frac{2\pi}{3} \bigl( b^3 -12 b - 16\bigr) t + 2\pi \int_0^b f(r) r dr. \] Thus, the heat energy in the annulus will be constant, which corresponds to the equilibrium solution, if the coefficient with $t$ in the formula for $E(t)$ is $0$, that is, \[ b^3 -12 b - 16 = 0, \] the same equation as in i.
- Yesterday we briefly derived the Heat Equation in three space dimensions. This is Section 1.5 in the book. The Heat Equation in three dimensions is \[ \rho \mkern 2mu c \mkern 2mu \frac{\partial u}{\partial t} = K_0 \mkern 0mu \left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}\right). \] Here $\rho, c, K_0$ are physical constants, $u(x,y,z,t)$ is a function of three space variables and time, and \[ \Delta u = \nabla^2 u = \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2} \] is the Laplacian in orthogonal coordinates.
- The Heat Equation in two dimensions is \[ \rho \mkern 2mu c \mkern 2mu \frac{\partial u}{\partial t} = K_0 \mkern 0mu \left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\right). \] Here $\rho, c, K_0$ are physical constants, $u(x,y,t)$ is a function of two space variables and time, and \[ (\Delta u)(x,y) = \bigl(\nabla^2\bigr) u (x,y) = \frac{\partial^2u}{\partial x^2}(x,y)+\frac{\partial^2u}{\partial y^2}(x,y) \] is the Laplacian in orthogonal coordinates.
- In two and three dimensions we have options for the choice of a coordinate system. An interesting feature here is that the expression for the Laplacian in different coordinate systems has different forms.
-
The webpage The Laplacian in Polar Coordinates contains a derivation of the Laplacian in polar coordinates: \[ \bigl(\nabla^2 w \bigr) (r,\theta) = \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,\theta) \right). \] Here $w(r,\theta)$ is a function of $r \in (0,+\infty)$ and $\theta \in [0, 2\pi)$.
For example, consider the following function in the orthogonal coordinate system $xy$, \[ u(x,y) = x y\bigl(x^2 + y^2\bigr) \quad x,y \in \mathbb{R}. \] One can think of this function as giving a temperature at each point in the $xy$-plane. So, at the point $P$ whose coordinates in $xy$ orthogonal coordinate system are $(2,2)$ the temperature is $32.$ The same temperature in polar coordinates is given by the function \[ w(r,\theta) = r^4 (\cos \theta)(\sin \theta) \quad r \in [0,+\infty), \quad \theta \in [0, 2 \pi). \] The same point $P$ from above has coordinates $r = 2\sqrt{2},$ $\theta = \pi/4.$ So, the temperature at the point $P$ is \[ w\bigl(2\sqrt{2}, \pi/4 \bigr) = 2^4 2^2 (\sqrt{2}/2) (\sqrt{2}/2) = 32. \] The Laplacian of the function $u(x,y)$ is \[ (\nabla^2 u)(x,u) = \frac{\partial^2 u}{\partial x^2}(x,y) + \frac{\partial^2 u}{\partial y^2}(x,y) = 12 x y. \] The Laplacian of the function $w(r,\theta)$ is \[ (\nabla^2 w)(r,\theta) = \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r}(r,\theta) \right) = 12 r^2 (\cos \theta)(\sin \theta). \] As expected, we have \[ 12xy \quad \text{in polar coordinates is} \quad 12 r^2 (\cos \theta)(\sin \theta). \]
- The Laplacian in spherical coordinates: \[ \Delta w = \nabla^2 w = {1 \over \rho^2} {\frac{\partial}{\partial \rho}} \left(\rho^2 {\frac{\partial w}{\partial \rho}} \right) + {\frac{1}{\rho^2 \sin \theta}} {\frac{\partial}{\partial \theta}} \left( (\sin \theta) {\frac{\partial w}{\partial \theta}} \right) + {1 \over \rho^2 (\sin \theta)^2} {\partial^2 w \over \partial \phi^2}. \] Here $w(\theta,\phi,\rho)$ is a function of $\rho \in (0,+\infty)$, $\theta \in [0, 2\pi),$ and $\phi \in [0,\pi]$.
- Problem 1. Let $a$ and $b$ be real numbers such that $a \lt b$. Determine the equilibrium temperature distribution inside a circular annulus \[ \bigl\{(r,\theta) \in (0,+\infty)\times[0,2\pi) : a \leq r \leq b \bigr\} \] if the outer radius is insulated and the inner radius is kept at the constant temperature $T_a$. Assume that the all other relevant physical constants are equal to $1$.
- For Problem 5 on Assignment 1 I ask you to illustrate the problem with a Mathematica plot. In that plot you need to graph a function of two variables and emphasize the boundary of that function. In Mathematica notebook Surface_and_Boundary.nb you can find sample plots of surfaces and their boundaries.
- This is a general remark about file names that I use. Instead of the blank space in a file name, I like using the underscore. Often my filename contain the underscore _. So, the name of the file linked above is Surface_and_Boundary.nb
-
One plot created in the above Mathematica file.
-
The plot below is the solution of Problem 4 on Assignmnt 1 with somewhat changed boundary data: the inner circle has radius $1$ and the temperature is constant $1$ along this circle, and the outer circle has radius $4$ and the temperature is constant $4$ along this circle.
Or, with a different viewpoint
-
Today we discussed the equilibrium solution for the heat equation. This is Section 1.4 in the book. There are six related problems posted yesterday. Below I solve Problem 2 posted yesterday. Problem 3 on Assignment 1 belongs to this circle of questions.
-
Problem 2.
Consider the equilibrium temperature distribution for a uniform one-dimensional rod with sources $Q/K_0 = x^2$ of thermal energy, subject to the boundary conditions $u(0) = 0$ and $u(2) = 0$.
- Determine the heat energy generated per unit time inside the entire rod.
- Determine the heat energy flowing out of the rod per unit time at the endpoints $0$ and $2$.
- What relationships should exist between the answers in parts a. and b.?
- Below is a solution of this problem.
- Recall the heat equation in a uniform rod, that is the specific heat \(c\), the mass density \(\rho\), the cross sectional area \(A\) and the conductivity \(K_0\) are all independent of position: \[ c \rho \frac{\partial u}{\partial t}(x,t) = K_0 \frac{\partial^2 u}{\partial x^2}(x,t) + Q(x), \qquad 0 \leq x \leq 2, \qquad t \geq 0. \] The boundary conditions are \[ u(0,t) = 0 \quad u(2,t) = 0 \qquad \text{for all} \quad t \geq 0. \] The units of the left-hand side of the equation are \[ \dfrac{\operatorname{cal}}{\operatorname{cm}^3 \cdot \operatorname{sec}} . \] So the units of the right-hand side are the same. In particular, the units of \(Q(x)\) are the same.
- Since we are looking for the equilibrium temperature distribution, we must find \(u(x,t)\) which does not depend on \(t.\) That is we must find \(u(x,t)\) for which \[ \frac{\partial u}{\partial t}(x,t) = 0 \qquad \text{for all} \quad x \in [0,2], \quad t \geq 0. \] Hence our PDE simplifies to \[ K_0 \frac{\partial^2 u}{\partial x^2}(x,t) + Q(x) = 0. \] It is given that \(Q(x) = K_0 x^2\), so we need to solve \[ K_0 \frac{\partial^2 u}{\partial x^2}(x,t) + K_0 x^2 = 0 \qquad u(0) = 0, \quad u(2) = 0, \] or, since $K_0 \gt 0$ simplified by dividing by $K_0$: \[ \frac{\partial^2 u}{\partial x^2}(x,t) + x^2 = 0 \qquad u(0) = 0, \quad u(2) = 0. \] The solution of the preceding differential equation together with the boundary conditions is \[ u(x) = - \frac{1}{12}x^4 + \frac{2}{3} x. \] This is the equilibrium temperature distribution. The heat energy of the rod is \[ c \rho A \int_0^2 \left( - \frac{1}{12}x^4 + \frac{2}{3} x \right) dx = c \rho A \frac{4}{5}. \]
- As we calculated in the preceding item, the heat energy of the rod is constant. So the change in time of the heat energy is \(0.\) However, due to the sources of the heat energy in the rod, the heat energy is constantly created at position \(x\) at the rate \[ K_0 x^2 \ \dfrac{\operatorname{cal}}{\operatorname{cm}^3 \cdot \operatorname{sec}}. \] Therefore, the total heat energy created in the rod (per one second) is \[ A \int_0^2 K_0 x^2 dx = A K_0 \frac{8}{3} \ \dfrac{\operatorname{cal}}{\operatorname{sec}}. \]
- So, the above amount of heat energy is created in the rod each second and the rod is staying at the constant total heat energy. How is this possible?
- The answer is the flux of the heat energy at the end-points of the rod. Based on the Fourier Law the flux of the heat energy is \[ \phi(0,t) = - A K_0 \frac{\partial u}{\partial x}(0,t), \qquad \phi(2,t) = - A K_0 \frac{\partial u}{\partial x}(2,t). \] Now calculate \[ \frac{\partial u}{\partial x}(0,t) = \frac{2}{3}, \qquad \frac{\partial u}{\partial x}(2,t) = -2 \] Thus the flux is \[ \phi(0,t) = - \frac{2}{3} A K_0 , \qquad \phi(2,t) = 2 A K_0. \] Recall that the total heat energy leaving the rod per second at both end-points is the following sum \[ -\phi(0,t) + \phi(2,t). \] Hence, the total heat energy leaving the rod per second at both end-points is \[ \frac{2}{3} A K_0 + 2 A K_0 = \frac{8}{3} A K_0 \ \dfrac{\operatorname{cal}}{\operatorname{sec}}. \]
- Now compare the the total heat energy created in the rod (per one second) \[ A \int_0^2 K_0 x^2 dx = A K_0 \frac{8}{3} \ \dfrac{\operatorname{cal}}{\operatorname{sec}} \] and the total heat energy leaving the rod per second at both end-points \[ \frac{2}{3} A K_0 + 2 A K_0 = \frac{8}{3} A K_0 \ \dfrac{\operatorname{cal}}{\operatorname{sec}}. \] Since the total amount of the heat energy in the rod remains constant, the last two numbers must be equal.
-
Problem 2.
Consider the equilibrium temperature distribution for a uniform one-dimensional rod with sources $Q/K_0 = x^2$ of thermal energy, subject to the boundary conditions $u(0) = 0$ and $u(2) = 0$.
- Tomorrow we will discuss the equilibrium solution for the heat equation. This is Section 1.4 in the book.
-
Below are several problems related to this topic.
-
Problem 1.
Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sources and boundary conditions:
- $Q = 0$, $u(0) = T$, $\frac{\partial u}{\partial x}(L) = \alpha$.
- $Q(x) = K_0$, $u(0) = T_1$, $u(L) = T_2$.
- $Q(x) = K_0 x^2$, $u(0) = T$, $\frac{\partial u}{\partial x}(L) = 0$.
-
Problem 2.
Consider the equilibrium temperature distribution for a uniform one-dimensional rod with sources $Q(x) = K_0 x^2$ of thermal energy, subject to the boundary conditions $u(0) = 0$ and $u(2) = 0$.
- Determine the heat energy generated per unit time inside the entire rod.
- Determine the heat energy flowing out of the rod per unit time at the endpoints $0$ and $2$.
- What relationships should exist between the answers in parts a. and b.?
- Problem 3. Let $\alpha$ be a real number. Consider a one-dimensional rod positioned over the interval $[0,2]$ that is composed of a uniform material with all relevant physical constants equal to $1.$ We assume that there are sources of heat energy in the rod given by $Q(x) = x-\alpha$. Assume that the endpoints of the rod are insulated. Find a condition on $\alpha$ which guarantees the existence of the equilibrium temperature distribution for this rod. Find a formula for equilibrium solution(s).
- Problem 4. Consider a one-dimensional rod positioned over the interval $[0,2]$ that is composed of two different materials. We assume that the different materials are in perfect thermal contact at the point $1$. For $x\in [0,1]$ all relevant physical constants are equal to $1$ and there is a variable source $Q(x) = x$ with $x\in [0,1].$ For $x\in [1,2]$ the specific heat of the material is $c = 1$, the mass density is $\rho = 1$ and the Fourier's constant is $K_0 = 3.$ We assume that there are no sources in this part of the rod. The boundary conditions are given to be $\frac{\partial u}{\partial x}(0) = 0$ and $u(2) = 0.$ Find the equilibrium temperature distribution for this rod if it exists, or explain why it cannot exist. There is a quick way of verifying that you solution is correct. What is it?
-
Problem 5.
Suppose the concentration $u(x,t)$ of a dye satisfies Fick's law with the constant $k$ and the initial concentration is given $u(x,0) = f(x)$. Consider a region $0 \leq x \leq L$ in which the flow is specified at both ends: $-k\frac{\partial u}{\partial x}(0,t) = \alpha$ and $-k\frac{\partial u}{\partial x}(L,t) = \beta$. Assume that $\alpha$ and $\beta$ are constants.
- Express the conservation law for the entire region.
- Determine the total amount of chemical in the region as a function of time (using the initial condition).
- Under which conditions is there an equilibrium chemical concentration and what is it?
- Problem 6. Do the preceding exercise if $\alpha$ and $\beta$ are functions of time.
-
Problem 1.
Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sources and boundary conditions:
- The derivation of the Heat Equation naturally leads to the concept of Boundary Conditions for the Heat Equation. This is Section 1.3 in the book. Among several other boundary conditions (see the Summary in Section 1.3), we introduced so called periodic boundary conditions which correspond to viewing a heated rod of length $L$ that is wrapped in a circle, so that points corresponding to $x = 0$ and $x = L$ coincide. We also need to assume that the physical parameters of the rod are the same at $0$ and $L$, that is $c(0) = c(L)$, $\rho(0) = \rho(L)$ and $K_0(0)= K_0(L).$ In this case the boundary conditions for the temperature become \[ u(0,t) = u(L,t), \quad \frac{\partial u}{\partial x}(0,t) = \frac{\partial u}{\partial x}(L,t) \quad \text{for all} \quad t\geq 0. \]
-
For homework do the following exercise:
- Problem 1. Consider a bath of volume $V$ containing a fluid of specific heat $c_b$ and mass density $\rho_b$ that surrounds the endpoint at $L$ of a one-dimensional rod. Suppose that the bath is ideally stirred such that the bath temperature is uniform throughout, equaling the temperature at $u(L, t).$ Assume that the bath is thermally insulated except at its perfect thermal contact with the rod, where the bath may be heated or cooled by the rod. Determine an equation for the temperature in the bath. (This will be a boundary condition at the end x = L.) (Hint: Apply the conservation of heat energy for the bath.) (This is 1.3.3 in the book.)
- There is a very important question asked in Discussions on Canvas about Problem 2 on Assignmnt 1.
- Today we derived the Heat Equation. This is Section 1.2 in the book. Understanding the derivation of the Diffusion Equation helps in understanding the derivation of the Heat Equation. Notice that both derivations result in the same partial differential equation.
-
For homework do the following exercise:
-
Problem 1.
Two one-dimensional rods of different materials joined at $x_0$
are said to be in perfect thermal contact if the following two conditions are satisfied:
(1) the temperature is continuous at $x_0$, that is, the one-sided limits
\[
\lim_{x\uparrow x_0} u(x, t) = \lim_{x\downarrow x_0} u(x, t)
\]
coincide for all $t \geq 0$, and (2) no heat energy is lost at $x_0$, that is, all the heat energy
flowing out of one rod flows into the other.
- What mathematical equation represents condition (2) at $x_0$?
- Under what special condition is the partial derivative $\displaystyle \frac{\partial u}{\partial x}(x,t)$ continuous at $x_0$? (This is 1.3.2 in the book.)
-
Problem 1.
Two one-dimensional rods of different materials joined at $x_0$
are said to be in perfect thermal contact if the following two conditions are satisfied:
(1) the temperature is continuous at $x_0$, that is, the one-sided limits
\[
\lim_{x\uparrow x_0} u(x, t) = \lim_{x\downarrow x_0} u(x, t)
\]
coincide for all $t \geq 0$, and (2) no heat energy is lost at $x_0$, that is, all the heat energy
flowing out of one rod flows into the other.
-
In class, you requested a sample LaTeX file showing how to include pictures in a LaTeX file. I posted this sample file and accompanying files in Discussions on Canvas. For convenience, I am posting them here as well.
- This is a general remark about file names that I use. Instead of the blank space in a file name, I like using the underscore. Often my filename contain the underscore _. So, the names of the files linked below are Sample_Picture_in_LaTeX.pdf, Sample_Picture_in_LaTeX.TeX
- Mathematica notebook that I used to make the pictures. Here you can see how to export Mathematica pictures as files of your choice, PNG, PDF, JPG, SVG, and many more PiecewiseFunctions3D.nb
- PDF file produced by the LaTeX file linked in the next item Sample_Picture_in_LaTeX.pdf
- LaTeX sample file explaining how to include pictures in a LaTeX file Sample_Picture_in_LaTeX.TeX
- Three pictures that are used in the LaTeX file linke in the preceding item myPlotfh.png myPlotfh.pdf myPlotfh.jpg
- I moved the HTML record of my thinking about Burgers' equation from the Method of Characteristics webpage to its dedicated webpage Burgers' equation. Smaller pages load faster and are easier to read.
- Mathematica notebook PiecewiseFunctions3D.nb. I also created PiecewiseFunctions3D.pdf and PiecewiseFunctions3D.html
-
The Mathematica notebook posted in the preceding item is partly written by ChatGPT with Mathematica plugin. Here is the original GhatGPT written notebook PiecewiseFunctions.nb.
-
If you want ChatGPT to generate a Mathematica notebook for you, then you need to ask a question like this: "Can you provide me with Wolfram Language code that, when executed in Mathematica, will generate a notebook with [specific content or topic]?" ChatGPT will generate code that will create the content that you requested. I did this in OpenAI's GPT-4, but the free version of OpenAI's GPT-3.5 generated answer would likely provide a similar answer, but the exact phrasing, depth, or structure might differ slightly.
Notice that ChatGPT generated code will produce a notebook in the directory $UserDocumentsDirectory. Mathematica will tell you this directory if you execute (CTRL+Ener) $UserDocumentsDirectory in an input cell.
To use the ChatGPT code you need to create a Mathematica notebook with the code given by ChatGPT. I have done it in this Mathematica notebook Test_GPT.nb. There is only one command in this notebook. You just execute that command and you will find a notebook created in your $UserDocumentsDirectory.
- I updated my presentation of the solution of Burgers' equation. For the context, see The Method of Characteristics webpage. I added a calculation of the maximum domain of the solution with a specific initial condition.
- Today we discussed the derivation of the Diffusion Equation. The derivation of the Heat Equation is similar.
- In the textbook, Problems 1.2.4 and 1.2.5 relate to the derivation of the diffusion equation. Section 1.2 presents the derivation of the heat equation.
- Yesterday we discussed Burgers' equation. I wrote about it on the page Burgers' equation. For the context, see The Method of Characteristics webpage.
- On Monday and yesterday in class I created a small Mathematica notebook 20231009.nb with basic illustrations related to Burgers' equation. A more extensive Mathematica notebook with Burgers' equation explorations in the link in the preceding item.
- Yesterday I posted on Canvas that Problem 13 and Problem 14 on The Method of Characteristics page will be on the first assignment.
- Today we discussed special partial differential equations of the first-order which can be solved using The Method of Characteristics.
- Example 1, Example 2, and Example 3 at The Method of Characteristics page are completely solved. Then, there are 12 problems of different level of difficulty. Try any or all of these problems. Post questions at the class Canvas page in Discussions.
-
One more example. Consider the following linear nonhomogeneous first-order PDE
\begin{equation*}
u_x + y^2 u_y = y^2 + 2 y\mkern 2mu u \quad \text{in} \quad \mathbb R^2
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(0,y) = y, \quad \text{for all} \quad y \in \mathbb R.
\end{equation*}
- The characteristic equations are: \begin{equation*} \begin{array}{l} \dfrac{dX}{ds} = 1 \\ \dfrac{dY}{ds} = Y^2 \\ \dfrac{dZ}{ds} = y^2 + 2 y\mkern 2mu Z \end{array} \end{equation*} subject to the initial conditions \begin{equation*} \begin{array}{l} X(0) = 0 \\ Y(0) = \xi \\ Z(0) = \xi \end{array} \end{equation*}
- The solution of the above system is \begin{equation*} \begin{array}{l} X(s,\xi) = s \\ Y(s,\xi) = \dfrac{\xi }{1-\xi s} \\ Z(s,\xi) = \dfrac{\xi (1 +\xi s)}{(1-\xi s)^2} \end{array} \end{equation*}
- Now solve the system \begin{equation*} \begin{array}{l} x = s \\ y = \dfrac{\xi }{1-\xi s} \end{array} \end{equation*} for $s$ and $\xi$. Substitute $x$ for $s$ in the second equation and solve for $\xi$ \begin{equation*} \begin{array}{l} s = x \\ \xi = \dfrac{y}{x y+1}. \end{array} \end{equation*}
- Finally, substitute the formulas for $s$ and $xi$ into $Z(s,\xi) = \dfrac{\xi (1 +\xi s)}{(1-\xi s)^2}$ to get \[ u(x,y) =\dfrac{\dfrac{y}{x y+1} \left(1 +\dfrac{y}{x y+1} x \right)}{\left(1-\dfrac{y}{x y+1} y \right)^2} = y (1 + 2 x y) \] Clearly the domain of $u$ is the entire plane ${\mathbb R}^2$.
- It might be prudent to verify whether the function $u(x,y)$ found in the preceding item is really a solution. Calculate \begin{align*} u_x(x,y) + y^2 u_y(x,y) & = 2 y^2 +y^2 \bigl( 1 + 4 x y \bigr) \\ & = 2 y ^2 + y^2 + 4 x y^3 \\ & = y^2 + 2 y \bigl( y + 2 x y^2\bigr) \\ & = y^2 + 2 y u(x,y) \end{align*}
- I need to write a good narrative for the whiteboard pictures below. Today was the first time I told this story in a class. It fascinated me when I saw it for the first time as an undergraduate student almost a half-century ago.
- I did not take a picture of the first white-board with the scalar initial value problem: Let $a\in \mathbb{R}\setminus\{0\}$. Find the unique solution of the Initial Value Problem: \[ y'(t) = a y(t), \qquad y(0) = 1. \]
-
In the picture below, we moved on to the vector Initial Value Problem, with a little sad face, and then to the matrix Initial Value Problem. On this whiteboard, $A$ is an $n\times n$ matrix, $\overrightarrow{y}(t)$ is an $n\times 1$ vector function, and $Y(t)$ is an $n\times n$ matrix function. The main point is the definition of the $n\times n$ matrix exponential function
\[
e^{At} \quad \text{for all} \quad t\in \mathbb{R}.
\]
.jpg)
-
On the left-hand side of the whiteboard below, we used the $n\times n$ matrix exponential function $e^{At}$ to solve the vector initial value problem
\[
(\overrightarrow{y})'(t) = A \overrightarrow{y}(t), \qquad \overrightarrow{y}(0) = \overrightarrow{v},
\]
where $\overrightarrow{v} \in \mathbb{R}^n$ is a given vector.
On the right-hand side of the whiteboard below we ask the main question of the day: If $D_x$ denotes differentiation with respect to $x$ on the space of differentiable functions in the variable $x \in \mathbb{R}$, can we assign a meaning to the expression
\[
e^{t D_x} \quad \text{for all} \quad t\in \mathbb{R}.
\]
_to_exp(tDx).jpg)
-
Finally, we solve a partial differential equation to explain that, for a given differentiable function $f(x)$ defined on $\mathbb{R}$ we have
\[
\bigl(e^{t D_x}\bigr) f(x) = f(x+t) \quad \text{for all} \quad x, t\in \mathbb{R}.
\]
.jpg)
The method presented on the whiteboard above is one example of the Method of Characteristics, that we will discuss in more detail tomorrow.
- Computer algebra system Mathematica will be very useful for the assignments in this class. To get started with Mathematica see my Mathematica page. A convenient way to get started is to watch the videos which are linked on my Mathematica page. Watching the movies is very helpful to get started with Mathematica efficiently! Mathematica is available in the computer labs in BH 215 and BH 209.
- During the first class I provide an overview of the course. As a guide to an overview I like to offer my explanations of the meaning of the words in the course title from the catalog. Today I offered the following explanation of the word equation: An equation is a mathematical expression which involves the equality sign and in which one component is considered unknown and the remaining components are considered known. To solve an equation means to find a formula for the unknown component in terms of the known quantities.
- Notice that in the explanation in the preceding item I use the concept of mathematical expression without explaining the meaning. As the Wikipedia page shows, an explanation of the concept mathematical expression is not simple.
- Today I offered the following definition of a Partial Differential Equation: A partial differential equation is an equation in which the unknown quantity is a function of several variables, and the equation involves this function, its partial derivatives, and other terms, called coefficients, which are specific or considered known. The coefficients can be constants or functions in the same independent variables as the unknown function.
- Tomorrow we will discuss special partial differential equations of the first order which can be solved using The Method of Characteristics. Reading the webpage The Method of Characteristics you will encounter several Partial Differential Equations. This pages also contains many ordinary differential equations.
- My presentation of the Method of Characteristics differs from the presentation in the textbook. For comparison read Section 12.2.2 Method of Characteristics in the textbook. Suggested problems are 2, 3, 4, 5, 6, 7, 8 in Exercises 12.2, see Chapter 12 from the textbook.
- The information sheet.
- We will start the class by studying the Method of characteristics for quasi-linear first order partial differential equations.
- My presentation of the Method of Characteristics differs from the presentation in the textbook. For comparison read Section 12.2.2 Method of Characteristics in the textbook. Suggested problems are 2, 3, 4, 5, 6, 7, 8 in Exercises 12.2, see Chapter 12 from the textbook.
- We will continue with the heat (or diffusion) equation. A derivation of the diffusion equation is here.
The animation below is obtained by solving that equation.
Place the cursor over the image to see the diffusion of the dye.
- Here is the complete list of all the webpages that I created for this class:
- Computer algebra system Mathematica will be very useful for the assignments in this class. You can start getting familiar with it. To get started with Mathematica see my Mathematica page. Please watch the videos that are on my Mathematica page. Watching the movies is very helpful to get started with Mathematica efficiently! Mathematica is available in the computer labs in BH 215 and BH 209.
- My one-page on the Power of Math.
- My one-page on Rigorous Reasoning.