-
Below I present two animations which are the final result of solving Problem 3 on Assignment 3. The first string is set in motion by the initial displacement which can be seen in the first scene of the animation. The second string is set in motion by the initial velocity which equal to the initial displacement of the first string.
Place the cursor over the image to start vibrations.
Place the cursor over the image to start vibrations.
-
Below I present an animation which is the final result of solving Problem 4 on Assignment 3. A string with two rigid ends is set in motion by the given initial displacement which can be seen in the first scene of the animation.
Place the cursor over the image to start vibrations.
-
Below I present two animations which are the final result of solving Problem 6 on Assignment 3. The first string is set in motion by the initial displacement which can be seen in the first scene of the animation. The second string is set in motion by the given initial velocity \(\sin\bigl(2\pi x/5\bigr)\).
Place the cursor over the image to start vibrations.
Place the cursor over the image to start vibrations.
-
Here I give a hint for Problem 6 on Assignment 3. In part (a) I ask you to establish a mathematical model for the vibrations of a string with the rigid part over the interval \([1,2]\). The string is placed over the interval \([0,5]\). The value of \(x\) in a mathematical model for the vibrations of this string should be restricted to the union \begin{equation*} [0,1] \cup [2,5]. \end{equation*} The fact that the string is rigid over \([1,2]\) should be reflected by boundary conditions involving the values of the function \(u(x,t)\) and its derivative at the points \(x=1\) and \(x=2\). Stating all the conditions carefully, you end up with four boundary conditions: one at \(x=0\), two involving \(x=1\) and \(x=2\), and one at \(x=5\).
Dealing with four boundary conditions is not easy, so to find eigenvalues and eigenfunctions I suggest that you start with the "general" solution on \([0,1] \) of the form \begin{equation*} C_1 \sin(\mu x) \end{equation*} which automatically satisfies the boundary condition at \(x=0\) and with the "general" solution on \([2,5] \) of the form \begin{equation*} C_2 \sin\bigl(\mu (5-x)\bigr) \end{equation*} which automatically satisfies the boundary condition at \(x=5\). In this setting we are left with only two remaining boundary conditions that we can use to set up a linear system for the unknowns \(C_1\) and \(C_2\) which will lead towards an equation which will determine \(\mu\)-s.
What I presented in the preceding paragraph applies only to positive eigenvalues. You need to make necessary adjustments for the negative eigenvalues and zero.
- Today we discussed how to use the natural modes of vibrations to solve the wave equation with given initial conditions. In this Mathematica notebook I illustrate the natural modes of vibration of a rectangular membrane and I illustrate vibrations of a rectangular membrane the prescribed specific initial displacement and prescribed initial velocity. 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 displacement 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 displacement 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 displacement or velocities were not symmetric. I thought that would result in interesting unpredictable vibrations. However, vibrations of membranes with symmetric initial displacement are also interesting. The initial displacement 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.
- What we discussed today is based on Sections 7.1, 7.2 and 7.3.
- In this Mathematica notebook I illustrate the natural modes of vibration of a rectangular membrane. I need to export animations of the natural modes of vibration of a rectangular membrane and post them.
- The notebook Super_glued_ends_solution_v12.nb contains a Mathematica implementation of finding a solution of the vibrating string equation with a Robin boundary condition. This Mathematica file is essential for the Mathematica part of Problem 4 on Assignment 3. A pdf printout of this notebook is Super_glued_ends_solution_v12.pdf. You can download the Mathematica notebook here Super_glued_ends_solution_v12.nb
-
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 Monday and Tuesday 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 the animation below.
Place the cursor over the image to start vibrations.
However, if we set the initial velocity $g(x)= 0$ and we make a special choice of the initial shape $f(x)$ that is orthogonal to the eigenfunction corresponding to the negative eigenvalue than the string does not break. See the animation below.
Place the cursor over the image to start vibrations.
-
Case 2. $h = -L$ In this case, $0$ is an eigenvalue and there are countably many positive eigenvalues. For a general choice of the initial shape $f(x)$ and no initial velocity, that is $g(x) = 0,$ the string will not break. This is illustrated in the animation below with $L = \pi$ and $h=-\pi$.
Place the cursor over the image to start vibrations.
For a general choice of the initial velocity $g(x)$ the string breaks. We did not illustrate this case.
-
Case 3. $h \geq 0$ or $h \leq -L$ In this case, there are no negative eigenvalues, $0$ is not an eigenvalue and there are countably many positive eigenvalues. See the animation below with $L =\pi$ and $h=-4$.
Place the cursor over the image to start vibrations.
In the animation below we used $L =\pi$ and $h=1$.
Place the cursor over the image to start vibrations.
-
- These are the complete notes of what we did on Monday and Tuesday and will continue the next week. The relevant part from the textbook is Section 5.8.
- Today we studied the boundary conditions of the "third kind" or Robin boundary conditions. I wrote the notes of what we did today and will continue tomorrow. The relevant part from the textbook is Section 5.8.
-
Today 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.
-
Below I present the first six natural modes of vibration of a string with fixed endpoints. For each picture hover the cursor over the image to start vibrations.
the first harmonic or fundamental
the second harmonic
the third harmonic
the fourth harmonic
the fifth harmonic
the sixth harmonic
-
Below is a string with fixed endpoints set in motion by initial displacement only, no initial velocity.
Place the cursor over the image to start vibrations.
-
The above animations are created in the Mathematica notebook linked below
- Natural_modes_of_vibration.nb Here is the pdf printout of the same notebook.
- This is in Sections 4.4 in the book.
- We talked about complex Fourier series; the relevant section of the book is Section 3.6.
-
The cornerstone in the theory of complex Fourier series is
-
And a note about the terminology: I prefer the phrase Euler's identity to Euler's formula used by Wikipedia. The noun "formula" is used very broadly in mathematics, while the noun "identity" according to Merriam-Webster means: an equation that is satisfied for all values of the symbols.
-
Seeing $e$, I could not resist talking about wonders of $e$. I remembered how I got inspired to create the $e$-shirt:
- I created this small Mathematica notebook in which I explain how to create animations and export them as common animated files. Here is the pdf printout of the same notebook.
-
Below is the animated gif that I produced
Below is the animated png that I produced
There is not much difference in quality. For more complicated pictures I think that png format is better.
- Unfortunately I did not find a simple way to include animated files in pdf documents. As you can see from my website, it is easy to include animated gifs and animated pngs in html files. These files show nicely in browsers.
-
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.
-
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}{k \pi} b_k \end{align*} Let us calculate $B_k$. Since $F(x)\sin\bigl(k\pi x/L\bigr)$ is continuous and $f$ is piecewise smooth the Integration by Parts Theorem applies and it yields \begin{align*} B_k & = \frac{1}{L} \int_{-L}^L F(\xi) \sin\left(\frac{k \pi}{L} \xi \right) d\xi \\ & = \frac{1}{L} \left( - \frac{L}{k\pi} F(L) \cos\bigl(k\pi\bigr) + \frac{L}{k\pi} F(-L) \cos\bigl(k\pi\bigr) + \frac{L}{k\pi} \int_{-L}^L f(\xi) \cos\left(\frac{k \pi}{L} \xi\right) d\xi \right) \\ & = \frac{L}{k \pi} a_k \end{align*} Hence, for all $x\in \mathbb{R}$ we have \[ {\widetilde F}(x) = A_0 - \frac{L}{\pi} \sum_{k=1}^{\infty} \frac{b_k}{k} \cos\left(\frac{k\pi}{L} x \right) + \frac{L}{\pi} \sum_{k=1}^{\infty} \frac{a_k}{k} \sin\left(\frac{k\pi}{L} x \right). \] In particular the preceding equality holds for $x=0.$ Since $F(0) = 0$ we have \[ 0 = A_0 - \frac{L}{\pi} \sum_{k=1}^{\infty} \frac{b_k}{k}. \] This completes the proof. - The preceding theorem states that if $a_0 = 0$ in a Fourier series of a piecewise smooth function $f,$ then the Fourier series of $F$ can be obtained by term-by-term integration of the Fourier series of $f.$ This claim is the consequence of the fact that integrating $k$-th terms we obtain \[ \int_{0}^{x} a_k \cos\left(\frac{k\pi}{L} \xi \right) d\xi = \frac{La_k}{k\pi} \sin\left(\frac{k\pi}{L} x \right) \] and \[ \int_{0}^{x} b_k \sin\left(\frac{k\pi}{L} \xi \right) d\xi = -\frac{Lb_k}{k\pi} \cos\left(\frac{k\pi}{L} x \right) + \frac{Lb_n}{k\pi}, \] which are exactly the terms of the Fourier series obtained in the preceding Theorem.
-
We can eliminate the condition that $a_0 = 0$ in the preceding theorem in the following way:
Integration term-by-term Theorem. Let $L \gt 0$ and let $f:[-L,L] \to \mathbb R$ be a function. Assume that
- $f$ is piecewise smooth function on $[-L,L],$
- $\displaystyle a_0 = \frac{1}{2L} \int_{-L}^{L} f(\xi) d\xi,$
- $\displaystyle F(x) = \int_{0}^{x} f(\xi) d\xi.$
-
Today we discussed Fourier series of a piecewise smooth function. The relevant section is
Section 3.2: the assigned problems are 3.2.1, 3.2.2, 3.2.3, 3.2.4.
- Definition. Let $L \gt 0$ and let $f:[-L, L] \to \mathbb R$ be a piecewise continuous function. The series \[ a_0 + \sum_{k=1}^{+\infty} \biggl( a_k \cos\Bigl(\!\tfrac{k \pi}{L} x\!\Bigr) + b_k \sin\Bigl(\!\tfrac{k \pi}{L} x\!\Bigr) \biggr) \] where \[ a_0 = \frac{1}{2L} \int_{-L}^L f(\xi) d\xi \] and, for $k \in {\mathbb N}$, \[ a_k = \frac{1}{L}\int_{-L}^L f(\xi) \cos\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \quad b_k = \frac{1}{L}\int_{-L}^L f(\xi) \sin\Bigl(\!\tfrac{k \pi}{L} \xi\!\Bigr) d\xi, \] 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 \[ x - \left\lfloor\!\tfrac{x-a}{b-a} \right\rfloor(b-a) \in [a,b). \]
-
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 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 that I wrote about what we did in class today.
-
An important question came up in class yesterday: Why the integrals of squares of positive integer multiples of cosines over $[-\pi,\pi]$ are all equal to $\pi$? That is
\[
\int_{-\pi}^{\pi} \bigl(\cos (n x) \bigr)^2 dx = \pi \qquad \text{whenever} \qquad n \in \mathbb{N}.
\]
Below are pictures and animations that justify this fact. In fact, there are no integrals in these
visual and colorful proofs. These questions could be explored in any trigonometry class. For example, one could print colored pictures from below and ask students to use scissors to perform a proof. That is well known method of proofs: Scissors Method. -
A trigonometry question: Given that the equation of the red curve in the picture below is
\[
y = (\cos x)^2,
\]
calculate the orange area?
-
The following picture could be regarded as a hint. Or, you can print out the picture below hand it out to trigonometry students, and ask them to use scissors to do the calculation. Colors help.
-
Another task for trigonometry students:
-
Animations, always helpful, hopefully.
-
Try larger domains.
- Today we talked further about the equilibrium temperature distribution in the unit disk. This leads to the next topic: Fourier series. This is a small notebook (version 12) that I wrote today in class. And the pdf printout of this notebook.
- Today we discussed Laplace's equation on a disk, or more general in polar coordinates friendly regions. This is Subsection 2.5.2 in the textbook.
- We applied the Method of Separation of Variables applied to the homogeneous Laplace's in a unit disk \[ \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial w}{\partial r}(r,\theta) + \frac{\partial^2 w}{\partial r^2}(r,\theta) = 0, \quad r \in [0,+\infty), \quad \theta \in [-\pi, \pi], \] subject to the Boundary Condition: \[ w(1,\theta) = f(\theta) \quad \theta \in [-\pi, \pi]. \] The separation of variables (ignoring the boundary condition) leads to the sequence of solutions \[ 1, \qquad r^n \cos (n \theta ), \quad r^n \sin (n \theta ), \quad \text{where} \quad n \in \mathbb{N}. \]
- It might be more convenient to write the Laplacian in polar coordinates in its compacted form \[ \frac{1}{r^2} \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r} (r,\theta) \right) = 0, \quad r \in [0,+\infty), \quad \theta \in [-\pi, \pi], \] or, equivalently, \[ \frac{\partial^2 w}{\partial \theta^2}(r,\theta) + r \frac{\partial}{\partial r} \left( r \frac{\partial w}{\partial r} (r,\theta) \right) = 0, \quad r \in [0,+\infty), \quad \theta \in [-\pi, \pi], \]
- It is always fun to verify that solutions that we found really satisfy this equation. For example $r^n \sin (n \theta )$: \begin{align*} \frac{\partial^2}{\partial \theta^2}\bigl(r^n \sin (n \theta )\bigr) + r \frac{\partial}{\partial r} \left( r \frac{\partial}{\partial r} \bigl(r^n \sin (n \theta )\bigr) \right) & = - n^2 r^n \sin (n \theta ) + r \frac{\partial}{\partial r}\bigl( r n r^{n-1} \sin (n \theta ) \bigr) \\ & = - n^2 r^n \sin (n \theta ) + r \frac{\partial}{\partial r}\bigl( n r^{n} \sin (n \theta ) \bigr) \\ & = - n^2 r^n \sin (n \theta ) + r n^2 r^{n-1} \sin (n \theta ) \\ & = 0. \end{align*}
- To satisfy the boundary condition \[ w(1,\theta) = f(\theta) \quad \theta \in [-\pi, \pi], \] we look for the solution in the form of a Fourier series \[ w(r,\theta) = a_0 + \sum_{n=1}^{\infty} a_n r^n \cos(n\theta) + \sum_{n=1}^{\infty} b_n r^n \sin(n \theta), \] where $a_0$, $a_n$, and $b_n$ are real numbers to be determined from the boundary condition: \[ f(\theta) = w(1,\theta) = a_0 + \sum_{n=1}^{\infty} a_n \cos(n\theta) + \sum_{n=1}^{\infty} b_n \sin(n \theta), \quad \theta \in [-\pi, \pi]. \] Next we use the fact that the functions \[ 1, \qquad \cos (n \theta ), \quad \sin (n \theta ), \quad \text{where} \quad n \in \mathbb{N}, \] are mutually orthogonal relative to the inner product \[ \bigl\langle u(\theta), v(\theta) \bigr\rangle = \int_{-\pi}^{\pi} u(\theta) v(\theta) d \theta \] and \[ \int_{-\pi}^{\pi} 1 d \theta = 2\pi, \quad \int_{-\pi}^{\pi} \bigl(\cos(n\theta)\bigr)^2 d \theta = \pi, \quad \int_{-\pi}^{\pi} \bigl(\sin(n\theta)\bigr)^2 d \theta = \pi, \] to calculate \[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\theta) d \theta, \quad a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \cos(n\theta) d \theta, \quad b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \sin(n\theta) d \theta, \quad n \in \mathbb{N}. \] With these coefficients in the above Fourier series we have the solution of the boundary value problem on the unit disk.
- This method is particularly rewarding when the boundary function $f(\theta)$ is a finite linear combination of the functions \[ 1, \qquad \cos (n \theta ), \quad \sin (n \theta ), \quad \text{where} \quad n \in \mathbb{N}. \] I listed several such functions on Saturday, October 29. Here are several more \begin{align*} (\cos x)^2 & = \frac{1}{2}+\frac{1}{2} \cos (2 x) \\ (\cos x)^4 & = \frac{3}{8} +\frac{1}{2} \cos (2 x)+\frac{1}{8} \cos (4 x)\\ (\cos x)^6 & = \frac{5}{16} + \frac{15}{32} \cos (2 x)+\frac{3}{16} \cos (4x)+\frac{1}{32} \cos (6x) \end{align*} These are trigonometric identities that can be deduced by the repeated application of the double-angle identities and angle sum and difference identities.
- Let \[ f(\theta) = 2 \bigl(\cos(\theta/2) \bigr)^6, \quad \theta \in [-\pi, \pi], \] in the boundary condition for Laplace's equation on a disk. Based on the last identity in the preceding item we have the following identity for the given boundary condition: \[ 2 \bigl(\cos(\theta/2) \bigr)^6 = \frac{5}{8}+\frac{15}{16} \cos (\theta ) +\frac{3}{8}\cos (2 \theta ) + \frac{1}{16} \cos (3 \theta ). \] From this formula we can read out the coefficients \[ a_0 = \frac{5}{8}, \quad a_1= \frac{15}{16}, \quad a_2 = \frac{3}{8}, \quad a_3 = \frac{1}{16}, \] and all other coefficients are zero. Surprisingly, from these trigonometric identities we can deduce that \begin{align*} \frac{5}{8} & = \frac{1}{2\pi} \int_{-\pi}^{\pi} 2 \bigl(\cos(\theta/2) \bigr)^6 d \theta \\ \frac{15}{16} & = \frac{1}{\pi} \int_{-\pi}^{\pi} 2 \bigl(\cos(\theta/2) \bigr)^6 \cos(\theta) d \theta \\ \frac{3}{8} & = \frac{1}{\pi} \int_{-\pi}^{\pi} 2 \bigl(\cos(\theta/2) \bigr)^6 \cos(2 \theta) d \theta \\ \frac{1}{16} & = \frac{1}{\pi} \int_{-\pi}^{\pi} 2 \bigl(\cos(\theta/2) \bigr)^6 \cos(3 \theta) d \theta \end{align*}
-
Hence, the exact solution of the boundary value problem on the unit disk with this specific boundary condition is
\[
w(r,\theta) = \frac{5}{8}+\frac{15}{16} r \cos (\theta )
+\frac{3}{8} r^2 \cos (2 \theta ) + \frac{1}{16} r^3 \cos (3 \theta ).
\]
- I updated the webpage Laplace's equation on a Rectangle with the section An Example with the exact solution written as a finite sum. This content is in Section 2.5.1 in the textbook.
- Since we work with trigonometric functions a lot, it is helpful to review beautiful world of trigonometric identities. The Wikipedia page List_of_trigonometric_identities is a good place for such a review. Among many, you can find the following formulas \begin{align*} (\sin x)^2 & = \frac{1}{2}-\frac{1}{2} \cos (2 x) \\ (\cos x)^3 & = \frac{3}{4}\cos (x)+\frac{1}{4} \cos (3 x)\\ (\sin x)^4 & = \frac{3}{8} -\frac{1}{2} \cos (2 x)+\frac{1}{8} \cos (4 x)\\ (\cos x)^5 & = \frac{5}{8}\cos (x)+\frac{5}{16} \cos (3 x)+\frac{1}{16} \cos (5 x)\\ (\sin x)^6 & = \frac{5}{16} - \frac{15}{32} \cos (2 x)+\frac{3}{16} \cos (4x)-\frac{1}{32} \cos (6x) \end{align*} \begin{align*} (\sin x)(\cos x) & = \frac{1}{2} \sin (2 x) \\ (\sin x)(\cos x)^2 & = \frac{1}{4} \sin (x) + \frac{1}{4} \sin (3 x) \\ (\sin x)(\cos x)^3 & = \frac{1}{4} \sin (2 x) + \frac{1}{8} \sin (4 x) \\ (\sin x)^3 & = \frac{3}{4} \sin (x) - \frac{1}{4} \sin (3 x) \\ (\sin x)^3(\cos x) & = \frac{1}{4} \sin (2 x) - \frac{1}{8} \sin (4 x) \\ (\sin x)^3 (\cos x)^2 & = \frac{1}{8} \sin (x) + \frac{1}{16} \sin (3 x) - \frac{1}{16} \sin (5 x) \\ (\sin x)^3(\cos x)^3 & = \frac{3}{32} \sin (2 x) - \frac{1}{32} \sin (6 x) \\ \end{align*} In the context of this class, each of these formulas gives us a (finite) Fourier series of the function on the left-hand side of the equality sign. Therefore, if a function on a left-hand side of an equality above is an initial temperature distribution in a heated rod of length $\pi,$ then, based on the above formulas we can write a solution of the heat equation with that initial condition and an appropriate choice of the boundary conditions: the Neumann boundary conditions for the first group of functions and the Dirichlet boundary conditions for the second group of functions.
- The preceding item is a hint for the problem posted on Thursday. Notice that the solutions of the specific problems posted on Thursday are exact solutions given as finite sums, not infinite series. Also, for me, the preceding item is a source of many heat equation problems which have exact solutions. I will post such problems related to equilibrium temperature distribution in a rectangular plate.
- 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
- We are working with trigonometric functions a lot.
-
Consider the one-dimensional diffusion of dye in a narrow glass tube of length $\pi$. Let the environment substance be uniform throughout the tube and let all the relevant diffusion constants be equal to $1.$ Assume that all the lateral sides, as well as the endpoints of the tube are insulated, so that no dye leaves the tube. This situation is modeled by the following partial differential equation with the boundary and initial conditions:
\begin{equation*}
\frac{\partial u}{\partial t}(x,t) = \frac{\partial^2 u}{\partial x^2}(x,t) \quad \text{where} \quad x \in [0, \pi], \quad t \geq 0,
\end{equation*}
subject to the boundary conditions
\[
\frac{\partial u}{\partial x}(0,t) = 0, \qquad \frac{\partial u}{\partial x}(\pi,t) = 0 \quad \text{for all} \quad t \geq 0,
\]
and the initial condition
\[
u(x,0) = f(x) \quad \text{for all} \quad x \in [0, \pi].
\]
The function $f:[0,\pi]\to \mathbb{R}$ is a differentiable function such that $f'(0) = f'(\pi) = 0$.
- Express the solution of the above problem as an infinite series. Give formulas for the coefficients of the series.
- Find the solution of the above problem for the specific function $f(x) = (\sin x)^2.$ In this case the exact solution is a sum of two terms whose coefficients you can find using relevant trigonometric formulas. No integration is needed.
- Find the solution of the above problem for the specific function $f(x) = (\cos x)^2.$ In this case the exact solution is a sum of two terms whose coefficients you can find using relevant trigonometric formulas. No integration is needed.
- What is a relationship between the solutions found in the preceding two items?
- Find the solution of the above problem for the specific function $f(x) = (\sin x)^4.$ In this case the exact solution is a sum of three terms whose coefficients you can find using relevant trigonometric formulas. No integration is needed.
- Find the solution of the above problem for the specific function $f(x) = (\cos x)^4.$ In this case the exact solution is a sum of three terms whose coefficients you can find using relevant trigonometric formulas. No integration is needed.
- Find the solution of the above problem for the specific function $f(x) = (\cos x)^2 (\sin x)^2.$ In this case the exact solution is a sum of two terms whose coefficients you can find using relevant trigonometric formulas. No integration is needed.
- What is a relationship between the solutions found in the preceding three items?
-
Today I wanted to celebrate the function that you obtained as a solution of Problem 2 on Assignment 1
\[
u(x,y) = \frac{\sin(x y)}{y}.
\]
It is not surprising to make a claim that the domain of this function is the set
\[
\bigl\{ (x,y) \in \mathbb{R}^2 \, : \, y \neq 0\bigr\}.
\]
However, the plot of a graph of this function indicates that this function is defined on the entire plane $\mathbb{R}^2$.
-
The explanation for the fact that the function behaves well along the $x$-axis is the existence of the following limit
\[
\lim_{y\to 0} \frac{\sin(xy)}{y} = x \quad \text{for all} \quad x \in \mathbb{R}.
\]
The above limit is a consequence of the most important trigonometric limit
\[
\lim_{x \to 0} \frac{\sin(x)}{x} = 1.
\]
The preceding limit is a motivation for the definition of the following special function
\[
\operatorname{sinc}(x) = \begin{cases}
\dfrac{\sin(x)}{x} & \quad \text{if} \quad x \in \mathbb{R}\setminus\{0\}, \\
1 & \quad \text{if} \quad x = 0.
\end{cases}
\]
The function $\operatorname{sinc}(x)$ is usually not studied in undergraduate math classes. However, it appears in many engineering applications. Look it up on Wikipedia. It is important to emphasise that the function $\operatorname{sinc}(x)$ is smooth on $\mathbb{R}$, that is, it has continuous derivatives of all orders. Below is a graph of $\operatorname{sinc}(x)$.
-
In the rest of the post I show pictures only; a narrative to follow.
- 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:
- To get started with Mathematica read my Mathematica page. On this page, there are links to several videos that might be helpful to efficiently get started with Mathematica. Mathematica is available in the computer labs in BH 215 and BH 209.
- Today 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.
-
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.
- Today we discussed the boundary conditions for the heat equation. This is Section 1.3 in the book. Among several other boundary conditions, we introduced so called periodic boundary conditions which correspond to viewing a heated rod of length $L$ that is wrapped in a circle, so that points corresponding to $x = 0$ and $x = L$ coincide. We also need to assume that the physical parameters of the rod are the same at $0$ and $L$, that is $c(0) = c(L)$, $\rho(0) = \rho(L)$ and $K_0(0)= K_0(L).$ In this case the boundary conditions for the temperature become \[ u(0,t) = u(L,t), \quad \frac{\partial u}{\partial x}(0,t) = \frac{\partial u}{\partial x}(L,t) \quad \text{for all} \quad t\geq 0. \]
-
For homework do the following 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.
-
For homework do the following two exercises:
-
Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sources and boundary conditions:
- $Q = 0$, $u(0) = T$, $\frac{\partial u}{\partial x}(L) = \alpha$.
- $\frac{Q}{K_0} = 1$, $u(0) = T_1$, $u(L) = T_2$.
- $\frac{Q}{K_0} = x^2$, $u(0) = T$, $\frac{\partial u}{\partial x}(L) = 0$.
-
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?
-
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.
-
Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sources and boundary conditions:
-
I started the following problem in class today.
-
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.?
- In class, I got somewhat confused by the problem. Below is a proper solution.
- Recall the heat equation in a uniform rod, that is the specific heat \(c\), the mass density \(\rho\), the cross sectional area \(A\) and the conductivity \(K_0\) are all independent of position: \[ c \rho \frac{\partial u}{\partial t}(x,t) = K_0 \frac{\partial^2 u}{\partial x^2}(x,t) + Q(x), \qquad 0 \leq x \leq 2, \qquad t \geq 0. \] The boundary conditions are \[ u(0,t) = 0 \quad u(2,t) = 0 \qquad \text{for all} \quad t \geq 0. \] The units of the left-hand side of the equation are \[ \dfrac{\operatorname{cal}}{\operatorname{cm}^3 \cdot \operatorname{sec}} . \] So the units of the right-hand side are the same. In particular, the units of \(Q(x)\) are the same.
- Since we are looking for the equilibrium temperature distribution, we must find \(u(x,t)\) which does not depend on \(t.\) That is we must find \(u(x,t)\) for which \[ \frac{\partial u}{\partial t}(x,t) = 0 \qquad \text{for all} \quad x \in [0,2], \quad t \geq 0. \] Hence our PDE simplifies to \[ K_0 \frac{\partial^2 u}{\partial x^2}(x,t) + Q(x) = 0. \] It is given that \(Q(x) = K_0 x^2\), so we need to solve \[ K_0 \frac{\partial^2 u}{\partial x^2}(x,t) + K_0 x^2 = 0 \qquad u(0) = 0, \quad u(2) = 0, \] or, simplified \[ \frac{\partial^2 u}{\partial x^2}(x,t) + x^2 = 0 \qquad u(0) = 0, \quad u(2) = 0. \] The solution of the preceding equation with the boundary conditions is \[ u(x) = - \frac{1}{12}x^4 + \frac{2}{3} x. \] This is the equilibrium temperature distribution. The heat energy of the rod is \[ c \rho A \int_0^2 \left( - \frac{1}{12}x^4 + \frac{2}{3} x \right) dx = c \rho A \frac{4}{5}. \]
- As we calculated in the preceding item, the heat energy of the rod is constant. So the change in time of the heat energy is \(0.\) However, due to the sources of the heat energy in the rod, the heat energy is constantly created at position \(x\) at the rate \[ K_0 x^2 \ \dfrac{\operatorname{cal}}{\operatorname{cm}^3 \cdot \operatorname{sec}}. \] Therefore, the total heat energy created in the rod (per one second) is \[ A \int_0^2 K_0 x^2 dx = A K_0 \frac{8}{3} \ \dfrac{\operatorname{cal}}{\operatorname{sec}}. \]
- So, the above amount of heat energy is created in the rod each second and the rod is staying at the constant total heat energy. How is this possible?
- The answer is the flux of the heat energy at the end-points of the rod. Based on the Fourier Law the flux of the heat energy is \[ \phi(0,t) = - A K_0 \frac{\partial u}{\partial x}(0,t), \qquad \phi(2,t) = - A K_0 \frac{\partial u}{\partial x}(2,t). \] Now calculate \[ \frac{\partial u}{\partial x}(0,t) = \frac{2}{3}, \qquad \frac{\partial u}{\partial x}(2,t) = -2 \] Thus the flux is \[ \phi(0,t) = - \frac{2}{3} A K_0 , \qquad \phi(2,t) = 2 A K_0. \] Hence, heat energy is leaving the rod at both end-points in the total amount of \[ \frac{2}{3} A K_0 + 2 A K_0 = \frac{8}{3} A K_0 \ \dfrac{\operatorname{cal}}{\operatorname{sec}}. \]
-
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 $u(2) = 0$.
- On Friday and today I derived the Diffusion Equation and the Heat Equation. My derivation of the heat equation today was more detailed than what is presented on the website. I included internal sources of heat and I provided detailed account of the units involved. I should update the website to be more like my presentation today. The corresponding sections in the textbook are Section 1.2 and 1.5. Do Exercises 1.2.1, 1.2.2, 1.2.3, 1.2.4.
- Below I present pictures and animations related to Burgers' equation.
-
Recall Burgers' partial differential equation:
\begin{equation*}
u_t(x,t) + u(x,t)\, u_x(x,t) = 0 \quad \text{in} \quad U \subseteq \bigl\{(x,t) \in \mathbb R^2 : t \gt 0 \bigr\}
\end{equation*}
subject to the initial condition
\begin{equation*}
u(x,0) = f(x), \quad x \in \mathbb R.
\end{equation*}
For illustrations we will use $f(x) = \exp(-x^2)$.
- The corresponding characteristic equations are \begin{equation*} \begin{array}{l} \color{red}{X}'(s) = \color{red}{Z}(s) \\ \color{red}{T}'(s) = 1 \\ \color{red}{Z}'(s) = 0 \end{array} \end{equation*} subject to the initial conditions \begin{equation*} \begin{array}{l} \color{red}{X}(0) = \color{green}{\xi} \\ \color{red}{T}(0) = 0 \\ \color{red}{Z}(0) = \color{green}{f}(\color{green}{\xi}) \end{array} \end{equation*} where $\xi \in \mathbb{R}$ and $s \geq 0.$
- The solution of the Initial Value Problem from the preceding item is \[ \color{red}{X}(s,\xi) = \color{green}{f}(\xi) \, s + \xi, \quad \color{red}{T}(s,\xi) = s, \quad \color{red}{Z}(s,\xi) = \color{green}{f}(\xi), \quad \text{with} \quad s \geq 0, \quad \xi \in \mathbb{R}. \]
-
The following vector equation
\[
\bigl\langle \color{green}{f}(\xi) \, s + \xi, s, \color{green}{f}(\xi) \bigr\rangle, \quad \text{with} \quad s \geq 0, \quad \xi \in \mathbb{R},
\]
represents a surface in \(\mathbb{R}^3.\) In the figure below we show a part of this surface with \(\xi \in [-3,3]\) and \(s\in [0,2.8]\)
-
Another way to see that the surface determined by the vector equation
\[
\bigl\langle \color{green}{f}(\xi) \, s + \xi, s, \color{green}{f}(\xi) \bigr\rangle, \quad \text{with} \quad s \geq 0, \quad \xi \in \mathbb{R},
\]
is not a graph of the function is to look at the characteristics of Burgurs' equation that we found by solving the characteristic equations. For a fixed $\xi \in \mathbb{R}$ the vector equation
\[
\bigl\langle \color{green}{f}(\xi) \, s + \xi, s, \color{green}{f}(\xi) \bigr\rangle, \quad \text{with} \quad s \geq 0,
\]
represents a line parallel with \(xt\)-plane. These lines are the characteristics of Burgers' equation. Since the characteristics are parallel with \(xt\)-plane, we can deduce properties of the surface that we found just by looking at the projected characteristics; for a fixed $\xi \in \mathbb{R}$ we consider a line determined by the vector equation
\[
\bigl\langle \color{green}{f}(\xi) \, s + \xi, s \bigr\rangle, \quad \text{with} \quad s \geq 0.
\]
These lines are the projected characteristics of Burgers' equation. We show \(121\) projected characteristics in the picture below.
- The items below are dedicated to finding a domain on which the surface that we found is a graph of a function. We approach this question first by looking at graphs, then we proceed with relevant calculations.
-
From the picture below we visually confirm that the surface presented is a graph of a function when we limit the domain for the variable $t = s.$
- In the preceding item we concluded that on restricted domain the surface determined by Burgers' equation represents a graph of a function. Since with $f(x) = \exp(x^{-2})$ we cannot find a closed form solution of the system \[ x = \exp(-\xi^2) \, s + \xi, \qquad t = s, \] we do not have a closed form expression for this function.
-
The first step towards finding formulas for the domain in which the surface that we found determines a function is to look at the evolution of the initial condition in time. That is, we study the curve
\[
\bigl\langle \color{green}{f}(\xi) \, s + \xi, s, \color{green}{f}(\xi) \bigr\rangle, \quad \text{with} \quad \xi \in \mathbb{R}.
\]
with fixed \(s \geq 0.\) The animation below illustrates this evolution:
-
The curve we found at the end of the preceding item is shown as the red curve in the picture below. In words, the vertical lines at the points on the red curve in the picture below are tangent to the surface.
-
An interesting phenomena occurs when we project the red curve one \(xt\)-plane:
\[
\left\langle - \frac{\color{green}{f}(\xi)}{\color{green}{f}'(\xi)} + \xi, - \frac{1}{\color{green}{f}'(\xi)} \right\rangle, \quad \text{with} \qquad \xi \in \mathbb{R}.
\]
The picture below shows the projection of the red curve from the preceding item together with projected characteristics. As you can see, the projected rec curve bounds the area in which the projected characteristics intersect each other. Therefore, I claim that outside of the region bounded by the red curve in the picture below, the surface that we found above represents a graph of a function.
-
Next we want to find the exact values of the coordinates of the lowest point on the red curve in the preceding item. Below is a graph of the function
\[
s = - \frac{1}{\color{green}{f}'(\xi)}, \quad \text{in our case} \quad s = \frac{\exp(\xi^2)}{2\xi}.
\]
-
Finally, to celebrate all the calculations so far, we calculate all the points at which vertical tangent lines occur and show them in the animation below
- I post two more Mathematica sample files. 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.
- I post the Mathematica notebook that I created today. In this notebook I explore the solution of Burgers' equation with the initial condition $u(x,0) = \exp(-x^2)$ with $x\in\mathbb{R}.$ You can download this Mathematica notebook named 20221003_MoC_Burgers.nb. As before, I printed this notebook as a pdf file 20221003_MoC_Burgers.pdf
- On Friday I got somewhat confused solving the following linear nonhomogeneous PDE: This known vector field will be central to solving the following quasi-linear first order PDE: \begin{equation*} y \, \color{red}{u}_x(x,y) - x \, \color{red}{u}_y(x,y) = \color{red}{u}(x,y) \qquad \text{in} \qquad \bigl\{(x,y) \in\mathbb{R}^2 \, : \, y \geq 0 \bigr\} \end{equation*} subject to the initial condition \begin{equation*} \color{red}{u}(x,0) = \color{green}{f}(x) \qquad \text{where} \qquad x \geq 0. \end{equation*}
- On Friday, I did not notice that our work below will simplify considerably if I multiply the given equation by $-1.$ Multiplication by a constant does not change the equation, so we consider the following problem \begin{equation*} - y \, \color{red}{u}_x(x,y) + x \, \color{red}{u}_y(x,y) = - \color{red}{u}(x,y) \qquad \text{in} \qquad \bigl\{(x,y) \in\mathbb{R}^2 \, : \, y \geq 0 \bigr\} \end{equation*} subject to the initial condition \begin{equation*} \color{red}{u}(x,0) = \color{green}{f}(x) \qquad \text{where} \qquad x \geq 0. \end{equation*} The vector field in $\mathbb{R}^3$ associated with this equation is \[ (x,y,z) \mapsto \langle - y , x, -z \rangle, \qquad (x,y,z) \in \mathbb{R}^3. \]
-
To find the characteristics of the given PDE, that is of the given Initial Value Problem we need to solve the following initial value problem for the system of ODEs (so called characteristic equations)
\begin{equation*}
\begin{array}{l}
\color{red}{X}'(s) = -\color{red}{Y}(s) \\
\color{red}{Y}'(s) = \phantom{-} \color{red}{X}(s) \\
\color{red}{Z}'(s) = -\color{red}{Z}(s)
\end{array}
\end{equation*}
subject to the initial conditions
\begin{equation*}
\begin{array}{l}
\color{red}{X}(0) = \color{green}{\xi} \\ \color{red}{Y}(0) = 0 \\ \color{red}{Z}(0) = \color{green}{f}(\color{green}{\xi})
\end{array}
\end{equation*}
where $\xi \geq 0$ and $s \in [0,\pi).$ Now, one is asking why this restriction on $s$? That is a consequence of the restriction $y \geq 0$ which is given in the problem.
- The Initial Value Problem \begin{equation*} \begin{array}{l} \color{red}{X}'(s) = -\color{red}{Y}(s) \\ \color{red}{Y}'(s) = \phantom{-} \color{red}{X}(s) \end{array} \end{equation*} subject to the initial conditions \begin{equation*} \begin{array}{l} \color{red}{X}(0) = \color{green}{\xi} \\ \color{red}{Y}(0) = 0 \end{array} \end{equation*} in particular with $\xi = 1$ is probably the most famous IVP for a system of two ODEs. It should be as famous as the IVP of an ODE: $g'(t) = g(t)$ with $g(0) = 1.$
- The solution of the IVP for ODEs in the previous item is (please enjoy the balance of colors below) \begin{equation*} \begin{array}{l} \color{red}{X}(s) = \color{green}{\xi} \color{green}{\cos}(s) \\ \color{red}{Y}(s) = \color{green}{\xi} \color{green}{\sin}(s) \\ \color{red}{Z}(s) = \color{green}{f}(\color{green}{\xi})\color{green}{\exp}(-s) \end{array} \end{equation*} where $\xi \geq 0$ and $s \in [0,\pi].$ Notice that for $s \in [0,\pi]$ and a fixed $\xi \gt 0$, the vector equation $\langle \xi \cos(s), \xi \sin(s) \rangle$ represents a half-circle with radius $\xi$ in $xy$-plane. This is a projected characteristic. Thus, for $s \in [0,\pi]$ and a fixed $\xi \gt 0$, the vector equation $\langle \xi \cos(s), \xi \sin(s), f(\xi)\exp(-s) \rangle$ represents a helix-like curve in $xyz$-plane. This is a characteristic of the given Initial Value Problem for the given PDE. These for all $\xi \gt 0$ these curves form a surface, which is the graph of our solution $u(x,y).$
-
In the figure below I show six characteristics of the given Initial Value Problem for the given PDE together with their projected characteristics. The green curve is the initial condition $\color{green}{f}(x) = x^2$ with $x\geq 0.$
- To find a formula for $u(x,y)$ we need to solve the system \begin{equation*} \begin{array}{l} x = \xi \cos(s) \\ y = \xi \sin(s) \end{array} \end{equation*} for $s$ and $\xi$ such that $\xi \geq 0$ and $s \in [0,\pi].$ Since the above equations are in fact the polar coordinates in a plane, the solution that we are seeking is available from the theory of polar coordinates, see Converting between polar and Cartesian coordinates at the Wikipedia page polar coordinates: \begin{equation*} \begin{array}{l} \xi = \sqrt{x^2 + y^2} \\ s = \arccos\left(\frac{x}{\sqrt{x^2 + y^2}}\right) \end{array} \end{equation*}
-
Finally, the solution $u(x,y)$ is
\[
\color{red}{u}(x,y)
=
\color{green}{f}\mkern-3mu\left(\sqrt{x^2 + y^2}\right) \mkern 2mu \color{green}{\exp}\mkern -5mu\left(- \color{green}{\arccos}\mkern-2mu\left(\frac{x}{\sqrt{x^2 + y^2}}\right) \right) \quad
\text{where} \quad x \in \mathbb{R} \quad \text{and} \quad y \geq 0,
\]
and do not forget to enjoy the balance of colors. In the figure below I show the graph of the function $u.$ The only difference from the graph of the parametric equation is in the mesh lines on the surface.
-
A natural question occurs: Is it possible to extend the definition of the solution in such a way that its domain is the entire plane $\mathbb{R}^2$? The answer is a conditional yes; not the entire plane, we have to exclude the negative part of the $y$-axis. In the figure below I show the graph of the surface with the parametric vector equation
\[
\bigl\langle \xi \cos(s), \, \xi \sin(s), \, \xi^2 \exp(-s) \bigr\rangle \quad \xi \geq 0, \quad s \in \left(-\frac{\pi}{2}, \frac{3\pi}{2}\right).
\]
Notice that the view-point is different from the view-point that was used for the figures in which we showed characteristics.
-
Below are the corresponding figures for the Initial Value
\[
u(x,0) = (\sin x)^2, \qquad x \geq 0.
\]
Below I show five characteristics
- I post the Mathematica notebook that I used to produce these pictures. You can download the notebook and explore it on the computers in Bond Hall 215 and Bond Hall 209.
- I printed the notebook from the previous item as pdf file MoC_w_circle.pdf.
-
The material that we discussed in the second part of the class today, is presented on the webpage Method of characteristics. Below I provide a summary.
-
- Let $\color{green}{A}$, $\color{green}{B}$, and $\color{green}{C}$ be functions of three variables defined on $\mathbb{R}^3.$ These functions can be defined on a subset of $\mathbb{R}^3.$ In this case we just need to take the common domain in the considerations. We consider the vector field \[ (x,y,z) \mapsto \bigl\langle \color{green}{A}(x,y,z),\color{green}{B}(x,y,z),\color{green}{C}(x,y,z) \bigr\rangle, \qquad (x,y,z) \in \mathbb{R}^3. \] This known vector field will be central to solving the following quasi-linear first order PDE: \begin{equation*} \color{green}{A}\bigl(x,y,\color{red}{u}(x,y)\bigr) \, \color{red}{u}_x(x,y) + \color{green}{B}\bigl(x,y, \color{red}{u}(x,y)\bigr) \, \color{red}{u}_y(x,y) = \color{green}{C}\bigl(x,y,\color{red}{u}(x,y)\bigr) \end{equation*} subject to the initial condition, for example, \begin{equation*} \color{red}{u}(x,0) = \color{green}{f}(x). \end{equation*} In last two displayed equations we assume that $x \in \mathbb{R}$ and $y \geq 0.$
- Now we will discover the geometric meaning of the partial differential equation \begin{equation*} \color{green}{A}\bigl(x,y,\color{red}{u}\bigr) \, \color{red}{u}_x + \color{green}{B}\bigl(x,y, \color{red}{u}\bigr) \, \color{red}{u}_y = \color{green}{C}\bigl(x,y,\color{red}{u}\bigr). \end{equation*} Assume that the preceding PDE is satisfied. Next we use this PDE to decompose the vector field \[ (x,y,z) \mapsto \bigl\langle \color{green}{A}(x,y,z),\color{green}{B}(x,y,z),\color{green}{C}(x,y,z) \bigr\rangle, \qquad (x,y,z) \in \mathbb{R}^3. \] on the surface of the graph of $z = \color{red}{u}(x,y)$ \[ \bigl\{ \bigl( x,y, \color{red}{u}(x,y) \bigr) \, : \, (x,y) \in \mathbb{R}^2 \bigr\}. \] Here is the promised decomposition. Since the expressions are long, we drop the variables $(x,y,\color{red}{u})$. At the first step we use the equality given by the PDE. After that we use vector algebra: \begin{align*} \bigl\langle \color{green}{A},\color{green}{B},\color{green}{C} \bigr\rangle & = \bigl\langle \color{green}{A},\color{green}{B}, \color{green}{A}\, \color{red}{u}_x + \color{green}{B} \, \color{red}{u}_y \bigr\rangle \\ & = \bigl\langle \color{green}{A},0, \color{green}{A}\, \color{red}{u}_x \bigr\rangle + \bigl\langle 0,\color{green}{B}, \color{green}{B} \, \color{red}{u}_y \bigr\rangle \\ & = \color{green}{A} \bigl\langle 1,0, \color{red}{u}_x \bigr\rangle + \color{green}{B} \bigl\langle 0, 1, \color{red}{u}_y \bigr\rangle \\ \end{align*} Thus we deduced \begin{equation*} \bigl\langle \color{green}{A},\color{green}{B},\color{green}{C} \bigr\rangle = \color{green}{A} \bigl\langle 1,0, \color{red}{u}_x \bigr\rangle + \color{green}{B} \bigl\langle 0, 1, \color{red}{u}_y \bigr\rangle \end{equation*} at each point $\bigl( x,y, \color{red}{u}(x,y) \bigr)$ on the graph of the function $\color{red}{u}.$ Thus, on the surface of the graph of $\color{red}{u},$ each vector of the vector field $\bigl\langle \color{green}{A},\color{green}{B},\color{green}{C} \bigr\rangle$ is a linear combination of the vectors \[ \bigl\langle 1,0, \color{red}{u}_x \bigr\rangle \qquad \text{and} \qquad \bigl\langle 0, 1, \color{red}{u}_y \bigr\rangle. \] As we know from a Multivariable Class the last two vectors are canonical tangent vectors to the surface of the graph of $\color{red}{u}.$ Since we assume that the function $\color{red}{u}$ is smooth, a linear combination of tangent vectors is also a tangent vector. Hence, the vector field $\bigl\langle \color{green}{A},\color{green}{B},\color{green}{C} \bigr\rangle$ is tangent to the graph of the function $\color{red}{u}.$
- I rephrase the last sentence: The geometric interpretation of the given quasi-linear first order PDE is that the vectors \begin{equation*} \bigl\langle \color{green}{A}\bigl(x, y, \color{red}{u}(x,y)\bigr) , \color{green}{B}\bigl(x, y, \color{red}{u}(x,y)\bigr) , \color{green}{C}\bigl(x, y, \color{red}{u}(x,y)\bigr) \bigr\rangle \end{equation*} are tangent to the graph of the function $z = \color{red}{u}(x,y)$, that is to the surface \begin{equation*} \bigl\{ \bigl( \, x, \, y, \, \color{red}{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, \, \color{green}{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 \color{green}{A}\bigl(x,y,z\bigr) , \color{green}{B}\bigl(x,y,z\bigr) , \color{green}{C}\bigl(x,y,z\bigr) \bigr\rangle. \end{equation*} Recalling what we know from Math 331 about the systems of ordinary differential equations, from each point \begin{equation*} \bigl( \, \color{green}{\xi}, \, 0, \, \color{green}{f}(\color{green}{\xi}) \bigr) \quad \text{for a fixed} \quad \xi \in \mathbb{R} \end{equation*} we can find the curve \begin{equation*} \bigl\{ \bigl( \, \color{red}{X}(s), \, \color{red}{Y}(s), \, \color{red}{Z}(s) \bigr) \, \bigl| \bigr. \, s \geq 0 \bigr\} \end{equation*} which is tangent to the given vector field \begin{equation*} \bigl\langle \color{green}{A}\bigl(x,y,z\bigr) , \color{green}{B}\bigl(x,y,z\bigr) , \color{green}{C}\bigl(x,y,z\bigr) \bigr\rangle. \end{equation*} To find such a curve we solve the system of differential equations \begin{equation*} \begin{array}{l} \color{red}{X}'(s) = \color{green}{A}\bigl(\color{red}{X}(s),\color{red}{Y}(s),\color{red}{Z}(s) \bigr) \\ \color{red}{Y}'(s) = \color{green}{B}\bigl(\color{red}{X}(s),\color{red}{Y}(s),\color{red}{Z}(s) \bigr) \\ \color{red}{Z}'(s) = \color{green}{C}\bigl(\color{red}{X}(s),\color{red}{Y}(s),\color{red}{Z}(s) \bigr) \end{array} \end{equation*} subject to the initial conditions \begin{equation*} \begin{array}{l} \color{red}{X}(0) = \color{green}{\xi} \\ \color{red}{Y}(0) = 0 \\ \color{red}{Z}(0) = \color{green}{f}(\color{green}{\xi}) \end{array} \end{equation*} For simple functions $\color{green}{A},$ $\color{green}{B}$ and $\color{green}{C}$ this system is solvable and the solutions of this system will lead to the solution of the given quasi-linear PDE.
- What is the solution of the Initial Value Problem at the end of the preceding item? Assuming that we have solved this IVP we have three functions $X,$ $Y,$ and $Z.$ The independent variable in each of these functions is $s.$ But, the solutions of the IVP also depend on $\xi.$ Hence, each of these functions is in fact a function of two independent variables $s$ and $\xi$. Hence the set of points \[ \bigl( X(s,\xi), Y(s,\xi), Z(s,\xi)\bigr), \quad s \geq 0, \quad \xi \in \mathbb{R}, \] is a surface in $\mathbb{R}^3.$ That is, the equations \begin{equation*} \begin{array}{l} x = X(s,\xi) \\ y = Y(s,\xi)\\ z = Z(s,\xi) \end{array} \end{equation*} are parametric equations of a surface in $\mathbb{R}^3.$ This surface is the graph of the solution $z = u(x,y)$ of the given PDE.
- Once we have the equations \begin{equation*} \begin{array}{l} x = X(s,\xi) \\ y = Y(s,\xi)\\ z = Z(s,\xi) \end{array} \end{equation*} to find the formula for the solution $u(x,y)$ we need to solve, if possible, the system of equations \begin{equation*} \begin{array}{l} x = X(s,\xi) \\ y = Y(s,\xi) \end{array} \end{equation*} for unknowns $s$ and $\xi.$ Ideally we want to find formulas \begin{equation*} \begin{array}{l} s = S(x,y) \\ \xi = \Xi(x,y). \end{array} \end{equation*} Once we have the last two formulas, the solution of the PDE is \[ u(x,y) = Z\bigl(S(x,y), \Xi(x,y) \bigr) \]
- I will update the post of Sunday, September 25, 2022 by adding more details of the solutions of the PDEs that we considered in class.
- From the textbook you can do 2, 3, 4, 5, 6, 7, 8 in Exercises 12.2 and the problems on the webpage Method of characteristics.
-
On Friday we solved the following initial value problem: Find a function $u(x,y)$ (defined on a domain that you need to determine) such that
\[
-3 u_x + u_y = 0, \qquad u(x,0) = \cos(x) \quad \text{for all} \quad x\in \mathbb{R}.
\]
- The first step is to find a geometric meaning of the given PDE. Let $\bigl\langle x(s), y(s) \bigr\rangle$ with $s \in \mathbb{R}$ be a curve in the $xy$-plane. Now find the derivative of the composite function \[ \frac{d}{ds} u\bigl( x(s), y(s) \bigr) = x'(s) u_x\bigl( x(s), y(s) \bigr) + y'(s) u_y\bigl( x(s), y(s) \bigr). \] Thus, if $x'(s) = -3$ and $y'(s) = 1$ for all $s \in \mathbb{R}$, then the PDE $-3 u_x + u_y = 0$ tells us that $u(x,y)$ is constant along the curve $\bigl\langle x(s), y(s) \bigr\rangle$.
-
Let us fix a point $(\xi,0)$ on the $x$-axis. We know that the value of $u$ at this point is $u(\xi,0) = \cos(\xi).$ Find the parametric curve $\bigl\langle x(s), y(s) \bigr\rangle$ which goes through $(\xi,0)$ and satisfies $x'(s) = -3$ and $y'(s) = 1$. That is, solve the initial value problem for the system of differential equations
\begin{align*}
x'(s) & = -3 \\
y'(s) & = 1 \\
x(0) & = \xi \\
y(0) & = 0.
\end{align*}
The solution is $x(s) = -3 s + \xi$ and $y(s) = s$. These equations represent a family of lines. These lines are called (projected) characteristics of the PDE $-3 u_x + u_y = 0.$ The picture below shows the (projected) characteristics in the $xy$-plane.
-
But the last formula for $u$ is not the complete solution to the initial value problem stated. We need $u$ as a function of $(x,y).$ For that we solve
\begin{align*}
x & = -3 s + \xi \\
y & = s
\end{align*}
for $\xi$ and $s.$ The solution is $\xi = x + 3 y$ and $s = y.$ Hence,
\[
u(x,y) = \cos(x + 3 y) \qquad \text{for all} \quad (x,y) \in \mathbb{R}^2.
\]
The solution is defined for all $(x,y) \in \mathbb{R}^2.$ The picture below illustrates the solution.
- The family of parallel lines $x + 3 y = \xi$ with $\xi \in \mathbb{R}$ is important for the given PDE since the solution is constant along each of these lines. These lines are called characteristics of the PDE $-3 u_x + u_y = 0.$
-
Let us make a small change in the given PDE. Find a function $u(x,y)$ (defined on a domain that you need to determine) such that
\[
-3 u_x + x u_y = 0, \qquad u(x,0) = \cos(x) \quad \text{for all} \quad x\in \mathbb{R}.
\]
- The first step is to find a geometric meaning of the given PDE. Let $\bigl\langle x(s), y(s) \bigr\rangle$ with $s \in \mathbb{R}$ be a curve in the $xy$-plane. Now find the derivative of the composite function \[ \frac{d}{ds} u\bigl( x(s), y(s) \bigr) = x'(s) u_x\bigl( x(s), y(s) \bigr) + y'(s) u_y\bigl( x(s), y(s) \bigr). \] Thus, if $x'(s) = -3$ and $y'(s) = x(s)$ for all $s \in \mathbb{R}$, then the PDE $-3 u_x + x u_y = 0$ tells us that $u(x,y)$ is constant along the curve $\bigl\langle x(s), y(s) \bigr\rangle$.
- Let us fix a point $(\xi,0)$ on the $x$-axis. We know that the value of $u$ at this point is $u(\xi,0) = \cos(\xi).$ Find the parametric curve $\bigl\langle x(s), y(s) \bigr\rangle$ which goes through $(\xi,0)$ and satisfies $x'(s) = -3$ and $y'(s) = x(s)$. That is, solve the initial value problem for the system of differential equations \begin{align*} x'(s) & = -3 \\ y'(s) & = x(s) \\ x(0) & = \xi \\ y(0) & = 0. \end{align*} The solution is $x(s) = -3 s + \xi$ and $y(s) = -\frac{3}{2} s^2 + \xi s.$ Since we deduced that $u$ is constant along these curves, we have \[ u\bigl(-3 s + \xi, -\tfrac{3}{2} s^2 + \xi s\bigr) = \cos(\xi) \qquad \text{for all} \quad s\in \mathbb{R} \quad \text{and} \quad \text{for all} \quad \xi \in \mathbb{R}. \]
- But the last formula for $u$ is not the complete solution to the initial value problem stated. We need $u$ as a function of $(x,y).$ For that we solve \begin{align*} x & = -3 s + \xi \\ y & = -\frac{3}{2} s^2 + \xi s \end{align*} for $\xi$ and $s.$
- Before solving this system it is helpful to look at the vector field $\langle -3, x \rangle.$ Also, we need to think that we are looking for the solution defined for all positive values of $y$ (think of $y$ as time). Therefore we need $s$ and $\xi$ such that $2 \xi - 3 s \geq 0.$ a picture needed
- The solution is $\xi =\sqrt{x^2 + 6 y}$ and $s = \bigl(-x + \sqrt{x^2 + 6 y} \bigr)/3.$ Hence, \[ u(x,y) = \cos\bigl(\sqrt{x^2 + 6 y}\bigr) \qquad \text{for all} \quad (x,y) \in ???. \] The domain needs to be discussed here. Clearly this function is defined on the set \[ \bigl\{ (x,y) \in \mathbb{R} : y \geq 0 \bigr\}. \] However this function is defined on a larger set: \[ \bigl\{ (x,y) \in \mathbb{R} : x^2 \geq 6 y \bigr\}. \]
- The family of parabolas $y = (-x^2 + \xi^2)/6$ with $\xi \geq 0$ is the family of characteristics of the PDE $-3 u_x + x u_y = 0.$
- Computer algebra system Mathematica will be very useful for the assignments in this class. You can start getting familiar with it. To get started with Mathematica see my Mathematica page. Please watch the videos that are on my Mathematica page. Watching the movies is very helpful to get started with Mathematica efficiently! Mathematica is available in the computer labs in BH 215 and BH 209.
- During the first class I provide an overview of the course. As a guide to an overview I like to offer my explanations of the meaning of the words in the course title from the catalog. Today I offered the following explanation of the word equation: An equation is a mathematical expression which involves the equality sign and in which one component is considered unknown and the remaining components are considered known. To solve an equation means to find a formula for the unknown component in terms of the known quantities.
- Notice that in the explanation in the preceding item I use the concept of mathematical expression without explaining the meaning. As the Wikipedia page shows, an explanation of the concept mathematical expression is not simple.
- Tomorrow we will discuss special partial differential equations of the first order which can be solved using The Method of Characteristics.
- The information sheet.
- We will start the class by studying the Method of characteristics for quasi-linear first-order partial differential equations.
- My presentation of the Method of Characteristics differs from the presentation in the textbook. For comparison read Section 12.2.2 Method of Characteristics in the textbook. Suggested problems are 2, 3, 4, 5, 6, 7, 8 in Exercises 12.2, see Chapter 12 from the textbook.
-
We will continue with the heat (or diffusion) equation. A derivation of the diffusion equation is here.
The animation below is obtained by solving that equation.
Place the cursor over the image to see the diffusion of the dye.
- Here is the complete list of all the webpages that I created for this class:
- Computer algebra system Mathematica will be very useful for the assignments in this class. You can start getting familiar with it. To get started with Mathematica see my Mathematica page. Please watch the videos that are on my Mathematica page. Watching the movies is very helpful to get started with Mathematica efficiently! Mathematica is available in the computer labs in BH 215 and BH 209.