- #1
mathmari
Gold Member
MHB
- 5,049
- 7
Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I am looking at the proof of the strong maximum principle:
If a function $u$ satisfies the Laplace's equation at the open space $D$ and is continuous at the boundary $\partial{D}$ and achieves its maximum at $\partial{D}$ and at a point of $D$ then the function is a constant.
which is the following:
View attachment 2666
$x_M$ : the point where the function achieves its maximum
$x_M \in D$
$u(x) \leq u(x_M)=M, \forall x \in D$
$u(x)=u(x_M), \forall \text{ choice of circle }$
(Thinking)
I am facing some difficulties understanding this proof.. (Worried)
So, we take a circle with center $x_M$.
Then from the mean value property we have that:
$$u(x_M)=\frac{1}{2 \pi a} \int_{| \overrightarrow{r'}|=a}{u(\overrightarrow{r'})}ds$$
We consider that $x_M \in D$ is the point at which $u$ achieves its maximum, so
$$u(x) \leq u(x_M)=M, \ \ \ \forall x \in D$$
So, we have the following:
$$M=u(x_M)=\frac{1}{2 \pi a} \int_{| \overrightarrow{r'}|=a}{u(\overrightarrow{r'})}ds$$
But since the integral is at the boundary $\partial{D}$, and we have supposed that $M$ is the maximum of all points at $D$, we cannot say that
$$M=u(x_M)=\frac{1}{2 \pi a} \int_{| \overrightarrow{r'}|=a}{u(\overrightarrow{r'})}ds \leq \frac{1}{2 \pi a} \int_{| \overrightarrow{r'}|=a}{M}ds$$
right??
How can I continue then to show that the function is constant?? (Wondering)
I am looking at the proof of the strong maximum principle:
If a function $u$ satisfies the Laplace's equation at the open space $D$ and is continuous at the boundary $\partial{D}$ and achieves its maximum at $\partial{D}$ and at a point of $D$ then the function is a constant.
which is the following:
View attachment 2666
$x_M$ : the point where the function achieves its maximum
$x_M \in D$
$u(x) \leq u(x_M)=M, \forall x \in D$
$u(x)=u(x_M), \forall \text{ choice of circle }$
(Thinking)
I am facing some difficulties understanding this proof.. (Worried)
So, we take a circle with center $x_M$.
Then from the mean value property we have that:
$$u(x_M)=\frac{1}{2 \pi a} \int_{| \overrightarrow{r'}|=a}{u(\overrightarrow{r'})}ds$$
We consider that $x_M \in D$ is the point at which $u$ achieves its maximum, so
$$u(x) \leq u(x_M)=M, \ \ \ \forall x \in D$$
So, we have the following:
$$M=u(x_M)=\frac{1}{2 \pi a} \int_{| \overrightarrow{r'}|=a}{u(\overrightarrow{r'})}ds$$
But since the integral is at the boundary $\partial{D}$, and we have supposed that $M$ is the maximum of all points at $D$, we cannot say that
$$M=u(x_M)=\frac{1}{2 \pi a} \int_{| \overrightarrow{r'}|=a}{u(\overrightarrow{r'})}ds \leq \frac{1}{2 \pi a} \int_{| \overrightarrow{r'}|=a}{M}ds$$
right??
How can I continue then to show that the function is constant?? (Wondering)