- The animations posted today are based on Sections 7.7 and 7.8.
- We study a circular drum of radius 1. Here is the Mathematica notebook which I used to produce the animations below. Here is a pdf file of this Mathematica notebook so that you can view it without access to Mathematica.
- In the above notebook we derived the formulas for normal modes of vibrations of a circular drum with the radius 1 and the constant $c = 1$. Below I show vibrations of 30 normal modes with $m=0,1,2,3,4,5$ and $n=1,2,3,4,5$.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=1$.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=2$.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=3$.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=4$.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=5$.
-
The initial velocity of the drum below is given by the following equation:
\[
g(r,\theta,0) = 2(1-r) \exp\bigl(-100(r-.5)^2 \bigr) \, \sin\bigl((\pi/2)\exp\bigl(-50(\theta-\pi)^2 \bigr) \bigr)
\]
Place the cursor over the image to start vibrations.
- When solving the natural modes of vibrations of the unit drum above only the Bessel functions of the first kind appear, since the Bessel functions of the second kind are not defined at $0$. However, if we consider a circular drum with a hole in the middle we need to consider both the Bessel functions of the first and of the second kind. I thought that it would be interesting to work out the details of this case in Mathematica.
- Below I study a circular drum with a hole in the middle. It is a drum of radius 3 with a hole of radius 1. I show vibrations of 30 normal modes with $m=0,1,2,3,4,5$ and $n=1,2,3,4,5$ of the drum.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=1$.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=2$.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=3$.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=4$.
- In the modes below we have $m=0,1,2,3,4,5$ and $n=5$.
-
The initial velocity of the drum below is given by the following polynomial equation:
\[
g(r,\theta,0) = (r-1)(3-r) \exp\bigl(-20(r-2)^2 \bigr) \, \sin\bigl((\pi/2)\exp\bigl(-30(\theta-\pi)^2 \bigr) \bigr)
\]
Place the cursor over the image to start vibrations.
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- What we discussed today is based on Sections 7.1, 7.2 and 7.3.
- In this Mathematica notebook I illustrate vibrations of a rectangular membrane. Three animations that follow below are produced using this notebook. One needs to be patient with this notebook. Manipulations with complicated 3D functions do not perform satisfactory. Therefore, I exported Mathematica images into animated gifs which show more realistic vibrations.
-
The initial shape of the membrane below is given by the following polynomial equation:
\[
u(x,y,0) = 10 x y (1 - x) (1 - y)^2, \quad 0\leq x \leq 1, \quad 0\leq y \leq 1.
\]
Place the cursor over the image to start vibrations.
-
The initial shape of the membrane below is given by a slightly different polynomial equation:
\[
u(x,y,0) = 10 x y (1 - x)^2 (1 - y)^2, \quad 0\leq x \leq 1, \quad 0\leq y \leq 1.
\]
Place the cursor over the image to start vibrations.
-
The initial velocity is given to the membrane below; in fact the membrane has been hit from below by a baton.
Place the cursor over the image to start vibrations.
-
Above I posted membranes whose initial shapes or velocities were not symmetric. I thought that would result in interesting unpredictable vibrations. However, vibrations of membranes with symmetric initial shape are also interesting. The initial shape of the membrane below is given by the following polynomial expression:
\[
u(x,y,0) = \bigl(16 x y (1 - x) (1 - y)\bigr)^{10}, \quad 0\leq x \leq 1, \quad 0\leq y \leq 1.
\]
Place the cursor over the image to start vibrations.
-
Here is a guide for Chapter 5:
- Read Section 5.1 and Section 5.2.
- Read Section 5.3. Pay special attention to Boundary conditions and 5.3.2: Regular Sturm-Liouville problems, in particular Statements of theorems (skip Rayleigh quotient). Read 5.3.3.
- Skip Section 5.4.
- We covered Section 5.5 in detail. Pay special attention to the integral form of Lagrange's identity (book calls it Green's formula). My approach to this part, see Symmetry of Sturm-Liouville eigenvalue problems, differs from the approach in the book. Notice that I will give a simple way of answering 5.5.1. Suggested problems: 5.5.1 and 5.5.4.
- Read Appendix to 5.5. This was covered in Math 304. It is good to read it for comparison.
- Skip Section 5.6 and Section 5.7.
- We will cover Section 5.8 in detail. Recommended problems: 5.8.1, 5.8.2, 5.8.3 in (a) ignore the comment in parenthesis and ignore question (c), 5.8.4, 5.8.5, 5.8.6, 5.8.8 solve (a) directly, 5.8.9, 5.8.11.
- Skip Section 5.9 and Section 5.10.
- I wrote a webpage on Symmetry of Sturm-Liouville eigenvalue problems
-
Let $L \gt 0$ 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 \geq 0 \\
&\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 \geq 0, \\
&\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,\pi].
\end{alignat*}
On Friday and today 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:
- 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 first illustration with $L =\pi$ and $h=-1$ posted on Saturday 20, 2021. For a very special choice of $f(x)$ and $g(x)= 0$, see the second illustration with $L =\pi$ and $h=1,$ we show a special $f(x)$ for which the spring does not break.
- 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. We illustrate this with $L = \pi$ and $h=-\pi$ in the next item. For a general choice of the initial velocity $g(x)$ the string will break. We did not illustrate this case.
- 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 with $L =\pi$ and $h=1$ and the animation with $L =\pi$ and $h=-4$ posted on Saturday 20, 2021.
-
Below is the animation of a string of length $\pi$ and with a Robin boundary condition with $h = -\pi$, that is,
\[
u(0,t) + \pi \frac{\partial u}{\partial x}(0,t) = 0 \quad \text{for all} \quad t \geq 0.
\]
Place the cursor over the image to start vibrations.
- What I presented on Friday is essential for what we will do for the rest of the quarter. So, I wrote the notes of what we did on Friday. The relevant part from the textbook is Section 5.8.
-
On Friday we did the first step towards the solving the vibrating string equation with the boundary conditions describing 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.
-
This is my interpretation of, what the book calls in Section 5.8, the "non-physical boundary conditions of the third kind". However, below I present a physical situation in which such boundary conditions occur.
The boundary condition at the endpoint $0$ in the animation below is
\[
u(0,t) + 1 \frac{\partial u}{\partial x}(0,t) = 0.
\]
As you can see in the animation below the string breaks in this case.
Place the cursor over the image to start vibrations.
In the above animation one end of the orange string is attached to an end of 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-point of the spring and the slope of the string at that end.
-
The setting shown in the next animation is the same as in the preceding one. This string does not break since the initial shape of the string is orthogonal to the eigenfunctions corresponding to the negative eigenvalue.
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 there are no negative eigenvalues. Consequently, for an arbitrary initial shape of the string the string keeps oscillating without breaking. The boundary condition at the endpoint $0$ is
\[
u(0,t) + 4 \frac{\partial u}{\partial x}(0,t) = 0.
\]
Place the cursor over the image to start vibrations.
- 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. If you shared your Dropbox folder with me, you can also find this notebook in Dropbox\430_Files. 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 is here Super_glued_ends_solution_v12.nb
-
Simple string with fixed endpoints set in motion by initial displacement only, no initial velocity.
Place the cursor over the image to start vibrations.
-
Below I present the first six natural modes of vibrations of a string with fixed endpoints.
Place the cursor over the image to start vibrations.
- The above animations are created in the Mathematica notebook linked below If you shared your Dropbox folder with me, you can also find this notebook in Dropbox\430_Files.
- This is in Sections 4.4 in the book.
- On Friday and today we covered vibrating string equation. Read Sections 4.1, 4.2, and 4.3 in the textbook, do the problems Section 4.2: 4.2.1, 4.2.2, Section 4.3: 4.3.1, 4.3.2.
-
Animations below illustrate my interpretation of the Robin boundary conditions (in the book these are called "third kind" bcs) which 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.
-
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.
-
We discussed further complex Fourier series. The cornerstone in the theory of complex Fourier series is
-
Seeing $e$, I could not resist talking about wonders of $e$. I remembered how I got inspired to create the $e$-shirt:
- We talked about complex Fourier series; the relevant section of the book is Section 3.6.
-
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 $[0,L].$
- 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 \[ a_k = \frac{2 (-1)^k \sinh(1)}{1+(k \pi)^2}, \quad b_k = -\frac{2 (-1)^k k \pi \sinh(1) }{1+(k \pi)^2} \quad \text{for all} \quad k\in\mathbb{N}. \] In class we confirmed this by plotting the Fourier periodic extension of $\exp(x)$ on [-1,1]$ and its approximation by its Fourier series in this is a small notebook (version 12), which we used before to define the periodic extension of a function in Mathematica.
-
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 $[-L,L],$
- $\displaystyle a_0 = \frac{1}{2L} \int_{-L}^{L} f(\xi) d\xi,$
- $\displaystyle F(x) = \int_{0}^{x} f(\xi) d\xi.$
- Today we found some basic Fourier series "by hand". It is certainly fun to find Fourier Series by hand, figuring out challenging integrals, finding ways of expressing series in a most compact way. However, it is very useful to finish explorations in Mathematica. In this Mathematica notebook (version 12) I did some basic series in Mathematica, just to check calculations and illustrate the convergence with plots. Here is the pdf printout of the same notebook. And the earlier version of this notebook Mathematica notebook (version 8).
- The relevant section is Section 3.3. In this Mathematica notebook v12 I solved some exercises from Chapter 3 in Mathematica. Here is the pdf printout of the same notebook.
-
Today we discussed Fourier series of a piecewise continuous 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 $k \in {\mathbb N}$, \[ a_k = \frac{1}{L}\int_{-L}^L f(\xi) \cos\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \quad b_k = \frac{1}{L}\int_{-L}^L f(\xi) \sin\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \] 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. 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 \[ \lim_{x\downarrow a} f(x), \quad \lim_{x\uparrow x_k} f(x), \quad \lim_{x\downarrow x_k} f(x), \ k \in \{1,\ldots,n\}, \quad \lim_{x\uparrow b} f(x). \]
-
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 \[ x - \left\lfloor\!\tfrac{x-a}{b-a} \right\rfloor(b-a) \in [a,b). \]
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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 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 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.
- Read Section 3.1 in the book. This is a small notebook (version 12) that I wrote today in class.
- Today we discussed Laplace's equation in a disk. That is Section 2.5.2 in the textbook. 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. A pdf printout of this notebook is Equilibrium_temperature_2D_disk_v12.pdf. You can download this notebook here Equilibrium_temperature_2D_disk_v12.nb
- 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
- 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 immediatelly, 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_dbc_v12.nb contains a Mathematica implementation of finding a solution of the heat equation under the Dirichlet boundary conditions. The Mathematica commands that I use in this notebook are somewhat different from what is in the notebook in the item below. In this notebook I tried to make a universal function that will work for most initial conditions, but this made the function complicated. A pdf printout of this notebook is Heat_dbc_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 three Mathematica notebooks at the following links:
- Last few days we discussed 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 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: \[ u(0,t) = 0 \quad \text{and} \quad u(L,t) = 0 \quad \text{for all} \quad t \geq 0. \] 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 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.
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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 $t \geq 0.$
- 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 \[ 1, \qquad \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right) \cos\left(\frac{m \pi}{L} x\right), \qquad \exp\left(\!-\frac{m^2 \pi^2}{L^2} \kappa\, t \right) \sin\left(\frac{m \pi}{L} x\right) \quad \text{where} \quad m \in \mathbb{N}. \]
- The relevant sections of the book are 2.1, 2.2, 2.3, 2.4.
- Last week we discussed the boundary conditions for the heat equation. This is Section 1.3 in the book. 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 \[ u(0,t) = u(L,t), \quad \frac{\partial u}{\partial x}(0,t) = \frac{\partial u}{\partial x}(L,t) \quad \text{for all} \quad t\geq 0. \]
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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: $u(x_0−, t) = u(x_0+, t)$ 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 heat energy for the bath.) (This is 1.3.3 in the book.)
- We also discussed equilibrium solution for the heat equation. This is Section 1.4 in the book.
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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 finished the derivation of the heat equation. In class I gave a detailed presentation of the units for each quantity that we used. On the Heat Equation webpage I did not do that. It seems to me that following the units gives some quidence to the derivation process. 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 $u(0) = u_0.$
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I will try again to give a better summary of the Method of Characteristics (I will ignore domains. That is important issue, but at this stage it might confuse things. You need to look closer into the domains when you have the parametric equations of the characteristics.):
- We are given a vector field in $\mathbb{R}^3$: \[ (x,y,z) \mapsto \Bigl\langle A(x,y,z), B(x,y,z), C(x,y,z) \Bigr\rangle. \]
- Next we consider the corresponding quasi-linear first order partial differential equation \[ A\bigl(x,y,u(x,y)\bigr) u_x(x,y) + B\bigl(x,y,u(x,y)\bigr) u_y(x,y) = C\bigl(x,y,u(x,y)\bigr). \] This PDE is accompanied with an ``initial condition,'' for example \[ u(x,0) = f(x) \]
- The PDE from the previous item is solved by solving the system of ordinary differential equations accompanied with initial conditions \[ \begin{array}{l} \dfrac{dX}{ds}(s,\xi) = A(X,Y,Z) \\ \dfrac{dY}{ds} = B(X,Y,Z) \\ \dfrac{dZ}{ds} = C(X,Y,Z) \end{array} \] subject to the initial conditions \[ \begin{array}{l} X(0) = \xi \\ Y(0) = 0 \\ Z(0) = f(\xi) \end{array} \] These equations are called the characteristics equations for the given PDE with the given initial condition.
- The solutions of the characteristic equations form a family of parametric curves \begin{equation*} \begin{split} x &= X(s,\xi) \\ y & = Y(s,\xi) \\ z &= Z(s,\xi) \\ \end{split} \end{equation*} For a fixed $\xi$ with $s$ being the independent variable the preceding three equations represent a curve in $xyz$-space. These curves are called the characteristics of the given PDE subject to the given initial condition.
- Another set of curves that plays important role for the given PDE is the set of projected characteristics. The projected characteristics are the curves in $xy$-plane. The significance of the projected characteristics is as follows: From the equations of the characteristics, it is often "easy" to deduce how the solution $u(x,y)$ behaves along the projected characteristics.
- There is one more step to find the formula for the function $z= u(x,y)$. We need to solve the system \begin{equation*} \begin{split} x &= X(s,\xi) \\ y & = Y(s,\xi) \end{split} \end{equation*} for $s$ and $\xi$ in terms of $x$ and $y$. Solving this system will lead to two functions \begin{equation*} \begin{split} s &= S(x,y) \\ \xi & = \Xi(x,y). \end{split} \end{equation*} Now the desired function $u(x,y)$ is \[ u(x,y) = Z\bigl( S(x,y), \Xi(x,y) \bigr). \]
- We finished the derivation of the diffusion equation in one and three dimensions. The 3rd edition of the book 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; although, it makes more sense to me to derive the diffusion equation before deriving the heat equation.
- The theory of quasi-linear first order PDEs is on the website Method of characteristics. On this website there are some worked out examples and several problems to be solved.
- The Mathematica file (notebook) posted on the website Method of characteristics is for Mathematica version 8. Mathematica v8 is installed on all computers where Mathematica is installed. I was not successful in converting this v8 file to v12. I am writing smaller v12 notebooks for each example separately.
- You can download the Mathematica notebooks MoC_Transport_eq_v12.nb and MoC_Burgers_eq1_v12.nb. Then you can open these notebooks in Mathematica v12.
- I printed the notebooks from the previous item as pdf files MoC_Transport_eq_v12.pdf and MoC_Burgers_eq1_v12.pdf.
- What we discussed in class is presented on the webpage Method of characteristics for quasi-linear first order partial differential equations. As in class, I first present prerequisites and then the geometric interpretation of a quasi-linear first order PDE: \begin{equation*} A\bigl(x,y,u(x,y)\bigr) \, u_x(x,y) + B\bigl(x,y,u(x,y)\bigr) \, u_y(x,y) = C\bigl(x,y,u(x,y)\bigr) \end{equation*} subject to the initial condition, for example, \begin{equation*} u(x,0) = f(x). \end{equation*} In both displayed equations we assume that $x \in \mathbb{R}$ and $y \geq 0.$ We explained that the geometric interpretation of the given quasi-linear first order PDE is that the vectors \begin{equation*} \Bigl\langle A\bigl(x,y,u(x,y)\bigr) , B\bigl(x,y,u(x,y)\bigr) , C\bigl(x,y,u(x,y)\bigr) \Bigr\rangle \end{equation*} are tangent to the graph of the function $z = u(x,y)$, that is to the surface \begin{equation*} \Bigl\{ \bigl( \, x, \, y, \, u(x,y) \bigr) \, \bigl| \bigr. \, x \in \mathbb{R}, \, y \geq 0 \Bigr\}. \end{equation*} The only part of this surface that we know is the curve determined by the initial condition: \begin{equation*} \Bigl\{ \bigl( \, x, \, 0, \, f(x) \bigr) \, \bigl| \bigr. \, x \in \mathbb{R} \Bigr\}. \end{equation*} And we know that the surface is tangent to the vector field \begin{equation*} \Bigl\langle A\bigl(x,y,z\bigr) , B\bigl(x,y,z\bigr) , C\bigl(x,y,z\bigr) \Bigr\rangle. \end{equation*} Recalling what we know from Math 331, from each point \begin{equation*} \bigl( \, \xi, \, 0, \, f(\xi) \bigr) \quad \text{for a fixed} \quad \xi \in \mathbb{R} \end{equation*} we can find the curve \begin{equation*} \Bigl\{ \bigl( \, X(s), \, Y(s), \, Z(s) \bigr) \, \bigl| \bigr. \, s \geq 0 \Bigr\} \end{equation*} which is tangent to the given vector field, that is satisfies the system of differential equations \begin{equation*} \begin{array}{l} X'(s) = A\bigl(X(s),Y(s),Z(s) \bigr) \\ Y'(s) = B\bigl(X(s),Y(s),Z(s) \bigr) \\ Z'(s) = C\bigl(X(s),Y(s),Z(s)\bigr) \end{array} \end{equation*} subject to the initial conditions \begin{equation*} \begin{array}{l} X(0) = \xi \\ Y(0) = 0 \\ Z(0) = f(\xi) \end{array} \end{equation*} For simple functions $A,$ $B$ and $C$ this system is solvable and the solutions of this system will lead to the solution of the given quasi-linear equation.
- Since Mathematica will be very useful for the assignments, 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!
- 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. Until the books that you ordered arrive I post Chapter 12 from the textbook.
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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: