1D wave equation $u_{t t} - c^2 u_{x x} = 0$ with forcing at one point

Question

I want to solve the 1-dimensional wave equation $$u_{t t} - c^2 u_{x x} = 0$$ for $u(x,t)$ in the domain $-\infty \leq x \leq \infty, t\geq 0$ with initial data $$ u(x,0) = g(x) \hspace{4mm} , \hspace{4mm} u_t(x,0) = 0$$ and forcing at location $x=0$ by a prescribed value of $$u(0,t) = h(t) \; .$$ The condition $$ h(0) = g(0) $$ is satisfied. The situation is shown in the following figure.

As is well known, in the case of zero forcing, the solution is $$ u = \frac{1}{2} [g(x+c t) + g(x-c t) ] \; . $$ How do I deal with the forcing? Can you show me a suitable approach or recommend literature that explains this? Thank you.

Answer 1

I found a general solution in the book "Handbook of linear partial differential equations for engineers and scientists, 2nd edition", section 6.1.2, for the domain $0 \leq x < \infty$. In my case, the solution is $$u(x,t) = \frac{1}{2} [g(x+c t) + g(x-c t)] + \operatorname{H}\left( t-\frac{x}{c} \right) h\left( t - \frac{x}{c} \right) $$ where $\operatorname{H}\left( \cdot \right)$ is the Heaviside step function.

Answer 2

In the case of the semi-infinite string with fixed end. You have ( these are some notes)

$$ \begin{align}\begin{cases} \frac{\partial^{2} u}{\partial t^{2}} = c^{2}\frac{\partial^{2} u}{\partial x^{2}} & x > 0 , t > 0 \\ u(x,0) = f(x) & x \geq 0 \\ \frac{\partial u}{\partial x}(x,0) = g(x) & x \geq 0 \\ u(0,t) = h(t) & t \geq 0 \end{cases} \end{align} \tag{1}$$

Then d'Alembert's Formula is

$$u(x,t) = \phi(x+ct) +\psi(x-ct) \tag{2} $$

you plug in $x=0$ to get $h(t)$

$$ u(0,t) = h(t) = \phi(ct) +\psi(-ct) \tag{3}$$

you then introduce some constant $\alpha = -ct$

$$ \psi(\alpha) = h(\frac{-\alpha}{c}) - \phi(-\alpha) \tag{4} $$

you then replace $\alpha$ with $x-ct$

$$\psi(x-ct) = h(t-\frac{x}{c}) - \phi(ct-x) \tag{5} $$

then you get for $ 0 \leq x \leq ct$

$$ u(x,t) = h(t-\frac{x}{c}) + \frac{1}{2}\big[f(x+ct) -f(ct-x) \big] + \frac{1}{2c} \int_{ct-x}^{x+ct} g(\tau) d \tau \tag{6} $$

and for $ x > ct$ we have

$$ u(x,t) = \frac{1}{2}\big[f(x+ct) -f(x-ct) \big] + \frac{1}{2c} \int_{x-ct}^{x+ct} g(\tau) d \tau \tag{7} $$

now popping some stuff in..

$$ \begin{align}\begin{cases} \frac{\partial^{2} u}{\partial t^{2}} = c^{2}\frac{\partial^{2} u}{\partial x^{2}} & x > 0 , t > 0 \\ u(x,0) = g(x) & x \geq 0 \\ \frac{\partial u}{\partial x}(x,0) = 0 & x \geq 0 \\ u(0,t) = h(t) & t \geq 0 \end{cases} \end{align} \tag{8}$$

we get the following answer

$$ u(x,t) = \begin{align}\begin{cases} h(t-\frac{x}{c}) + \frac{1}{2}\big[g(x+ct) -g(ct-x) \big] & 0 \leq x \leq ct \\ \frac{1}{2}\big[g(x+ct) -g(x-ct) \big] & x > ct \end{cases} \end{align} \tag{9}$$

you can enforce your last condition fairly easily because it is $u(0,0)$