The transverse displacement $u(x, t)$ of a semi-infinite elastic string satisfies: $$\frac{\partial^2u}{\partial t^2} = c^2 \frac{\partial^2u}{\partial x^2}$$
$ x > 0, t > 0$ with initial conditions $$u(x,0) = \frac{\partial}{\partial t}u(x,0) = 0$$ $x > 0$
and boundary condition $$\beta \frac{\partial}{\partial x} u(0,t) = f(t)$$ $t > 0$. Show, using Laplace transforms that the solution can be written as: ($H(x)$ is the Heaviside function)
$$u(x,t) = \frac{c}{\beta}H\left(t - \frac{x}{c}\right) \int_{0}^{t - x/c}f(u)du$$
Can you interpret this result physically?
I'm struggling to get the answer and need some help. Let me know if there are any typos.
Let $L$ be the Laplace transform. Take the Laplace transform of both sides of the PDE to get:
$L({\frac{\partial^2u}{\partial t^2}})$ = $L(c^2 \frac{\partial^2u}{\partial x^2})$ $\implies$ $s^2U(s,x) - su(x,0)-u_t(x,0)$ = $c^2 \frac{\partial^2U}{\partial x^2}$
Apply the initial conditions above to get:
$s^2U(s,x)$ = $c^2 \frac{\partial^2U}{\partial x^2}$ $\implies$ $\frac{s^2}{c^2}U(s,x)$ = $ \frac{\partial^2U}{\partial x^2}$
Since we only have $x$, we can think of this as an ODE (eliminate dependency on time variable):
$\frac{s^2}{c^2}U(x)$ = $ \frac{\partial^2U(x)}{\partial x^2}$
From here, you should be able to solve the above as a BVP, with the boundary condition given as $-\beta u_x(0,t) = f(t)$
Take Laplace transform of the BC:
$L(-\beta u_x(0,t)) = -\beta L(u_x(0,t)) = L(f(t)) \implies -\beta \frac{dU(0,s)}{dx} =F(s) \implies \frac{dU(0,s)}{dx}= -\frac{F(s)}{\beta}$
Use the general solution you got to get that:
$\frac{dU(0,s)}{dx} = A \frac{s}{c} -B \frac{s}{c} = -\frac{F(s)}{\beta} \implies A -B = -\frac{cF(s)}{s\beta}$
By definition of the Laplace transform, we may write:
$$\lim_{x\to\infty} \int_{0}^{\infty}e^{-st}u(x,t)dt = \int_{0}^{\infty} \lim_{x\to\infty}e^{-st}u(x,t)dt =0$$
From this, you may deduce that the first coefficient is zero, since $c>0$ and for fixed $s>0$, the function $e^{\frac{sx}{c}}$ increases as $x$ increases.
From here, it follows that $U(x,s) = Be^{\frac{-sx}{c}} = \frac{cF(s)}{s\beta} e^{\frac{-sx}{c}}$ and so we arrive at the answer.
So we needed a further assumption on the solution $U(x,s)$, which says that as $x \rightarrow \infty$, then $U(x,s) \rightarrow 0$.