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Andrey Rekalo
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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 sphere which is contained in $\Omega$$x\in\Omega$ and any $r>0$ such that $\{y\in\mathbb R^3:\ |x-y|\leq r \}\subset\Omega$, we have $$u(x)=\frac {p(R)}{4\pi R^2}\int_{|x-y|=R} u(y)dS_y,$$$$u(x)=\frac {p(r)}{4\pi r^2}\int_{|x-y|=r} u(y)dS_y,\qquad\qquad\qquad(1)$$ where $$p(R)=\frac{Rk}{\sin Rk}$$$$p(r)=\frac{rk}{\sin rk}.$$

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

The formulaEdit added: relation (1) is valid for all $r_1\leq r$. If we multiply it by $4\pi r^2/p(r)$ and itsintegrate 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).

If $u$ is a solution to the equation $\triangle u +k^2 u=0$ in a 3D domain $\Omega$, then for any sphere which is contained in $\Omega$, we have $$u(x)=\frac {p(R)}{4\pi R^2}\int_{|x-y|=R} u(y)dS_y,$$ where $$p(R)=\frac{Rk}{\sin Rk}$$.

The formula and its derivation can be found in chapter IV of Methods of Mathematical Physics (Vol. 2) by Courant and Hilbert.

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).

Source Link
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122

If $u$ is a solution to the equation $\triangle u +k^2 u=0$ in a 3D domain $\Omega$, then for any sphere which is contained in $\Omega$, we have $$u(x)=\frac {p(R)}{4\pi R^2}\int_{|x-y|=R} u(y)dS_y,$$ where $$p(R)=\frac{Rk}{\sin Rk}$$.

The formula and its derivation can be found in chapter IV of Methods of Mathematical Physics (Vol. 2) by Courant and Hilbert.