Partial Differential Equations/Poisson's equation

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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

Definition 6.5:

The Gamma function   is defined by


The Gamma function satisfies the following equation:

Theorem 6.6:




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



Theorem 6.8:

The volume of the  -dimensional ball with radius  ,   is given by



Theorem 6.9:

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).



Green's kernelEdit

We recall a fact from integration theory:

Lemma 6.11:   is integrable     is integrable.

We omit the proof.

Theorem 6.12:

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


For  ,  

For  ,


, 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

Theorem 6.12:

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:


Recalling that


, the claim follows using the definition of the surface integral.

Theorem 6.13:

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,


Harmonic functionsEdit

Definition 6.14: Let   be open and let   be a function. If   and


  is called a harmonic function.

Theorem 6.15:

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


, that


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.

Definition 6.16:

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.

Theorem 6.17:

Let   and let   be a function. If


then   for almost every  .

Theorem 6.18:

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.

Theorem 6.19:

Let   be a domain and let   be harmonic. If there exists an   such that


, then   is constant.


We choose


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  .


Theorem 6.18:

Let   be a domain and let   be harmonic. If there exists an   such that


, then   is constant.

Proof: See exercise 7.

Corollary 6.20:

Let   be a bounded domain and let   be harmonic on   and continuous on  . Then



Theorem 6.20:

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  



Definition 6.26:

Let   be a sequence of harmonic functions, and let   be a function.   converges locally uniformly to   iff

Theorem 6.27:

Let   be open and let   be harmonic functions such that the sequence   converges locally uniformly to a function  . Then also   is harmonic.


Definition 6.28:

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:

  1.   (the closure of  ) is compact
  2.   is bounded and equicontinuous


Definition 6.30:

Theorem 6.31:

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  .

Representation formulaEdit

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

Let's choose




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.

Solution formulaEdit

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:

  1.   is continuous
  2.   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.

Harmonic loweringEdit

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.

Definition 3.1Edit

Let  . Then we define the following set:


Lemma 3.2Edit

  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.

Lemma 3.3Edit

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.

Lemma 3.4Edit

If   is bounded and  , then the function


is harmonic.


Lemma 3.5Edit

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.

Corollary 6.last:

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.


  1. Prove theorem 6.3 using theorem 6.2 (Hint: Choose   in theorem 6.2).
  2. Prove that  , where   is the factorial of  .
  3. Calculate  . Have you seen the obtained function before?
  4. 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).
  5. For all   and  , calculate   and  .
  6. Let   be open and   be continuous. Prove that   almost everywhere in   implies   everywhere in  .
  7. Prove theorem 6.20 by modelling your proof on the proof of theorem 6.19.
  8. For all dimensions  , give an example for vectors   such that neither   nor  .