Partial Differential Equations/Poisson's equation< Partial Differential Equations
This chapter deals with Poisson's equation
Provided that , we will through distribution theory prove a solution formula, and for domains with boundaries satisfying a certain property we will even show a solution formula for the boundary value problem. We will also study solutions of the homogenous Poisson's equation
The solutions to the homogenous Poisson's equation are called harmonic functions.
Important theorems from multi-dimensional integrationEdit
In section 2, we had seen Leibniz' integral rule, and in section 4, Fubini's theorem. In this section, we repeat the other theorems from multi-dimensional integration which we need in order to carry on with applying the theory of distributions to partial differential equations. Proofs will not be given, since understanding the proofs of these theorems is not very important for the understanding of this wikibook. The only exception will be theorem 6.3, which follows from theorem 6.2. The proof of this theorem is an exercise.
Theorem 6.2: (Divergence theorem)
Let a compact set with smooth boundary. If is a vector field, then
, where is the outward normal vector.
Theorem 6.3: (Multi-dimensional integration by parts)
Let a compact set with smooth boundary. If is a function and is a vector field, then
, where is the outward normal vector.
Proof: See exercise 1.
The volume and surface area of d-dimensional spheresEdit
The Gamma function is defined by
The Gamma function satisfies the following equation:
If the Gamma function is shifted by 1, it is an interpolation of the factorial (see exercise 2):
As you can see, in the above plot the Gamma function also has values on negative numbers. This is because what is plotted above is some sort of a natural continuation of the Gamma function which one can construct using complex analysis.
Definition and theorem 6.7:
The -dimensional spherical coordinates, given by
are a diffeomorphism. The determinant of the Jacobian matrix of , , is given by
The volume of the -dimensional ball with radius , is given by
The area of the surface of the -dimensional ball with radius (i. e. the area of ) is given by
The surface area and the volume of the -dimensional ball with radius are related to each other "in a differential way" (see exercise 3).
We recall a fact from integration theory:
Lemma 6.11: is integrable is integrable.
We omit the proof.
The function , given by
is a Green's kernel for Poisson's equation.
We only prove the theorem for . For see exercise 4.
We show that is locally integrable. Let be compact. We have to show that
is a real number, which by lemma 6.11 is equivalent to
is a real number. As compact in is equivalent to bounded and closed, we may choose an such that . Without loss of generality we choose , since if it turns out that the chosen is , any will do as well. Then we have
, where we applied integration by substitution using spherical coordinates from the first to the second line.
We calculate some derivatives of (see exercise 5):
For , we have
For , we have
For all , we have
We show that
Let and be arbitrary. In this last step of the proof, we will only manipulate the term . Since , has compact support. Let's define
Since the support of
, where is the characteristic function of .
The last integral is taken over (which is bounded and as the intersection of the closed sets and closed and thus compact as well). In this area, due to the above second part of this proof, is continuously differentiable. Therefore, we are allowed to integrate by parts. Thus, noting that is the outward normal vector in of , we obtain
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 .
Integration over spheresEdit
Let be a function.
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.
Let be a function. Then
We have , where are the spherical coordinates. Therefore, by integration by substitution, Fubini's theorem and the above formula for integration over the unit sphere,
Definition 6.14: Let be open and let be a function. If and
is called a harmonic function.
Let be open and let . The following conditions are equivalent:
- 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 grows at , 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.
A domain is an open and connected subset of .
For the proof of the next theorem, we need two theorems from other subjects, the first from integration theory and the second from topology.
Let and let be a function. If
then for almost every .
In a connected topological space, the only simultaneously open and closed sets are the whole space and the empty set.
We will omit the proofs.
Let be a domain and let be harmonic. If there exists an such that
, then is constant.
Since is open by assumption and , for every exists an such that
By theorem 6.15, we obtain in this case:
, which is why
, we have even
By theorem 6.17 we conclude that
almost everywhere in , and since
is continuous, even
really everywhere in (see exercise 6). Therefore , and since was arbitrary, is open.
and is continuous. Thus, as a one-point set is closed, lemma 3.13 says is closed in . Thus is simultaneously open and closed. By theorem 6.18, we obtain that either or . And since by assumtion is not empty, we have .
Let be a domain and let be harmonic. If there exists an such that
, then is constant.
Proof: See exercise 7.
Let be a bounded domain and let be harmonic on and continuous on . Then
Let be open and be a harmonic function, let and let such that . Then
What we will do next is showing that every harmonic function is in fact automatically contained in .
Theorem 6.25: Let be open, and let be harmonic. Then . Furthermore, for all , there is a constant depending only on the dimension and such that for all and such that
Let be a sequence of harmonic functions, and let be a function. converges locally uniformly to iff
Let be open and let be harmonic functions such that the sequence converges locally uniformly to a function . Then also is harmonic.
Theorem 6.29: (Arzelà-Ascoli) Let be a set of continuous functions, which are defined on a compact set . Then the following two statements are equivalent:
- (the closure of ) is compact
- is bounded and equicontinuous
Let be a locally uniformly bounded sequence of harmonic functions. Then it has a locally uniformly convergent subsequence.
Boundary value problemEdit
The dirichlet problem for the Poisson equation is to find a solution for
Uniqueness of solutionsEdit
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 kindEdit
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 problemEdit
Green's function of the first kind for the ballEdit
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:
Here shall be continuous on . Then 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.
To do this, we notice first that
This follows due to the fact that if , then solves the problem
and the application of the representation formula.
Furthermore, if and , we have due to the second triangle inequality:
In addition, another application of the second triangle inequality gives:
Let then be arbitrary, and let . Then, due to the continuity of , we are allowed to choose such that
In the end, with the help of all the previous estimations we have made, we may unleash the last chain of inequalities which shows that the representation formula is true:
Since implies , we might choose close enough to such that
- . Since was arbitrary, this finishes the proof.
Let be a domain. A function is called a barrier with respect to if and only if the following properties are satisfied:
- is continuous
- is superharmonic on
Exterior sphere conditionEdit
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 functionsEdit
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 functionsEdit
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 an exercise, you might try to prove this minimum principle yourself.
Let , and let . If we define
, then .
Proof: For this proof, the very important thing to notice is that the formula for inside is nothing but the solution formula for the Dirichlet problem on the ball. Therefore, we immediately obtain that is superharmonic, and furthermore, the values on don't change, which is why . This was to show.
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.
If is bounded and , then the function
If satisfies the exterior sphere condition, then for all there is a barrier function.
Existence theorem of PerronEdit
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 .
Let's summarise the results of this section.
Let be a domain satisfying the exterior sphere condition, let , let be continuous and let be a Green's function of the first kind for . Then
is the unique continuous solution to the boundary value problem
In the next chapter, we will have a look at the heat equation.
- Prove theorem 6.3 using theorem 6.2 (Hint: Choose in theorem 6.2).
- Prove that , where is the factorial of .
- Calculate . Have you seen the obtained function before?
- Prove that for , the function as defined in theorem 6.11 is a Green's kernel for Poisson's equation (hint: use integration by parts twice).
- For all and , calculate and .
- Let be open and be continuous. Prove that almost everywhere in implies everywhere in .
- Prove theorem 6.20 by modelling your proof on the proof of theorem 6.19.
- For all dimensions , give an example for vectors such that neither nor .