Elliptic eigenvalues need not be simple

Membership in $H_0^1$. For $\epsilon >0$ small let ${\chi }_{\epsilon }\in C_c^{\mathrm{\infty }}(0,\pi )$ satisfy $0\le {\chi }_{\epsilon }\le 1$, ${\chi }_{\epsilon }=1$ on $[2\epsilon ,\pi -2\epsilon ]$, ${\chi }_{\epsilon }=0$ outside $[\epsilon ,\pi -\epsilon ]$ and $\mathrm{\mid }{\chi }_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }\le C\mathrm{/}\epsilon$; put $w_{\epsilon }(x,y):={\chi }_{\epsilon }(x){\chi }_{\epsilon }(y)w(x,y)$ for $w=u$ or $w=v$. Then $w_{\epsilon }\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$. Let $A_{\epsilon }$ be the union of the boundary strips where either cutoff differs from $1$; its area is $O(\epsilon )$. The sine factors give $\mathrm{\mid }w\mathrm{\mid }\le C_1\epsilon$ on $A_{\epsilon }$, while $\mathrm{\mid }\mathrm{\nabla }w\mathrm{\mid }\le C_2$ on the whole square. Thus $\mathrm{\parallel }{w}_{\epsilon }-w{\mathrm{\parallel }}_{L^2}^2=O({\epsilon }^3)$. For the gradient, $\mathrm{\nabla }(w_{\epsilon }-w)=({\chi }_{\epsilon }(x){\chi }_{\epsilon }(y)-1)\mathrm{\nabla }w+w\mathrm{\nabla }({\chi }_{\epsilon }(x){\chi }_{\epsilon }(y))$; the first term has squared $L^2$ norm $O(\epsilon )$, and on the support of the cutoff derivatives $\mathrm{\mid }w\mathrm{\mid }\le C_1\epsilon$, so the second term is bounded pointwise and supported on area $O(\epsilon )$, also giving squared $L^2$ norm $O(\epsilon )$. Hence $w_{\epsilon }\to w$ in $H^1(\mathrm{\Omega })$, and $u,v\in H_0^1(\mathrm{\Omega })$ by the closure definition [F1].

Weak eigenidentity. For every $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, integration by parts on the square has no boundary term and $-\mathrm{\Delta }u=5u$, $-\mathrm{\Delta }v=5v$ by [F1], so $a(w,\phi )={\int }_{\mathrm{\Omega }}\mathrm{\nabla }w\cdot \overline{\mathrm{\nabla }\phi }={\int }_{\mathrm{\Omega }}(-\mathrm{\Delta }w)\bar{\phi }=5{\int }_{\mathrm{\Omega }}w\bar{\phi }\mathrm{.}$ Given arbitrary $\phi \in H_0^1(\mathrm{\Omega })$, choose ${\phi }_k\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ with ${\phi }_k\to \phi$ in $H^1$; both pairings are continuous in the $H^1$ norm, so passing to the limit extends the identity to all $\phi \in H_0^1(\mathrm{\Omega })$. By [F2], $u$ and $v$ are weak Dirichlet eigenfunctions with the common eigenvalue $5=1^2+2^2=2^2+1^2$.

Orthogonality and dimension. The inner product factors: $(u,v{)}_{L^2}=({\int }_0^{\pi }\sin x\sin 2xdx\left)\right({\int }_0^{\pi }\sin 2y\sin ydy)=0$ by [F5] (both one-dimensional integrals vanish), while the same factorization gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2=\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}^2=(\pi \mathrm{/}2{)}^2>0$, so both are nonzero classes. Hence $u$ and $v$ are $L^2$-orthogonal nonzero classes, so they are linearly independent by [F3], and the eigenspace of the eigenvalue $5$ has dimension at least two.

Conclusion. The eigenspace of dimension at least two established in step 2.1 contains the linearly independent eigenfunctions $u,v$, so the eigenvalue $5$ occurs with multiplicity at least two in the list of [F4]; the refuted statement, that all eigenvalues of the Dirichlet Laplacian on a bounded domain are simple (no repetitions in the list with multiplicity), therefore fails on the square.