4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ looks right to me. $\endgroup$ – Mortified Through Math Jun 12 '17 at 19:39
  • $\begingroup$ It seems correct. $\endgroup$ – VanDerWarden Jun 12 '17 at 19:48
  • $\begingroup$ what disturb me is that $x(l-x)$ does not respect $v_x(0,0)=v_x(l,0)=0$ $\endgroup$ – enzotib Jun 12 '17 at 21:15
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.