— 1. Definition of Newtonian Potential —
Definition 1 Let be a bounded open set, we define the Newtonian potential of is the function on by
where is the fundamental solution of Laplace’s equation.
We need to check if is well-defined. If , then one can easily see . We can further relax to be in , where . Since the Holder conjugate , simply by Holder’s inequality, for any
Therefore is well-defined in the whole space. If we only require to be defined in , we only need to be integrable plus a little bit local boundedness near the singularity , for example, local Holder continuity or so. In this case, we can see is also well defined outside , but not defined on . So this case will become less interesting when we try to solve Dirichlet problem through Newtonian potential, since the function on boundary needs to be defined.
— 2. Regularity of Newtonian Potential —
Theorem 2 (Representation for first derivative) Let be bounded and integrable in . Then and for any ,
Proof: Define a function
Since , we know is well-defined. We want to show .We fix a function in satisfying for , for . And define for small,
where . Clearly since we make the singularity vanished. And we have
Therefore uniformly in compact subsets of as . One can also easily see that uniformly as . Therefore, and
Theorem 3 (Representation for second derivative) Let , with , and also bounded. Then , in , and for any ,
where is any domain containing for which the divergence theorem holds and is extended to be 0 outside .
Proof: The proof follows from the same strategy as the previous theorem. We define to be the RHS of the equation stated in theorem. Since is locally Holder continuous, the integrand is integrable near the singularity, hence is well-defined. Let be the same as defined in last theorem, then define
We will have
provided . Hence, by subtraction
Therefore converges to uniformly on any compact subset of as . One can also easily see that converges uniformly to in . Therefore and .Finally, setting , we have for sufficiently large ,
But we can estimate
for any . Therefore and hence .
Remark 1 If the local Holder continuity of is replaced by Dini continuity, we still have the same result. And this theorem suggests that we can use Newtonian potential to get a solution of Poisson equation.
Theorem 4 (interior Schauder estimate) If , then for any . Furthermore,
Proof: Without loss of generality we can assume and . Take , from the representation we have, we know
If we set , We can split the RHS into 6 parts, namely,
We will estimate all these 6 terms. Since , we have
And should have a same estimate by this argument. To estimate , notice that if , , and if , . And we will have by divergent theorem
To estimate , note that is smooth when , by mean value theorem, we know for some between and . And for , we always have . Then we will have
Since on , we have trivial estimate for :
To estimate , we again have mean value theorem as in estimate of , and since and are bounded, we have
Combining these 6 estimates we can conclude . From the previous theorems we can bounded the norm of and its gradient by . Therefore we have the inequality
This tells us if and , then . And the following example tells us if only continuous, then is not necessarily ( cannot be 0).
Remark 2 If is only continuous, is , but not necessarily .
An example: For with , let be a homogeneous harmonic polynomial of degree 2 with . Choose with when . Define Notice that , when or . Therefore
And since vanishes on the boundary of each annulus and as , we can easily see that is continuous.
\noindent Now for such that define . Then by a similar argument we can see is well-defined and smooth away from . For any , we can find an such that , and Notice that
Since is not , as , we can see is not around . Noticing that
since is homogeneous. We can see as , Also as , therefore stays bounded when . Now suppose is a solution of . Then for . Since is bounded around 0, is a removable singularity for , which makes it a harmonic function. But this implies , which leads to a contradiction. So there is no solutions in any neighborhood of the origin to .
— 3. Solution to Dirichlet problem —
Theorem 5 Let be a bounded domain and suppose each point of is regular (in the sense that Perron’s method will give a harmonic function prescribed continuous boundary value, e.g. is .) Then if is bounded and locally Holder continuous, the classical Dirichlet problem : in , on is uniquely solvable for any continuous boundary values .
Proof: From Theorem 2 and 3, we know that , in . Consider , then the Dirichlet problem is equivalent to in , on , which has a unique solution by Perron’s method.
The regularity of can be relaxed as following theorem.
Theorem 6 Let be as described above, for some and is local Holder continuous then the classical Dirichlet problem can by uniquely solved.
Proof: From section 1 we know is well-defined on . And we need to check . We fix a function as in Theorem 2. And define for small,
where . Clearly . Extending to be 0 outside , then
Hence uniformly as , and thus . Then follow the proof of Theorem 2 one can see . Note that we might not have anymore since we lose boundedness of near the boundary of . Then we can follow the exact same proof of Theorem 3 and Theorem 8 to get the result.
Remark 3 In order to solve Dirichlet problem, we can’t assume for some , otherwise might not be well-defined on boundary.