Explication on how obtaining $\int \langle \nabla w, \nabla w\rangle = \lambda \int \langle w, w\rangle$

Question

Could anyone is able to explain to me how to obtain $\int \langle \nabla w, \nabla w\rangle = \lambda \int \langle w, w\rangle$ related to user7530's comment in the question : Rayleigh quotient $Q=(\frac{||\triangledown w||}{||w||})^2$ in using the eigenfunction $\sin(x)$ on the segment $(0,\pi)$ ?

Answer 1

This follows from integration by parts. If $\Delta w = \lambda w$ and $w$ satisfies Dirichlet's boundary conditions then

$$ \lambda \int_{\Omega} \left< w(\mathbf{x}), w(\mathbf{x}) \right>_{\mathbb{C}^n} \, d\mathbf{x} = \lambda \left <w, w \right>_{L^2(\Omega)} = \left <\Delta w, w \right>_{L^2(\Omega)} = \sum_{i=1}^n \int_{\Omega} \frac{\partial^2 w}{\partial x_i^2}(\mathbf{x})\overline{w}(\mathbf{x}) \, dx_1 \dots dx_n \\ = -\sum_{i=1}^n \int_{\Omega} \frac{\partial w}{\partial x_i}(\mathbf{x}) \overline{\frac{\partial w}{\partial x_i}}(\mathbf{x}) \, dx_1 \dots dx_n \\ = -\int_{\Omega} \left<\nabla w, \nabla w \right>_{\mathbb{C}^n} \, d\mathbf{x} = - \left< \nabla w, \nabla w \right>_{L^2(\Omega;\mathbb{C}^n)}. $$

Answer 2

Suppose that $\lambda$ is an eigenvalue of the Laplacian and $w$ is a corresponding eigenfunction so that $-\Delta u = \lambda u$. Integration by parts (using the Dirichlet boundary condition) gives you $$\int \langle \nabla w,\nabla w \rangle = \int \sum_{k=1}^n \frac{\partial w}{\partial x_k} \frac{\partial w}{\partial x_k} = - \int \left( \sum_{k=1}^n \frac{\partial^2 w}{\partial x_k^2} \right) w = \int (-\Delta w) w = \lambda \int w^2$$

Answer 3

Since $\lambda$ is a eigenvalue of $-\Delta$, we have:

$-\Delta w= \lambda w$ with $w=0$ in $\partial \Omega$ (using the Dirichlet boundary condition) or $\frac{\partial w}{\partial \eta} = 0$ in $\partial \Omega$ (using the Neumann boundary condition) . Then, multiplying both sides by $w$ and appliyng the Green's identity (my comment above), we obtain:

$\displaystyle\int_{\Omega} \nabla w \nabla w = \displaystyle \int \lambda w^{2}$

And so,

$||\nabla w||^{2}_{L^{2}} = \lambda ||w||^{2}_{L^{2}}$