Skip to main content
edited title
Link

Functions orthogonal Orthogonality to harmonic functions

Notice removed Draw attention by A random mathematician
Bounty Ended with Mateusz Kwaśnicki's answer chosen by A random mathematician
edited body
Source Link

Let $a_0$ and $b_0$ be smooth compactly supported functions in $\Omega \subset R^3$$B \subset R^3$, $f\in C^1(\Omega)$, and define

$a_n=f\Delta^{-1}(a_{n-1})=-f(x)\int_{\Omega}a_{n-1}(y)\Phi(x-y)dy$$a_n=f\Delta^{-1}(a_{n-1})=-f(x)\int_{B}a_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

$b_n=f\Delta^{-1}(b_{n-1})=-f(x)\int_{\Omega}b_{n-1}(y)\Phi(x-y)dy$$b_n=f\Delta^{-1}(b_{n-1})=-f(x)\int_{B}b_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

where $\Phi$ is the fundamental solution of the laplacian. Suppose

$\int_{\Omega}(a_n-b_n)\varphi=0$$\int_{B}(a_n-b_n)\varphi=0$ for every harmonic function $\varphi$ and every $n\geq 0$. Is $a_0=b_0$? When $f$ is a constant, the above means that $a_0-b_0$ is orthogonal to all functions $\varphi$ with $\Delta^n \varphi=0$ for some $n\in N$. This would imply $a_0=b_0$. Can this be generalized for a general coefficient $f$?

Let $a_0$ and $b_0$ be smooth compactly supported functions in $\Omega \subset R^3$, $f\in C^1(\Omega)$, and define

$a_n=f\Delta^{-1}(a_{n-1})=-f(x)\int_{\Omega}a_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

$b_n=f\Delta^{-1}(b_{n-1})=-f(x)\int_{\Omega}b_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

where $\Phi$ is the fundamental solution of the laplacian. Suppose

$\int_{\Omega}(a_n-b_n)\varphi=0$ for every harmonic function $\varphi$ and every $n\geq 0$. Is $a_0=b_0$? When $f$ is a constant, the above means that $a_0-b_0$ is orthogonal to all functions $\varphi$ with $\Delta^n \varphi=0$ for some $n\in N$. This would imply $a_0=b_0$. Can this be generalized for a general coefficient $f$?

Let $a_0$ and $b_0$ be smooth compactly supported functions in $B \subset R^3$, $f\in C^1(\Omega)$, and define

$a_n=f\Delta^{-1}(a_{n-1})=-f(x)\int_{B}a_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

$b_n=f\Delta^{-1}(b_{n-1})=-f(x)\int_{B}b_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

where $\Phi$ is the fundamental solution of the laplacian. Suppose

$\int_{B}(a_n-b_n)\varphi=0$ for every harmonic function $\varphi$ and every $n\geq 0$. Is $a_0=b_0$? When $f$ is a constant, the above means that $a_0-b_0$ is orthogonal to all functions $\varphi$ with $\Delta^n \varphi=0$ for some $n\in N$. This would imply $a_0=b_0$. Can this be generalized for a general coefficient $f$?

Notice added Draw attention by A random mathematician
Bounty Started worth 50 reputation by A random mathematician
added 229 characters in body
Source Link

Let $a_0$ and $b_0$ be smooth compactly supported functions in $\Omega \subset R^3$, $f\in C^1(\Omega)$, and define

$a_n=f\Delta^{-1}(a_{n-1})=-f(x)\int_{\Omega}a_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

$b_n=f\Delta^{-1}(b_{n-1})=-f(x)\int_{\Omega}b_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

where $\Phi$ is the fundamental solution of the laplacian. Suppose

$\int_{\Omega}(a_n-b_n)\varphi=0$ for every harmonic function $\varphi$ and every $n\geq 0$. Is $a_0=b_0$? When $f$ is a constant, the above means that $a_0-b_0$ is orthogonal to all functions $\varphi$ with $\Delta^n \varphi=0$ for some $n\in N$. This would imply $a_0=b_0$. Can this be generalized for a general coefficient $f$?

Let $a_0$ and $b_0$ be smooth compactly supported functions in $\Omega \subset R^3$, $f\in C^1(\Omega)$, and define

$a_n=f\Delta^{-1}(a_{n-1})=-f(x)\int_{\Omega}a_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

$b_n=f\Delta^{-1}(b_{n-1})=-f(x)\int_{\Omega}b_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

where $\Phi$ is the fundamental solution of the laplacian. Suppose

$\int_{\Omega}(a_n-b_n)\varphi=0$ for every harmonic function $\varphi$ and every $n\geq 0$. Is $a_0=b_0$?

Let $a_0$ and $b_0$ be smooth compactly supported functions in $\Omega \subset R^3$, $f\in C^1(\Omega)$, and define

$a_n=f\Delta^{-1}(a_{n-1})=-f(x)\int_{\Omega}a_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

$b_n=f\Delta^{-1}(b_{n-1})=-f(x)\int_{\Omega}b_{n-1}(y)\Phi(x-y)dy$, $n\geq 1$

where $\Phi$ is the fundamental solution of the laplacian. Suppose

$\int_{\Omega}(a_n-b_n)\varphi=0$ for every harmonic function $\varphi$ and every $n\geq 0$. Is $a_0=b_0$? When $f$ is a constant, the above means that $a_0-b_0$ is orthogonal to all functions $\varphi$ with $\Delta^n \varphi=0$ for some $n\in N$. This would imply $a_0=b_0$. Can this be generalized for a general coefficient $f$?

added 1 character in body
Source Link
Loading
Source Link
Loading