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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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    $\begingroup$ My guess would either separating variables or doing Fourier transform in space variable would work, try it. $\endgroup$
    – Shuhao Cao
    Commented Sep 7, 2012 at 17:41
  • $\begingroup$ Sorry, I mistyped the original question, I really meant $\frac{\partial^4y}{\partial x^4}=c^2\frac{\partial^2y}{\partial t^2}$. Sorry about that! $\endgroup$ Commented Sep 9, 2012 at 11:35
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    $\begingroup$ @anegligibleperson: Because people have already answered the question you asked, you should simply start a new question with the correction. Otherwise, the answers that were already given for the question you asked will no longer appear to match the question. I edited this question to restore the original wording. It's perfectly fine to ask a second question, there is no limitation of resources or anything like that to worry about. $\endgroup$ Commented Sep 10, 2012 at 1:22
  • $\begingroup$ Ok sure. Thanks Carl! $\endgroup$ Commented Sep 11, 2012 at 4:04

3 Answers 3

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Since all terms of the PDE are in same order and constant coefficient, you can apply the similar technique that solving the wave equation:

$\dfrac{\partial^4y}{\partial x^4}=c^2\dfrac{\partial^4y}{\partial t^4}$

$\dfrac{\partial^4y}{\partial x^4}-c^2\dfrac{\partial^4y}{\partial t^4}=0$

$\left(\dfrac{\partial^4}{\partial x^4}-c^2\dfrac{\partial^4}{\partial t^4}\right)y=0$

$\left(\dfrac{\partial}{\partial x}-\sqrt{c}\dfrac{\partial}{\partial t}\right)\left(\dfrac{\partial}{\partial x}+\sqrt{c}\dfrac{\partial}{\partial t}\right)\left(\dfrac{\partial}{\partial x}-i\sqrt{c}\dfrac{\partial}{\partial t}\right)\left(\dfrac{\partial}{\partial x}+i\sqrt{c}\dfrac{\partial}{\partial t}\right)y=0$

$y(x,t)=f_1(t+\sqrt{c}x)+f_2(t-\sqrt{c}x)+f_3(t+i\sqrt{c}x)+f_4(t-i\sqrt{c}x)$

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  • $\begingroup$ I congratulate you for the hard work in typing all of this, but perhaps you should show computations in less detail and highlight more the important points. To be honest, this post is almost unreadable. $\endgroup$ Commented Sep 9, 2012 at 21:59
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    $\begingroup$ $Giuseppe Negro: I like answer for the OP in detail. This is my style. Sorry about that it takes quite long time to load my answer. $\endgroup$ Commented Sep 9, 2012 at 22:43
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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.

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  • $\begingroup$ How do you find the eigenvalues for this problem? $\endgroup$ Commented Nov 1, 2015 at 18:21
  • $\begingroup$ Where in this problem do you see an equation of the form $M {\bf v} = m {\bf v}$?, with $M$ a matrix, $\bf v$ a vector, and $m$ a scalar? $\endgroup$
    – Eric Angle
    Commented Nov 2, 2015 at 14:10
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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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  • $\begingroup$ Note that using separation of variables can also take integration and not only can take summation. Since integration is the continuous superposition while summation is the discrete superposition. $\endgroup$ Commented Sep 7, 2012 at 22:40
  • $\begingroup$ Using separation of variables should be better than using integral transforms, since you will not miss the special cases. $\endgroup$ Commented Sep 7, 2012 at 22:46
  • $\begingroup$ @doraemonpaul I am not certain about cases you can just use eigenfunctions of the operator because I am not sure about conditions needed for orthogonality and inversion. For example, eigenfunctions of the derivative operator is $e^{sx}$, but $s$ must be allowed to take on complex values even when you're expanding a real function. That's why I suggest Fourier and Laplace transforms. They have been well studied. (I don't understand what you mean by missing special cases.) $\endgroup$
    – Tunococ
    Commented Sep 8, 2012 at 0:29

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