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edited Sep 18, 2022 at 16:07

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Take the eigenbasis $\phi_n$ you exhibited for $a=0$. It is still an eigenbasis for $-\Delta + a$, with shifted eigenvalues $\lambda_n + a$. Write $$ E_a=\{n\in\mathbb N : \lambda_n =- a\}, $$ which is either an empty set or a finite set. If $E_a=\emptyset$ (case 1.) then $$ u = \sum_{n\in\mathbb N} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n, $$ and this sum is well defined. Note that only finitely many terms correspond to the case $\lambda_n+a<0$, since $\lambda_n\to\infty$.

If $E_a\neq\emptyset$ (case 2.) note that if $\langle\phi_n,f\rangle\neq0$ for some $n\in E_a$, there isn't a solution, as an integration by parts against $\phi_n$ will show.

If $\langle\phi_n,f\rangle=0$ for all $n\in E_a$, then any solution writes $$ u=\sum_{n\in E_a} \alpha_n \phi_n + \sum_{n\not\in E_a} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n, $$ where the $\alpha_n$ are arbitrary. So to obtain uniqueness, you need to set $\langle\phi_n,u\rangle$ for all $n \in E_a$. When $a=0$, there is a unique eigensolution, corresponding to $\lambda=0$, which is the constant eigensolution $\phi_0=1$. So usually one sets $\langle\phi_0,u\rangle=\int_{[0,1]^2} u dxdy=0$ to ensure uniqueness (but you could choose another constant if you so wished).

Take the eigenbasis $\phi_n$ you exhibited for $a=0$. It is still an eigenbasis for $-\Delta + a$, with shifted eigenvalues $\lambda_n + a$. Write $$ E_a=\{n\in\mathbb N : \lambda_n =- a\}, $$ which is either an empty set or a finite set. If $E_a=\emptyset$ (case 1.) then $$ u = \sum_{n\in\mathbb N} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n, $$ and this sum is well defined. Note that only finitely many terms correspond to the case $\lambda_n+a<0$, since $\lambda_n\to\infty$.

If $E_a\neq\emptyset$ (case 2.) note that if $\langle\phi_n,f\rangle\neq0$ for some $n\in E_a$, there isn't a solution, as an integration by parts against $\phi_n$ will show.

If $\langle\phi_n,f\rangle=0$ for all $n\in E_a$, then any solution writes $$ u=\sum_{n\in E_a} \alpha_n \phi_n + \sum_{n\not\in E_a} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n, $$ where the $\alpha_n$ are arbitrary. So to obtain uniqueness, you need to set $\langle\phi_n,u\rangle$ for all $n \in E_a$.

Take the eigenbasis $\phi_n$ you exhibited for $a=0$. It is still an eigenbasis for $-\Delta + a$, with shifted eigenvalues $\lambda_n + a$. Write $$ E_a=\{n\in\mathbb N : \lambda_n =- a\}, $$ which is either an empty set or a finite set. If $E_a=\emptyset$ (case 1.) then $$ u = \sum_{n\in\mathbb N} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n, $$ and this sum is well defined. Note that only finitely many terms correspond to the case $\lambda_n+a<0$, since $\lambda_n\to\infty$.

If $E_a\neq\emptyset$ (case 2.) note that if $\langle\phi_n,f\rangle\neq0$ for some $n\in E_a$, there isn't a solution, as an integration by parts against $\phi_n$ will show.

If $\langle\phi_n,f\rangle=0$ for all $n\in E_a$, then any solution writes $$ u=\sum_{n\in E_a} \alpha_n \phi_n + \sum_{n\not\in E_a} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n, $$ where the $\alpha_n$ are arbitrary. So to obtain uniqueness, you need to set $\langle\phi_n,u\rangle$ for all $n \in E_a$. When $a=0$, there is a unique eigensolution, corresponding to $\lambda=0$, which is the constant eigensolution $\phi_0=1$. So usually one sets $\langle\phi_0,u\rangle=\int_{[0,1]^2} u dxdy=0$ to ensure uniqueness (but you could choose another constant if you so wished).

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created Sep 18, 2022 at 16:00

2.5k
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36

Take the eigenbasis $\phi_n$ you exhibited for $a=0$. It is still an eigenbasis for $-\Delta + a$, with shifted eigenvalues $\lambda_n + a$. Write $$ E_a=\{n\in\mathbb N : \lambda_n =- a\}, $$ which is either an empty set or a finite set. If $E_a=\emptyset$ (case 1.) then $$ u = \sum_{n\in\mathbb N} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n, $$ and this sum is well defined. Note that only finitely many terms correspond to the case $\lambda_n+a<0$, since $\lambda_n\to\infty$.

If $E_a\neq\emptyset$ (case 2.) note that if $\langle\phi_n,f\rangle\neq0$ for some $n\in E_a$, there isn't a solution, as an integration by parts against $\phi_n$ will show.

If $\langle\phi_n,f\rangle=0$ for all $n\in E_a$, then any solution writes $$ u=\sum_{n\in E_a} \alpha_n \phi_n + \sum_{n\not\in E_a} \frac{1}{\lambda_n + a} \langle\phi_n,f\rangle \phi_n, $$ where the $\alpha_n$ are arbitrary. So to obtain uniqueness, you need to set $\langle\phi_n,u\rangle$ for all $n \in E_a$.