I'm trying to solve the 1D heat equation on a semi-infinite domain with no heat sources and inhomogeneous boundary conditions:
$$\frac{\partial T}{\partial t}=\alpha\frac{\partial^2T}{\partial x^2}$$ $$T(0,t)=\phi(t)$$ $$T(x\rightarrow\infty,t) = T_i$$ $$T(x,0)=T_i$$
for which the Green's Function is:
$$G(x,t|x',\tau)=\frac{1}{\sqrt{4\pi\alpha(t-\tau)}}\left(e^{-\frac{(x-x')^2}{4\alpha(t-\tau)}}-e^{-\frac{(x+x')^2}{4\alpha(t-\tau)}}\right)$$
My questions are:
- When writing the solution: $$T(x,t)=\int_{x'=0}^{\infty}\left.G(x,t|x',\tau)\right|_{\tau=0}T_idx'+\alpha\int_{\tau=0}^{t}\left.\frac{\partial G}{\partial x'}\right|_{x'=0}\phi(\tau)d\tau$$ why is the boundary condition, $\phi(t)$, not recovered when $x=0$?
- When numerically evaluating: $$\alpha\int_{\tau=0}^t\left.\frac{\partial^2 G}{\partial x\partial x'}\right|_{x'=0,x=0}\phi(\tau)d\tau$$ where $$\alpha\left.\frac{\partial^2 G}{\partial x\partial x'}\right|_{x'=0}=\frac{1}{\sqrt{4\pi\alpha}}e^{-\frac{x^2}{4\alpha(t-\tau)}}\left((t-\tau)^{-3/2}-\frac{x^2}{2\alpha}(t-\tau)^{-5/2}\right)$$ what happens when $t=\tau$ at $x=0$? Performing the integral just gives $\infty$ for all time. I imagine the integration as a convolution of $\phi(t)$ with $1/t$.
- Can this be solved by splitting $T(x,t)=U(x,t)+V(x,t)$ where $U$ satisfies the inhomogeneous BCs and $V$ satisfies the inhomogeneous BCs? If so, does the form of $V(x,t)$ matter? Working it through requires evaluation of $\dot{\phi}(t)$? Or am I just completely wrong on all counts?