I am trying to solve the following boundary value problem:

$\frac{\partial^2{u}}{\partial t^2} = \frac{\partial^2{u}}{\partial x^2} + Q(x,t)$ subject to the time dependent boundary conditions:

$u(0,t) = A(t)$ and $u(L,t) = B(t)$ with initial conditions $u(x,0) = f(x)$,$\frac{\partial{}}{\partial t}(x,0) = g(x) $.

I know how to solve the problem when the boundary conditions are homogeneous by the method of eigen-function expansion, however I am not sure how to do it when the BCs are not homogeneous. The book I am using vaguely says that it is not necessary to reduce to homogeneous BCs.


Similar to ordinary linear autonomous differential equations, you can always add any solution to the homogeneous differential equation, to the solution to the inhomogeneous differential equation, often called the particular solution. The same can be said for a linear partial differential equation.

So if you have found a particular solution, lets call it $u_Q(x,t)$, then you can subtract it from the boundary conditions $A(t)$ and $B(t)$, such that the remaining equations can be used as new boundary conditions for a homogeneous equation and thus can be solved by the method of eigen-function expansion.

Writing this in terms of equation; if you define the solution as $u(x,t) = u_Q(x,t) + u_h(x,t)$, such that

$$ \frac{\partial^2 u_Q(x,t)}{\partial t^2} = \frac{\partial^2 u_Q(x,t)}{\partial x^2} + Q(x,t), $$

$$ \frac{\partial^2 u_h(x,t)}{\partial t^2} = \frac{\partial^2 u_h(x,t)}{\partial x^2}, $$

then $u(0,t) = u_Q(0,t) + u_h(0,t) = A(t)$ and $u(L,t) = u_Q(L,t) + u_h(L,t) = B(t)$. This can be rewritten to constraints for $u_h(x,t)$, namely

$$ u_h(0,t) = A(t) - u_Q(0,t), $$

$$ u_h(L,t) = B(t) - u_Q(L,t). $$

Finding $u_h(x,t)$ should be trivial to the method you know, however finding a $u_Q(x,t)$ (since you can add any combination of solutions of the homogeneous equation to it) in general might be harder, but since you have not defined $Q(x,t)$ I got not say anything about that.

  • $\begingroup$ can you elaborate on this please? $\endgroup$ – continental Oct 29 '15 at 8:44
  • $\begingroup$ @continental I have edited my answer. $\endgroup$ – Kwin van der Veen Oct 29 '15 at 20:06

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