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Solve: $\frac{\partial u}{\partial t}=\frac{\partial^2u}{\partial x^2}+4\frac{\partial u}{\partial x}+2u$ for $0<x<\pi$, $t>0$ with boundary conditions $u(0,t)=u(\pi,t)=0$ for $t>0$ and initial condition $u(x,0)=x(\pi-x)$ for $0<x<\pi$.

The way I am solving it is by reducing it to one involving the standard heat equation by setting $u(x,t)=(e^{\alpha x+\beta t})U(x,t)$. Then substituting this into the problem and choosing $\alpha$ and $\beta$ to obtain a standard problem for $U(x,t)$ for a bar with ends kept at zero temperature. I have gotten to the point where I differentiated the equation and plugged it back into the initial PDE and took out $e^{\alpha x+\beta t}$. It's after here that I am unsure of what to do. What I dont understand is choosing alpha and beta or the following steps to finish the overall problem. Also, I apologize for the syntax, I don't have Latex or anything on my computer and the derivative notation is supposed to be partial derivatives.

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You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". Here's a basic tutorial and quick reference. There's an "edit" link under the question. –  joriki Sep 22 '12 at 0:06
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Just directly use separation of variables is OK, letting $u(x,t)=e^{\alpha x+\beta t}U(x,t)$ should be superfluous. –  doraemonpaul Sep 22 '12 at 0:40
    
The point of the question is to solve it this way and not use separation of variables. Sorry I should have clarified on that. Thanks for the info on Tex as well. –  Sprock Sep 22 '12 at 3:47

2 Answers 2

up vote 1 down vote accepted

Setting $u(x,t)=e^{\alpha x+\beta t}U(x,t)$ we have for $(x,t) \in (0,\pi)\times(0,\infty)$: $$ \frac{\partial U}{\partial t}(x,t)=\frac{\partial^2U}{\partial x^2}(x,t)+2(\alpha+2)\frac{\partial U}{\partial x}(x,t)+[(\alpha+2)^2-\beta-2]U(x,t). $$ Choosing $$ \alpha=\beta=-2, $$ the problem reduces to the Heat Equation $$ \begin{cases} \frac{\partial U}{\partial t}(x,t)=\frac{\partial^2U}{\partial x^2}(x,t) &\text{ for } (x,t) \in (0,\pi)\times(0,\infty),\\ U(0,t)=U(\pi,t)=0 &\text{ for } t \in (0,\infty),\\ U(x,0)=x(\pi-x)e^{2x} &\text{ for } x \in (0,\pi), \end{cases} $$ whose solution is given by $$\tag{1} U(x,t)=\sum_{n=1}^\infty D_n\sin\left(\frac{n\pi x}{\pi}\right)\exp\left(-\frac{n^2\pi^2t}{\pi^2}\right)=\sum_{n=1}^\infty D_n\sin(nx)e^{-n^2t}, $$ where $$\tag{2} D_n :=\frac{2}{\pi}\int_0^\pi U(x,0)\sin\left(\frac{n\pi x}{\pi}\right)dx =\frac{2}{\pi}\Im\int_0^\pi x(\pi-x)e^{(2+in)x}dx. $$ Your solution is given by $u(x,t)=e^{-2(x+t)}U(x,t)$, with $U$ defined by (1) and (2).

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@ Mercy-how did you get beta=0? I got -2. Also where does the $e^(2x)$ come from in $U(x,o)=x(pi-x)e^(2x)$ as well as in the final solution of $u(x,t)$? –  Sprock Sep 24 '12 at 13:15
    
Those are supposed to be ${e^2}^x$. By the time I figured out how to do it right it wouldnt let me edit anymore. –  Sprock Sep 24 '12 at 13:24
    
1) It was a mistake! $\beta$ is in fact equal to $-2$. 2) Since $U(x,t)=e^{-\alpha x-\beta t}u(x,t)$, it follows that $U(x,0)=e^{-\alpha x}u(x,0)=x(\pi-x)e^{2x}$. –  Mercy Sep 24 '12 at 13:54
    
Duh, when t=o that gets rid of the $\beta$ and $\alpha$ is -2. I should have been able to see that. Thanks. –  Sprock Sep 24 '12 at 17:21

Note that when I made the derivations, I considered ${{\rm e}^{\alpha\,t+\beta\,x}}$ not $ {{\rm e}^{\alpha\,x+\beta\,t}} $ which does not change the result.

The heat equation has the form $$ \frac{\partial u}{\partial t}=C\frac{\partial^2u}{\partial x^2}\,, $$ where $C$ is a constant.

As you mentioned above, you have reached with the problem at the following stage

$$ \left( \alpha-4\,\beta-{\beta}^{2}-2 \right) U \left( x,t \right) + \left( -4-2\,\beta \right) {\frac {\partial }{\partial x}}U \left( x, t \right) -{\frac {\partial ^{2}}{\partial {x}^{2}}}U \left( x,t \right) +{\frac {\partial }{\partial t}}U \left( x,t \right) =0$$ So, you can see from the above equation that you have to get rid of the first two terms on the left of the above equation which implies that

$$ \left( \alpha-4\,\beta-{\beta}^{2}-2 \right)=0 \,,$$ $$ \left( -4-2\,\beta \right) = 0 $$

Solving the above system for $\alpha$ and $\beta$, you should have $\alpha=-2$ and $\beta = -2$.

You solve the resulting heat equation in $U(x,t)$ and then your final solution should be

$$ u(x,t) = {{\rm e}^{-2\,(t+x)}}U(x,t) \,.$$

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