I'm currently working on solving the Heat Equation in a one dimensional rod of length $L$. However, instead of the 'usual' singular condition $u(x,0)=f(x)$ for all $0\leq x\leq L$, I am given
$u(x,0)=f(x)$ for $0\leq x \leq L/2$.
$u(x,0)=g(x)$ for $L/2 \leq x \leq L$.
I can find the solutions individually for each of these conditions, but what do I do with these two solutions to solve the PDE? Or am I supposed to have two separate solutions, one for each half of the rod?
My gut feeling is telling me to sum the two, but I could be mistaken.
You still have a single function specifying the heat distribution, it just happens to be given by a function defined piecewise:
$$u_0(x) = \left\{ \begin{array}{cl} f(x), & 0 \leq x \leq \tfrac{L}{2} \\ g(x), & \tfrac{L}{2} \leq x \leq L \end{array} \right. .$$
Assuming you're using Fourier series to solve the evolution problem, use that
$$\int_0^L h(x) u_0(x) \,dx = \int_0^{\frac{L}{2}} h(x) f(x) \,dx + \int_{\frac{L}{2}}^L h(x) g(x).$$