# Find a formula for the inhomogenous wave equation

Find a formula to the solution to $$\begin{cases} w_{tt} - c^2 w_{xx} = f(x,t) \ \ &\text{for} \ \ x > 0, t > 0\\ w(x,0) = \phi(x) \ \ &\text{for} \ \ x > 0\\ w_t(x,0) = \psi(x) \ \ &\text{for} \ \ x > 0\\ w_x(0,t) = h(t) \ \ &\text{for} \ \ t > 0 \end{cases}$$

I normally don't post a question without somewhat of an attempted solution but I am a little lost here as to how to approach this. Any suggestions are greatly appreciated.

• I do not understand the fourth equation. Shouldn't it be $w_x(0,t)=h(t)$ or something like that ? Commented Jun 8, 2017 at 20:35
• @Tom-Tom Yes my apologies, just edited that Commented Jun 8, 2017 at 21:08

Both the spatial and temporal domains are semi-infinite. You can leverage this and use, for instance, Laplace transform in $t$ to find

$$s^2 W(x;s) - s \phi(x)-\psi(x)-c^2 W_{xx}(x;s) = F(x;s)$$

where $W(x;s)$ and $F(x;s)$ are the Laplace transforms of $w(x,t)$ and $f(x,t)$, respectively, defined as

$$W(x;s) = \int^\infty_0 e^{-st}w(x,t)\,\mathrm{d}t$$

Now you can see this equation as a constant-coefficient inhomogenous second order ODE for $W(x;s)$ as a function of $x$

$$W_{xx} - (s/c)^2 W = - (F+s\phi + \psi)/c^2 \equiv Q(x;s),$$

which is easy to work with, given the initial conditions in $x$. Now we have to work out the solution for $W$. Let $W_1 = e^{sx/c}$ and $W_2 = e^{-sx/c}$ be the solution of the homogenous part of the equation. Then, according to Bender & Orszag (p15), variation of parameters gives the complete solution as

$$W(x;s) = A_1 W_1 + A_2 W_2 + W_p$$

where the particular solution $W_p$ turns out to be

$$W_p(x;s) = -W_1 \int^{x}_0 \frac{Q(\xi;s) W_2(\xi;s)}{R(\xi)}\,\mathrm{d}\xi + W_2 \int^{x}_0 \frac{Q(\xi;s) W_1(\xi;s)}{R(\xi)}\,\mathrm{d}\xi$$

where $R = W_1 \, \partial_x W_2 - W_2 \, \partial_x W_1$ is the Wronksian of $(W_1,W_2)$. You now may impose the initial conditions on $x$ and inverse Laplace transform your solution, which, of course depends on the shape of $f$ through the definition of $Q$.

An alternative

View the PDE as

$$(\partial_t + c \partial_x)(\partial_t - c \partial_x) w = f(x,t)$$

and define $u = (\partial_t - c \partial_x) w$ such that $(\partial_t + c \partial_x) u = f$, the latter being amenable for the use of the method of characteristics, which reads

$$\mathrm{d}x/c = \mathrm{d}t = \mathrm{d}u/f,$$

which tells us that $x - ct = c_1$ is a characteristic line. On the other hand, one must integrate the relation $\mathrm{d}t = \mathrm{d}u/f(ct + c_1, t)$, or equivalently:

$$u = c_2 + \int f(c t + c_1, t) \, \mathrm{d}t$$

put $c_2$ as a function of $c_1$ to have $u = H(x-ct) + \int f(x, t) \, \mathrm{d}t$ where $H$ is an arbitrary function of its argument. Similarly, according to the definition of $u$, one gets the 1st-order PDE for w:

$$u = H(x-ct) + \int f(x, t) \, \mathrm{d}t = (\partial_t - c \partial_x) w$$ from which the method of characteristics provides

$$-\mathrm{d}x/c = \mathrm{d}t = \frac{\mathrm{d}w}{H(x-ct) + \int f(x, t) \, \mathrm{d}t}$$ which provides $w = M(x-ct) + N(x+ct) + P$ is as solution, where $M$ and $N$ are arbitrary functions and $P$ is a particular solution depending on $f(x,t)$. Initial conditions should be imposed and convenient limits should be considered for the integrals (but I'm not sure about satisfying the conditions at $x = 0$).

I hope you find this helpful.

• I am still stuck, could you provide the full solution?> Commented Jun 9, 2017 at 3:24
• I'll try to elaborate on the first part Commented Jun 9, 2017 at 15:27