# My first partial differential equation attempt

have I solved this correctly? My textbook is asking for the relation between $\alpha$ and $\beta$:

$$\frac{\partial{u}}{\partial{t}}=\frac{\partial^2{u}}{\partial{x^2}}$$ Textbook's proposed solution: $$u\left(x,t\right)=\sin\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}$$ My work: $$\frac{\partial}{\partial{t}}\left(\sin\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}\right)=-\beta\cdot{\sin}\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}$$ $$\frac{\partial^2{u}}{\partial{x^2}}=\frac{\partial^2}{\partial{x^2}}\left(\sin\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}\right)=\frac{\partial}{\partial{x}}\left(\alpha\cdot{\cos}\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}\right)=-\alpha^2\cdot{\sin}\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}$$ $$\therefore-\beta\cdot{\sin}\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}=-\alpha^2\cdot{\sin}\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}$$ $$\implies\boxed{\beta=\alpha^2}$$

My textbook states that it's some kind of a heat equation relating the distribution of heat at a specific time t. I've looked at the Wikipedia page on heat equations and I find it fascinating. I've spent every spare moment this summer in-between semesters at RIT trying to get a better grasp on the calc. II content and I couldn't help but take a look at the chapters I'm bound to study in multivariable calc. this fall.

• Do you mean to say you want to solve $\frac{\partial f(x,t)}{\partial t} = \frac{\partial^2 f(x,t)}{\partial x^2}$? The notation for taking a derivative of $f = \sin(\alpha x) \cdot e^{- \beta t}$ with respect to $x$ would be $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( \sin(\alpha x) \cdot e^{- \beta t} \right)$ as opposed to repeating the $f$ in the second statement as you do Aug 3, 2014 at 22:47
• Yes, that's exactly right, I shouldn't have included the $f$ in the second one, so it should read $\frac{\partial{f}}{\partial{x}}=\frac{\partial}{\partial{x}}\left({sin}\left(\alpha{x}\right){e}^{-\beta{t}} \right)$ Aug 3, 2014 at 22:50
• I have edited your question to remove some of the confusion you have placed in the text. To answer your question you have found the relationship between $\alpha$ and $\beta$ correctly. Aug 5, 2014 at 22:06
• Is this solution unique? What happens if I tell you that $u(x,t)= A x + B$ is also a solution? ($A$ and $B$ are constants). Aug 6, 2014 at 14:05

As you've correctly figured out, $$u\left(x,t\right)=\sin\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}$$ satisfies the heat equation, $$\partial u/\partial t = \partial^2 u / \partial x^2$$, provided that $$\beta=\alpha^2$$. As pointed out in the comments, this solution is far from unique. Let's try to provide a little more context.
Often, the heat equation is accompanied by one or more boundary conditions. For example, we might specify that $$u(0,t) = u(1,t) = 0$$. From a physical perspective, this means that $$u$$ describes the temperature along the line segment $$0 \leq x \leq 1$$ and that the temperature at the two endpoints is fixed to be zero. If we want to ensure that your book's proposed solution $$u\left(x,t\right)=\sin\left(\alpha{x}\right)\cdot{e}^{-\beta{t}}$$ satisfies the boundary conditions, as well as the heat equation, then from just the right hand endpoint we need $$u(1,t) = \sin(\alpha)e^{-\alpha^2} = 0.$$ This only happens when $$\alpha$$ is an integer multiple of $$\pi$$. Thus, one solution is $$u\left(x,t\right)=\sin\left(\pi{x}\right)\cdot{e}^{-\pi^2{t}}.$$ Let's take a look at how the graph of this functions evolves over $$7/10$$ of a second.
At the next step, you start to consider other possible initial distributions of heat. At that point, you take into account the fact that any function of the form $$u_n\left(x,t\right)=\sin\left(n\pi{x}\right)\cdot{e}^{-n^2\pi^2{t}}$$ solves the equation and the boundary conditions. At time $$t=0$$, we have $$u_n\left(x,0\right)=\sin\left(n\pi{x}\right)$$ and, in fact, any sum of the $$u_n$$s satisfies the heat equation and the boundary conditions. You're left to ask whether, given a more or less arbitrary initial temperature distribution, is there a sum of the $$c_n\sin(n\pi x)$$ terms that matches that initial distribution. This is how the topic of Fourier series arises.
• Nice explanation, @Mark. Could you please tell me what software you used for the graphics? It's pretty cool (never better said). Furthermore it's a nice exercise to show that this parabolic problem subjected to homogenous Dirichlet boundary conditions is given by: $$u(x,t) = \sum_{n=0}^{\infty} G_n e^{- n^2 \pi^2 t} \sin{ n \pi x},$$ where $G_n$ is a constant related to the initial condition. Aug 8, 2014 at 9:11
• @Dmoreno I'm glad you liked it. I generate the image with Mathematica. I generated a list of images and then exported that to an animated GIF. The command to generate the list of images was something like so: Table[Plot[Exp[-Pi^2 t] Sin[Pi*x], {x, 0, 1}, PlotRange -> {0, 1}], {t, 0, 0.7, 0.01}]. Aug 10, 2014 at 0:43