I need help checking my solution to a PDE. The book I am using is PDE for Scientists and Engineers(exponentially increasing functions are excluded) and the chapter is on Laplace transform. Furthermore, the Laplace transform is with respect to time. Here is the problem:
PDE: $$u_t = u_{xx}$$
BC: $$u(0,t) = sin(t)$$
IC: $$u(x, 0) = 0$$
Here is my work:
$$\mathcal{L}[u_t] = \mathcal{L}[u_{xx}]$$ $$sU(x,s)-u(x,0) = \frac{d^2 U}{dx^2} = sU$$
Assuming that the $U = e^{rx}$,
$$\frac{d^2U}{dx^2} = r^2e^{rx} = sU = se^{rx}$$ $$r^2 = s$$ $$r = \pm \sqrt{s}$$ Which leads to a general solution of, $$U = c_1e^{x\sqrt{s}} + c_2e^{-x\sqrt{s}}$$
Converting the BC, $$\mathcal{L}[u(0,t)] = \mathcal{L}[sin(t)] = \frac{1}{s^2+1}$$ $$U(0) = \frac{1}{s^2+1} = c_1+c_2$$ $$\frac{1}{s^2+1} = \frac{A}{s+i}+ \frac{B}{s-i} = \frac{A(s-i)+B(s+i)}{s^2+1}$$ Noticing that $i$ appears on one side and not the other,($A=B$) then $$As-Ai +Bs+Bi = 1 = 2As$$ $$A = \frac{1}{2s}$$ $$c_1 + c_2 = \frac{A}{s+i}+ \frac{B}{s-i}= \frac{1}{2s(s+i)}+\frac{1}{2s(s-i)}$$
$$U = \frac{1}{2s(s+i)}e^{x\sqrt{s}}+\frac{1}{2s(s-i)}e^{-x\sqrt{s}}$$ Here I eliminate the exponentially increasing solution leading to, $$U = \frac{1}{2s(s-i)}e^{-x\sqrt{s}}$$
Unfortunately I do not have a way to apply the Laplace transform, so what I did next, which doesn't make much sense to me, was to drop the exponentially increasing solution before applying the BC. Thus the work here goes as follows $$U = ce^{-x\sqrt{s}}$$ Applying BC's,
$$U(0) = \frac{1}{s^2+1} = c$$ $$U = \frac{s}{s} \frac{1}{s^2+1}e^{-x\sqrt{s}} = \frac{s}{s^2+1} \frac{e^{-x\sqrt{s}}}{s}$$ Here I use the convolution property of transforms, $$\mathcal{L^{-1}}[f*g] = \mathcal{L^{-1}}[f] \mathcal{L^{-1}}[g]$$ $$\mathcal{L^{-1}}\left[\frac{s}{s^2+1}\right] \mathcal{L^{-1}} \left[ \frac{e^{-x\sqrt{s}}}{s} \right]$$ $$[cos(t)]*\left[ erfc \left(\frac{x}{2\sqrt{t}}\right) \right]$$ Note I am not required to simplify a convolution.