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Let $$\Delta^2 u-\lambda u =0$$ where $\lambda>0$ and

$$\Delta^2 u = \frac{\partial ^4 u }{\partial x^4} + 2 \frac{\partial^4 u }{\partial x^2 \partial y^2} + \frac{\partial ^4 u }{\partial y^4}$$

Is there a maximum principle for this equation in a rectangular domain $[-a,a]\times[-b,b]$ or any other domain?

Fact: it is known that there is a maximum principle for $$\Delta^2 u =0$$ but not in every domain.

ps: this question arises from another question, i.e. that question is solved if the maximum principle exists.

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Michael Medvinsky
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  • I think I heard somewhere that those problems lose the maximum principle. This is a thing that I can loosely understand by observing that the solutions to $f''''(x)=0$ in one dimension clearly do not have a maximum principle (unlike $f''(x)=0$ which are affine functions). But you require $\lambda > 0$, so mine is not a counterexample. – Giuseppe Negro Sep 29 '15 at 23:40

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I would be surprised if this equation had a maximum principle.

My intuition comes from the corresponding problem in one dimension, which is $$\tag{1}f^{(4)}(x) =f(x), \qquad x \in (0, 2\pi).$$ This problem fails to have a maximum principle, because it supports the oscillating solution $f(x)=\sin x$.

This solution satisfies Dirichlet's boundary conditions. If you want an example with Neumann's boundary conditions, consider the solution $f(x)=\cos x$.

Also, by translation and rescaling, all problems $$ u^{(4)}=\lambda u,\qquad x \in (a, b)$$ can be reduced to (1). So, in one dimension one never has a maximum principle.

Giuseppe Negro
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  • you don't need to go that far as higher dimensions...in your example if you go for $x\in (0,\frac{\pi}2)$ the maximum principle holds so the dependence on the domain exists even in such simple example... – Michael Medvinsky Sep 30 '15 at 00:12
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    @MichaelMedvinsky: I disagree. The function $f(x)=\sin\left(x+\frac\pi4\right)$ solves the equation and has a maximum at $x=\frac\pi4\in\left(0, \frac\pi2\right)$. – Giuseppe Negro Sep 30 '15 at 00:22
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    Can you please post a reference to this domain dependent maximum principle to $\Delta^2 u =0$? – Giuseppe Negro Sep 30 '15 at 00:23
  • Note the both solution satisfies different BC's I think it was here: https://www.google.co.il/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&sqi=2&ved=0CDMQFjACahUKEwiE_bCkv53IAhUHnRoKHVE1B9Y&url=http%3A%2F%2Fwww.mathematics-and-its-applications.com%2Fpreview%2Fdec2011%2Fdata%2Fdanet_c.pdf&usg=AFQjCNGSZt763Hkl9Z1rhTV1-fP-6jvCkg&sig2=9TktqLsfiep3WyiAk7eP7g – Michael Medvinsky Sep 30 '15 at 00:34
  • or here https://www.google.co.il/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&sqi=2&ved=0CDMQFjACahUKEwiE_bCkv53IAhUHnRoKHVE1B9Y&url=http%3A%2F%2Fwww.mathematics-and-its-applications.com%2Fpreview%2Fdec2011%2Fdata%2Fdanet_c.pdf&usg=AFQjCNGSZt763Hkl9Z1rhTV1-fP-6jvCkg&sig2=9TktqLsfiep3WyiAk7eP7g – Michael Medvinsky Sep 30 '15 at 00:37
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    @MichaelMedvinsky: In those papers, the functions satisfying a maximum principle are not the solutions themselves, but functions of them. For the equation $\Delta^2 u=0$, it is the function $P_9=\lvert\nabla u\rvert^2 -u\Delta u$ that has a maximum principle (note: notations are taken from the linked paper). By the way, if $u(x)=\sin x$ in one dimension, like in our example in this post, then $P_9\equiv 1$. – Giuseppe Negro Sep 30 '15 at 16:28
  • I see that I've sent the same link twice, there were another one, can't find it atm....sorry – Michael Medvinsky Sep 30 '15 at 20:26
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Let $f$ be the eigenfunction for Laplacian with eigenvalue $\mu$. Then

$$\Delta f + \mu f= 0 \Rightarrow \Delta ^2 f + \mu \Delta f =0 \Rightarrow \Delta^2 f - \mu^2 f = 0.$$

Thus if the biharmonic equation satisfy maximum principle with $\lambda >0$, then so is the Laplace equation with arbitrary $\mu \neq 0$.

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The first eigenvalue of the clamped square plate is sign-changing, for a reference see my answer here.

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tks
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What do you mean by a maximum principle here? There are lots of versions of this rule. I assume you mean that the absolute maximum and minimum of all functions defined on the same domain and satisfying the same differential equation will happen on the boundary of the domain. I don't think that such a thing exist for this eigenvalue problem. To give some supporting examples for my claim I challenge you to prove such a maximum principle exist for the following simple eigenvalue problems.

Eigenvalue Problem 1. Let $V$ be the set of all continuous functions $f(x)$ on $[a\,\,\,b]$ satisfying ${{{d^2}f} \over {d{x^2}}}(x) = \lambda f(x)$ on $(a\,\,\,b)$. Can one prove that for every function belonging to $V$, the absolute maximum and minimum will occur at $a$ or $b$? I don't think so!

Eigenvalue Problem 2. Let $V$ be the set of all continuous functions $f(x)$ on $[a\,\,\,b]$ satisfying ${{{d^4}f} \over {d{x^4}}}(x) = \lambda f(x)$ on $(a\,\,\,b)$. Can one prove that for every function belonging to $V$, the absolute maximum and minimum will occur at $a$ or $b$? I am afraid NO!

Also, I may say that you need such a maximum principle to truly prove the only solution to the biharmonic eigenvalue problem mentioned in this question is zero. Consequently, I can say that you cannot prove the only solution to the aforementioned biharmonic eigenvalue problem is zero since there is no maximum principle (with the definition I mentioned above) for this problem.

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Hosein Rahnama
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