Heat Equation With Boundary Value Problem

Question

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?

Answer 1

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 $w(0,x)=u(0,x)-v(0,x)=200-300\left(e^{k_+x}-e^{k_{-}x}\right)/\left(k_+e^{k_+L}-k_{-}e^{k_{-}L}\right)$.