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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.

Thank you for your time,

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    $\begingroup$ 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 $\endgroup$
    – DanZimm
    Aug 3, 2014 at 22:47
  • $\begingroup$ 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) $ $\endgroup$
    – bjd2385
    Aug 3, 2014 at 22:50
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    $\begingroup$ 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. $\endgroup$
    – Chinny84
    Aug 5, 2014 at 22:06
  • $\begingroup$ 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). $\endgroup$
    – Dmoreno
    Aug 6, 2014 at 14:05

1 Answer 1

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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.

enter image description here

I guess the point is that the object is warmest in the middle and, since the endpoints are immersed in an ice bath to hold their temperature at zero, the heat diffuses out of the bar.

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.

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    $\begingroup$ 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. $\endgroup$
    – Dmoreno
    Aug 8, 2014 at 9:11
  • $\begingroup$ @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}]. $\endgroup$ Aug 10, 2014 at 0:43
  • $\begingroup$ Wow, I'm sorry I didn't see your answer earlier, this makes a lot of sense now! $\endgroup$
    – bjd2385
    Aug 21, 2014 at 0:21

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