I am attempting to solve a PDE $$y_{tt} = y_{xx}, -\infty < x < 0,\ t > 0$$ with boundary conditions $$ y_x(0,t) = k(t),\ y(x,t) \rightarrow 0\ \mbox{as}\ x \rightarrow -\infty,\ y(x,0) = 0,\ y_t(x,0) = 0$$.
Let $Y(x,s)$ be the Laplace Transform $\mathcal{L}[y(x,t)]$.
I arrive at $$Y_x(x,s) = Ase^{sx}$$ with the condition $$y_x(0,t) = k(t)$$
therefore $$Y_x(0,s) = As = \mathcal{L}[y_x(0,1)] = \mathcal{L}[k(t)]$$
therefore $$A = \frac{1}{s}\mathcal{L}[k(t)]$$
so $$Y(x,s) = \mathcal{L}[k(t)]\left(\frac{1}{s}\right)e^{sx}$$
and so $$ y(x,t) = \mathcal{L}^{-1}[\mathcal{L}[k(t)]\left(\frac{1}{s}\right)e^{sx}]$$.
I know that the time delay (or t-shift) is: $$\mathcal{L}^{-1}[\mathcal{L}[h(t)]e^{-sx}] = h(t-x)H(t-x)$$ where $H$ is the step (or Heaviside) function.
And that is my problem, how do I deal with the extra $-\frac{1}{s}$ term?
Best Answer
Hint:
$$\mathscr{L}\left(\int_0^t f(t')dt'\right)(s)=\int_0^{\infty}e^{-st}\left(\int_0^t f(t')dt'\right)dt=\frac{1}{s}\,\mathscr{L}\{f\}(s)$$
So, the Laplace Transform of the integral of a function $f$ is equal to $1/s$ times the Laplace Transform of $f$.