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How do we solve the folllowing diffusion problem?

$u_t=4u_{xx}$

$u(0,t)=0$

$u(3,t) = 0$

$u(x,0)=\sin(2\pi x/3)-2\sin(\pi x)+7\sin(5\pi x/3)$

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    I don't really understand what either of the two of you are talking about. It's OK to do calculations in the beginning without fully understanding. This doesn't mean you will never understand the method. Why is it disgusting to a apply a formula that has been fully derived in class? I actually find it disgusting to make students repeatedly re-derive the same thing over and over again. This is the beauty of math. Once you've derived something it is true forever.2012-12-01

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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: $\sum\limits_{n=1}^\infty C_t(n,t)\sin\dfrac{n\pi x}{3}=-\dfrac{4n^2\pi^2}{9}\sum\limits_{n=1}^\infty C(n,t)\sin\dfrac{n\pi x}{3}$

$\therefore C_t(n,t)=-\dfrac{4n^2\pi^2}{9}C(n,t)$ And:

$\dfrac{C_t(n,t)}{C(n,t)}=-\dfrac{4n^2\pi^2}{9}\rightarrow\int\dfrac{C_t(n,t)}{C(n,t)}dt=\int-\dfrac{4n^2\pi^2}{9}dt$ $\ln C(n,t)=-\dfrac{4n^2\pi^2t}{9}+f(n)\rightarrow C(n,t)=F(n)e^{-\frac{4n^2\pi^2t}{9}}$

$\therefore u(x,t)=\sum\limits_{n=1}^\infty F(n)e^{-\frac{4n^2\pi^2t}{9}}\sin\dfrac{n\pi x}{3}$ Given that: $u(x,0)=\sin\dfrac{2\pi x}{3}-2\sin\pi x+7\sin\dfrac{5\pi x}{3}$ We can derive: $\sum\limits_{n=1}^\infty F(n)\sin\dfrac{n\pi x}{3}=\sin\dfrac{2\pi x}{3}-2\sin\pi x+7\sin\dfrac{5\pi x}{3}$

$F(n)=\begin{cases}1&\text{when}~n=2\\-2&\text{when}~n=3\\7&\text{when}~n=5\\0&\text{otherwise}\end{cases}$

$\therefore u(x,t)=e^{-\frac{16\pi^2t}{9}}\sin\dfrac{2\pi x}{3}-2e^{-4\pi^2t}\sin\pi x+7e^{-\frac{100\pi^2t}{9}}\sin\dfrac{5\pi x}{3}$

Note that this solution suitable for $x,t\in\mathbb{C}$ .