- Today we did Problem 8 in The Method of Characteristics. (I will post details later.)
- I updated the post from Monday, January 13, 2025.
- Today we presented a general strategy for solving quasilinear partial differential equation: \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 \begin{equation*} u(x,0) = f(x), \quad x \in \mathbb R, \end{equation*} where $A, B, C$ are given real functions of three variables $x,y,z$, $f$ is a given real function of one variable $x$ and $u$ is an unknown function of two variables $x,y$. The objective is to find a formula for $u$ in terms of $A, B, C$ and $f$. For simplicity we assume that the given functions $A$, $B$ and $C$ are smooth functions defined on ${\mathbb R}^3$ with the values in $\mathbb R$ and that the given function $f$ is a smooth function defined on $\mathbb R$ with the values in $\mathbb R$. Read more at my webpage The Method of Characteristics.
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Today we solved the following first order PDE
\begin{equation*}
\frac{\partial u}{\partial x}(x,y) - \frac{\partial u}{\partial y}(x,y) = 0
\quad \text{where} \quad (x,y) \in \mathbb R^2
\end{equation*}
subject to the boundary condition
\begin{equation*}
u(x,0) = f(x), \quad \text{for} \quad x \in \mathbb R.
\end{equation*}
- Since the Partial Differential Equation is homogeneous, that is the expression with the first order derivatives equals \(0\), it is reasonable to expect that along some curves in \(xy\)-plane there is no change in values of \(u(x,y)\). Let us look for those curves.
- Let \[ x = X(s), \qquad y = Y(s), \] be parametric equations of a curve in \(xy\)-plane. Now we focus on the values of \(u(x,y)\) along this curve. We consider the following function of \(s\): \[ u\bigl(X(s), Y(s) \bigr). \]
- Using the chain rule for multivariable functions, the derivative of this function of \(s\) is \[ \frac{d}{ds} u\bigl(X(s), Y(s) \bigr) = \frac{\partial u}{\partial x}\bigl(X(s), Y(s) \bigr) X'(s) + \frac{\partial u}{\partial y}\bigl(X(s), Y(s) \bigr) Y'(s). \] Compare the last expression to the given Partial Differential Equation. Recall that we are just experimenting with the functions \(X(s)\) and \(Y(s)\); we can choose these functions to be any differentiable functions. One observes that choosing \[ X'(s) = 1 \qquad \text{and} \qquad Y'(s) = -1 \] we obtain that \begin{align*} \frac{d}{ds} u\bigl(X(s), Y(s) \bigr) & = \frac{\partial u}{\partial x}\bigl(X(s), Y(s) \bigr) X'(s) + \frac{\partial u}{\partial y}\bigl(X(s), Y(s) \bigr) Y'(s) \\ & = \frac{\partial u}{\partial x}\bigl(X(s), Y(s) \bigr) - \frac{\partial u}{\partial y}\bigl(X(s), Y(s) \bigr) = 0. \end{align*}
- Thus, there is no change along the curve determined by the parametric equations \(x = X(s)\), \(y = Y(s) \). To get useful solutions for \(X(s)\) and \(Y(s)\) we choose the initial conditions: \(X(0) = \xi\) and \(Y(0) = 0\). The reason for this choice is that from the initial values for the unknown \(u(x,y)\) we know the values \(u(\xi,0) = f(\xi)\). (A comment about my choice of notation: Instead of \(x\), I use \(\xi\) because \(x\) is often used as a general variable. The Greek lowercase letter \(\xi\) serves as a Greek counterpart to \(x\).) The nature of \(\xi\) here is that it is arbitrary real number, but fixed for now.
- The solution of the initial value problem \[ X'(s) = 1, \quad X(0) = \xi, \] is \[ X(s) = s + \xi. \] The solution of the initial value problem \[ Y'(s) = -1, \quad Y(0) = 0, \] is \[ Y(s) = - s. \]
- Thus, along the line \(x=s+\xi, \ y= -s\), where \(s\in\mathbb{R}\) and \(\xi\) is fixed, the function \(u(x,y)\) is constant. In other words, with fixed \(\xi\), if \(x+y = \xi\), we have that \(u(x,y)\) is constant. Therefore, if \(x+y = \xi\), we have \[ u(x,y) = u(\xi,0) = f(\xi) = f(x+y). \] Thus, the solution to the given Partial Differential Equation with the given Initial Values is \[ u(x,y) = f(x+y). \]
- Recall the post of Saturday, January 11, 2025. We introduced the unknown function \(u(x,t)\) as follows \[ u(x,t) = \biggl({\huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\biggr) f(x), \quad \normalsize \text{where} \quad \large x, t \in \mathbb{R}, \] Today, we solved the partial differential equation which gave us the function \(u(x,t)\): \[ u(x,y) = f(x+y). \] Combining the last two displayed equalities, we have the formula for the exponential function with \(t \frac{d}{dx}\) in the exponent: \[ \biggl({\huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\biggr) f(x) = f(x+t), \quad \text{where} \quad \large x, t \in \mathbb{R}. \] Thus, for each \(t\in\mathbb{R}\) the transformation \(\displaystyle {\huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\) acts on a function \(f(x)\) by shifting it's graph horizontally. The shift is to the left if \(t \gt 0\) and the shift is to the right if \(t \lt 0\).
- Now that we have established the formula \[ \biggl({\huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\biggr) f(x) = f(x+t), \quad \text{where} \quad x, t \in \mathbb{R}, \] we can draw a fascinating analogy with the standard exponential function from calculus. For real numbers \(a, c \in \mathbb{R}\), the standard exponential function satisfies: \[ {\Large c} \mkern 2mu {\huge e}^{^{\Large a \mkern 0.75mu t}} = \Bigl( {\huge e}^{^{\Large t \mkern 0.75mu a}} \Bigr) {\Large c}. \] Here, the expression \(\displaystyle \bigl( {\Large e}^{^{\Large t \mkern 0.75mu a}} \bigr) c\) is analogous to \[ \biggl({\huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\biggr) f(x), \] where \(\frac{d}{dx}\) plays the role of \(a\), and \(f(x)\) corresponds to \(c\).
- This analogy extends to the derivatives at \(t=0\). Consider the derivative of \(\displaystyle \bigl( {\Large e}^{^{\Large t \mkern 0.75mu a}} \bigr) {\large c}\) at \(t = 0\): \[ \lim_{t\to 0} \frac{\bigl(e^{t \mkern 0.75mu a} \bigr) c - c}{t} = c \lim_{t\to 0} \frac{e^{a \mkern 0.75mu t} - 1}{t} = c \mkern 2mu a = a \mkern 2mu c. \] The analogous computation for \(\displaystyle \Bigl({\Large e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\Bigr) f(x)\) gives: \[ \lim_{t\to 0} \frac{\Bigl({\Large e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\Bigr) f(x) - f(x)}{t} = \lim_{t\to 0} \frac{f(x+t) - f(x)}{t} = \frac{d}{dx} f(x). \] Notice the structure of the final terms in the last two displayed formulas: \(\frac{d}{dx}\) plays the role of \(a\), and \(f(x)\) corresponds to \(c\). This shows how the operator exponential \({\huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\) parallels the familiar exponential function in calculus.
- The analogy presented here has inspired a significant amount of research in mathematical analysis. This subfield is known as the theory of operator semigroups.
- The story that I am telling in this post fascinated me when I saw it for the first time as an undergraduate student a half-century ago.
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In this item we review the most fundamental Initial Value Problem for Ordinary Differential Equations. Let \(a \in \mathbb{R}\) be an arbitrary real number.
- In calculus we studied the following problem: Find a function \(y(t)\) that satisfies the initial value problem \[ \Large \frac{dy}{dt}(t) = a \mkern 2mu y(t), \qquad y(0) = 1. \] A key topic in calculus is that this problem has a unique solution, given by the exponential function \[ \Large y(t) = e^{a \mkern 1.5mu t}, \quad \normalsize \text{where} \quad \Large t \in \mathbb{R}. \]
- More generally, one can consider the initial value problem for any \(c \in \mathbb{R}\): Find \(y(t)\) that solves \[ \Large \frac{dy}{dt}(t) = a \mkern 2mu y(t), \qquad y(0) = c. \] Again, in calculus we established that the unique solution to this problem is \[ \Large y(t) = c \mkern 2mu e^{a \mkern 1.5mu t}, \quad \normalsize \text{where} \quad \Large t \in \mathbb{R}. \]
- The problems stated in the preceding items are the simplest initial value problems for ordinary differential equations. Their importance lies in the fact that these problems describe the exponential growth or decay, a phenomenon commonly observed in nature.
- Given the importance and fundamental nature of the Initial Value Problems presented in the preceding item, courses on Ordinary Differential Equations or the second quarter course of Linear Algebra often explore a generalized version of these Initial Value Problems for square matrices. This exploration typically occurs during an in-depth study of matrix diagonalization. I do this in my Math 304, see the end of the post on Wednesday, October 9, 2024, starting with Problem 5 on Assignment 1 and the posts on Thursday, October 10, 2024, and Friday, October 11, 2024.
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In this item we state the Matrix Version of the Initial Value Problem presented in the above item. Let \(n \in \mathbb{N}\) be a positive integer and \(A \in \mathbb{R}^{n\times n}\) is a square matrix.
- Find an $n\times n$ matrix valued function \(Y(t)\) that satisfies the initial value problem \[ \Large \frac{dY}{dt}(t) = A \mkern 2mu Y(t), \qquad Y(0) = I_n. \] Here, \(I_n\) is the $n\times n$ identity matrix. By definition, the solution of this Matrix Initial Value Problem is the Matrix Exponential Function \[ \Large Y(t) = {\huge e}^{^{{\Large A} \mkern 1.5mu t}}, \quad \normalsize \text{where} \quad \Large t \in \mathbb{R}. \]
- For example, you can verify that \[ {\Huge e}^{^{ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} {\Large t}}} = \left[\!\begin{array}{rr} \cos(t) & -\sin(t) \\[5pt]\sin(t) & \cos(t) \end{array}\!\right], \] \[ {\Huge e}^{^{\begin{bmatrix} -1 & 2 \\ 0 & 1 \end{bmatrix}{\Large t}}} = \left[\!\begin{array}{cc} e^{-t} & e^{t} - e^{-t} \\[5pt] 0 & e^{t} \end{array}\!\right], \] \[ {\Huge e}^{^{\left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \\ \end{array}\right] {\Large t}}} = \left[\!\begin{array}{ccc} 1 & t & t^2 \\ 0 & 1 & 2t \\ 0 & 0 & 1 \\ \end{array}\!\right]. \]
- Instead of matrix valued functions, we can consider vector valued functions and formulate the Vector Initial Value Problem. Let \(\vec{\mathbf{v}}\) be a vector in \(\mathbb{R}^n\). Find a vector valued function \(\mathbf{y}(t)\) that solves the following vector initial value problem: \[ \Large \frac{d\vec{\mathbf{y}}}{dt}(t) = A \mkern 2mu \vec{\mathbf{y}}(t), \qquad \vec{\mathbf{y}}(0) = \vec{\mathbf{v}}. \] The unique solution of this initial value problem is \[ \Large \vec{\mathbf{y}}(t) = {\huge e}^{^{t \mkern 1mu {\Large A} }} \vec{\mathbf{v}}, \quad \normalsize \text{where} \quad \Large t \in \mathbb{R}. \]
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Inspired by what is presented in Matrix Initial Value Problem we can answer the question: What is the exponential function of the derivative \(\frac{d}{dx}\)? That is we will give a meaning to
\[
{\Huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}, \quad \normalsize \text{where} \quad \Large t \in \mathbb{R}.
\]
We proceed by analogy.
- Instead of an \(n\times n\) matrix \(A\) in Matrix Initial Value Problem we have the derivative \(\frac{d}{dx}\). The matrix \(A\) acts on vectors \(\vec{\mathbf{v}}\) in the vector space \(\mathbb{R}^n\). The derivative \(\frac{d}{dx}\) acts on the space of differentiable functions, for example on \(f(x)\).
- In Matrix Initial Value Problem we established that, for \(t\in \mathbb{R}\), \[ {\huge e}^{^{ {\Large t \mkern 1mu A} }} \] is an \(n\times n\) matrix. Then, in Vector Initial Value Problem, for a given vector \(\vec{\mathbf{v}} \in \mathbb{R}^n\), we established that \[ \Large \vec{\mathbf{y}}(t) = {\huge e}^{^{t \mkern 1mu {\Large A} }} \vec{\mathbf{v}} \] is the solution of the initial value problem \[ \large \frac{d\vec{\mathbf{y}}}{dt}(t) = A \mkern 2mu \vec{\mathbf{y}}(t), \qquad \vec{\mathbf{y}}(0) = \vec{\mathbf{v}}. \]
- To understand \[ {\Huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}, \quad \normalsize \text{where} \quad \Large t \in \mathbb{R}, \] instead of a vector \(\vec{\mathbf{v}} \in \mathbb{R}^n\) we need to consider a differentiable function \(f(x)\) defined on \(\mathbb{R}\), that is defined and differentiable for all real numbers. Thus, instead of \[ \Large \vec{\mathbf{y}}(t) = {\huge e}^{^{t \mkern 1mu {\Large A} }} \vec{\mathbf{v}} \] we need to study \[ u(x,t) =\mkern-16mu \underbrace{\biggl({\huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\biggr) f(x)}_{ \begin{array}{l} \text{this is a function of $x$ and $t$;} \\[-12pt] \text{we call it $u(x,t)$} \end{array}}, \quad \normalsize \text{where} \quad \large x, t \in \mathbb{R}, \]
- Now, in the initial value problem \[ \large \frac{d\vec{\mathbf{y}}}{dt}(t) = A \mkern 2mu \vec{\mathbf{y}}(t), \qquad \vec{\mathbf{y}}(0) = \vec{\mathbf{v}}, \] instead of \(\vec{\mathbf{y}}(t)\) we have \(u(x,t)\), instead of \(A\) we have \(\frac{d}{dx}\), and instead if \(\vec{\mathbf{v}}\) we have \(f(x)\). Thus, to understand the expression \[ u(x,t) = \biggl({\huge e}^{^{\Large t \mkern 0.5mu \frac{d}{dx}}}\biggr) f(x), \] we need to solve the Initial Value Problem for the Partial Differential Equation \[ \Large \frac{\partial u}{\partial t}(x,t) = \frac{\partial u}{\partial x}(x,t), \qquad u(x,0) = f(x). \]
- Computer algebra system Mathematica will be very useful for the assignments in this class. To get started with Mathematica see my Mathematica page. A convenient way to get started is to watch the videos which are linked on my Mathematica page. Watching the movies is very helpful to get started with Mathematica efficiently! Mathematica is available in the computer labs in BH 215, HH 233 and BH 209/211A.
- 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 its meaning. As the Wikipedia page shows, an explanation of the concept mathematical expression is not simple.
- Today I offered the following definition of a Partial Differential Equation: A partial differential equation is an equation in which the unknown quantity is a function of several variables, and the equation involves this function, its partial derivatives, and other terms, called coefficients, which are specific or considered known. The coefficients can be constants or functions in the same independent variables as the unknown function.
- On Thursday, we will discuss special partial differential equations of the first order which can be solved using The Method of Characteristics. Reading the webpage The Method of Characteristics you will encounter several Partial Differential Equations. This pages also contains many ordinary differential equations; in fact systems of ordinary differential equations.
- My presentation of the Method of Characteristics differs from the presentation in the textbook. For comparison read Section 12.2.2 Method of Characteristics in the textbook. Suggested problems are 2, 3, 4, 5, 6, 7, 8 in Exercises 12.2, see Chapter 12 from the textbook.
- If you have any questions about the content or the organization of the class, please post them on Discussions on Canvas.
- 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, HH 233 and BH 209.
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I recommend that you look into learning
LaTeX which is a superior typesetting system for writing mathematics. Not only that LaTeX is the best way to write mathematical content on computer, but it is the best way to display math on the Internet. All the mathematics that you see on my website is written in LaTeX. All the mathematics that you see on Wikipedia is written in LaTeX. In fact, if you go to a Wikipedia page with mathematics, on the line below the title, next to the Read button, you will find the Edit button. Clicking this Edit button will take you to the code for that webpage. You will see a lot of LaTeX code there. In fact, if you need exactly that formula, you can copy and paste it in your document.
- I created Getting Started with LaTeX page to help you with this.
- Here is a simple LaTeX sample document in which I prove an interesting inequality.
- If you need help starting with LaTeX, feel free to ask me for help. It is as important as learning one nice piece of mathematics. And I want to help you with that, not only because that is my job, but because I deeply believe that both math - through learning rigorous thinking - and professional writing skills - supported by rigorous thinking - will help you in your life more than anything else.
- I have noticed that many students use the online LaTeX editor Overleaf. I do not have any experience with Overleaf, except that I have seen that many students use it with nice results.
- LaTeX is just like a computer program that compiles into your mathematical document. As you probably know, ChatGPT is exceptionally good at writing code in all kinds of different languages. So, it is good at writing LaTeX code. If you are confused on how to do a particular task in LaTeX, you can ask ChatGPT for advise. All LaTeX related advise that I obtained from ChatGPT was superb. ChatGPT can also help you with Wolfram Mathematica code. Sometimes ChatGPT is sloppy, and makes mistakes, parenthesis not matching and stuff like that. If that happens, ask for corrections. It always helps to write very detailed user prompts.
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Here is one example of my conversation with ChatGPT:
You:
Can you please write a complete LaTeX file with instructions on using basic mathematical operations, like fractions, sums, integrals, basic functions, like cosine, sine, and exponential function, and how to structure a document and similar features? Please explain the difference between the inline and displayed mathematical formulas. Please include examples of different ways of formatting displayed mathematical formulas. Please include what you think would be useful to a mathematics student. Also, can you please include your favorite somewhat complicated mathematical formula as an example of the power of LaTeX? I emphasize I want a complete file that I can copy into the LaTeX compiler and compile into a pdf file. Please ensure that your document contains the code for the formulas you are writing, which displays both as code separately from compiled formulas. Also, please double-check that your code compiles correctly. Remember that I am a beginner and cannot fix the errors. Please act as a concerned teacher would do.
This is the LaTeX document that ChatGPT produced base on the above prompt. Here is the compiled PDF document.
You can ask ChatGPT for specific LaTeX advise. To get a good response, think carefully about your prompt. Also, you can offer to ChatGPT a sample of short mathematical writing from the web or a book as a PNG file and it convert that writing to LaTeX. You can even try with neat handwriting. The results will of course depend on the clarity of the file, ChatGPT makes mistakes, but I found it incredibly useful.
- It is tempting to make LaTeX a mandatory tool for creating submissions in this class. However, I do not appreciate having mandates imposed on me; I prefer the freedom to think my own thoughts. An important part of a teacher's role is to encourage students to think their own thoughts, and imposing a mandate could undermine that principle.
- My one-page on the Power of Math.
- My one-page on Rigorous Reasoning.