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a friend and i are working on a research problem and we just hit a wall:

if $u$ is a solution of the 3-dimensional helmholtz operator $\Delta + k^2$ in the ball $B_r(x)$ of radius $r > 0$, then does $u$ satisfy anything like the mean value for harmonic functions? basically, we are wondering what an analogous "mean value property" would be for a nonzero $k$ term. we can't find anything obvious like a harmonic function, where $k = 0$.

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If $u$ is a solution to the equation $\triangle u +k^2 u=0$ in a 3D domain $\Omega$, then for any $x\in\Omega$ and any $r>0$ such that $\{y\in\mathbb R^3:\ |x-y|\leq r \}\subset\Omega$, we have $$u(x)=\frac {p(r)}{4\pi r^2}\int_{|x-y|=r} u(y)dS_y,\qquad\qquad\qquad(1)$$ where $$p(r)=\frac{rk}{\sin rk}.$$

Formula (1) is an analogue of the mean value theorem for harmonic functions (in the case of spherical means).

Edit added: relation (1) is valid for all $r_1\leq r$. If we multiply it by $4\pi r^2/p(r)$ and integrate between $0$ and $r$ we will obtain that $$u(x)=\frac{k^3}{4\pi(\sin rk-rk\cos rk)}\int_{|x-y|\leq r} u(y)dy.$$ The latter formula generalizes the property that the value of a harmonic function at $x\in\Omega$ is equal to function's average value over a ball with the center at $x$.

A short derivation of formula (1) can be found in chapter IV of Methods of Mathematical Physics (Vol. 2) by Courant and Hilbert (or see Harald's comment below).

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This is clearly related to mathoverflow.net/questions/24039/maximum-decay-rate and for a similar reason: In the present case, if $\bar u(r)$ is the mean value over the sphere of radius $r$ then $v(r)=r\bar u(r)$ satisfies $v''+k^2v=0$. Also $v(0)=0$ and $v'(0)=u(x)$, so $v(r)=u(x)\sin kr$. –  Harald Hanche-Olsen May 10 '10 at 4:01
    
it came up when we were researching for the same topic, something about distributions solving certain equations =) –  chris May 10 '10 at 4:07
    
what are some of the best books for studying distributions from a very applied perspective? –  chris May 10 '10 at 4:09
    
Typo in my previous comment: It should be $v(r)=k^{-1}u(x)\sin kr$. –  Harald Hanche-Olsen May 10 '10 at 4:20
    
@chris: you will probably get more attention if you ask for references in a separate question. –  Andrey Rekalo May 10 '10 at 17:55
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