I am computing the eigenvalue of the Laplace problem on an unit square.
Consider the eigenvalue problem as follows. $$-L u(x, y) = \lambda u(x, y)$$ where $$L = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$ and the Dirichlet boundary condition is $u = 0$ at the boundary of the unit square.
Now I am interested to calculate $N(t)$ that is, the number of eigenvalues less than or equal to $t$.
I have a formula, namely Weyl's law to calculate $N(t)$. Now Weyl's law is as follows:
Consider $\Omega$ be a bounded region in $\mathbb{R}^d$ and $\omega_{d}$ is the volume of the unit ball in $\mathbb{R}^d$, then Weyl's law gives
$${\displaystyle N(\lambda )=(2\pi )^{-d}\lambda ^{d/2}\omega _{d}\mathrm {vol} (\Omega )\mp {\frac {1}{4}}(2\pi )^{1-d}\omega _{d-1}\lambda ^{(d-1)/2}\mathrm {area} (\partial \Omega )+o(\lambda ^{(d-1)/2})}$$
My question is how to unwind the formula for $2$ dimensional case, i.e for the unit square in $\mathbb{R}^2$.
Please help me. Thanking in advanced.
Best Answer
The notation $\mathrm{vol}(\Omega)$ and $\mathrm{area}(\partial \Omega)$ can be misleading. The term $\mathrm{vol}(\Omega)$ is the $n$-Lebesgue measure of $\Omega$ which is the area of $\Omega$ when $n=2$. Also, $\mathrm{area}(\partial \Omega)$ is the $(n-1)$-Hausdorff measure of $\partial \Omega$ which is the length of $\partial \Omega$ (i.e the perimeter of $\Omega$) when $n=2$. Hence, when $\Omega$ is the unit square in $\mathbb{R}^2$, $\mathrm{vol}(\Omega)=1$ and $\mathrm{area}(\partial \Omega)=4$.
To calculate $\omega_1$ remember that the unit ball in $\mathbb{R}$ is just the interval $(-1,1)$ which has length 2, so $\omega_1=2$. Similarly, the unit ball in $\mathbb{R}^2$ is just the region enclosed by a circle of radius 1, so $\omega_1=\pi$. Plugging all of this into Weyl's Law gives \begin{align*} N(\lambda) &= (2 \pi)^{-2}\cdot \pi \lambda \mp \frac 1 4 (2 \pi)^{-1} \cdot 2 \cdot 4 \lambda^{\frac 1 2} + o (\lambda^{\frac 1 2 }) \\ &= \frac \lambda {4 \pi} \mp \frac{\lambda^{\frac 1 2 }}{\pi}+ o (\lambda^{\frac 1 2 }) \end{align*} as $\lambda \to \infty$.