Find the steady state temperature of the rod A rod occupying the interval $0 \leq x \leq l$ is subject to the heat source
$f(x) =0, $ for $ 0 < x < L/2$,
$f(x) =H $ for $  L/2 <x <L ,H>0$
(1)The rod satisﬁes the heat equation $$u_t = u_{xx} + f(x)$$ and its ends are kept at zero temperature. Find the steady-state temperature of the rod. 
(2)Which point is the hottest, and what is the temperature there?
Attempt: I only know that since, its ends are kept at zero temperature, $$u_t=0$$. Where should I go from here. 
 A: For the steady state, you are solving
$$u_t=0 \implies u_{xx}=-f(x)$$
The general solution to this, given $f$, is pretty straightforward:
$$u(x) = \begin{cases} c_1 x+c_2 & x \in \left (0,\frac{L}{2}\right)\\-\frac12 H x^2 + d_1 x+d_2 & x \in \left (\frac{L}{2},L\right) \end{cases} $$
Yes, we have different constants in each branch of the integration interval.  We thus need four conditions.  Two come from the boundary conditions:
$$u(0)=u(L)=0$$
The other two come from continuity requirements for $u$ and its derivative (otherwise, the heat flow is discontinuous, not what should happen in a steady-state solution).  Thus,
$$c_1 \frac{L}{2}+c_2 = -\frac18 H L^2 + d_1 \frac{L}{2} + d_2$$
$$c_1 = -\frac12 H L + d_1$$
Four equations, four unknowns, solve.  It's not as bad as you may think.  For example, $c_2=0$.  The rest fall out from simple substitutions.  The result I get is
$$u(x) = \begin{cases} \frac18 H L x & x \in \left (0,\frac{L}{2}\right)\\-\frac12 H x^2 + \frac{5}{8} H L x-\frac18 H L^2 & x \in \left (\frac{L}{2},L\right) \end{cases} $$
EDIT
To see what the above temperature looks like, rewrite the above result in the following form:
$$f(y) = y \,\theta \left ( \frac12-y\right) - (4 y-1)(y-1)\, \theta \left (y- \frac12\right) $$
where $\theta$ is the Heaviside step function, $y=x/L$, and $f(y) = 8 u(L y)/(H L^2)$.  
Here is a plot of $f(y)$:

