Use the Laplace transform to solve $ u_{tt}(x, t) − c^2u_{xx}(x, t) = 0$

Question

Use the Laplace transform to solve the following initial boundary value problem for the wave equation $ u_{tt}(x, t) − c^2u_{xx}(x, t) = 0$

$u(x, 0) = 0$, $u_{t}(x, 0) = 0 ∀x > 0$ , and

$u(0, t) = \sin\left(2\pi t\right)$ $ 0 ≤ t < 1$, and $0$ otherwise. (with $lim_{x→∞}u(x, t) = 0 $). You may use standard results for the Laplace transform without proof.

Then, sketch the solution profile at times $t = 0, t = 1$ and $ t = 2$.

I'm a bit stuck on this question. Here's what I've done so far:

$ u_{tt}(x, t) − c^2u_{xx}(x, t) = 0$ Taking the laplace transform with respect to $t$ i got: $s^2\tilde u - su(x,0)-u_{t}(x,0)- c^2u_{xx} = 0$

Substituting in the conditions = $s^2\tilde u (x,s) - c^2\tilde u_{xx}(x,s) = \tilde u_{xx} - s^2/c^2 \tilde u= 0$ which is a 2nd order constant coefficient ODE in x . $\tilde u (x,s) = A(s)e^{sx/c} + B(s)e^{-sx/c}$.

Am i correct so far? Im not sure what to do with the initial conditions $u(0,t) = sin(2πt)$. My guess is to the the Laplace transform of that.

Would greatly appreciate the help.

Edit: I managed to complete the question, but im not certain it's correct:

Taking the Laplace transform of the 2 conditions:

(1) $u(0,t) = sin(2πt)$ which is $\tilde u (0,s) = 2π/(4π^2 +t^2)$

(2) $lim_{x→∞}u(x, t) = 0 $ gives $lim_{x→∞} \tilde u (x,s) =0$

So we have 2 constants A,B and 2 conditions. 2nd condition gives $A (s) = 0 $ and 1st condition gives $B(s)= 2π/(4π^2 +t^2)$ which gives $\tilde u (x,s) = 2π/(4π^2 +t^2).e^{-sx/c}$

Therefore the solution is $u(x,t) = H(t-x/c)sin2π(t-x/c)$. Where H is the heavy side function. For $x/c>t$ we have $0$ and for $t<x/c$ we have no solution.

Answer 1

For the boundary function, write

\begin{align} u(0,t) &= \sin(2\pi t)u(1-t) \\ & = \sin(2\pi t)\big[1 - u(t-1)\big] \\ &= \sin(2\pi t) - \sin(2\pi(t-1) + 2\pi)u(t-1) \\ &= \sin(2\pi t) - \sin(2\pi(t-1))u(t-1) \end{align} where $u(t)$ is the unit step function

Using the time-shift property of the Laplace transform

$$ \tilde u(0,s) = \frac{2\pi}{s^2+4\pi^2} - e^{-s}\frac{2\pi}{s^2+4\pi^2} = \frac{2\pi(1-e^{-s})}{s^2+4\pi^2} $$

Hence

$$ \tilde u(x,s) = \frac{2\pi(1-e^{-s})}{s^2+4\pi^2}e^{-xs/c} $$

To transform back, write

$$ \tilde u(x,s) = \frac{2\pi}{s^2+4\pi^2}e^{-(x/c)s} - \frac{2\pi}{s^2+4\pi^2}e^{-(x/c+1)s} $$

Then use the shift theorem in reverse.

\begin{align} u(x,t) &= \sin\big(2\pi(t-x/c)\big)u(t-x/c) - \sin\big(2\pi(t-x/c-1)\big)u(t-x/c-1) \\ &= \sin\big(2\pi(t-x/c)\big)u(t-x/c) - \sin\big(2\pi(t-x/c)\big)u(t-x/c-1) \\ &= \sin\big(2\pi(t-x/c)\big)\big[u(t-x/c) - u(t-x/c-1)\big] \end{align}

$$ \implies u(x,t) = \begin{cases} \sin\big(2\pi(t-x/c)\big), & x/c < t < x/c+1 \\ 0, & t < x/c, \ t > x/c+1 \end{cases} $$

Edit: We can rewrite in terms of $x$

$$ u(x,t) = \begin{cases} -\sin\big(2\pi(x/c-t)\big), & c(t-1) < x < ct \\ 0, & x < c(t-1), \ x > ct \end{cases} $$

For $t=0$ $u$ is non-zero for $-c < x < 0$ but this isn't in the domain, so $u(x,0)=0, \forall x > 0$

For the other cases

\begin{align} u(x,1) &= \begin{cases} -\sin\big(2\pi x/c\big), & 0 < x < c \\ 0, & x > c \end{cases} \\ u(x,2) &= \begin{cases} -\sin\big(2\pi x/c\big), & c < x < 2c \\ 0, & x < c, \ x > 2c \end{cases} \end{align}