Last modified on 28 February 2014, at 21:56

Partial Differential Equations/Fundamental Solutions, Green's functions, Green's kernels and Dirichlet Problems

This chapter will show how you can apply distributions to find solutions for certain PDEs (partial differential equations).

Definition: Solutions in the sense of distributionsEdit

Let L = \sum_{|\alpha| \le k} a_\alpha (x) \frac{\partial^\alpha}{\partial x^\alpha}. Consider the equation:

Lu = f ~ (*)

We call a distribution T \in \mathcal D'(\Omega) a solution in the sense of distributions for (*), if and only if:

LT = T_f

Fundamental SolutionsEdit

Definition: Fundamendal SolutionsEdit

We call a function K_{(\cdot)} : \Omega \to \mathcal D'(\Omega) a fundamental solution for L, if and only if

  • it is continuous[1], and
  • LK_\xi = \delta_\xi, where \delta_\xi(x) is the Dirac delta distribution.

Lemma 2.1Edit

Let T_n be a sequence of distributions in \mathcal A', and let \mathcal A be a function space, which is a locally convex space. Let us furthermore have that \lim_{n \to \infty} T_n(\phi) = T(\phi) for all \phi. Then

T: \mathcal A \to \R, T(\varphi) = \lim_{n \to \infty} T_n(\varphi)

defines a distribution.

Proof: Very difficult; The appendix will have a section about this.

Lemma 2.2Edit

Let \mathcal A be a function space, which is a locally convex normed space. For \Lambda \subseteq \R^d, let \{T_\lambda : \lambda \in \Lambda\} \subseteq \mathcal A' be a family of distributions on that function space. Let's further assume that for all \varphi \in \mathcal A, the function \lambda \mapsto T_\lambda(\varphi) is continuous on \Lambda and bounded, and let f \in L^1(\Lambda). Then

T(\varphi) := \int_\Lambda f(\lambda) T_\lambda(\varphi) d \lambda

defines a distribution.

Proof: Due to the truncation of L^p-functions, we have that there are radii R_i \in \R_+ such that

\int_{\Lambda \setminus B_{R_i}(0)} | f(\lambda) | d\lambda < \frac{1}{2 i \|T_\lambda(\varphi)\|_\infty}

, where \|T_\lambda(\varphi)\|_\infty is the supremum of the function \lambda \mapsto T_\lambda(\varphi).

B_{R_i}(0) is a compact set, since it is bounded as well as closed. Therefore, we may divide B_{R_i}(0) into finitely many (let's say n_i) squares d_{m_i} with diameter at most \delta_i, such that

\forall \nu \in B_{R_i}(0) : \lambda \in B_{\delta_i} (\nu) \Rightarrow |T_\lambda(\varphi) - T_\nu(\varphi)| < \frac{1}{2i \|f\|_{L^1}}

. This we may do because continuous functions are uniformly continuous on compact sets. At the border, we just round the squares so that they fit in with the sphere. Furthermore, we choose for each square a \lambda_{m_i} inside this square.

We choose now

T_i(\varphi) := \sum_{m=0}^{n_i} \int_{d_{m_i}} f(\lambda) T_{\lambda_{m_i}}(\varphi) d \lambda

, which is a finite linear combination of distributions and therefore a distribution. Due to the normal triangle inequality for the absolute value, the triangle inequality for the Lebegue integral, our first calculation and the fundamental integral estimation, we obtain:

|T_i(\varphi) - T(\varphi)| < \sum_{m=0}^{n_i} \int_{d_{m_i}} | f(\lambda) (T_{\lambda_{m_i}}(\varphi) - T_\lambda(\varphi))| d \lambda + \frac{\|T_\lambda(\varphi)\|_\infty}{2 i \|T_\lambda(\varphi)\|_\infty} \le \frac{1}{i}

This obviously goes to zero, and this lemma follows with Lemma 2.1.

Theorem 2.3Edit

Let's assume that in equation (*), f is integrable. Let K_{(\cdot)} be a fundamental solution for (*) with respect to the locally convex normed function space \mathcal A, such that \forall \phi \in \mathcal A, the function \xi \mapsto K_\xi(\phi) is bounded. Then we can know, that:

T(\varphi) = \int_{\R^d} f(\xi) K_\xi(\varphi) d\xi

is well-defined and solves (*) in the sense of distributions.

Proof: Since by the definition of fundamental solutions, the function \xi \mapsto K_\xi(\phi) is continuous, we may apply lemma 2.2, which gives us that T is indeed well-defined.

To show that it really solves (*) in the sense of distributions, we need the following calculation:

LT(\varphi) = T(L^*\varphi) = \int_{\R^d} f(\xi) K_\xi(L^* \varphi) d\xi = \int_{\R^d} f(\xi) LK_\xi(\varphi) d\xi
 = \int_{\R^d} f(\xi) \delta_\xi(\varphi) d\xi = \int_{\R^d} f(\xi) \varphi(\xi) d\xi = T_f(\varphi)

, which is what we wanted to show.

Green's functions and Green's kernelsEdit

Definition: Green's functionEdit

Assume that for each \xi, the fundamental solution K_\xi is a regular distribution, i. e. for each \xi \in \Omega, there is an integrable function G( \cdot| \xi) such that K_\xi = T_{G(\cdot | \xi)}. Then we call this function G: \R^d \times \Omega \to \R a Green's function for L.

Definition: Green's kernelEdit

Let's assume that L has the Green's function G(\cdot|\xi). If there exists a function \tilde G: \R^d \to \R such that

G(\cdot|\xi) = \tilde G(\cdot - \xi)

, then we call \tilde G a Green's kernel for L.

Lemma 2.4Edit

Let \tilde G be a locally integrable function, and \Omega \subseteq \R^d be a domain. Then the family of distributions K_\xi := T_{\tilde G(\cdot - \xi)} \in \mathcal D'(\Omega) is well-defined and depends continuously on \xi. Furthermore, for each \phi \in \mathcal D(\Omega), the function \xi \mapsto K_\xi(\phi) is bounded.

Proof: Well-definedness follows from Lemma 1.3.

Let \phi \in \mathcal D(\Omega), and let \xi_n \to \xi. Then we can calculate the following:

T_{\tilde G(\cdot - \xi_n)}(\phi) - T_{\tilde G(\cdot - \xi)}(\phi) = \int_{\R^d} \tilde G(x - \xi_n) \phi(x) dx - \int_{\R^d} \tilde G(x - \xi) \phi(x) dx = \int_{\R^d} \tilde  G(x) (\phi(x + \xi_n) - \phi(x + \xi)) dx
\le \max_{x \in \R^d} |\phi(x + \xi_n) - \phi(x + \xi)| \underbrace{\int_{\text{supp } \phi + B_{2\xi}(0)} \tilde  G(x) dx}_\text{constant}

for sufficiently large n, where the last expression goes to 0 as n \to \infty, since the support of \phi(x) is compact and therefore the function is (even uniformly) continuous.

Furthermore, we have

T_{\tilde G(\cdot - \xi)}(\phi) = \int_{\R^d} \tilde G(x - \xi) \phi(x) dx = \int_{\text{supp } \phi} \tilde G(x) \phi(x + \xi) dx

, which is zero for \|\xi\| sufficiently large, which is why the function \xi \mapsto K_\xi(\phi) has compact support. But since the function is also continuous, we know that it obtains a maximum and a minimum and is therefore bounded.

RemarkEdit

This lemma shows that if we have found a locally integrable function \tilde G such that LT_{\tilde G(\cdot - \xi)} = \delta_\xi, we already know that it is a Green's kernel, and don't need to check the continuity property.

Theorem 2.5Edit

Now this theorem finally shows us why distributions are useful:

Let \tilde G be a Green's kernel for L, and let f \in L^\infty(\R^d). If

u(x) = (f * \tilde G)(x)

is sufficiently often differentiable such that L u is continuous, then it is a solution for (*) in the classical sense.

Proof: From a case of Hölder's inequality (namely p = 1, q = \infty, i. e. \|f \cdot \tilde G\|_{L^1} \le \|\tilde G\|_{L^1} \cdot \|f\|_{L^\infty}), we obtain that u is locally integrable, which is why T_u is a distribution in \mathcal D'(\R^d).

Furthermore, due to the theorem of Fubini, we have for \varphi \in \mathcal D(\R^d), that

T_u(\varphi) = \int_{\R^d} (f * \tilde G)(x) \varphi(x) dx = \int_{\R^d} \int_{\R^d} f(y) \tilde G(x - y) \varphi(x) dy dx
 = \int_{\R^d} \int_{\R^d} \tilde G(x - y) \varphi(x) dx ~ f(y) dy = \int_{\R^d} T_{\tilde G(\cdot - \xi)}(\varphi) f(y) dy

, which is why T_u solves (*) in the sense of distributions (this is due to theorem 2.3).

Thus, for all \varphi \in \mathcal D(\R^d), we can calculate the following:

\int_{\R^d} (Lu)(x) \varphi(x) dx = T_{Lu} (\varphi) = LT_u(\varphi) = T_f(\varphi) = \int_{\R^d} f(x) \varphi(x) dx

and therefore

\int_{\R^d} ((Lu)(x) - f(x)) \varphi(x) dx = 0.

From this follows that Lu = f almost everywhere. But since Lu and f are both continuous, they must be equal everywhere. This is what we wanted to prove.

Definition: Dirichlet problemsEdit

If one looks at a domain \Omega \subset \R^d with boundary \partial \Omega, then finding a function u which satisfies

\begin{cases}
Lu(x) = f(x) & x \in \Omega \\
u(x) = g(x) & x \in \partial  \Omega
\end{cases}

is called a Dirichlet Problem.

NotesEdit

  1. This means here that \xi \mapsto K_\xi(\phi) as a function from \R^d to \R is continuous for every \phi \in \mathcal D(\Omega).