A biharmonic converse to Krein–Rutman: a maximum principle near a positive eigenfunction

Abstract

The Green function $G_0(x,y)$ for the biharmonic Dirichlet problem on a smooth domain $\Omega$, that is $\Delta ^{2}u=f$ in $\Omega$ with $u=u_{n}=0$ on $\partial \Omega$, can be written as the difference of a positive function, which bears the singularity at $x=y$, and a rank-one positive function, both of which satisfy the boundary conditions. See Grunau et al. (Proc Am Math Soc 139:2151–2161, 2011). More precisely $G_0(x,y)= H(x,y)- c\, d(x)^2 d(y)^2$ holds, where $d(\cdot )$ is the distance to the boundary $\partial \Omega$ and where H contains the singularity and is positive. We will extend the corresponding estimates to $G_{\lambda }(x,y)$ for the differential operator $\Delta ^{2}-\lambda$ with an optimal dependence on $\lambda$. As a consequence, strict positivity of an eigenfunction with a simple eigenvalue $\lambda _{i}$ implies a positivity preserving property for $\left( \Delta ^{2}-\lambda \right) u=f$ in $\Omega$ with $u=u_{n}=0$ on $\partial \Omega$ for $\lambda$ in a left neighbourhood of $\lambda _{i}$. This result can be viewed as a converse to the Krein–Rutman theorem.

Computation of the first eigenfunctions for the spherical shell

As for the annulus in 2D the first eigenvalues appear as zeroes of explicit transcendental functions as follows. The eigenfunction $\varphi _{{\textcircled {1}}}:\Omega _{\varepsilon }\subset {\mathbb {R}}^{3}\rightarrow {\mathbb {R}}$ is radially symmetric. Therefore, we write $\varphi _{{\textcircled {1}} }(x)={\widetilde{\varphi }}_{{\textcircled {1}}}(|x|)={\widetilde{\varphi }}_{ {\textcircled {1}}}(r)$ and find that the eigenfunction is of the form

$$\begin{aligned} {\widetilde{\varphi }}_{{\textcircled {1}}}(r)=c_{1}\tfrac{\exp (\root 4 \of { \lambda _{{\textcircled {1}}}}r)}{\root 4 \of {\lambda _{{\textcircled {1}}}}r} +c_{2}\tfrac{\exp (-\,\root 4 \of {\lambda _{{\textcircled {1}}}}r)}{\root 4 \of { \lambda _{{\textcircled {1}}}}r}+c_{3}\tfrac{\sin \left( \root 4 \of {\lambda _{ {\textcircled {1}}}}r\right) }{\root 4 \of {\lambda _{{\textcircled {1}}}}r}+c_{4} \tfrac{\cos \left( \root 4 \of {\lambda _{{\textcircled {1}}}}r\right) }{\root 4 \of { \lambda _{{\textcircled {1}}}}}. \end{aligned}$$

Setting $u_{1}(t)=\frac{\exp (t)}{t}$, $u_{2}(t)=\frac{\exp (-\,t)}{t}$, $u_{3}(t)=\frac{\sin (t)}{t}$ and $u_{4}(t)=\frac{\cos (t)}{t}$ one obtains $\lambda _{{\textcircled {1}}}$ as the first zero of

$$\begin{aligned} \lambda&\mapsto \det \begin{pmatrix} u_{1}(\varepsilon \root 4 \of {\lambda }) &{} u_{2}(\varepsilon \root 4 \of {\lambda }) &{} u_{3}(\varepsilon \root 4 \of {\lambda }) &{} u_{4}(\varepsilon \root 4 \of {\lambda }) \\ u_{1}^{\prime }(\varepsilon \root 4 \of {\lambda }) &{} u_{2}^{\prime }(\varepsilon \root 4 \of {\lambda }) &{} u_{3}^{\prime }(\varepsilon \root 4 \of {\lambda }) &{} u_{4}^{\prime }(\varepsilon \root 4 \of {\lambda }) \\ u_{1}(\root 4 \of {\lambda }) &{} u_{2}(\root 4 \of {\lambda }) &{} u_{3}(\root 4 \of { \lambda }) &{} u_{4}(\root 4 \of {\lambda }) \\ u_{1}^{\prime }(\root 4 \of {\lambda }) &{} u_{2}^{\prime }(\root 4 \of {\lambda }) &{} u_{3}^{\prime }(\root 4 \of {\lambda }) &{} u_{4}^{\prime }(\root 4 \of {\lambda }) \end{pmatrix} \\&=\tfrac{4}{\varepsilon ^{2}\lambda }\left( 1-\cosh \left( \left( 1-\varepsilon \right) \root 4 \of {\lambda }\right) \cos \left( \left( 1-\varepsilon \right) \root 4 \of {\lambda }\right) \right) . \end{aligned}$$

Assuming that $\varphi _{{\textcircled {2}}}:\Omega _{\varepsilon }\subset {\mathbb {R}}^{3}\rightarrow {\mathbb {R}}$ can be written as $\varphi _{{\textcircled {2}}}(x)=x_{1}\widetilde{\varphi }(\left| x\right| )$ we find that ${\widetilde{\varphi }}$ is of the form

$$\begin{aligned} {\widetilde{\varphi }}(r)&=c_{5}\frac{\left( 1-\root 4 \of {\lambda _{{\textcircled {2}}}}r\right) \exp \left( \root 4 \of {\lambda _{{\textcircled {2}}}}r\right) }{\left( \root 4 \of {\lambda _{{\textcircled {2}}}}r\right) ^{3}}+c_{6}\frac{\left( 1+\root 4 \of {\lambda _{{\textcircled {2}}}}r\right) \exp \left( -\,\root 4 \of {\lambda _{{\textcircled {2}}}}r\right) }{\left( \root 4 \of {\lambda _{{\textcircled {2}}}}r\right) ^{3}} \\&\quad +c_{7}\frac{\sin \left( \root 4 \of {\lambda _{{\textcircled {2}}}}r\right) -\root 4 \of { \lambda _{{\textcircled {2}}}}r\cos \left( \root 4 \of {\lambda _{{\textcircled {2}}}}r\right) }{\left( \root 4 \of {\lambda _{{\textcircled {2}}}}r\right) ^{3}} \\&\quad +c_{8}\frac{\cos \left( \root 4 \of { \lambda _{{\textcircled {2}}}}r\right) +\root 4 \of {\lambda _{{\textcircled {2}}}}r\sin \left( \root 4 \of {\lambda _{{\textcircled {2}}}}r\right) }{\left( \root 4 \of {\lambda _{{\textcircled {2}}}}r\right) ^{3}}. \end{aligned}$$

As above we define $v_{1}(t)=\frac{(1-t)\exp (t)}{t^{3}}$, $v_{2}(t)=\frac{ (1+t)\exp (-\,t)}{t^{3}}$, $v_{3}(t)=\frac{\sin (t)-t\cos (t)}{t^{3}}$ and $v_{4}(t)=\frac{\cos (t)+t\sin (t)}{t^{3}}$. The eigenvalue $\lambda _{{\textcircled {2}}}$ is the first zero of the function one obtains by replacing $u_{i}$ by $v_{i}$ in the determinant above. Numerical approximations of $\lambda _{{\textcircled {1}}}$ and of $\lambda _{{\textcircled {2}}}$ as a function of $\varepsilon$ show that for all $\varepsilon \in \left( 0,0.3\right)$:

$$\begin{aligned} \lambda _{1,\varepsilon }=\lambda _{{\textcircled {1}}}<\lambda _{{\textcircled {2}}}=\lambda _{2,\varepsilon }. \end{aligned}$$

The sketches of $\varepsilon \mapsto \lambda _{i,\varepsilon }$ with $i\in \{1,2\}$ are found in Fig. 5.