# Solve wave equation with Neumann boundary conditions

Solve the wave equation with Neumann boundary conditions: $$\begin{cases} v_{tt} = c^2 v_{xx} \ \ \text{for} \ \ 0 < x < l\\ v_x(0,t) = v_x(l,t) = 0\\ v(x,0) = x(l - x)\\ v_t(x,0) = x\\ \end{cases}$$ Your final answer should remain in series form. You can use, without proof, that, for $0 < x < l$, $$x(l - x) = \frac{l^2}{6} + \sum_{n=1}^{\infty}\frac{2 l ^2}{n^2 \pi ^2}((-1)^{n+1} - 1)\cos\left(\frac{n\pi x}{l}\right); \ \ \ x = \frac{l}{2} + \sum_{n=1}^{\infty}\frac{2l}{n^2 \pi^2}((-1)^{n+1} - 1)\cos\left(\frac{n\pi x}{l}\right)$$

Attempted solution - Using separation of variables we have $$v(x,t) = X(x)T(t)$$ This PDE yields $$X(x)T^{\prime \prime} = c^2 X^{\prime\prime}T(t)$$ Rearranging we have $$\frac{1}{c^2}\frac{T^{\prime\prime}}{T(t)} = \frac{X^{\prime\prime}}{X(x)} = -\lambda$$ Rearranging we get \begin{equation} X^{\prime\prime}(x) + \lambda X(x) = 0 \end{equation} and \begin{equation} T^{\prime\prime}(t) + \lambda c^2 T(t) = 0 \end{equation} ODE $(1)$ is a Strum-Lioville problem when coupled with boundary conditions $$v_x(0,t) = X'(0)T(t) = 0 \Rightarrow X'(0) = 0$$ and $$v_x(l,t) = X'(l)T(t) = 0 \Rightarrow X'(l) = 0$$ For $\lambda = 0$ we get $X^{\prime\prime}(x) = 0$. The solution is $X(x) = c_1 + c_2 x$. Applying boundary conditions \begin{align*} X'(0) &= c_2 = 0\\ X'(l) &= c_2 = 0\\ \end{align*} So we have $\lambda_0 = 0$ and $X_0(x) = 1$ as the first eigenvalue and eigenfunction of the system.\ For $\lambda < 0$, say $\lambda = -\mu^2$, $(1)$ becomes $$X^{\prime\prime}(x) - \mu^2 X(x) = 0$$ $$X'(0) = 0, \ \ X'(l) = 0$$ The characteristic equaiton is $$m^2 - \mu^2 = 0\Rightarrow m = \pm \mu$$ So $X(x) = c_1 e^{-\mu x} + c_2 e^{\mu x}$. Applying boundary conditions. $$X'(0) = -c_1 + c_2 = 0$$ $$X^{\prime}(l) = -c_1 e^{-\mu l} + c_2 e^{\mu l} = 0$$ Solving for $c_1,c_2$, we get $c_1 = c_2 = 0$. Thus $X(x) = 0$.\ Now for $\lambda > 0$, say $\lambda = \mu^2$ $(1)$ becomes $$X^{\prime\prime}(x) + \mu^2 X(x) = 0$$ $$X'(0) = 0, \ \ X'(l) = 0$$ The characteristic equation is $$m^2 + \mu^2 = 0 \Rightarrow m = \pm \mu i$$ So $X(x) = c_1 \cos(\mu x) + c_2 \sin(\mu x)$. Applying boundary conditions, \begin{align*} X'(0) &= c_2 = 0\\ X'(l) &= -\mu c_1 \sin(\mu l) = 0 \end{align*} For $c_1 \neq 0$, \begin{align*} \sin(\mu l) &= 0\\ \sin(\mu l) &= \sin(n\pi) \ \ n = 1,2,\ldots\\ \Rightarrow \mu l &= n \pi\\ \mu &= \frac{n\pi}{l} \end{align*} Thus we have eigenvalues $\lambda_n = \frac{n^2 \pi^2}{l^2}$, and corresponding eigenfunction $X_n(x) = \cos\left(\frac{n\pi x}{l} \right) n = 1,2,\ldots$. Noew we can solve $(2)$ for $T(t)$ using $\lambda_n = \frac{n^2 \pi^2}{l^2}$ $$T^{\prime\prime}(t) + \frac{n^2 \pi^2}{l^2}c^2 T(t) = 0$$ The characteristic equation being $$m^2 + \frac{n^2\pi^2 c^2}{l^2} = 0 \Rightarrow m = \pm \frac{n\pi c}{l}i$$ So $$T_n(t) = a_n\cos\left( \frac{n\pi c}{l} t\right) + b_n \sin\left( \frac{n\pi c}{l}t \right) \ \ n = 1,2,\ldots$$ For $n = 0$, we get $T^{\prime\prime}(t) = 0$ $$T_0(t) = a_0 + b_0 t$$ The solution is now ready to be expressed in series form $$u(x,t) = (a_0 + b_0 t) + \sum_{n=1}^{\infty}\left[a_n \cos\left(\frac{n\pi c}{l}t \right) + b_n \sin\left(\frac{n\pi c}{l}t \right) \right]\cos\left(\frac{n\pi x}{l} \right)$$ All is left is to determine the coefficients $a_n$,$b_n$. This can be done by the initial conditions $$v(x,0) = a_0 + \sum_{n=1}^{\infty}a_n\cos\left(\frac{n\pi x}{l} \right) = x(l-x)$$ In other words, the Fourier cosine series of $x(l-x)$ which is given by $$x(l-x) = \frac{l^2}{6} + \sum_{n=1}^{\infty}\frac{2 l^2}{n^2 \pi^2}((-1)^{n+1} - 1)\cos\left(\frac{n\pi x}{l} \right)$$ Equating coefficients, we get $$a_0 = \frac{l^2}{6}, \ \ a_n = \frac{2 l^2}{n^2 \pi^2}((-1)^{n+} - 1)$$ Next, $$v_t(x,0) = b_0 + \sum_{n=1}^{\infty}\frac{n \pi c}{l} b_n \cos\left(\frac{n\pi x}{l} \right) = x$$ Again a Fourier cosine series, which is given by $$x = \frac{l}{2} + \sum_{n=1}^{\infty}\frac{2 l}{n^2 \pi^2}((-1)^n - 1)\cos\left(\frac{n\pi x}{l} \right)$$ Equating coefficients we get $$b_0 = \frac{l}{2}, b_n = \frac{2l}{n^2 \pi^2}((-1)^n - 1)$$ Thus finally, $$v(x,t) = \frac{l^2}{6} + \frac{l}{2}t + \sum_{n=1}^{\infty}\left[\frac{2 l^2}{n^2 \pi^2}((-1)^{n+1} - 1)\cos\left(\frac{n\pi c}{l}t \right) + \frac{2l}{n^2 \pi^2}((-1)^{n} - 1)\sin\left(\frac{n\pi c}{l}t \right) \right]\cos\left(\frac{n \pi x}{l} \right)$$

I know this is long, but I just wanted to check if this is correct. Any suggestions are greatly appreciated.

• looks right to me. – Mortified Through Math Jun 12 '17 at 19:39
• It seems correct. – VanDerWarden Jun 12 '17 at 19:48
• what disturb me is that $x(l-x)$ does not respect $v_x(0,0)=v_x(l,0)=0$ – enzotib Jun 12 '17 at 21:15

Solve the wave equation with initial and boundary conditions: \begin{cases} v_{tt} = c^2 v_{xx} \ \ \text{for} \ \ 0 < x < l\\ v(x,0) = \varphi(x)\qquad\qquad\qquad\qquad (1)\\ v_t(x,0) = \psi(x)\\ +\text{"boundary conditions"}\\ \end{cases}

1. Solve eigenvalue problem $$-X''_k=\lambda_k X_k$$ with "boundary conditions".
2. Find eigenvalue expansions $\varphi(x)=\sum\varphi_kX_k,\quad \psi(x)=\sum\psi_kX_k$, $$\varphi_k=\frac{1}{\|X_k\|^2}\langle \varphi,X_k\rangle,\quad \psi_k=\frac{1}{\|X_k\|^2}\langle \psi,X_k\rangle.$$ Here $\|u\|^2=\int_0^lu^2dx,\quad \langle u,v\rangle=\int_0^luvdx$.
3. Solve ode problems $$T''_k(t)+c^2\lambda_kT_k(t)=0,\;T_k(0)=\varphi_k,\;T'_k(0)=\psi_k.$$
4. Then solution of problem $(1)$ is $$v=\sum T_k(t)X_k$$

In our case

1. $$\lambda_k=(\pi k)^2,\quad X_k=\cos{\left( \frac{k\pi x}{l}\right) },\;k=0,\,1,\,2,\,\ldots$$
2. $$\varphi_0=\frac{l^2}{6},\quad\psi_0=\frac l2$$ $$\varphi_k=-\frac{2 \left( {{\left( -1\right) }^{k}}+1\right) \, {{l}^{2}}}{{{\pi }^{2}}\, {{k}^{2}}},\quad\psi_k=\frac{2 \left( {{\left( -1\right) }^{k}}-1\right) l}{{{\pi }^{2}}\, {{k}^{2}}},\;k=1,\,2,\,\ldots$$
3. if $k=0$ solve ode $T''(t)=0,\;T(0)=\frac{l^2}{6},\;T'(0)=\frac l2$, $\quad\Longrightarrow\quad$ $T_0(t)=\frac{l t}{2}+\frac{{{l}^{2}}}{6}$

if $k>0$ solve ode $$T''(t)+c^2(\pi k)^2\,T(t)=0,\;T(0)=\varphi_k,\;T'(0)=\psi_k$$

$\quad\Longrightarrow\quad$

$$T_k(t)=\varphi_k \cos{\left( \frac{\pi c k t}{l}\right)+\psi_k\frac{ l \sin{\left( \frac{\pi c k t}{l}\right) }}{\pi c k} }$$

Final solution is $$v=\sum_{k=0}^\infty T_k(t)X_k=\frac{l t}{2}+\frac{{{l}^{2}}}{6}\\ +\sum_{k=1}^{\infty }{\left. \left( \frac{2 \left( {{\left( -1\right) }^{k}}-1\right) \, {{l}^{2}} \sin{\left( \frac{\pi c k t}{l}\right) }}{{{\pi }^{3}} c\, {{k}^{3}}}-\frac{2 \left( {{\left( -1\right) }^{k}}+1\right) \, {{l}^{2}} \cos{\left( \frac{\pi c k t}{l}\right) }}{{\pi^{2}}\, {{k}^{2}}}\right) \cos{\left( \frac{\pi k x}{l}\right) }\right.}.$$