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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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Welcome to Math.SE! A few guidelines on getting the best help from the community possible: What have you done so far? What do you know? If you can support your questions with answers to these two questions, you will be more likely to receive the best help. – Emily Nov 28 '12 at 16:46
Use separation of variables. – dcs24 Nov 28 '12 at 16:46
Do you know the general form of the solution to the heat equation? If so, then it will be very quick to read off the Fourier coefficients and get a full exact solution. – Matt Nov 28 '12 at 17:50
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. – doraemonpaul Apr 21 '13 at 2:13

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

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



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


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

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

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


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

Note that this solution suitable for $x,t\in\mathbb{C}$ , not only suitable for $0<x<3$ and $t>0$ .

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