The initial condition for a heat equation with stationary solution subtracted

I am presented with the following question for exam revision:

Heat is supplied at a prescribed rate $Q(x) > 0$ (per unit volume) to an isotropic conducting rod that occupies the region $0≤x≤L$. The rod has density $\rho$, specific heat $c$ and thermal conductivity $k$, all of which are constant. The faces at $x = 0, L$ are kept at zero temperature. The initial temperature at time $t = 0$ is zero.

a) Derive the heat equation $$\rho c\frac{\partial T}{\partial t} = k \frac{\partial ^2 T}{\partial x^2} + Q(x), \text{ for } 0<x<L, t>0$$ b) State the differential equation and boundary conditions satisfied by the steady-state solution $T = T_s(x)$. Hence, state the differential equation, boundary conditions and initial condition satisfied by $U(x, t) = T(x, t) − T_s(x)$.

I'm quite sure I've done part a) correctly. For part b) I'm quite sure $T_s$ has boundary conditions $T_s(0) = 0, T_s(L) = 0$ and that $U(x,t)$ satisfies $$\rho c\frac{\partial U}{\partial t} = k \frac{\partial ^2 U}{\partial x^2} + Q(x), \text{ for } 0<x<L, t>0$$ with boundary conditions $U(0,t)=0, U(L,t) = 0$

What I'm unsure about is the initial condition because of the presence of the $Q(x)$ term. How do I fit this into my initial condition for $U$?

• @muaddib No. The problem states $T(x,0) = 0 \text{ } \forall x$ – elDin0 Jun 9 '15 at 12:41
• I see. And it isn't sufficient to simply state: $U(x, 0) = - T_s(x)$? – muaddib Jun 9 '15 at 12:46

I think you are expected to state: $U(x,0) = -T_s(x)$, where $T_s$ is the steady-state solution.
In principle, you could express $T_s$ as an integral involving $Q$ and Green's function for the interval $[a,b]$: $$T_s(x) = \frac{-1/k}{b-a} \int_a^b \max\Big((t-a)(x-b),(t-b)(x-a)\Big) Q(t)\,dt$$ But I don't think this was the intent of the question.