How does one solve a fourth-order PDE of the form $\frac{\partial^4y}{\partial x^4}=c^2\frac{\partial^4y}{\partial t^4}$? It looks like a one dimensional wave equation, but I'm unfortunately very bad at PDEs. Would separating variables and using Fourier series work if I try hard enough?
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$\begingroup$Since all terms of the PDE are in same order and constant coefficient, you can apply the similar technique that solving the wave equation:
$\dfrac{\partial^4y}{\partial x^4}=c^2\dfrac{\partial^4y}{\partial t^4}$
$\dfrac{\partial^4y}{\partial x^4}-c^2\dfrac{\partial^4y}{\partial t^4}=0$
$\left(\dfrac{\partial^4}{\partial x^4}-c^2\dfrac{\partial^4}{\partial t^4}\right)y=0$
$\left(\dfrac{\partial}{\partial x}-\sqrt{c}\dfrac{\partial}{\partial t}\right)\left(\dfrac{\partial}{\partial x}+\sqrt{c}\dfrac{\partial}{\partial t}\right)\left(\dfrac{\partial}{\partial x}-i\sqrt{c}\dfrac{\partial}{\partial t}\right)\left(\dfrac{\partial}{\partial x}+i\sqrt{c}\dfrac{\partial}{\partial t}\right)y=0$
$y(x,t)=f_1(t+\sqrt{c}x)+f_2(t-\sqrt{c}x)+f_3(t+i\sqrt{c}x)+f_4(t-i\sqrt{c}x)$
$\endgroup$ 2 $\begingroup$I would try separation of variables, $y\left(x,t\right) = X\left(x\right) T\left(t\right)$, to give $$ \frac{1}{X} X^{\left(4\right)} = c^2 \frac{1}{T} T^{\left(4\right)} = - \lambda^4 \Rightarrow X^{\left(4\right)} + \lambda^4 X = T^{\left(4\right)} + \left(\frac{\lambda}{\sqrt{c}}\right)^4 T = 0, $$ where $\lambda$ is a constant. Assume solutions $X \propto \exp\left(\mu x\right)$ and $T \propto \exp\left(\rho t\right)$ and you will find that $\mu^4 + \lambda^4 = \rho^4 + \left(\lambda / \sqrt{c}\right)^4 = 0$, or $$ \mu_{1,2,3,4} = \pm \lambda \sqrt{\pm i} $$ and $$ \rho_{1,2,3,4} = \pm \frac{\lambda}{\sqrt{c}} \sqrt{\pm i}, $$ so that your general solution is $$ y\left(x,t\right) = \left[\sum_{i=1}^4 a_i \exp\left(\mu_i x\right)\right] \left[\sum_{j=1}^4 b_j \exp\left(\rho_j t\right)\right]. $$ Now start applying initial and boundary conditions.
$\endgroup$ 2 $\begingroup$Usually there is no theory like Sturm-Liouville for this, but depending on your boundary conditions, you might get a complete orthogonal basis. For example, if $x \in [0, L]$ and boundary conditions are $y(0, t) = \frac{\partial^2 y}{\partial x^2}(0, t) = y(L, t) = \frac{\partial^2 y}{\partial x^2}(L, t) = 0$, separation of variables will eventually give $\sin\left(\frac{n\pi x}L\right)$ as your basis. Then you can solve for the corresponding $t$-dependent part, giving the general solution
$$ \begin{align*} y(x, t) = & \sum_{n=1}^\infty \sin\left(\frac{n\pi x}L\right)\left(A_n\cos\left(\frac{n\pi t}{L\sqrt c}\right) + B_n\sin\left(\frac{n\pi t}{L\sqrt c}\right) + \right.\\ & \left.C_n \cosh\left(\frac{n\pi t}{L\sqrt c}\right) + D_n \sinh\left(\frac{n\pi t}{L\sqrt c}\right)\right) \end{align*} $$
Use initial conditions to find out what $A_n, B_n, C_n$ and $D_n$ are.
If you have different boundary conditions, your eigenfunctions will be different. If your domain is not finite, you might need Laplace or Fourier transform instead of separation of variables.
I'm not sure about how to use method of characteristics with complex variables. It might work out.
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