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Solve $\frac{\partial u}{\partial t} = 2\frac{\partial^2 u}{\partial x^2}$ with $0 < x < 3, t > 0$, given that $u(0,t) = u(3,t) = 0$, and $u(x,0) = 5\sin 4\pi x - 3\sin 8\pi x + 2\sin 10 \pi x.$ Note that $u(x,t)$ is bounded.

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    This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date.2013-04-21

1 Answers 1

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Let $u(x,t)=\sum\limits_{n=1}^\infty C(n,t)\sin\dfrac{n\pi x}{3}$ so that it automatically satisfies $u(0,t)=u(3,t)=0$ ,

Then $\sum\limits_{n=1}^\infty\dfrac{\partial C(n,t)}{\partial t}\sin\dfrac{n\pi x}{3}=-\dfrac{2n^2\pi^2}{9}\sum\limits_{n=1}^\infty C(n,t)\sin\dfrac{n\pi x}{3}$

$\therefore\dfrac{\partial C(n,t)}{\partial t}=-\dfrac{2n^2\pi^2}{9}C(n,t)$

$\dfrac{dC(n,t)}{C(n,t)}=-\dfrac{2n^2\pi^2}{9}dt$

$\int\dfrac{dC(n,t)}{C(n,t)}=\int-\dfrac{2n^2\pi^2}{9}dt$

$\ln C(n,t)=-\dfrac{2n^2\pi^2t}{9}+f(n)$

$C(n,t)=F(n)e^{-\frac{2n^2\pi^2t}{9}}$

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

$u(x,0)=5\sin4\pi x-3\sin8\pi x+2\sin10\pi x$ :

$\sum\limits_{n=1}^\infty F(n)\sin\dfrac{n\pi x}{3}=5\sin4\pi x-3\sin8\pi x+2\sin10\pi x$

$F(n)=\begin{cases}5&\text{when}~n=12\\-3&\text{when}~n=24\\2&\text{when}~n=30\\0&\text{otherwise}\end{cases}$

$\therefore u(x,t)=5e^{-32\pi^2t}\sin4\pi x-3e^{-128\pi^2t}\sin8\pi x+2e^{-200\pi^2t}\sin10\pi x$

Note that this solution suitable for $x,t\in\mathbb{C}$ , not only suitable for $0 and $t>0$ .