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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

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1 Answer 1

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.

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