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I'm not a mathematician, I may be inaccurate ; sorry for that ! I have two questions.

First, I need some help to clarify the method about solving and homogeneous PDE. I have read somewhere (and my advisor said) that finding the eigenvalues of an differential operator can be helpful to construct the solutions of the homogeneous PDE associated to this operator.

What does that mean ? Is it true for all order of differential operators ? Or only for second order ?

On another hand, I'm working on solving the biharmonic equation $\nabla^4 f = 0$ on some strange subspaces in $\mathbb{R}^2$. My advisor said that on the whole plane, the smaller eigenvalue associated to $\nabla^4$ will be zero. Is it true ? Why ?

Thanks for your answer. I already search references about all of that on Internet but didn't find anything suitable for my problem.

Have a nice day ! :-)

Edit : Solving $\nabla^4 f = 0$ on the plane and on the sphere will have different solutions ? Is it relevant to study the comportment of the solution on the plane to infer something about the solution on the sphere ? And why is my first sentence "Hello everyone" deleted when I post my question ?

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    Basic fact from linear algebra: if $A$ is a linear operator and $X$ some vector space, then you can solve $Ax = 0$ for $x\in X$ where $x$ is non-zero if and only if 0 is an eigenvalue of $A$.2012-03-12
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    Secondly, what function space are you working on from the biharmonic equation? If $\nabla^4f = 0$ then the function $g = \nabla^2f$ solves $\nabla^2 g = 0$. But in most reasonable function spaces over Euclidean space, the Laplacian has no embedded eigenvalues in its spectrum, hence $g \equiv 0$ necessarily. Applying this again you get that $f\equiv 0$, hence $\nabla^4$ has no eigenvalue at 0. However, if you are asking about the spectrum of the operator, it is a different issue.2012-03-12
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    The functions I consider are from the functional space of $\mathcal{C}^2$ functions at least, and when I work on other space than the euclidean one, with null border conditions ($f=0$ on $\partial \Omega$ and $\frac{\partial f}{\partial n} = 0$ on $\partial \Omega$). So the $\nabla^4$ do not have null eigenvalue, I see. What will be the smallest positive eigenvalue of this operator ?2012-03-12
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    On the plane, if you take the Fourier transform, you see that under suitable decay conditions there cannot be **any** eigenvalues. On a bounded domain for the Dirichlet problem you are considering, it is known that there are a countable number of positive eigenvalues. But the exact values of the eigenvalues will depend on your domain. [This reference](http://www.springerlink.com/content/u64186k5v68111u6/) may help.2012-03-12
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    Okay, thanks for the comments and for the reference ! :-)2012-03-12

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