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How do we solve the folllowing diffusion problem?



$u(3,t) = 0$

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

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Are you familiar with the method of separation of variables? – Christopher A. Wong Nov 30 '12 at 2:17
This must be a homework. What have you tried? There is just a formula for the general solution involving Fourier coefficients ... and you don't even need to compute those for this problem because you can just read them from the initial distribution! – Matt Nov 30 '12 at 2:48
Well, I think this is very straightforward one. If someone says more than use "separation of variables" will be helping more than expected since you doesn't show your work. Maybe is a good idea to make a question over separation of variables method, since I think that is your problem: that you are missing something about it. – elessartelkontar Nov 30 '12 at 3:41
@elessartelkontar I disagree. I'm teaching a class that does this right now, and if this is a homework problem, then they have probably derived the general solution in class and are not expected to run through separation of variables everytime they solve it. – Matt Nov 30 '12 at 5:36
However, if one is supposed to solve a problem is because he knows the subtleties behind solution methods, if not they are not seeing the important part of the method and it is nonsense to solve a problem. – elessartelkontar Nov 30 '12 at 6:09

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: $$\sum\limits_{n=1}^\infty F(n)\sin\dfrac{n\pi x}{3}=\sin\dfrac{2\pi x}{3}-2\sin\pi x+7\sin\dfrac{5\pi x}{3}$$


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

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

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