I'm trying to solve a fourth-order linear non-homogeneous PDE

$$ \frac{\partial u}{\partial t} + \frac{\partial^2 u}{\partial \theta^2} + \frac{\partial^4 u}{\partial \theta^4} = - \sin{\theta}. $$

This PDE is periodic in $\theta$ and has domain $\theta \in [0, 2 \pi)$, $t\geq 0$, with boundary conditions

$$ \frac{\partial u}{\partial \theta} = 0 $$

on $\theta = \pi/2$ and $\theta = 3 \pi/2$, with initial condition $u(\theta , t=0) = 0$.

I'm not sure of the best way to approach solving this PDE, so any help would be much appreciated!

I know that in the non-homogeneous case with a constant initial condition, then there is a constant solution, but I'm not sure how to proceed in the non-homogeneous case with the $-\sin{\theta}$ term on the RHS.

  • $\begingroup$ Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$ Jul 15, 2023 at 11:54
  • $\begingroup$ What's the domain? $\endgroup$
    – tommy1996q
    Jul 15, 2023 at 12:55
  • $\begingroup$ @tommy1996q I've edited the question to add the domain. $\endgroup$
    – James
    Jul 15, 2023 at 13:08
  • $\begingroup$ Since it's fourth order the solution is gonna be non unique, if you're ok with stationary solutions you basically reduce to an ODE $\endgroup$
    – tommy1996q
    Jul 15, 2023 at 13:13
  • 2
    $\begingroup$ By simple inspection one see that $u(t,\theta)=-t\,\sin(\theta)$ satisfies the PDE and the conditions. $\endgroup$
    – JJacquelin
    Jul 15, 2023 at 13:37

1 Answer 1


You can, for each time $t$, project the solution $u$ on the $\sin$ and $\cos$ basis: $$u(t,\theta) = \sum_{n=1}^\infty a_n(t) \sin(n \theta) + \sum_{n=0}^\infty b_n(t) \cos(n \theta)$$ You can project the PDE to get: $$a_n' + (n^4-n^2) a_n = - \delta_{1,n}$$ $$b_n' + (n^4-n^2) b_n = 0$$ and solve these simple ODE The $a_n(t)$, $n\neq 1$ and $b_n(t)$ will be exponentials, and I beleieve we can find equations on their coefficients with the boundary conditions. I Hope this approach helps.


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