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It is well known that every harmonic function satisfy the mean value property. I was wondering if there exists some similar property for functions $u \colon \Omega \subset \mathbb{R}^n \to \mathbb{R}$ satisfying the equation $$ \Delta u + \nabla u \cdot v = 0 $$ where $v$ is a constant vector. Maybe there is a kind of mean value property, where the "mean" is taken over some possibly more complicated geometric object (such as the heat balls in the case of the heat equation).

Any hint or reference would be very useful! Thanks!

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If, $u : \Omega (\underset{\text{open}}{\subset} \mathbb{R}^n) \to \mathbb{R}$ satisfies $\Delta u + v \cdot \nabla u = 0 \tag{1}$

Since, $$\displaystyle \Delta \left(u(x)e^{\frac{1}{2}v\cdot x}\right) = e^{\frac{1}{2}v\cdot x}\Delta u(x) + e^{\frac{1}{2}v\cdot x}v\cdot \nabla u(x) + \frac{|v|^2}{4}e^{\frac{1}{2}v\cdot x}u(x)$$ writing, $\displaystyle w(x) = u(x)e^{\frac{1}{2}v\cdot x}$, the equation $(1)$ can be written as: $$\Delta w(x) - c^2w(x) = 0 \tag{$1^{*}$}$$ where, $\displaystyle c = |v|/2 > 0$.

If, $B(x_0,r) \subset\subset \Omega$, then for a solution with $ w \in C^{2m}(\Omega)$, expanding in Taylor series leads to,

\begin{align*}w(x) = \sum\limits_{0 \le |\alpha| \le 2m} \frac{1}{\alpha !}D^{\alpha}w(x_0)(x - x_0)^{\alpha} + O(|x-x_0|^{2m})\end{align*}

Now, call a multi-index $\alpha = (\alpha_1, \cdots , \alpha_n)$ even if each coordinate is even.

Then using $$\int_{B(x_0,r)} (x-x_0)^{\alpha}\,dx = \begin{cases} \dfrac{\Gamma\left(\frac{\alpha_1 +1}{2}\right) \cdots \Gamma\left(\frac{\alpha_n +1}{2}\right)r^{|\alpha|+n}}{\Gamma\left(\frac{|\alpha| + n}{2}+1\right)} &\text{ when, } \alpha \text{ (even) } \\ 0 &\text{ otherwise }\end{cases} \tag{2}$$ integrating both sides of Taylor expansion we have,

\begin{align}\int_{B(x_0,r)} w(x)\,dx &= \sum\limits_{0 \le |\alpha| \le 2m} \frac{D^{\alpha}w(x_0)}{\alpha !}\int_{B(x_0,r)} (x-x_0)^{\alpha}\,dx + O(r^{2m}) \\&= \sum\limits_{0 \le |\alpha| \le m} \frac{D^{2\alpha}w(x_0)}{(2\alpha) !}\int_{B(x_0,r)} (x-x_0)^{2\alpha}\,dx + O(r^{2m})\tag{3}\\&= \sum\limits_{0 \le |\alpha| \le m} \frac{D^{2\alpha}w(x_0)}{(2\alpha) !}\frac{\left(\alpha-\frac{1}{2}\right)!}{\left(|\alpha|+\frac{n}{2}\right)!}r^{2|\alpha|+n} + O(r^{2m}) \tag{4}\\&= \sum\limits_{0 \le |\alpha| \le m} \frac{D^{2\alpha}w(x_0)}{\alpha !}\frac{\pi^{n/2}r^n}{\left(|\alpha|+\frac{n}{2}\right)!}\left(\frac{r}{2}\right)^{2|\alpha|} + O(r^{2m}) \tag{5}\\&= \sum\limits_{0 \le k \le m} \frac{\Delta^{k} w(x_0)}{k!}\frac{\pi^{n/2}r^n}{\left(k+\frac{n}{2}\right)!}\left(\frac{r}{2}\right)^{2k} + O(r^{2m}) \tag{6}\\&= |B(x_0,r)|\left(\frac{n}{2}\right)!\sum\limits_{0 \le k \le m} \frac{\Delta^{k} w(x_0)}{k!\left(k+\frac{n}{2}\right)!}\left(\frac{r}{2}\right)^{2k} + O(r^{2m}) \tag{7}\end{align}

where, In line $(3)$ the sum only survives over even indices. In line $(4)$ we used the Legendre Duplication identity, $$(2\alpha)! = \prod\limits_{j=1}^{n} \Gamma\left(2\alpha_j+1\right) = \prod\limits_{j=1}^{n} \frac{2^{2\alpha_j}\Gamma\left(\alpha_j+\frac{1}{2}\right)\Gamma\left(\alpha_j+1\right)}{\sqrt{\pi}} = \frac{2^{2|\alpha|}\alpha!\left(\alpha-\frac{1}{2}\right)!}{\pi^{n/2}}$$. In line $(5)$ we used the identity, $\displaystyle \Delta^{k}w = \sum\limits_{|\alpha| = k} \frac{k!}{\alpha!}D^{2\alpha}w$. Finally in line $(6)$ we used the volume of ball $\displaystyle |B(x_0,r)| = \frac{\pi^{n/2}r^n}{(n/2)!}$.

Now, if $w \in C^{\infty} (\Omega)$, from $(1^{*})$ we have, $\displaystyle \Delta^{k}w = c^{2k}w$ in $\Omega$ and hence from $(7)$ we have, $$\frac{1}{|B(x_0,r)|}\int_{B(x_0,r)} w(x)\,dx = w(x_0)\left(\frac{n}{2}\right)!\sum\limits_{k=0}^{\infty} \frac{(cr/2)^{2k}}{k!\left(k+\frac{n}{2}\right)!} = \frac{\left(\frac{n}{2}\right)!I_{n/2}\left(cr\right)}{(cr/2)^{n/2}}w(x_0) \tag{8}$$

where, $I_{n/2}$ is the Modified Bessel Function of first kind.

Now, $u(x) = 1$ also satisfies $(1)$, hence, we may rewrite $(8)$ in terms of the mean value formula,

$$u(x_0)\int_{B(x_0,r)} e^{\frac{1}{2}v.x}\,dx = \int_{B(x_0,r)} u(x)e^{\frac{1}{2}v.x}\,dx \tag{9}$$

Differentiating both sides w.r.t., $r$ we get the corresponding mean value formula for spheres,

$$u(x_0)\int_{\partial B(x_0,r)} e^{\frac{1}{2}v.x}\,dS(x) = \int_{\partial B(x_0,r)} u(x)e^{\frac{1}{2}v.x}\,dS(x)\tag{10}$$

Similar to the Laplace equation the converse to the mean value property should also hold in this case (I'll try and add more details if needed).

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