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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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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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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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