- I post a pdf file of the notes that I wrote today during the class. The topics were from Sections 7.1, 7.2 and 7.3 of the textbook.
- I post a pdf file of the notes that I wrote today during the class. We discussed Problem 2 on Assignment 2 and Problem 5 on Assignment 3 (a very interesting problem, I am very proud for creating it). I saved Problem 2 on Assignment 2 in a separate file.
- I post a pdf file of the notes that I wrote today during the class.
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 5.8.
- The notebook Super_glued_ends_solution_v12.nb contains a Mathematica implementation of finding a solution of the vibrating string equation with a boundary condition of the third kind. This 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
- If you have problems accessing this Mathematica notebook on K:\Math\Curgus\430_Files of Dropbox\430_Files you can download it:
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 5.8.
-
Today we solved "on paper" the vibrating string equation with the boundary conditions describing a string with a rigid end as in the animation below. in this animation the parameter $h=1$. The boundary condition at the endpoint $0$ is
\[
\frac{\partial u}{\partial x}(0,t) - 1 \cdot u(0,t) = 0.
\]
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.
- The notebook Super_glued_ends_solution_v12.nb contains a Mathematica implementation of finding a solution of the vibrating string equation with a boundary condition of the third kind. This 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
- If you have problems accessing this Mathematica notebook on K:\Math\Curgus\430_Files of Dropbox\430_Files you can download it:
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 5.3 and the beginning of Section 5.8.
- I wrote a webpage about some of the theory of The Symmetry of the Sturm-Liouville Eigenvalue Problem. On this page I present the general form of the symmetric boundary conditions.
-
Towards the end of the class we started solving the vibrating string equation with the boundary conditions describing a string with a rigid end as in the animation below.
Place the cursor over the image to start vibrations.
Notice that the red part of the string is rigid, while the blue part is governed by the vibrating string equation.
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 5.2.
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 4.4.
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 4.3.
-
Animations below illustrate my interpretation of the boundary conditions that involve both the function and its derivative. To motivate these boundary conditions the book used the massless spring-string system. These animation illustrate that such boundary conditions can be interpreted as one portion of the string being rigid (for example soaked in super-glue).
Place the cursor over the image to start vibrations.
Notice that the red part of the string is rigid, while the blue part is governed by the vibrating string equation.
-
Sometimes the boundary conditions illustrated below are called "non-physical boundary conditions." This name might be coming from the fact that the string breaks after a while.
Place the cursor over the image to start vibrations.
In the above animation one end of a blue string is attached to an end a rigid red bar which is free to move up and down. The other end of the bar is fixed. Mathematically the red bar establishes a relation between the position of the end of the spring and the slope of the string at that end.
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The setting shown in this item is the same as in the previous animation. However, the string does not break since we carefully chosen our initial conditions. We will be explaining what this means for the rest of the quarter.
Place the cursor over the image to start vibrations.
-
The setting shown below is also termed "non-physical" in the book. However, in this case for an arbitrary initial shape of the string the string keeps oscillating without breaking.
Place the cursor over the image to start vibrations.
-
An interesting situation arises when a part of a string is rigid (dipped in super-glue, for example).
Place the cursor over the image to start vibrations.
Notice that the red part of the string is rigid, while the blue part is governed by the vibrating string equation.
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 3.6 and 4.2.
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 3.5.
-
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 $[-L,L],$
- $\displaystyle a_0 = \frac{1}{2L} \int_{-L}^{L} f(\xi) d\xi = 0,$
- $\displaystyle F(x) = \int_{0}^{x} f(\xi) d\xi.$
To toggle the proof of the Integration term-by-term Theorem clickSince 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 b_n}{k \pi} \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 a_n}{k \pi} \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 $[-L,L],$
- $\displaystyle a_0 = \frac{1}{2L} \int_{-L}^{L} f(\xi) d\xi,$
- $\displaystyle F(x) = \int_{0}^{x} f(\xi) d\xi.$
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 3.4.
-
We first prove a Differentiation term-by-term 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 $[-L,L],$
To toggle the proof of the Differentiation term-by-term Theorem clickSince $f'$ is piecewise smooth its Fourier series converges pointwise to the Fourier periodic extension of $f'.$ Denote the coefficients of this Fourier series by $A_0,$ $A_k$ and $B_k.$ Let us calculate $A_0.$ Since $f$ is continuous and $f'$ is piecewise continuous the Fundamental Theorem of Calculus applies and it yields \[ A_0 = \frac{1}{2L} \int_{-L}^L f'(x) dx = \frac{1}{2L} \bigl( f(L) - f(-L) \bigr). \] 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*} A_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*} B_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 term-by-term Theorem for Fourier sine series.
Differentiation term-by-term Theorem. Let $L \gt 0$ and let $f:[0,L] \to \mathbb R$ be a function. Assume that
- $f$ is continuous on the closed interval $[0,L],$
- $f$ is piecewise smooth function on $[0,L],$
- $f'$ is piecewise smooth function on $[0,L].$
- 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 an even extension of $\sin(x)$ and similar.
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Section 3.3.
- In this Mathematica notebook v12 and similar Mathematica notebook v8 I did some basic series in Mathematica, just to check calculations. Here is the pdf printout of the version 8 file.
- I post a pdf file of the notes that I wrote today during the class. The relevant part from the textbook is Sections 3.1 and 3.2.
- Our next topic is to find basic Fourier series "by hand". In this Mathematica notebook v12 and similar Mathematica notebook v8 I did some basic series in Mathematica, just to check calculations. Here is the pdf printout of the version 8 file. I needed to do many small modifications for v12, so the last three Examples might not work properly in v12 file.
-
Yesterday we defined the concept of a periodic extension of a function. Today we will introduce a modification of this concept which I call the Fourier periodic extension of a piecewise continuous function. First recall the definitions of periodic extensions.
- 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 \[ \tilde{f}(x) = f\left(\!x- \Bigl(\left\lceil\!\tfrac{x-b}{b-a}\! \right\rceil \Bigr)(b-a)\!\right), \quad x \in {\mathbb R}. \] 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 \[ \tilde{f}(x) = f\left(\!x- \left\lfloor\!\tfrac{x-a}{b-a}\! \right\rfloor (b-a)\!\right), \quad x \in {\mathbb R}. \] is called the periodic extension of $f.$
-
Next we illustrate this concept on a simple example.Consider the restriction of the function $x \to x$ to the interval $[1,4).$ A detailed graph of this function is below.
In the figure below the function $f$ is the restriction of the function $x \mapsto x$ (in blue) to the interval $[1,4).$
-
It is clear that the above function has a periodic extension with period $3.$ Below I show the given function in blue and its periodic extension in red. Since the periodic extension has jumps, I emphasize its values at the jumps.
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This brings us to a new concept: the Fourier periodic extension of a piecewise continuous function. It is not difficult to see that a periodic extension of a piecewise continuous function is itself piecewise continuous. The Fourier periodic extension of a piecewise continuous function is a modification of the periodic extension. The modification is done at the jumps: the value of the Fourier periodic extension at a jump $x$ of $\tilde{f}$ is
\[
\frac{1}{2} \bigl(\tilde{f}(x+)+\tilde{f}(x-)\bigr)
\]
The concept of the Fourier periodic extension is not explicitly defined in our textbook. In fact most textbooks on this topic do not define this concept, although it is implicitly present in each. Since there is no standard notation for the Fourier periodic extension, I will denote it by $\tilde{f}_{\!\!\rm Fourier}$. Here is a formal definition:
- Definition. Let $a$ and $b$ be real numbers such that $a \lt b$ and let $f:[a,b) \to \mathbb R$ be a piecewise continuous function. Let $\tilde{f}:{\mathbb R} \to {\mathbb R}$ be the periodic extension of $f$. The Fourier periodic extension of $f$ is the following function \[ \tilde{f}_{\!\!\rm Fourier}(x) = \begin{cases} \tilde{f}(x) & \text{if $\tilde{f}$ is continuous at $x$} \\[10pt] \tfrac{1}{2}\!\! \bigl(\tilde{f}(x+)+\tilde{f}(x-)\bigr) & \text{if $\tilde{f}$ is not continuous at $x$} \end{cases}, \qquad x\in\mathbb R. \] Here \[ \tilde{f}(x+) = \lim_{\xi\downarrow x}\tilde{f}(\xi), \quad \tilde{f}(x-) = \lim_{\xi\uparrow x}\tilde{f}(\xi), \] are one-sided limits of $\tilde{f}$ at $x.$
-
Here is the Fourier periodic extension of the blue function above:
-
One more example. The given function is in blue, its periodic extension in red and its Fourier periodic extension in green.
- You can find several examples of piecewise smooth functions, their periodic extensions and their Fourier periodic extensions on this webpage.
-
Next I present two important convergence theorems for Fourier series.
- 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 $k \in {\mathbb N}$, \[ a_k = \frac{1}{L}\int_{-L}^L f(\xi) \cos\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \quad b_k = \frac{1}{L}\int_{-L}^L f(\xi) \sin\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \] 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 \[ \bigl|S_n^f(x) - \tilde{f}(x) \bigr| < \varepsilon. \]
-
The book does not have formal definitions of the concepts that are studied in Chapter 3. Therefore, I give few formal definitions here.
-
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 \[ \lim_{x\downarrow a} f(x), \quad \lim_{x\uparrow x_k} f(x), \quad \lim_{x\downarrow x_k} f(x), \ k \in \{1,\ldots,n\}, \quad \lim_{x\uparrow b} f(x). \]
-
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 \[ \tilde{f}(x) = f\left(\!x- \Bigl(\left\lceil\!\tfrac{x-b}{b-a}\! \right\rceil \Bigr)(b-a)\!\right), \quad x \in {\mathbb R}. \] 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 \[ \tilde{f}(x) = f\left(\!x- \left\lfloor\!\tfrac{x-a}{b-a}\! \right\rfloor (b-a)\!\right), \quad x \in {\mathbb R}. \] 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 \[ x - \left\lfloor\!\tfrac{x-a}{b-a} \right\rfloor(b-a) \in [a,b). \]
-
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:
-
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 the fault of our curriculum. 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 (version 8) and
Mathematica notebook (version 12) 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.
- I post a pdf file of the notes that I wrote today during the class. The relevant section from the textbook is Sections 2.5.2 and 2.5.4.
- I post a pdf file of the notes that I wrote today during the class. We worked on solving the Laplace's Equation in a disk. The relevant section from the textbook is Section 2.5.2.
- The Mathematica notebook Equilibrium_temperature_2D_disk_v12.nb contains a Mathematica implementation of finding a solution to Laplace's equation in a disk with some specific examples. You can find this notebook in K-Drive\Math\Curgus\430_Files\ and in Dropbox\430_Files\. A pdf printout of this notebook is Equilibrium_temperature_2D_disk_v12.pdf. You can also download this notebook Equilibrium_temperature_2D_disk_v12.nb
- I have placed four Mathematica notebooks in K-Drive\Math\Curgus\430_Files\ and in the directory Dropbox\430_Files\ that I shared with you on Dropbox.
- 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. In Problem 1 on Assignment 2 I ask you to reproduce that animation. This notebook might be useful for creating that animation. A pdf printout of this notebook is Coloring_Functions_v12.pdf
- The notebook Equilibrium_temperature_2D_v12.nb contains a Mathematica implementation of finding a solution to Laplace's equation in a rectangle with some specific examples. In Problem 3 on Assignment 2 I ask you to improve that is in this notebook by using the method presented in Problem 5 on Assignment 1. This notebook might be useful for Problem 3 on Assignment 2. A pdf printout of this notebook is Equilibrium_temperature_2D_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. A pdf printout of this notebook is Heat_eq_3BCs_v12.pdf
- The notebook Heat_dbc_v12.nb contains a Mathematica implementation of finding a solution of the heat equation under Dirichlet boundary conditions. The Mathematica command that I use in this notebook is somewhat different from what is in the notebook in the preceding item. A pdf printout of this notebook is Heat_dbc_v12.pdf
- If you have problems accessing Mathematica notebooks on K:\Math\Curgus\430_Files you can download them:
- I post a pdf file of the notes that I wrote today during the class. We worked on solving the Laplace's Equation on a rectangle. The relevant section from the textbook is Section 2.5.
- I wrote a webpage The Laplace equation on a Rectangle with my approach to this problem.
- I will post more about new Mathematica files on K-drive tomorrow.
- I post a pdf file of the notes that I wrote today during the class. We worked on solving the heat equation subject to Periodic Boundary Conditions. The relevant section from the textbook is Section 2.4.
- 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 1.7 and in Remark 1.9.
- I post a pdf file of the notes that I wrote today during the class. We worked on solving the heat equation subject to Neumann Boundary Conditions and Periodic Boundary Conditions. The relevant section from the textbook is Section 2.4.
- I post a pdf file of the notes that I wrote today during the class. Basically it was how to solve the heat equation with Dirichlet boundary conditions and an initial condition.
- Today I continued to write the notes in the notebook that I started yesterday. The main message of today is: As a part of the implementation of the Separation of Variables Method we are solving the an eigenvalue problem for the second order ordinary differential operator with boundary conditions. Today we found the eigenvalues to be $\lambda_m = \bigl(m \pi /L\bigr)^2$ with $m\in \mathbb{N}$ and with the corresponding eigenfunctions $\sin\bigl(x m \pi /L\bigr).$ We just started talking about the orthogonality of these eigenfunctions.
- I post a pdf file of the notes that I wrote today during the class. The main message of today is: As a part of the implementation of the Separation of Variables Method we are solving the an eigenvalue problem for the second order ordinary differential operator with boundary conditions. This is like doing linear algebra with a differential operator instead of a matrix.
- I post a pdf file of the notes that I wrote today during the class. Today we did Fourier's Method of Separation of Variables. We obtained an important sequence of solutions of the homogeneous heat equation in a one-dimensional rod of length $L$ subject to 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: \[ u(0,t) = 0 \quad \text{and} \quad u(L,t) = 0 \quad \text{for all} \quad t \geq 0. \] The important sequence of special solutions is \[ \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right) \sin\left(\frac{m \pi}{L} x\right) \quad \text{where} \quad m \in \mathbb{N}. \] We use $\mathbb{N}$ to denote the set of all positive integers. 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 kappa letter, 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.
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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 $t \geq 0.$
- I post a pdf file of the notes that I wrote today during the class. Today we did the preparations for the Separation of Variables Method.
- Inspired by your question I came up with a general definition of a "characteristic equation": When we reduce solving a hard mathematical equation to solving of a significantly simpler equation, then we call the simpler equation the characteristic equation for the harder equation.
- I post a pdf file of the notes that I wrote today during the class. Today we did the derivation of the Laplacian in polar coordinates.
- This derivation is presented on my webpage The Laplacian in polar coordinates The formula for the Laplacian in polar coordinates is \[ \nabla^2 w = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}. \] Here $w$ is a function of $r \in [0,+\infty)$, and $\theta \in [0, 2\pi)$.
- The formula for the Laplacian in spherical coordinates is more complicated: \[ \nabla^2 w = {1 \over \rho^2} {\frac{\partial}{\partial \rho}} \left(\rho^2 {\frac{\partial w}{\partial \rho}} \right) + {\frac{1}{\rho^2 \sin \theta}} {\frac{\partial}{\partial \theta}} \left( (\sin \theta) {\frac{\partial w}{\partial \theta}} \right) + {1 \over \rho^2 (\sin \theta)^2} {\partial^2 w \over \partial \varphi^2}. \] Here $w$ is a function of $\rho \in [0,+\infty)$, $\theta \in [0, 2\pi)$ and $\varphi \in [0,\pi]$.
- The formula for the Laplacian in cylindrical coordinates is just a variation on the formula for the Laplacian in polar coordinates: \[ \nabla^2 w = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2} + \frac{\partial^2 w}{\partial z^2}. \] Here $w$ is a function of $r \in [0,+\infty)$, $\theta \in [0, 2\pi)$ and $z \in \mathbb R$.
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For homework do the following two problems:
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In the three items that follow below, we consider three equilibrium problems in three different dimensions. In each item, the physical constants of the uniform material involved are equal to $1$ and there are constant sources of heat energy that equal to $1.$
- Consider a one-dimensional rod placed over the interval $[-1,1]$ with heat sources equal to $1$ throughout the rod. The temperature at the boundary points of the rod is kept at the temperature $0.$ Find the equilibrium temperature distribution of this rod. State specifically what is the temperature at the midpoint of the rod.
- Consider a two-dimensional disk of radius $1.$ Assume that there are heat sources equal to $1$ throughout the disk. The temperature at the circular boundary of the disk is kept at the temperature $0.$ Find the equilibrium temperature distribution of this disk. State specifically what is the temperature at the center of the disk.
- Consider a three-dimensional ball of radius $1.$ Assume that there are heat sources equal to $1$ throughout the ball. The temperature at the spherical boundary of the ball is kept at the temperature $0.$ Find the equilibrium temperature distribution of this sphere. State specifically what is the temperature at the center of the sphere.
- It is probably just an interesting coincidence that there is a simple relationship between the temperature at the center and the dimension: the product of the dimension and the temperature at the center equals $1/2$.
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As in the preceding problem, in the three items that follow below, we consider three equilibrium problems in three different dimensions. In each item, the physical constants of the uniform material involved are equal to $1$ and there are sources of heat energy. The unifying theme is that the heat sources are concentrated near the center with the radius $s \in (0,1]$ and the intensity of the heat sources increases as $s$ becomes smaller. For $s=1$ this problem reduces to the preceding problem. (This problem is inspired by Zac Pontrantolfi's mention of the Dirac delta function the other day. In the items below we simulate the Dirac delta function as describing a heat source and study the resulting equilibrium temperature distribution.)
- Let $s\in(0,1].$ Consider a one-dimensional rod placed over the interval $[-1,1]$ with heat sources equal to $s^{-1}$ in the interval $[-s,s].$ The temperature at the boundary points $-1$ and $1$ of the rod is kept at the temperature $0.$ Find the formula for the equilibrium temperature distribution in this rod. Denote this temperature by $T_{1,s}(x)$ where $x\in [-1,1].$ State specifically what the temperature at the midpoint of the rod is and how it changes with $s,$ that is describe $T_{1,s}(0).$
- Consider a two-dimensional disk of radius $1$ and let $s\in(0,1].$ Assume that there are heat sources equal to $s^{-2}$ in the concentric disk of radius $s.$ The temperature at the circular boundary of the disk (that is along the unit circle) is kept at the temperature $0.$ Find the formula for the equilibrium temperature distribution in this disk. Denote this temperature by $T_{2,s}(r)$ where $r\in [0,1].$ State specifically what the temperature at the center of the disk is and how it changes with $s,$ that is describe $T_{2,s}(0).$
- Consider a three-dimensional ball of radius $1$ and let $s\in(0,1].$ Assume that there are heat sources equal to $s^{-3}$ in the concentric ball of radius $s.$ The temperature at the spherical boundary of the ball (that is over the unit sphere) is kept at the temperature $0.$ Find the formula for the equilibrium temperature distribution in this sphere. Denote this temperature by $T_{3,s}(\rho)$ where $\rho\in [0,1].$ State specifically what the temperature at the center of the sphere is and how it changes with $s,$ that is describe $T_{3,s}(0).$
- Describe the differences and similarities in the behaviour of the equilibrium temperature distribution in these three cases as $s\in (0,1]$ changes. Do you notice the special behaviour of the functions \[ \left(\frac{d T_{1,s}}{dx}\right)(1), \quad \left(\frac{d T_{2,s}}{dr}\right)(1), \quad \left(\frac{dT_{3,s}}{d\rho}\right) (1)? \] How do you explain this behaviour?
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In the three items that follow below, we consider three equilibrium problems in three different dimensions. In each item, the physical constants of the uniform material involved are equal to $1$ and there are constant sources of heat energy that equal to $1.$
- Today we discussed the equilibrium solution for the heat equation. This is Section 1.4 in the book. I post a pdf file of the notes that I wrote today during the class. The theme was an equilibrium solution of the heat equation.
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For homework do the following two exercises:
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Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sources and boundary conditions:
- $Q = 0$, $u(0) = T$, $\frac{\partial u}{\partial x}(L) = \alpha$.
- $\frac{Q}{K_0} = 1$, $u(0) = T_1$, $u(L) = T_2$.
- $\frac{Q}{K_0} = x^2$, $u(0) = T$, $\frac{\partial u}{\partial x}(L) = 0$.
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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.?
- 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).
- 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?
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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?
- Do the preceding exercise if $\alpha$ and $\beta$ are functions of time.
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Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sources and boundary conditions:
- Today we discussed the boundary conditions for the heat equation. This is Section 1.3 in the book. I post a pdf file of the notes that I wrote today during the class. In these notes I introduce 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 \[ u(0,t) = u(L,t), \quad \frac{\partial u}{\partial x}(0,t) = \frac{\partial u}{\partial x}(L,t) \quad \text{for all} \quad t\geq 0. \]
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For homework do the following two exercises:
- Two one-dimensional rods of different materials joined at $x_0$ are said to be in perfect thermal contact if the temperature is continuous at $x_0$ if the following conditions are satisfied: $u(x0−, t) = u(x0+, t)$ and no heat energy is lost at $x_0$, that is, the heat energy flowing out of one flows into the other. What mathematical equation represents the latter condition at $x_0$? Under what special condition is $(\partial u)/(\partial x)$ continuous at $x_0$? (This is 1.3.2 in the book.)
- 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 approximately 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 hear energy for the bath.) (This is 1.3.3 in the book.)
- Last week we derived the diffusion equation. We also derived the diffusion equation in 3 dimensions. The 3rd edition derives only the heat equation in Section 1.2. The 4th and the 5th edition derive the diffusion equation at the end of Section 1.2.
- Today we finished the derivation of the heat equation. This is Section 1.2 in the book.
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For homework do the following two exercises:
- Consider a rod of variable cross-sectional area $A(x), 0 \lt x \lt L$. Assume that all thermal quantities are constant across a section, the rod is made of a uniform material and that the rod is well insulated so that no heat energy can pass through the lateral surface. Assume no sources of heat energy in the rod. Derive the heat equation for this rod. (This is my version of Exercise 1.2.3 in the 4th and 5th ed.)
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Consider a thin one-dimensional rod of length $L$ positioned on the $x$-axis with the specific heat $c(x)$, mass density $\rho(x)$, Fourier coefficient $K_0(x)$ and without sources of thermal energy, but whose lateral surface area is not insulated. Assume that the cross-sectional area of the rod is independent of $x$ with the perimeter $P$ and the cross-sectional area $A$.
- Assume that the heat energy flowing out of the lateral sides per unit surface area per unit time is $w(x, t).$ Derive the partial differential equation for the temperature $u(x, t).$
- Assume that the heat energy flux $w(x, t)$ is governed by Newton's law of cooling. That is, $w(x, t)$ is proportional to the temperature difference between the rod temperature $u(x, t)$ and a known outside temperature $b(x, t)$ with the constant of proportionality $h(x).$ Derive the partial differential equation for $u(x,t):$ \begin{equation} \tag{*}\label{pde1} c(x) \rho(x) \frac{\partial u}{\partial t}(x,t) = \frac{\partial}{\partial x} \left( K_0(x) \frac{\partial u}{\partial x}(x,t) \right) - \frac{P}{A}h(x) \bigl( u(x,t) - b(x,t) \bigr). \end{equation}
- Write the partial differential equation $\eqref{pde1}$ from the preceding item for a rod of circular cross section with the radius $r$, with constant thermal properties and $0^\circ$C outside temperature.
- Consider the rod as in the preceding item and assume that the temperature of the rod is uniform along the rod. That is, $u(x,t)$ is a constant function of $x$. Determine a formula for $u(t)$ if the initial temperature of the rod is given to be $u(0) = u_0.$
- I post a pdf file of the notes that I wrote today during the class.
- I post a pdf file of the notes that I wrote today during the class. Today we did the derivation of the Diffusion equation in three dimensions.
- At the end of the class Garrett asked me a question about Problem 12 on the Method of characteristics webpage. There are some notes related to Garrett's question at the end of today's notes. Garrett asked: What is the domain of the solution of Problem 12. Look carefully at the domain of the solution indicated by the characteristics that are found by solving the characteristic equation you would thing that the domain is \[ \bigl\{ (x,y) \in \mathbb{R}^2 \, : \, y \gt 0 \bigr\}. \] However, one can find the closed form expression for the solution in the form $z = u(x,y).$ Interestingly, this function is defined on $\mathbb{R}^2.$ This example shows that the method of characteristics does not guarantee that the solution which is found by this method has in some sense maximal domain. It is also interesting to consider Problem 12 in which the initial condition is replaced by $u(x,0) = \exp(x^2)$ with $x\in \mathbb{R}$. Try the Method of Characteristics for this new problem. Reflect whether the solution of the original Problem 12 tells you something about this new problem.
- I post a pdf file of the notes that I wrote today during the class. Today we did the derivation of the Diffusion equation.
- This derivation is presented on my webpage Diffusion equation
- I post a pdf file of the notes that I wrote today during the class. Today we did two examples of the Method of Characteristics.
- I post a pdf file of the notes that I wrote today during the class. It is a discussion of the Burgers' equation that we continued in Mathematica.
- Please look for the K-drive on when you remote connect to a BH 215 computer. Now I have placed MoC_Burgers_eq1_v12.nb and MoC_Transport_eq_v12.nb in K:\Math\Curgus\430_Files.
- I printed the notebooks from the previous item as pdf files MoC_Burgers_eq1_v12.pdf and MoC_Transport_eq_v12.pdf.
- If you have problems accessing Mathematica notebooks on K:\Math\Curgus\430_Files you can download them MoC_Burgers_eq1_v12.nb and MoC_Transport_eq_v12.nb.
- I have rewritten the abstract part of this webpage: Method of characteristics. The link to a Mathematica file on this website gives you version 8 file. You can open it in version 8 rather than converting it to version 12. I am writing version 12 files in small pieces since I am having problems with version 12 on my laptop (and in general it might be a good idea to deal with smaller files).
- I post a pdf file of the notes that I wrote today during the class. It is a detailed discussion of the transport equation.
- Please look for the K-drive on when you remote connect to a BH 215 computer. Please watch the Mathematica videos that are on my Mathematica page. Watching the movies is very helpful to get started with Mathematica efficiently!
- I post a pdf file of the notes that I wrote today during the class. It is a review of the Method of Characteristics and an example.
- Today we talked about the abstract part of this file: Method of characteristics. There is a link to a Mathematica file on this website. That Mathematica file is written in version 8. You can open it in version 8 rather than converting it to version 12. Soon I will write a new file in version 12.
- I post a pdf file of the notes that I wrote today.
- Today I posted how to establish Remote access to a BH 215 workstation on Canvas.
- 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!
- We will start the class by studying the Method of characteristics for the first order partial differential equations.
- The information sheet
- To get you started I post some topics that we will study at the beginning of the class. If you need some mathematical entertainment (mathtertainment) this Summer you can engage with the webpages that I created for some topics that we will study in this class. These pages are work in progress.
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I believe in writing. I believe that improvement is possible. I feel that in my classes I am not encouraging you enough to write more and improve your writing. To promote my belief in writing and improving, inspired by "veni, vidi, vici", I wrote:
I came, I thought, I wrote, I taught, I thought more, I rewrote.
Verba volant, scripta manent.
- We will start the class by studding the Method of characteristics for the first order partial differential equations.
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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: