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The following Heat equation:

  • $u_t=4 u_{xx} $, $(t,x) \in (0, \infty)\times (0,1)$

  • $u(0,x)=x(1-x), x \in (0,1)$

  • $u(t,0)=0=u(t,1), t \in (0, \infty)$

should have the following solution:

$u(t,x)= \frac {8}{\pi^3} \sum_{n=1} ^{\infty} \frac {1}{(2n-1)^3} e^{-4(2n-1)^2 \pi^2 t} \sin((2n-1)\pi x) $

So to get this answer I used seperation of variables:

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

$\frac{X''}{X}=\frac{T'}{4T}=-\lambda$

Now I set ${\lambda = \alpha^2}$:

  • $(I): X''+\alpha^2 X=0$
  • $(II): T'+\alpha^2 T=0$

solving $(I)$:

$X(x)=c_1 \cos(\alpha x)+ c_2 \sin(\alpha x)$

Now using the boundary conditions $u(t,0)=0=u(t,1)$:

$c_1=0$ and $\sin(\alpha)=0$ so $\alpha= n \pi$

It follows that $X(x)=a_n \sin(n \pi x)$

Now I get the following solution, which is not identical to the given solution:

$u(t,x)=\sum_{n=0} ^{\infty} c_n e^{-4n^2 \pi^2 t} \sin(n\pi x)$

I know how to proceed from here on, but I don't understand how $\alpha = (2n-1) \pi $ in the given solution. We are asked to show that we get the given solution as a final solution for the equation.

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    $\begingroup$ Can you use the other boundary condition that $u(0,x)=x(1-x), x \in (0,1)$ and express $x(1-x)$ as a Fourier series and compare coefficients? $\endgroup$
    – user1174498
    Commented Jun 3, 2023 at 21:27
  • $\begingroup$ Can you please explain? I know how to do the Fourier series for x(1-x), but how to compare coefficients? If you mean how to get $c_n$, I can do that. But I can't proceed, since in my case $\alpha \neq (2n-1) \pi$ not like the expected solution, where $\alpha= (2n-1) \pi$ $\endgroup$ Commented Jun 3, 2023 at 22:10
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    $\begingroup$ Your $\alpha$ is totally correct, there is a different reason why in the solution there only appear the terms with $2n-1$. As @KevinPWalker suggested, if you express $x(1-x)$ in a Fourier series and compare coefficients, you probably get something like $c_k = \frac{8}{\pi^3} \frac{1}{(2n-1)^3}$, if $k=2n-1$ for some $n\in\mathbb{N}$, and $0$ otherwise. If you then plug these coefficients into your Fourier series $u(t,x)$, you get exactly the desired solution. $\endgroup$
    – stange
    Commented Jun 3, 2023 at 22:41
  • $\begingroup$ Ok, thank you, I understand now. $\endgroup$ Commented Jun 4, 2023 at 11:57

1 Answer 1

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We know that $u(x, 0) =x(1-x)$ and from the solution we Have that $u(x, 0) =\sum c_k\sin(k\pi x) $, so the unique posibility is that the coefficients $c_n$ are the same as the ones on the sine Fourier series of $x(1-x) $. If you compute that series you will get that $$x(1-x) = \frac{4}{\pi^3}\sum_{n=1}^{\infty}\frac{\left(1-\left(-1\right)^{n}\right)}{n^{3}}\sin\left(n\pi x\right),$$ since $1-(-1) ^n$ is zero for even $n$ and $2$ for odd $n$, we get that $c_k=0$ if $k=2n$ and $c_k=\frac{8}{\pi^3}\frac{1}{(2n-1) ^3}$ if $k=2n-1$.

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  • $\begingroup$ Thank you for the explanation $\endgroup$ Commented Jun 4, 2023 at 11:57

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