Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been trying to come up with a good way to get rid of an inhomogeneity in this PDE, but I have two different solutions. I am not sure if this is a question for the Math community. If not I'll ask the physics community about this

Consider this PDE which models the Wave Equation

$$u_{tt} = u_{xx} + c\sin(2\pi x) $$

With these conditions

$$u(0,t) = u(1,t) = u(x,0) = u_t(x,0) = 0$$

So I thought about making a substitution where $u_1 = \sum_{n\in \mathbb{N}} T(t)X(x)$ and to get rid of that sine function, I let $X(x) = \sin(n\pi x)$ and all the other $n$s to be $0$ for convenience and let $u(x,t) = v(x,t) + u_1$ where $u_1$ is my particular solution and $v(x,t)$ is my homogenous solution. After a while, I managed to solved that $v(x,t) = 0$ and there only exists $u_1$.

So I thought about making another substitution. Using the theory of the heat equation where I let

$$u(x,t) = w(x,t) + z(x)$$

where $z(x)$ is the steady state solution to the Wave Equation (if I can even use such a terminology) and $w(x,t)$ is the transient state (this is my "particular" solution).

I reapplied the techniques as I did for the heat equation where I first set all the time derivatives to $0$ and solved for the steady state and then the transient state. Neither the transient state nor steady state were $0$ and after computing, I got my solution to be the same for both PDE

So my question is, is it a coincidence they both worked? Does the Wave Equation have those steady/transient states solution?

Thank you

share|cite|improve this question
up vote 1 down vote accepted

Your equation is not "the Wave Equation", but it is an inhomogeneous wave equation. Inhomogeneous wave equations of the form $u_{tt} = u_{xx} + g(x)$, where the inhomogeneous term doesn't depend on $t$, do have particular solutions that don't depend on $t$, namely solutions of the ODE $u_{xx} + g(x) = 0$. If $v(x)$ is such a solution, by linearity $u(x,t)$ satisfies the (homogeneous) Wave equation if and only if $u(x,t) + v(x)$ satisfies this inhomogeneous wave equation.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.