I am asked to solve the following diffusion problem:
$$u'_t-au''_{xx} = 0, \quad x>0, \ t>0,$$ $$u(0,t) = 0, \quad t>0,$$ $$u(x,0)=1-\theta(x-1), \quad x>0.$$
I expand this problem to $x\in \mathbb{R}$ by setting $u^-=u(x,t)$ for $x>0$ and $u^-=-u(-x,t)$ for $x<0.$ I now obtain:
$$u^{-''}_t-au^{-''}_{xx} = 0, \quad x\in \mathbb{R}, \ t>0,$$ $$u^-(0,t) = 0, \quad t>0,$$ $$u^-(x,0)= -\theta(x+1)+2\theta(x)-\theta(x-1) \quad x\in\mathbb{R}.$$
I laplace-transform this in x-direction and end up with the following:
$$\frac{\partial U^-}{\partial t}=as^2U^- \iff U(s,t)=g(s)e^{as^2t}$$ with the conditions $$g(0) = 0,$$ $$g(s) = -\frac{e^s}{s}+\frac{2}{s}-\frac{e^{-s}}{s}.$$
I cannot figure out how to inverse laplace-transform this, if it is even possible, by both table and the definition. So my question is, is this the correct method to solve this type of problem, if it is, how do I perform the inverse laplace transformation, if it isn't, how would I solve it?