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Consider the $(d+1)$-Laplacian denoted as $\Delta_{d+1}$, ($p$-laplacian with $p=d+1$) and its eigenfunctions in $\mathbb{R}^d$.

I'd like to know whether we can say that any linear combination of eigenfunctions belonging to the same eigenvalue is also an eigenfunction with the same eigenvalue? ( despite the fact that not all eigenfunctions of same eigenvalue are necessarily scaled versions of one another). It is easily verified for the case $d=1$.

PS :

A function $u$ is eigen function corresponding to eigenvalue $\lambda$, if it satisfies the following equation

$$ 0= \lambda u |u|^{d-1} +\nabla \cdot( \nabla u |\nabla u|^{d-1}) $$

Rajesh D
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2 Answers2

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The desired properly fails when $n\geq2$ and $\lambda=0$. Probably it also fails for $\lambda\neq0$, but the special case is enough to show that the claim is false as stated.

If the eigenvalue is zero, the eigenfunctions are simply the $p$-harmonic functions. It is not important that $p=d+1$; the only thing that matters is $p\neq2$. Then the question is whether the sum of two $p$-harmonic functions is always $p$-harmonic.

The answer is no, and there are probably several ways to see it. Here is one method chosen at random. Perhaps the simplest way would be to choose two explicit $p$-harmonic functions and show that the sum is not $p$-harmonic. (One could argue that since the equation is not linear, the space of solutions shouldn't be linear, either. But that's not a sound argument as such.)

Suppose $u$ and $v$ are $p$-harmonic (and in $W^{1,p}_{\text{loc}}(\mathbb R^d)$) and the sum of any two such functions is also $p$-harmonic. Let $f_h=u+hv$ for any $h\in\mathbb R$. Assume that $u$ is not constant.

For convenience, consider the equation in a small open set $\Omega$ where $|\nabla u|$ is bounded away from zero. Such a set exists due to local regularity results unless $u$ is constant. We have that $$ \begin{split} 0 &= \Delta_p f_h \\&= \nabla\cdot(|\nabla f_h|^{p-2}\nabla f_h) \\&= \nabla\cdot[|\nabla u+h\nabla v|^{p-2}(\nabla u+h\nabla v)] \\&= \nabla\cdot[|\nabla u|^{p-2}(1+(p-2)h|\nabla u|^{-2}\nabla u\cdot\nabla v+O(h^2))(\nabla u+h\nabla v)] \\&= \Delta_pu + h \nabla\cdot[ |\nabla u|^{p-2} ((p-2)|\nabla u|^{-2}\nabla u\cdot\nabla v)\nabla u + \nabla v) ] + O(h^2). \end{split} $$ This vanishes for all $h$, so $$ \nabla\cdot[ |\nabla u|^{p-2} ((p-2)|\nabla u|^{-2}\nabla u\cdot\nabla v)\nabla u + \nabla v) ] = 0. $$ If we now choose $u(x)=x_i$, we get $$ (p-2)\partial_i^2 v + \Delta v = 0. $$ Summing this over all $i$ gives $(p+d-2)\Delta v=0$, implying that $\Delta v=0$. That is, we have concluded that any $p$-harmonic function $v$ is also harmonic. But this is not true. To show this, simply examine your favorite example of a non-affine $p$-harmonic function.

Joonas Ilmavirta
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To get started, and to throw more light on what I am seeking, I'd like to verify for the case $d=1$, that is $\mathbb{R}$.

One can verify the equation in $\mathbb{R}$ is $$u''+\lambda u = 0$$ and any function of the form $$u(x) = A\sin(\sqrt{\lambda} x + \theta), \forall A,\theta \in \mathbb{R}$$ is an eigenfunction (Please note that there are no boundary or any boundary restrictions in the problem considered in this Q&A). Note that not all these eigenfunctions are scaled versions of one another.

Proof basis: $$A_1\sin(\sqrt{\lambda} x + \theta_1) + A_2\sin(\sqrt{\lambda} x + \theta_2) = A_3\sin(\sqrt{\lambda} x + \theta_3)$$

Rajesh D
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