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Let $\Delta$ denote the Laplace operator in $\mathbb R^2$, and let $u_1$ and $u_2$ be two orthogonal eigenfunction of $-\Delta$, i.e., $-\Delta u_1=\alpha_1 u_1$ and $-\Delta u_2=\alpha_2 u_2$, where $\alpha_1$ and $\alpha_2$ are the corresponding eigenvalue. I am wondering do we have $$ \Delta^{-1}(u_1+u_2)=\Delta^{-1}u_1+\Delta^{-1}u_2\tag 1 $$ and if yes, how we proof it?

I know $(1)$ does not hold for general function, but I think it might work for orthogonal functions... PS: by orthogonal I mean $$ \int u_1 u_2dx=0. $$ PPS: by $\Delta^{-1}$ I mean the inverse of Laplace operator, i.e., $\Delta^{-1}\Delta=I$

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    $\begingroup$ The inverse operator is linear. If $\Delta \phi =u$, then $u=\int G\phi dS$ where $G$ is the Green's function. $\endgroup$
    – Mark Viola
    Aug 18, 2016 at 15:19
  • $\begingroup$ @Dr.MV Any reference? $\endgroup$
    – spatially
    Aug 18, 2016 at 15:21
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    $\begingroup$ The inverse is not unique unless you specify more condition. For example $\Delta^{-1}(u_1 + u_2) = \Delta^{-1}u_1 + \Delta^{-1}u_2 + u_3$ is also a valid inverse where $u_3$ is any function satisifying $\Delta u_3 = 0$. $\endgroup$
    – Kibble
    Aug 18, 2016 at 15:36
  • $\begingroup$ @Kibble Sure of course. I will define the boundary condition so that it will be clear. $\endgroup$
    – spatially
    Aug 18, 2016 at 15:40

1 Answer 1

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Note that if we have $\Delta \phi(x,y)=u(x,y)$, then

$$\phi(x,y)=\int_{\mathbb{R}^2} u(x',y')G(x,y|x',y')\,dx'\,dy' \tag 1$$

where $G(x,y|x',y')=\frac{1}{2\pi}\log\left(\sqrt{(x-x')^2+(y-y')^2}\right)$ is the "free-space" Green (or Green's) Function (SEE HERE).

Therefore, the inverse operator of the Laplacian is a linear operator and we assert that $\Delta^{-1}(u_1+u_2)=\Delta^{-1}(u_1)+\Delta^{-1}(u_2)$.

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    $\begingroup$ the link is broken $\endgroup$
    – Kai
    Nov 3, 2020 at 0:18

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