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Solve $$\frac{\partial u}{\partial t}=k \frac{\partial^2 u}{\partial x^2} + Q(x)\quad u(x,0)= f(x)$$ if an equilibrium solution exists. If no equilibrium solution exists reduce the problem to one with homogeneous boundary conditions(do not solve). Assume the following (separate) conditions

  1. $Q(x)=0$, $u(0,t)=A$, $\frac{\partial u(L,t)}{\partial x}=B$
  2. $Q(x)=k$, $u(0,t)=A$, $u(L,t)=B$

My work thus far: First I am unsure how to find the equilibrium solution in each case so below. I know that the equilibrium condition is that $\frac{\partial u}{\partial t}=0$. So \begin{align} 0&=k \frac{\partial^2 u}{\partial x^2} + Q(x) \\ \frac{\partial^2 u}{\partial x^2}&= -\frac{1}{k}Q(x) \\ &\text{Integrate twice assuming Q(x) is constant}\\ \frac{\partial^2 u}{\partial x^2}&= -\frac{1}{k}Q(x) \\ u_{eq}&=-\frac{Q}{2k}x^2+C_1x+C_2\\ \end{align} So I think the equilibrium solutions are (after applying the conditions)

  • Condition 1:$u_{eq}=C_1x+C_2$ applying boundary conditions $C_2=A$ and $C_1=B$ thus $u_{eq}=Bx+A$.
  • Condition 2: $u_{eq}=-\frac{1}{2}x^2+C_1x+C_2$ applying boundary conditions $C_2=A$ and $C_1=\frac{B-A+\frac{1}{2}L^2}{L}$ thus $u_{eq}=-\frac{1}{2}x^2+\frac{B-A+\frac{1}{2}L^2}{L}x+A$.

I am unsure if these equilibrium solutions are correct though. As I cannot solve the equation using separation of variables (non-homogeneous boundary conditions). So we now define a new function and find some derivatives so that we can solve the problem: $$v(x,t)=u(x,t)-u_{eq}(x,t) \quad \frac{\partial u}{\partial t}=\frac{\partial v}{\partial t}+\frac{\partial u_{eq}}{\partial t}=\frac{\partial v}{\partial t} \quad \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 u_{eq}}{\partial x^2}=\frac{\partial^2 v}{\partial x^2} $$ From there I am confused on how to use this newly defined $v$ to solve the PDE. Any hints into how to solve the problem is appreciated.

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    If you are unsure about the solutions you have found then check them explicitly. Compute $0 = ku_{xx} + Q$ and check if your solution satisfy it. Next check the boundary conditions: plug in $x=0$ and $x=L$ and see that you get the right result. If your solution passes these tests then you can be absolutely sure you have found the correct one. – 2017-01-17
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    I'm making many mistakes tonight, it's been a long day. I only checked that the boundary conditions was satisfied, however it seems you might have forgotten a factor of $\frac{1}{2}$ when integrating $x$. With $u = - x^2 + \ldots$ then $u_{xx} = -2$ so $u_{xx} + 1 \not = 0$. If I'm not mistaken (again) it should be $-\frac{x^2}{2}$. This mistake is also present in $u_{\rm eq} = - \frac{Q}{k} x^2 + \ldots$ which should be $-\frac{Q}{2k}x^2$. You will then have to recompute the constants derived from the BC. – 2017-01-17
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    @Winther Thanks for the correction. Any hints for how to proceed afterwords? – 2017-01-17
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    Take $u(x,t) = v (x,t)+ u_{\rm eq}(x)$ in the PDE $u_{t} = ku_{xx} + Q(x)$ and it will simplify. Then compute the new boundary values for $v(x,t)$ (which should also simplify so that you can apply separation of variables). – 2017-01-18

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