Toronto Math Forum
APM3462016F => APM346Tests => TT2 => Topic started by: Victor Ivrii on November 17, 2016, 03:22:16 AM

Solve by Fourier method
\begin{align}
& u_{tt}u_{xx}=0\qquad 0<x<\pi,\label{11}\\
& u_{x=0}=u_{x=\pi},\qquad u_x_{x=0}=u_x_{x=\pi},\label{12}\\
&u _{t=0}=1,\qquad u_t_{t=0}=0.\label{13}
\end{align}
Hint: $\lambda_n\ge 0$. Also remember how solution looks like in the case of double eigenvalues.

Introduce $u=X(x)T(t)$ and then separation of variables yield $X''+\lambda X=0$ and $T''+\lambda T=0$. Assume that $\lambda=\omega^2$ where $\omega\geq 0$. Then for $\omega\neq 0$, we have solutions $X_1=\sin \omega x$ and $X_2=\cos\omega x$. Applying the boundary conditions to $X_1$ we find
$$0=\sin \omega \pi$$
$$1=\cos\omega \pi$$
Applying the boundary conditions to $X_2$ yields the same thing. The first equation suggests that $\omega=n$ where $n=1,2,...$. The second equation suggests that $\omega=2n+1$, $n=0,1,2,...$. To satisfy both, we must take the second one. So we have $\omega_n=2n+1$, $n=0,1,2,...$. Now for $\omega=0$ the solution is $Ax+B$. But the second B.C. requires $B=B$ so $B=0$. Then applying the first B.C. we find $A=A$ so $A=0$. So we do not have this eigenvalue. Thus our solutions for the spacial part is
$$X_{1,n}=\sin((2n+1)x), X_{2,n}=\cos((2n+1)x)$$
With $\lambda_n=(2n+1)^2$, $n=0,1,2...$. Then the equation for time can be solved immediately: $T=A\sin((2n+1)t)+B\cos((2n+1)t)$. Applying $u_t_{t=0}=0$ we find $A=0$, and now we may write down the general solution:
$$u(x,t)=\sum_{n\geq 0}[A_n\sin((2n+1)x)+B_n\cos((2n+1)x)]\cos((2n+1)t)$$
Applying the condition $u_{t=0}=1$ we have
$$1=\sum_{n\geq 0}A_n\sin((2n+1)x)+B_n\cos((2n+1)x)$$
At this point the coefficients can be calculated via the standard method using orthogonality:
$$A_n=\frac{2}{\pi}\int_0^\pi \sin((2n+1)x)\,dx=\frac{2}{\pi(2n+1)}\cos((2n+1)x)_\pi^0=\frac{4}{\pi (2n+1)}$$
$$B_n=\frac{2}{\pi}\int_0^\pi \cos((2n+1)x)\,dx=0$$
Where the second integral is zero because we will be evaluating sine functions at integer values of $\pi$. So we have our solution:
$$u(x,t)=\frac{4}{\pi}\sum_{n\geq 0}\frac{1}{2n+1}\sin((2n+1)x)\cos((2n+1)t)$$