In this chapter, we will consider Poisson's equation:
The Gaussian integralEdit
The following integral formula holds:
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.
The Gamma functionEdit
The so-called Gamma function is defined as follows:
If one shifts the Gamma function by -1, it interpolates the factorial:
For , the Gamma function satisfies the functional equation of the Gamma function:
This functional equation can be proven in the following way (with integration by parts):
Integration over the surface of d-dimensional spheresEdit
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 spheresEdit
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 spheresEdit
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 skinsEdit
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 equationEdit
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.
Harmonic functions: Elementary propertiesEdit
Definitions: Laplace's equation and harmonic functionsEdit
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
for all below a certain constant, 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 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.
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
Minimum and maximum principlesEdit
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:
Dirichlet problem: Elementary propertiesEdit
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.
Harmonic functions: Advanced propertiesEdit
Multiindex binomial coefficient and multiindex orderEdit
In order to proceed, we need a new version of the binomial coefficient, and an order for multiindices. We start with the multi-index version of the binomial coefficient. We want to define a binomial coefficient, where we put in two multiindices, instead of two integers. We define it in the following way: If ,
We directly observe one property of these multi-index binomial coefficients: If is arbitrary, and , where the is at the -th place, we find:
This formula follows from the definition of and the formula
Next, we want to define an order for multi-indices: Let , then we say:
Notice, that there might be vectors such that neither nor . An example for this are ,
Another way to say this is the order is not total.
Generalized Leibniz product ruleEdit
Now, having defined these two things for multi-indices, we may prove the following generalized form of the Leibniz product rule:
Proof: We prove the claim by applying induction, where the induction variable is .
We start with the induction base . Then the formula just reads
, and this is true. Therefore, we have completed the induction base.
Next, we do the induction step. Let's assume the claim is true for all such that . Let now such that . Let's choose such that (we may do this because ). We define again , where the is at the -th place. Then we have, due to the the Schwarz theorem and the not-generalized Leibniz product rule:
Now we may use the linearity of derivation and the induction hypothesis to obtain:
Then, here comes a key ingredient for the proof: Noticing that
, we notice that we are allowed to shift indices in the first of the two above sums, and furthermore simplify both sums with the rule
Therefore, we obtain:
Now we just sort the sum differently, and then apply our observation
which we made immediately after we defined the binomial coefficients, as well as the observations that
- where in , and (these two rules may be checked from the definition of )
, to find in conclusion:
Let be a domain, and let , and let be a harmonic function on , i. e. . Then the following representation formula for holds:
Proof: 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.
Smoothness and boundedness of the derivativesEdit
Let be a domain. If is a harmonic function, then it is automatically infinitely often differentiable, i. e. . Furthermore, for every , there is a constant such that:
Proof: Due to our representation formula, we obtain:
Convergence theorem for harmonic functionsEdit
If locally uniformly, and all the are harmonic, then also the limit is harmonic.
Theorem of Arzela-AscoliEdit
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
Bolzano-Weierstrass-like theorem for harmonic functionsEdit
If is a sequence of harmonic functions which is locally uniformly bounded, we can find a subsequence which converges locally uniformly towards a harmonic function .
Dirichlet problem: Existence of solutionsEdit
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 .