Define
$$y(x,s) = \int_0^{\infty} dt \, Y(x,t) \, e^{-s t}$$
Then, integrating by parts:
$$\int_0^{\infty} dt \, Y_t(x,t) \, e^{-s t} = -Y(x,0) + s y(x,s)$$
$$\int_0^{\infty} dt \, Y_{tt}(x,t) \, e^{-s t} = -Y_t(x,0) + s Y(x,0) + s^2 y(x,s)$$
Then using the initial conditions $Y(x,0)=Y_y(x,0)=0$, the PDE becomes the following ODE:
$$y''-2 s y' + s^2 y = 0$$
where $y(0,s)=0$ and $y(1,s)=f(s)$, where the prime represents derivative with respect to $x$, and where
$$f(s) = \int_0^{\infty} dt \, F(t) \, e^{-s t} $$
The general solution of the ODE is (I will not derive here)
$$y(x,s) = (A + B x) \, e^{s x}$$
Using the boundary conditions, we may find $A$ and $B$ and therefore the LT of the solution to the PDE:
$$y(x,s) = x \, f(s) \, e^{-s (1-x)} $$
We may find the inverse LT by convolution, as we know the individual LT's. The ILT of $f(s)$ is obviously $F(t)$ by definition, and the ILT of $e^{-s (1-x)}$ is $\delta(t-(1-x))$. Therefore, the ILT, and the solution to the equation, is
$$y(x,t) = x \int_0^t dt' \, F(t') \delta(t-t'-(1-x)) = x F(t-(1-x)) \theta(t-(1-x))$$
where $\theta$ is the Heaviside step function, which is necessary because the contribution to the integral from the $\delta$ function for $t < 1-x$ is zero (i.e., $t' \gt 0$ in the integral.)
Best Answer
The equation formed by operating the Laplace transform on the given PDE is $$\frac{\partial U(x,s)}{\partial x}+ xsU(x,s) = \frac{x}{s}$$ Since the $U(x,s)$ no longer is a function of variable $t$ is can be written as $\frac{\partial U(x,s)}{\partial x}= \frac{dU(x,s)}{dx}$
The equation thus formed is a linear equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$ The integrating factor is $$I.F = e^{\int{xs}dx} = e^{\frac{sx^2}{2}}$$ And the solution $$U(x,s)e^{\frac{sx^2}{2}} = \int \left(\frac{x}{s}\right) e^{\frac{sx^2}{2}} dx +C$$ On solving $$U(x,s) = \frac{1}{s^2}+ Ce^{\frac{-sx^2}{2}}$$ inverting this $$\mathcal{L^{-1}}(U(x,s)) = \mathcal{L^{-1}}(\frac{1}{s^2}) + \mathcal{L^{-1}}(Ce^{\frac{-sx^2}{2}})$$ $$u(x,t) = t + CH\left(t-\frac{x^2}{2}\right)$$ Where $H(t-a)$ is Heaviside unit step function defined as $$ H(t-a) = \begin{cases} 0, & \text{t<a} \\ 1, & \text{t> a} \end{cases}$$ From the boundary condition $u(0,t)=0, t>0$ we can write, $u(0,t)=0= t+CH(t-0)$ since
from the defination $H(t) =1$ so the value of constant $C = -t$ The solution for the PDE is written as $$u(x,t) = t\left[1-H \left(t-\frac{x^2}{2}\right)\right]$$ Where $$H \left(t-\frac{-x^2}{2} \right) = \begin{cases} 0, & \text{t<$\frac{-x^2}{2}$} \\ 1, & \text{t $\geq \frac{-x^2}{2}$} \end{cases}$$