On page 115 of the third edition of Logan's Applied Partial Differential Equations, the reader is asked to solve the following:
$$u_{tt} = u_{xx}, \,\,\: x,t>0$$
$$ u(0,t) = \sin t ,\,\,\: t\ge 0 $$
$$ u(x,0) = 0;\,\,u_t(x,0) = 1 ,\,\,\: x \ge 0. $$
Since this problem is given immediately after a section on Laplace transforms, I figured that would be the place to start, but I only got as far as transforming the equation and substituting the boundary conditions like so:
$$ u_{tt} = u_{xx} \Longrightarrow s^2\,U(x,s) - s\,u(x,0) - u_t(x,0) = s^2 \,U(x,s) -1 = U_{xx}(x,s)$$
$$\Longrightarrow U_{xx} - s^2\,U = -1 \Longrightarrow U(x,s) = A(s)\exp(st) + B(s)\exp(-st) + \frac{1}{s^2}$$
with
$$ U(x,0) = \frac{1}{s^2+1},$$
where $U(x,s):= \mathscr{L}\{ u\}(x,s)$ and $A,B$ denote arbitrary bounded, twice continuously differentiable functions of $s$ alone with $a(t) := \mathscr{L}^{-1}\{ A(s)\}(t)$ and $b(t):= \mathscr{L}^{-1}\{ B(s)\}(t)$. The author's examples were pretty sparse, so I didn't know exactly where to go from here, but I assumed it would be a safe bet to just apply the inverse transform to get
$$u(x,t) = a(t) \,\ast \,\delta(t+x) + b(t) \,\ast\,\delta(t-x) + t = a(t+x) + b(t-x) +t .$$
The boundary and initial data give
$$ u(0,t) = a(t) + b(t) +t = \sin t $$
$$ u(x,0) = a(x) + b(-x) =0 ,\:\: u_t(x,0) = a'(x) + b'(-x) + 1 = 1$$
so $b(-x) = -a(x), b'(-x) = -a'(x)$. I have a feeling that I should end up with an expression involving multiples of $\sin( t\pm x)$ or $\cos(t\pm x)$, or maybe a convolution involving trig functions, but I was stumped. I later tried splitting the problem into one PDE with zero boundary data (which I do know how to solve) and another with homogeneous initial conditions, but again ran into a similar issue with the boundary condition. My question is the following: In general, how does one solve a semi-infinite hyperbolic BVP of the form
$$u_{tt} = c^2 \,u_{xx}, \,\,\: x,t>0$$
$$ u(0,t) = \phi(t) ,\,\,\: t\ge 0 $$
$$ u(x,0) = f(x);\,\,u_t(x,0) = g(x),\,\,\: x \ge 0,$$
where $u, \phi, f, g$ are all well-behaved, etc. using an integral transform method?
This resource gives the solution as
$$ w(x,t) = \left [ \frac{1}{2}f(x+ct) + \frac{1}{2}f(x-ct)\right ] + \frac{1}{2c}\int_{x-ct}^{x+ct} \!\!\! g(\xi)\,\mathrm{d}\xi ,\:\: x > ct$$
and
$$ w(x,t) = \left [ \frac{1}{2}f(x+ct) + \frac{1}{2}f(x-ct)\right ] + \frac{1}{2c}\int_{ct-x}^{ct+x} \!\!\! g(\xi)\,\mathrm{d}\xi + \phi\!\left ( t - \frac{x}{c}\right ),\:\: x <ct$$
(being the region of interest wherein the boundary data actually affects the solution), which in the case of the above textbook problem gives
$$ u(x,t) = \frac{1}{2}\int_{t-x}^{t+x} \!\!\mathrm{d}\xi + \sin(t-x) =\frac{1}{2}\big ( (t+x) - (t-x)\big ) + \sin(t-x) = x + \sin(t-x) $$
for $x<t$ and
$$u(x,t) = \frac{1}{2}\int_{x-t}^{x+t} \!\!\mathrm{d}\xi = t ,\:\: x > t,$$
which I'm pretty sure is correct, but how on earth was I supposed to get all that using the Laplace transform? Finally, if not from the Duhamel principle, change of variable, or transform methods, how is the form of solution on eqworld derived in the first place?