Consider the following wave equation: $$\begin{align} u_{tt}&=u_{xx},\quad x\in(0,\pi),\quad t>0,\\ u(x,0)&=0,\quad u_t(x,0)=0,\\ \color{red}{u(0,t)}&\color{red}{=\phi(t)},\quad u_x(\pi,t)=0. \end{align}$$ Since the highlighted boundary condition is non-homogeneous, I cannot apply the method of separation of variables. What can I do to solve this?

I am given a hint to use $u(x,t)=v(x,t)+\phi(t)$, but I have no idea how. Any other hint would be greatly appreciated.

Edit: Performing the hinted substitution, I managed to come up with

$$\begin{align} v_{tt}+\phi''&=v_{xx},\quad x\in(0,\pi),\quad t>0,\\ v(x,0)&=-\phi(0),\quad v_t(x,0)=-\phi'(0),\\ v(0,t)&=0,\quad v_x(\pi,t)=0. \end{align}$$

I now have homogeneous boundary conditions, but the PDE is non-homogeneous! I suppose it can be solved using Fourier transforms?

  • $\begingroup$ By the way, the hint implies that I can use that to obtain homogeneous boundary conditions. $\endgroup$ – wjm Nov 28 '12 at 20:38
  • $\begingroup$ To work according to the hint, substitute $v(x,t)=u(x,t)-/phi (t)$ to receive a non homogeneous pde of the function v with homogeneous boundary conditions. $\endgroup$ – nodwj Nov 28 '12 at 20:59
  • $\begingroup$ Since this problem also contains the conditions $u(x,0)=0$ and $u_t(x,0)=0$ , we do not necessarily have to consider the issues about the conditions $u(0,t)=\phi(t)$ and $u_x(\pi,t)=0$ . $\endgroup$ – doraemonpaul Nov 28 '12 at 22:57
  • $\begingroup$ You can now use Fourier transforms or D'Alembert's Formula. $\endgroup$ – Hans Engler Nov 29 '12 at 18:22
  • $\begingroup$ If I want to use Fourier transforms, wouldn't $\mathcal F(\phi'')$ be undefined? $\endgroup$ – wjm Dec 2 '12 at 20:07

Using Laplace Transform you have $$ s^2 \hat{u} - \hat{u}_{xx} = 0 $$ with boundary conditions $$ \hat{u}(0,s) = \hat{\phi}(s), \quad \hat{u}_x(\pi,s) = 0 $$ wich leads to the solution $$ \hat{u}(x,s) = \hat{\phi}(s) \frac{\cosh s(\pi-x)}{\cosh \pi s} $$This is fantastic modulo inverting the Laplace transform, which I think it can be done, but I'm still not clear on the details.

Another way would be taking $v(x,t) = u(x,t) - \phi(t)$ and using separation of variables. This is the battle horse and it's kind of fool proof, but it involves a lot of work.

The most intuitive and, in my opinion, beautiful way to solve the equation is using the fact that for any parallelogram $ABCD$ in the $xt$-plane bounded by four characteristic lines, the sums of the values of u at opposite vertices are equal, that is $$ u(A) + u(C) = u(B) + u(D) \tag{1} $$ If we divide the $xt$-plane in regions delimited by the characteristis as shown in the figure

enter image description here $\hskip1in$enter image description here

and take $A = (x,t) \in \mbox{II}$, $B = (0,t_B) \in \mbox{II}$, $C = (x_C,0) \in \mbox{I}$, $D=(x_D,t_D) \in \mbox{I}$, D'Alambert solution implies that $$ u(x,t) = 0, \quad (x,t) \in \mbox{I} $$ and using $(1)$, we have that $$ u(x,t) = \phi(t-x), \quad (x,t) \in \mbox{II} $$ For III, no wave can reach the region, hence $$ u(x,t) = 0, \quad (x,t) \in \mbox{III} $$ For region IV, take $A = (x,t) \in \mbox{IV}$, $B = (0,t_B) \in \mbox{II}$, $C = (x_C,t_C) \in \mbox{I}$ and $D = (x_D,t_D) \in \mbox{III}$ and equation $(1)$ implies that $$ u(x,t) = \phi(t-x), \quad (x,t) \in \mbox{IV} $$

For Region V we take $A = (x,t) \in \mbox{V}$, $B = (0,t_B) \in \mbox{V}$, $C = (x_C,t_C) \in \mbox{III}$ and $D = (x_D,t_D) \in \mbox{III}$ we have that $$ u(x,t) = \phi(t-x), \quad (x,t) \in \mbox{V} $$

Region VI is more interesting: taking $A = (x,t) \in \mbox{VI}$, $B = (0,t_B) \in \mbox{II}$, $C = (x_C,t_C) \in \mbox{II}$ and $D = (\pi,t_D) \in \mbox{VI}$, equation $(1)$ implies that $$ u(x,t) + u(x_C,t_C) = u(0,t_B) + u(\pi,t_D) \, \Longrightarrow $$ $$ u(x,t) = \phi(t_B) - \phi(t_C - x_C) + u(\pi, t_D) $$ but \begin{align} t_B &= t - x \\ t_D &= t + x - \pi\\ x_C &= \pi - x\\ t_C &= t - \pi \end{align} and then $$ u(x,t) = \phi(t-x) - \phi(t + x -2\pi) + u(\pi,t + x - \pi) $$ Now, using $u_x(\pi,t) = 0 = -2 \phi'(t - \pi) + u_t(\pi,t)$ we have $$ u(x,t) = \phi(t-x) + \phi(t + x - 2\pi), \quad (x,t) \in \mbox{VI} $$

In region VII, taking $A = (x,t) \in \mbox{VII}$, $B = (0,t_B) \in \mbox{V}$, $C = (x_C,t_C) \in \mbox{IV}$ and $D = (\pi,t_D) \in \mbox{VI}$, equation $(1)$ implies that $$ u(x,t) = \phi(t-x) + \phi(t + x - 2\pi), \quad (x,t) \in \mbox{VII} $$ So far, it's easy to understand the results of all regions. For I and III, there is no wave, for II, IV and V, there is only the wave originating from the boundary $x = 0$. In VI there are two waves, the one originated at II plus the reflection on the boundary $x = \pi$. In VII, there is the wave from $x=0$ and region VI. This logic tells us that in VIII there will be three waves: the one from the boundary $x = 0$, the one coming from VI and it's reflection. To see this, we take $A = (x,t) \in \mbox{VIII}$, $B = (0,t_B) \in \mbox{VIII}$, $C = (x_C,t_C) \in \mbox{VI}$ and $D = (\pi,t_D) \in \mbox{VI}$, \begin{align} u(x,t) &= \phi(t_B) - \phi(t_C - x_C) - \phi(t_C + x_C -2\pi) + 2\phi(t_D - \pi)\\ &= \phi(t-x) + \phi(t + x - 2\pi) - \phi(t - x - 2\pi), \quad (x,t) \in \mbox{VIII} \end{align}

Why the change of sign on the reflecting wave? one might ask. The answers is simple: the boundary $x = 0$ is hard (Dirichlet), while the boundary in $x = \pi$ is soft (Neumann).

For region IX, we take $A = (x,t) \in \mbox{IX}$, $B = (0,t_B) \in \mbox{V}$, $C = (x_C,t_C) \in \mbox{V}$ and $D = (\pi,t_D) \in \mbox{IX}$, \begin{align} u(x,t) &= \phi(t_B) - \phi(t_C-x_C) + u(\pi,t_D)\\ &= \phi(t-x) - \phi(t + x - 2\pi) + u(\pi, t + x - \pi) \end{align} Again, using the boundary condition in $x = \pi$ we have $u(\pi,t) = 2\phi(t - \pi)$ for $(x,t) \in \mbox{IX}$ and $$ u(x,t) = \phi(t-x) + \phi(t + x - 2\pi), \quad (x,t) \in \mbox{IX} $$

There is clearly a pattern arising in the triangular regions. In the parallelograms, a little more work must be performed, but all in all, the table is set to propose a general solution by induction.

Can you finish it off?

  • $\begingroup$ This is really the best method indeed, as this method can get the results that suitable for all $\phi(t)$ , unlike many other methods that suitable for only some $\phi(t)$ . I hope that you can show off this talent in the further overdetermining PDE problems. $\endgroup$ – doraemonpaul Nov 30 '12 at 0:59
  • $\begingroup$ @doraemonpaul Thanks! Props for D'Alambert and company that developed such beautiful theory for the 1-D wave equation :) $\endgroup$ – Pragabhava Nov 30 '12 at 1:02
  • $\begingroup$ I asked an extended question math.stackexchange.com/questions/266290, please have a look. $\endgroup$ – doraemonpaul Jan 3 '13 at 0:50
  • $\begingroup$ I am a little late to this party, but I do want to contribute by saying it is possible to compute that inverse laplace transform. In this link: math.stackexchange.com/a/2548022/17243 OP asked to solve the case when one side is forced by a sine function. I'm thinking it may be possible to write a general formula for suitable $\hat{\phi}$. $\endgroup$ – DaveNine Nov 15 '18 at 12:05

The function $v$ satisfies an inhomogeneous wave equation $v_{tt} = v_{xx} - \phi_{tt}$ with homogeneous boundary conditions and initial value $v(x,0) = -\phi(0)$. Can you solve such an equation?


I have a better approach.

Since this problem also contains the conditions $u(x,0)=0$ and $u_t(x,0)=0$ , we do not necessarily have to consider the issues about the conditions $u(0,t)=\phi(t)$ and $u_x(\pi,t)=0$ .

Because the PDE is an overdetermining problem and with the conditions $u(x,0)=0$ and $u_t(x,0)=0$ , So it is better to take Laplace transform on $t$ :

$\mathcal{L}_{t\to s}\{u_{tt}\}=\mathcal{L}_{t\to s}\{u_{xx}\}$





$u(0,t)=\phi(t)$ :


$\because t>0$

$\therefore f(t)+g(t)=\phi(t)$


$\therefore u(x,t)=f(t+x)H(t+x)+(\phi(t-x)-f(t-x))H(t-x)$


$u_x(\pi,t)=0$ :


$t\to t+\pi$ :


$\because t>0$

$\therefore f_x(t+2\pi)+f_x(t)-\phi_x(t)=0$


$\therefore u(x,t)=f(t+x)H(t+x)+(\phi(t-x)-f(t-x))H(t-x)$ , where $f(t)$ is the general solution of $f_x(t+2\pi)+f_x(t)=\phi_x(t)$


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