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Solve the boundary value problem $$\begin{cases} \displaystyle \frac{\partial u}{\partial t} = 2 \frac{\partial^2 u}{\partial x^2} \\ \ \\u(0,t) = 10 \\ u(3,t) = 40 \\ u(x, 0) = 25 \end{cases}$$

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This uses separation of variables and Fourier series. I am sure there are many resources you can find online telling you how to solve a problem like this. –  nayrb Nov 28 '12 at 16:36

2 Answers 2

HINT: Let's make the boundary condition homogenous by $v(x,t)=u(x,t)-10-10x$.
(How to see that?). The pde for $v(x,t)$ is thus: (Notice that $v_t=u_t$, $v_{xx}=u_{xx}$, and why?)

$$\begin{cases} v_t - 2 v_{xx} = 0, & 0<x<3, t>0, \\ v(0,t)=v(3,t)= 0, & t\geqslant 0, \\ v(x,0)=15-10x, & 0<x<l. \end{cases}$$

This question should be familiar to you now, as it's got homogenous b.c. and good i.c.. It shouldn't be hard once you find $v$, because you get $u$ the instant you get $v$.

Next: Familiar stuffs: Apply the method of separation of variables, i.e. $v(x,t)=X(x)T(t)$ and solve for the eigenvalues and eigenfunctions. Expand $f(x)=15-x$ into the Fourier series based on the eigenvalues, and solve for the specific result.

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Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=2X''(x)T(t)$

$\dfrac{T'(t)}{2T(t)}=\dfrac{X''(x)}{X(x)}=-\dfrac{n^2\pi^2}{9}$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-\dfrac{2n^2\pi^2}{9}\\X''(x)+\dfrac{n^2\pi^2}{9}X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(n)e^{-\frac{2n^2\pi^2t}{9}}\\X(x)=\begin{cases}c_1(n)\sin\dfrac{n\pi x}{3}+c_2(n)\cos\dfrac{n\pi x}{3}&\text{when}~n\neq0\\c_1x+c_2&\text{when}~n=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\sum\limits_{n=0}^\infty C_3(n)e^{-\frac{2n^2\pi^2t}{9}}\sin\dfrac{n\pi x}{3}+\sum\limits_{n=0}^\infty C_4(n)e^{-\frac{2n^2\pi^2t}{9}}\cos\dfrac{n\pi x}{3}$

$u(0,t)=10$ :

$C_2+\sum\limits_{n=0}^\infty C_4(n)e^{-\frac{2n^2\pi^2t}{9}}=10$

$\sum\limits_{n=0}^\infty C_4(n)e^{-\frac{2n^2\pi^2t}{9}}=10-C_2$

$C_4(n)=\begin{cases}10-C_2&\text{when}~n=0\\0&\text{when}~n\neq0\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\sum\limits_{n=0}^\infty C_3(n)e^{-\frac{2n^2\pi^2t}{9}}\sin\dfrac{n\pi x}{3}+10-C_2=C_1x+10+\sum\limits_{n=1}^\infty C_3(n)e^{-\frac{2n^2\pi^2t}{9}}\sin\dfrac{n\pi x}{3}$

$u(3,t)=40$ :

$3C_1+10=40$

$C_1=10$

$\therefore u(x,t)=10x+10+\sum\limits_{n=1}^\infty C_3(n)e^{-\frac{2n^2\pi^2t}{9}}\sin\dfrac{n\pi x}{3}$

$u(x,0)=25$ :

$10x+10+\sum\limits_{n=1}^\infty C_3(n)\sin\dfrac{n\pi x}{3}=25$

$\sum\limits_{n=1}^\infty C_3(n)\sin\dfrac{n\pi x}{3}=-10x+15$

$C_3(n)=\dfrac{2}{3}\int_{3k}^{3(k+1)}(-10x+15)\sin\dfrac{n\pi x}{3}dx$ , $\forall k\in\mathbb{Z}$ , $x\in(3k,3(k+1))$

$C_3(n)=\left[\dfrac{20x-30}{n\pi}\cos\dfrac{n\pi x}{3}-\dfrac{60}{n^2\pi^2}\sin\dfrac{n\pi x}{3}\right]_{3k}^{3(k+1)}$ , $\forall k\in\mathbb{Z}$ , $x\in(3k,3(k+1))$

$C_3(n)=\dfrac{(-1)^{n(k+1)}(60k+30)-(-1)^{nk}(60k-30)}{n\pi}$ , $\forall k\in\mathbb{Z}$ , $x\in(3k,3(k+1))$

$\therefore u(x,t)=10x+10+\sum\limits_{n=1}^\infty\dfrac{(-1)^{n(k+1)}(60k+30)-(-1)^{nk}(60k-30)}{n\pi}e^{-\frac{2n^2\pi^2t}{9}}\sin\dfrac{n\pi x}{3}$ , $\forall k\in\mathbb{Z}$ , $x\in(3k,3(k+1))$

Hence $u(x,t)=10x+10+\sum\limits_{k=-\infty}^\infty\sum\limits_{n=1}^\infty\dfrac{(-1)^{n(k+1)}(60k+30)-(-1)^{nk}(60k-30)}{n\pi}\prod_{3k,3(k+1)}(x)e^{-\frac{2n^2\pi^2t}{9}}\sin\dfrac{n\pi x}{3}~,~x\in\mathbb{R}$

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For homework (and, for some of us, suspected homework) we usually give hints instead of a full solution. –  Ross Millikan Dec 1 '12 at 3:18

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