Partial Differential Equations/Poisson's equation
In this chapter, we will consider Poisson's equation:
- 1 Green's kernel
- 1.1 Gaussian integral
- 1.2 Definition: Gamma function
- 1.3 Functional equation of the Gamma function
- 1.4 Integration over the surface of d-dimensional spheres
- 1.5 The surface area of d-dimensional spheres
- 1.6 The volume of d-dimensional spheres
- 1.7 Integration by onion skins
- 1.8 Green's kernel of Poisson's equation
- 2 Harmonic functions
- 3 Dirichlet problem
- 3.1 Uniqueness of solutions
- 3.2 Green's functions of the first kind
- 3.3 Representation formula
- 3.4 Harmonic functions on the ball: A special case of the Dirichlet problem
- 3.5 Exterior sphere condition
- 3.6 Subharmonic and superharmonic functions
- 3.7 Minimum principle for superharmonic functions
- 3.8 Definition 3.1
- 3.9 Lemma 3.2
- 3.10 Lemma 3.3
- 3.11 Existence theorem of Perron
Now we transform variables by using two dimensional polar coordinates, and then transform one-dimensional with :
Taking the square root on both sides of the equation finishes the proof.
Definition: Gamma function
The function has the following definition:
This function is called the Gamma function.
Functional equation of the Gamma function
The Gamma function satisfies the following functional equation:
Integration over the surface of d-dimensional spheres
be the spherical coordinates. Then we know:
Proof: We choose as an orientation the border orientation of the sphere. We know that for , an outward normal vector field is given by . As a parametrisation of , we only choose the identity function, obtaining that the basis for the tangent space there is the standard basis, which in turn means that the volume form of is
Now, we use the normal vector field to obtain the volume form of :
We insert the formula for and then use Laplace's determinant formula:
As a parametrisation of we choose spherical coordinates with constant radius .
We calculate the Jacobian matrix for the spherical coordinates:
We observe that in the first column, we have only the spherical coordinates divided by . If we fix , the first column disappears. Let's call the resulting matrix and our parametrisation, namely spherical coordinates with constant , . Then we have:
, the claim follows using the definition of the surface integral.
The surface area of d-dimensional spheres
Let . Then:
Proof: Due to the above formula for integration over , we obtain:
Let's also define
Through direct calculation, we easily obtain that and . With one-dimensional integration by parts, we can obtain the following result:
With the help of the equation , we obtain the result
From this we see that
, and therefore
From this we obtain that
This is already an expression, but it is still complicated. To bring it into the form stated in the claim of this lemma, we can use induction:
Let's evaluate first and :
We calculate , and :
Then we note that
which is why
Furthermore, we have for , that
And the important thing is now: has the same recursion equation; for :
Therefore, the two expressions must be equal for all , which proves the claim.
The volume of d-dimensional spheres
Let . Then:
Proof: With Gauss' theorem, we find:
We only need our formula for the surface of the sphere to finish the proof.
Integration by onion skins
Let f be an integrable function. Then:
Proof: Let again be the spherical coordinates. Due to transformation of variables, we obtain:
But due to the formula for integrating on the sphere surface, we also have that
Combining this formula with finishes the proof.
Green's kernel of Poisson's equation
The Poisson's equation has, depending on it's dimension, the following Green's kernel:
, where denotes the surface area of .
First, we show that is locally integrable. Let's choose an arbitrary compact and such that . For , we can see:
By transformation with polar coordinates, we obtain for :
This shows us that we are allowed to apply lemma 2.4, which shows us that is continuous. Well-definedness follows from theorem 1.3.
Furthermore, we calculate now the gradient and the laplacian of for , because we will need them later:
- , since is continuous and has constant absolute value for
Let first .
We define now
Due to the dominated convergence theorem, we have
Let's furthermore choose . Then
From Gauß' theorem, we obtain
, where the minus in the right hand side occurs because we need the inward normal vector. From this follows immediately that
We can now calculate the following, using the Cauchy-Schwartz inequality:
Now we define , which gives:
Applying Gauß' theorem on gives us therefore
, noting that .
We furthermore note that
Therefore, we have
due to the continuity of .
Thus we can conclude that
Therefore, is a Green's kernel for the Poisson's equation for .
For , we can calculate directly, using one-dimensional integration by parts:
, and dividing by 2 gives the result that we wanted.
Definitions: Laplace's equation and harmonic functions
The special case of the Poisson's equation where , i. e.
is called Laplace's equation. A function which satisfies this equation is called a harmonic function.
Let be a harmonic function, i. e. , and let be defined on a superset of . Then the following is true:
, where is the surface area and is the volume of the ball of radius . The two above formulas are average value formulas: They tell us that is equal to it's own average value on the border of a ball and equal to it's own average value on the whole ball.
Also, the following holds: If is a domain and is two times continuously differentiable on (i. e. ), and if satisfies one of the two formulas
, then is harmonic.
Proof: Let's define the following function:
From first coordinate transformation with the diffeomorphism and then applying our formula for integration on the unit sphere twice, we obtain:
From first differentiation under the integral sign and then Gauss' theorem, we know that
Case 1: If is harmonic, then we have
, which is why is constant. Now we can use the dominated convergence theorem for the following calculation:
Therefore for all .
With the relationship
, which is true because of our formula for , we obtain that
, which proves the first formula.
Furthermore, we can prove the second formula by first transformation of variables, then integrating by onion skins, then using the first formula of this theorem and then integration by onion skins again:
This shows that if is harmonic, then the two formulas for calculating , hold.
Case 2: Suppose that is not harmonic. Then there exists an such that . Without loss of generality, we assume that ; the proof for will be completely analoguous exept that the direction of the inequalities will interchange. Then, since as above, due to the dominated convergence theorem, we have
Since is continuous (by the dominated convergence theorem), this is why is not constant, which is a contradiction to the first formula.
The contradiction to the second formula can be obtained by observing that is continuous and therefore there exists a
This means that since
and therefore, by the same calculation as above,
This shows (by proof with contradiction) that if one of the two formulas hold, then is harmonic.
In the chapter about distributions, an example for a bump function was the standard mollifier, given by
, where .
We can also define mollifiers with different support sizes as follows:
With transformation of variables, we have that
Let be a domain. If is a harmonic function, then it is automatically infinitely often differentiable, i. e. . This is surprising, but it is really true.
Proof: By using the property of compact sets of being able to be covered by finitely many balls and looking at the sets , one can prove that there exists a sequence of subsets and radii such that:
Since is infinitely often differentiable on , we also know (due to the Leibniz rule of differentiation under the integral sign) that is infinitely often differentiable on .
We now calculate the following, using integration by onion skins, noticing that is radially symmetric, using one of the two mean-value formulas for and then integration by onion skins again:
This proves that is infinitely often differentiable on for every , and since , the theorem is proven.
Minimum and maximum principles
Let be a harmonic function on the connected domain such that or . If attains maximum or minimum in , i. e. if or for a , then is a constant function.
Proof: We prove the statement for the supremum, and the case for infimum is completely analoguous; it just reverses the only inequality in the proof.
Let's define . Let . Due to the assumption, we have that is not empty. Furthermore, since is open, there is an open ball around every such that . With one of the mean-value formulas (see above), we obtain the inequality
, which implies that on it holds almost everywhere. But since the function which is constantly and are both continuous, we even have on the whole ball . Thus , and therefore is open.
But since , and is continuous, we also have that is relatively closed in , and since is connected, the only possibility is that .
Let be a harmonic function on the connected and bounded domain and also continuous on . Then:
Let be a domain, and let , and let be a harmonic function on , i. e. . Then the following representation formula for holds:
Proof: For this proof, we unfortunately need some stuff of the next subchapter about the Dirichlet problem. So unfortunately, I have to recommend to read the second, third and fourth subsubchapters of this subchapter first.
The proof of this theorem is just calculating the solution formula for the Dirichlet problem
Below we calculated that a Green's function of the first kind for is given by
Furthermore, we have shown below that the representation formula for the general Dirichlet problem
. But since in our case, we have and for , we know that the first term vanishes, which leads to the expression
Calculating this expression explicitly gives the theorem.
The dirichlet problem for the Poisson equation is to find a solution for
Uniqueness of solutions
If is bounded, then we can know that if the problem
has a solution , then this solution is unique on .
Proof: Let be another solution. If we define , then obviously solves the problem
, since for and for .
Due to the above corollary from the minimum and maximum principle, we obtain that is constantly zero not only on the boundary, but on the whole domain . Therefore on . This is what we wanted to prove.
Green's functions of the first kind
Let be a domain. Let be the Green's kernel of Poisson's equation, which we have calculated above, i.e.
, where denotes the surface area of .
Suppose there is a function which satisfies
Then the Green's function of the first kind for for is defined as follows:
is automatically a Green's function for . This is verified exactly the same way as veryfying that is a Green's kernel. The only additional thing we need to know is that does not play any role in the limit processes because it is bounded.
A property of this function is that it satisfies
The second of these equations is clear from the definition, and the first follows recalling that we calculated above (where we calculated the Green's kernel), that for .
Let be a domain, and let be a solution to the Dirichlet problem
. Then the following representation formula for holds:
, where is a Green's function of the first kind for .
Proof: Let's define
. By the theorem of dominated convergence, we have that
Using multi-dimensional integration by parts, it can be obtained that:
When we proved the formula for the Green's kernel of Poisson's equation, we had already shown that
The only additional thing which is needed to verify this is that , which is why it stays bounded, while goes to infinity as , which is why doesn't play a role in the limit process.
This proves the formula.
Harmonic functions on the ball: A special case of the Dirichlet problem
Green's function of the first kind for the ball
is a Green's function of the first kind for .
Proof: Since and therefore
Furthermore, we obtain:
, which is why is a Green's function.
The property for the boundary comes from the following calculation:
, which is why , since is radially symmetric.
Let's consider the following problem:
The following holds: The unique solution for this problem is given by:
Proof: Uniqueness we have already proven; we have shown that for all Dirichlet problems for on bounded domains (and the unit ball is of course bounded), the solutions are unique.
Therefore, it only remains to show that the above function is a solution to the problem. To do so, we note first that
Let be arbitrary. Since is continuous in , we have that on it is bounded. Therefore, by the fundamental estimate, we know that the integral is bounded, since the sphere, the set over which is integrated, is a bounded set, and therefore the whole integral must be always below a certain constant. But this means, that we are allowed to differentiate under the integral sign on , and since was arbitrary, we can directly conclude that on ,
Furthermore, we have to show that , i. e. that is continuous on the boundary.
this i will try to show tomorrow
Exterior sphere condition
Let be a domain. We say that it satisfies the exterior sphere condition, if and only if for all there is a ball such that for some and .
Subharmonic and superharmonic functions
Let be a domain and .
We call subharmonic if and only if:
We call superharmonic if and only if:
From this definition we can see that a function is harmonic if and only if it is subharmonic and superharmonic.
Minimum principle for superharmonic functions
A superharmonic function on attains it's minimum on 's border .
Proof: Almost the same as the proof of the minimum and maximum principle for harmonic functions. As a nice exercise, you might build a proof for this minimum principle on your own :-)
Let . Then we define the following set:
is not empty and
Proof: The first part follows by choosing the constant function , which is harmonic and therefore superharmonic. The second part follows from the minimum principle for superharmonic functions.
Let . If we now define , then .
Proof: The condition on the border is satisfied, because
is superharmonic because, if we (without loss of generality) assume that , then it follows that
, due to the monotony of the integral. This argument is valid for all , and therefore is superharmonic.
Existence theorem of Perron
Let be a bounded domain which satisfies the exterior sphere condition. Then the Dirichlet problem for the Poisson equation, which is, writing it again:
has a solution .