# Prove the mean value property of harmonic functions

• Ryker
In summary: This means that G is constant, and since G(x) = g(x) for all x, we can conclude that g(x) must also be constant. And since g is continuous, we can take the limit r → 0 to show that g(x) = g(x), proving that g(x) is harmonic.In summary, we used Green's theorem to show that the function G(x) = \frac{1}{2 \pi r} \int_{\partial B_{r}(x)}gds is constant, and since G(x) = g(x), we can conclude that g(x) is also constant, proving that it is harmonic. I hope this helps
Ryker

## Homework Statement

Let $A \subset \mathbb{R}^{2}$ be an open connected set, and g: A → ℝ a C2 function. Show that if g is harmonic, i.e. $\frac{\partial ^{2} g}{\partial {x_{1}}^{2}} + \frac{\partial ^{2} g}{\partial {x_{2}}^{2}} = 0$, then $g(x) = \frac{1}{2 \pi r} \int_{\partial B_{r}(x)}gds$.

## Homework Equations

We're to use Green's theorem in the form $\int_{\partial A} G \cdot x = \int_{A} (\frac{\partial G_{2}}{\partial {x_{1}}} - \frac{\partial G_{1}}{\partial {x_{2}}})dx$.

## The Attempt at a Solution

OK, I've been struggling to wrap my head around this one for at least three hours now, and after failing to make any sense of it, looking for help online, as well. I've stumbled upon many proofs, but they all use different versions of Green's theorem, and I just really don't get what's going on.

All I gathered is that I need to somehow show that the derivative of the RHS is 0, implying it's constant, and then use that by taking the limit r → 0 and argue that due to continuity it is equal to the LHS, i.e. g(x).

Any help would be greatly appreciated, and I'm sorry I don't have more of an attempt to show, but I have zero idea on how to do this. I tried taking the derivative of the RHS myself, but I can't seem to do that, either...

First of all, don't worry, Green's theorem can be quite confusing at first. But once you understand it, it becomes a powerful tool in solving problems like this one. Let me try to guide you through it.

To start off, let's define the function G(x) = \frac{1}{2 \pi r} \int_{\partial B_{r}(x)}gds. This is the function we want to show is constant. To do this, we will take the derivative of G with respect to r, and show that it is equal to 0. This will imply that G is constant, and thus equal to g(x).

So, let's begin. Using Green's theorem, we can rewrite G as:

G(x) = \frac{1}{2 \pi r} \int_{\partial B_{r}(x)}gds = \frac{1}{2 \pi} \int_{B_{r}(x)} \frac{\partial g}{\partial x_{1}} - \frac{\partial g}{\partial x_{2}} dx

Notice that we have switched the order of the integrals, because we can do that according to Fubini's theorem. Now, we can take the derivative of G with respect to r:

\frac{\partial G}{\partial r}(x) = \frac{1}{2 \pi} \int_{B_{r}(x)} \frac{\partial}{\partial r}(\frac{\partial g}{\partial x_{1}} - \frac{\partial g}{\partial x_{2}}) dx

Using the chain rule, we can simplify this to:

\frac{\partial G}{\partial r}(x) = \frac{1}{2 \pi} \int_{B_{r}(x)} \frac{\partial^{2} g}{\partial x_{1}^{2}} + \frac{\partial^{2} g}{\partial x_{2}^{2}} dx

But wait, we know that g is harmonic, so \frac{\partial^{2} g}{\partial x_{1}^{2}} + \frac{\partial^{2} g}{\partial x_{2}^{2}} = 0. Therefore, we have:

\frac{\partial G}{\partial r}(x) = \frac{1}{2 \pi} \int_{B

## 1. What is the mean value property of harmonic functions?

The mean value property of harmonic functions states that the value of a harmonic function at any point inside a region is equal to the average of its values on the boundary of that region. In other words, the function at a point is the average of its values on all sides of that point.

## 2. How can the mean value property be used to prove the continuity of harmonic functions?

The mean value property can be used to prove the continuity of harmonic functions because it shows that the function's value at any point is determined by its values on the boundary. Therefore, if the function is continuous on the boundary, it must also be continuous at that point.

## 3. What is the significance of the mean value property in understanding the behavior of harmonic functions?

The mean value property is significant in understanding the behavior of harmonic functions because it allows us to make inferences about the function's behavior at any point based on its values on the boundary. This property is also important in various applications, such as in studying heat distribution and fluid flow.

## 4. Can the mean value property be extended to higher dimensions?

Yes, the mean value property can be extended to higher dimensions. In two dimensions, the average is taken over the boundary curve, while in three dimensions, it is taken over the boundary surface. This generalization is known as the mean value property for Laplace's equation.

## 5. How is the mean value property related to the Laplace equation?

The mean value property is closely related to the Laplace equation, which is a partial differential equation that describes the behavior of harmonic functions. In fact, the mean value property can be used as a tool to solve the Laplace equation, as it allows us to find the value of a harmonic function at any point based on its values on the boundary.

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