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Define the elliptic PDE operator $Lu:=-\partial_j(a_{ij}\partial_iu)+cu$ where $A=(a_{ij})$ is and uniformly elliptic matrix and $c\geq 0$, i.e., $A\xi\cdot\xi\geq \theta \lvert\xi\lvert^2$ for $\theta>0$. Hence the corresponding bilinear operate $B[u,v]$ is defined as $$ B[u,v]=\int_\Omega A\nabla u\nabla v\,dx+\int_\Omega cuv\,dx $$ More reference can be founded in Evans PDE book, chapter $6.5.1$. We define the eigenvalue $(\lambda_k)$ for operator $L$ and corresponding eigenvector $w_k\in C^\infty(\Omega)$ such that $$ \begin{cases} Lw_k=\lambda_kw_k&x\in\Omega\\ w_k=0 & x\in\partial\Omega \end{cases} $$ where $\|w_k\|_{L^2(\Omega)}=1$ and we sort $0<\lambda_1<\lambda_2\leq \lambda_3\leq\cdots$

Then the Rayleigh Quotient theorem states that we could find $\lambda$ by $$ \lambda_k=\min\{B[u,u],\,\, \|u\|_{L^2(\Omega)}=1,\,(u,w_l)_{L^2}=0\,\,\forall l\leq k-1\} $$ On Evans PDE book, chapter 6.5.1, it has a wonderful proof by using ONB. But now I want to prove it by using Direct method. i.e., formulate a CoV problem.

Define $E[u]=B[u,u]$ and set $$ M:= \{\|u\|_{L^2(\Omega)}=1,\,(u,w_l)_{L^2}=0\,\,\forall l\leq k-1\}$$ We can check $M$ is weakly closed w.r.t $H_0^1$ norm. Now the variation problem is to minimize $E[u]$ within set $M$. I will omit the details in applying directly method and jump to the end. After applying directly method, I obtained an minimizer $\bar u\in M$ such that $$ E[\bar u]=\min_{u\in M}E[u]:=\lambda_k' $$ and satisfies $$ B[\bar u,v]=\int_\Omega A\nabla \bar u\nabla v\,dx+\int_\Omega c\bar uv\,dx = \gamma \int_\Omega \bar uv\,dx$$ where $\gamma$ is the Largaronger multiplier.

Next, by taking $v=\bar u$ we could prove that $\gamma=\lambda_k'$

Now I need to prove $\lambda_k=\lambda_k'$. Since I have $w_k\in M$ so I quickly conclude that $\lambda_k'\leq \lambda_k$

But I got stuck on how to show $\lambda_k\leq \lambda_k'$... Any help is really welcome!

PS: Sorry for long background introduction... If there are any confusions, please let me know and I will write more details.

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Well, $\lambda_k'$ is an eigenvalue of your problem, with eigenvector $\bar{u}$. If $\lambda_k'< \lambda_k$ then we know that $\bar{u}$ must be in the span of $\{ w_1, \ldots, w_{k-1}\}$, but since $\bar{u}\in M$ we must have $\bar{u}=0$, a contradiction.

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  • $\begingroup$ Ha, I see I see. Thank you! $\endgroup$ – spatially Jan 3 '15 at 12:47

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