Heat Equation With Boundary Value Problem I have a heat equation
$$
\frac{\partial u}{\partial t} = a^2 \frac {\partial^2 u}{\partial x^2} + b \frac{\partial u}{\partial x}  \
$$
with boundary conditions
$$
u(0,x) = 300 \qquad u(t,0) = 100 \qquad  u_{x}(t, l) = 300e^{-t}
$$
I understand perfectly how to solve this equation with zero boundary conditions, but I can't figure out what to do with the boundary conditions. Can u help me please?
 A: In order to reduce the problem to one with homogeneous boundary conditions, we first try to find a particular solution $v(t,x)$ to the heat equation satisfying the boundary conditions $v(t,0)=100$ and $v_x(t,L)=300\,e^{-t}$. Once $v(t,x)$ is found, we solve the heat equation $w_t=a^2w_{xx}+bw_x$ with homogeneous boundary conditions $w(t,0)=0$, $w_x(t,L)=0$, and initial condition $w(0,x)=u(x,0)-v(0,x)$. The solution to the original problem is then given by $u(t,x)=v(t,x)+w(t,x)$.
Let's now find $v(t,x)$. Since the boundary condition at $x=0$ is independent of $t$, and the one at $x=L$ is an eigenfunction of $\partial_t$, it's natural to try an ansatz of the form $v(t,x)=v_0(x)+v_1(x)e^{-t}$. Plugging it in the heat equation yields
$$
a^2v_0''+bv_0'=0,\qquad a^2v_1''+bv_1'+v_1=0,
$$
whereas the boundary conditions $v(t,0)=100$ and $v_x(t,L)=300\,e^{-t}$ imply
$$
v_0(0)=100,\qquad v_1(0)=0, \qquad v_0'(L)=0, \qquad v_1'(L)=300.
$$
Solving for $v_0(x)$ and $v_1(x)$ we obtain
$$
v(t,x)=100+300\,\frac{e^{k_+x}-e^{k_{-}x}}{k_+e^{k_+L}-k_{-}e^{k_{-}L}}\,e^{-t},
$$
where $k_{\pm}=\frac{-b\pm\sqrt{b^2-4a^2}}{2a^2}$.
The original problem has now been reduced to the solution of the heat equation $w_t=a^2w_{xx}+bw_x$ with homogeneous boundary conditions $w(t,0)=w_x(t,L)=0$ and  initial condition $w(0,x)=u(0,x)-v(0,x)=200-300\left(e^{k_+x}-e^{k_{-}x}\right)/\left(k_+e^{k_+L}-k_{-}e^{k_{-}L}\right)$.
