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I need help with this exercise:

Given the PDE $$u_t=-10u_{xx}\tag{1},$$ with periodic boundary conditions in $[-1,1]$: $$u(-1,t)=u(1,t), \qquad u_x(-1,t)=u_x(1,t).$$

A) Obtain the solution $u(x,t)$ if $u(x,0)=u_0(x)=\sin(\frac{\pi}{10}x)$

B) If we consider as initial data $u_0(x)=x^2$ defined also in $[-1,1]$, does (1) have a solution in this case?

Thank you very much.

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2 Answers 2

up vote 3 down vote accepted

First of all, periodic boundary conditions look like a pair of boundary conditions, e.g. like $$u(-1,t)=u(1,t), \quad u_x(-1,t)=u_x(1,t).$$

Start by separating variables, finding the eigenvalues $\lambda_n$, eigenfunctions $X_n(x)$, and temporal solutions $T_n(t)$.

Superimpose those to get something of the form $u(x,t)=\sum_n c_n X_n(x)T_n(t)$.

To obtain the coefficients in the series solution above, evaluate $u(x,t)$ at $t=0$, apply the initial data, and recognize this as an appropriate Fourier series. Use your knowledge of Fourier series and/or orthogonality on $[-1,1]$ to deduce formulas for the coefficients.

Edit based on progress in the comments:

The superimposed solution should look like $$u(x,t)=\sum_{n=1}^\infty [a_n\cos(n\pi x)+b_n\sin(n\pi x)]e^{10(n\pi)^2 t}.$$

Then, the initial condition $u(x,0)=f(x)$, $-1<x<1$, leads to $$u(x,0)=\underbrace{\sum_{n=1}^\infty a_n\cos(n\pi x)+b_n\sin(n\pi x)=f(x)}, \quad -1<x<1,$$ which can be found as standard Fourier coefficients.

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Thanks, John. I could do A without problems, I think. But, in B, I follow the same method as in A, and I got what it seems a solution too. ¿Should I find any problem with $u_0(x)=x^2$? –  Mark_Hoffman Jan 21 '13 at 22:40
    
(For B, I got $u(x,t)=\sum_{n=1}^{\infty}\frac{4(-1)^n}{(n\pi)^2}cos(n\pi x)*e^{10n^2\pi^2t}$) This verifies the boundary and initial conditions, and I think that also verifies, $u_t=-10u_{xx}$. ¿It's ok, or are there any problems considering $x^2$ as an initial function in the pde? –  Mark_Hoffman Jan 21 '13 at 23:07
    
Check your eigenfunction calculations again. You should get sines and cosines for periodic BCs. –  JohnD Jan 21 '13 at 23:57
    
Sorry I misunderstood. Yes, you are correct for that specific initial data. –  JohnD Jan 22 '13 at 0:34
1  
What you have is fine: $b_n=0$ for all $n$, and $a_n=\int_{-1}^1 x^2\cos(n\pi x)\,dx$. There is what happens when you are computing the full Fourier series of an even function (such as $x^2$): the sine coefficients vanish leaving simply a Fourier cosine series! (Similarly, for an odd function, the cosine coefficients would vanish leaving a Fourier sine series. You should be able to prove it with just calculus.) –  JohnD Jan 22 '13 at 0:58

I doubt that whether you can really solve it even for A).

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

Then $X(x)T'(t)=-10X''(x)T(t)$

$\dfrac{T'(t)}{10T(t)}=-\dfrac{X''(x)}{X(x)}=n^2\pi^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=10n^2\pi^2\\X''(x)+n^2\pi^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(n)e^{10n^2\pi^2t}\\X(x)=\begin{cases}c_1(n)\sin n\pi x+c_2(n)\cos n\pi x&\text{when}~n\neq0\\c_1x+c_2&\text{when}~n=0\end{cases}\end{cases}$

$\therefore u(x,t)=\sum\limits_{n=0}^\infty C_1(n)e^{10n^2\pi^2t}\sin n\pi x+\sum\limits_{n=0}^\infty C_2(n)e^{10n^2\pi^2t}\cos n\pi x$

$u(-1,t)=u(1,t)$ :

$\sum\limits_{n=0}^\infty C_1(n)e^{10n^2\pi^2t}\sin(-n\pi)+\sum\limits_{n=0}^\infty C_2(n)e^{10n^2\pi^2t}\cos(-n\pi)=\sum\limits_{n=0}^\infty C_1(n)e^{10n^2\pi^2t}\sin n\pi+\sum\limits_{n=0}^\infty C_2(n)e^{10n^2\pi^2t}\cos n\pi$

$-\sum\limits_{n=0}^\infty C_1(n)e^{10n^2\pi^2t}\sin n\pi+\sum\limits_{n=0}^\infty C_2(n)e^{10n^2\pi^2t}\cos n\pi=\sum\limits_{n=0}^\infty C_1(n)e^{10n^2\pi^2t}\sin n\pi+\sum\limits_{n=0}^\infty C_2(n)e^{10n^2\pi^2t}\cos n\pi$

$2\sum\limits_{n=0}^\infty C_1(n)e^{10n^2\pi^2t}\sin n\pi=0$

$C_1(n)=0$

$\therefore u(x,t)=\sum\limits_{n=0}^\infty C_2(n)e^{10n^2\pi^2t}\cos n\pi x$

$u_x(x,t)=-\sum\limits_{n=0}^\infty n\pi C_2(n)e^{10n^2\pi^2t}\sin n\pi x$

But $u_x(-1,t)\neq u_x(1,t)$

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