Why does my solution to $\Delta u = \lambda u$ contradict the regularity theorem?

Question

I am confused about the regularity theorem for Laplacian. It states that if we take a weak solutions of $\Delta u = \lambda u$, then $u$ must be a smooth function. But I cannot understand this statement in the following example:

Consider two domains: $\Omega_1 = \{0 \leqslant x, y \leqslant 1\}$, and $\Omega_2 = \{1 \leqslant x \leqslant 2, 0 \leqslant y \leqslant 1\}$. Consider the functions $u_1(x,y) = \sin \pi x \sin \pi y$ and $u_2(x,y) = - \sin \pi x \sin \pi y$. They are both eigenfunctions of Laplacian on $\Omega_1$ and $\Omega_2$ with eigenvalue $2$.

Let's construct a function $u$ on $\Omega_1 \cup \Omega_2$

$u(x,y) = \begin{cases} u_1(x,y), \ (x,y) \in \Omega_1 \\ u_2(x,y), \ (x,y) \in \Omega_2\end{cases} $

This function is continuous on $\Omega_1 \cup \Omega_2$ and has first derivatives everywhere except for $\{(x,y) | x = 1\}$. This function is also a weak solution on $\Omega$. So, by the ergularity theorem, it must be smooth. But it is not. What am I missing?

Answer 1

Your $u$ is NOT a weak solution. First of all, it is direct to show that $$\partial_x^2u_1(x,y)=-\pi^2\sin\pi x\sin\pi y=-\pi^2 u_1(x,y)=\partial_y^2u_1(x,y)\Longrightarrow\Delta u_1=-2\pi^2 u_1, $$ and similarly for $u_2$, hence $\lambda=-2\pi^2$ (NOT $2$ as in OP).

$u$ is a weak solution to $\Delta u=\lambda u$ if and only if $$\int_\Omega \nabla u(x,y)\cdot \nabla\phi(x,y)+\lambda u(x,y)\phi(x,y)\,dxdy=0$$ for all $\phi\in C_c^\infty(\Omega)$. Now take an arbitrary $\phi\in C_c^\infty(\Omega)$, we have $$\int_0^1\partial_xu(x,y)\partial_x\phi(x,y)\,dx=\partial_xu_1(1,y)\phi(1,y)-\int_0^1 \partial_x^2u_1(x,y)\phi(x,y)\,dx, \qquad y\in[0,1],$$ hence $$\int_0^1\int_0^1\partial_xu(x,y)\partial_x\phi(x,y)\,dxdy=\int_0^1\partial_xu_1(1,y)\phi(1,y)\,dy-\int_0^1\int_0^1 \partial_x^2u_1(x,y)\phi(x,y)\,dxdy.$$ Similarly we have $$\int_0^1\int_0^1\partial_yu(x,y)\partial_y\phi(x,y)\,dxdy=-\int_0^1\int_0^1 \partial_y^2u_1(x,y)\phi(x,y)\,dxdy,$$ hence \begin{align*} \int_0^1\int_0^1\nabla u(x,y)\cdot \nabla\phi(x,y)\,dxdy&=\int_0^1\partial_xu_1(1,y)\phi(1,y)\,dy-\int_0^1\int_0^1\Delta u_1(x,y)\phi(x,y)\,dxdy\\ &=-\pi\int_0^1\sin(\pi y)\phi(1,y)\,dy-\lambda\int_0^1\int_0^1 u_1(x,y)\phi(x,y)\,dxdy. \end{align*} Similarly, \begin{align*} \int_1^2\int_0^1\nabla u(x,y)\cdot \nabla\phi(x,y)\,dxdy&=-\int_0^1\partial_xu_2(1,y)\phi(1,y)\,dy-\int_1^2\int_0^1\Delta u_2(x,y)\phi(x,y)\,dxdy\\ &=-\pi\int_0^1\sin(\pi y)\phi(1,y)\,dy-\lambda\int_1^2\int_0^1 u_2(x,y)\phi(x,y)\,dxdy. \end{align*} Hence $$\int_\Omega \nabla u(x,y)\cdot \nabla\phi(x,y)+\lambda u(x,y)\phi(x,y)\,dxdy=-2\pi\int_0^1\sin(\pi y)\phi(1,y)\,dy.$$ Taking $\phi\in C_c^\infty(\Omega)$ such that $\int_0^1\sin(\pi y)\phi(1,y)\,dy\neq0$ shows that $u$ is NOT a weak solution to $\Delta u=\lambda u$ in $\Omega$.

Answer 2

For $u_i$ to have eigenvalue $2$ your sign convention should be $-\Delta u = \lambda u$, and there should be no $\pi$s in the function. Then the weak form is $\int_\Omega \nabla u\cdot \nabla \phi - \lambda u \phi =0$.

Let $\Omega_s = \Omega \setminus \{ (x,y) : |x-1|<s\}$. Then $$\int_{\Omega} \nabla u \cdot \nabla \phi = \lim_{s\to 0}\int_{\Omega_s} \nabla u \cdot \nabla \phi.$$ Now $u\in C^\infty(\Omega_s)$ and we are free to integrate by parts (assume $\phi$ is compactly supported away from boundary of $\Omega$) $$\int_{\Omega_s} \nabla u \cdot \nabla \phi = -\int_{\Omega_s} \Delta u \cdot \phi + \int_{\Omega\cap\{x=1\pm s\}}\phi\nabla u\cdot n$$ By construction $-\int_{\Omega_s} \Delta u \cdot \phi\to +\int_{\Omega} \lambda u \cdot \phi$ so we focus on the boundary term. On $x=1\pm s$ the outward normal is horizontal and points towards $1$, i.e. $n=\binom{\mp1}0$. So $$\int_{\Omega\cap\{x=1\pm s\}}\phi\nabla u\cdot n= \left(\int_{y:(1-s,y)\in\Omega} - \int_{y:(1+s,y)\in\Omega} \right)\phi u_x dy$$ Now crucially, $u_x = (u_1)_x$ for $x<1$ and $u_x = (u_2)_x = -(u_1)_x$ for $x>1$. So \begin{align}\int_{\Omega\cap\{x=1\pm s\}}\phi\nabla u\cdot n& = \left(\int_{y:(1-s,y)\in\Omega} + \int_{y:(1+s,y)\in\Omega} \right)\phi (u_1)_x dy \\ &\to 2\int_{y:(1,y)\in\Omega} \phi (u_1)_x dy \end{align} which is not zero for all choices of $\phi$. So its not a weak solution.