Partial Differential Equations/Distributions

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To get solutions to the first more difficult partial differential equations (like, for example, Poisson's equation, the heat equation and a more general version of the transport equation), we will now set up the theory of distributions. Distributions are functions which map a function to a real number.

Distributions[edit]

Definition: Distributions[edit]

Let \mathcal A be a function space with a notion of convergence. A distribution T is a mapping T: \mathcal A \to \R with two properties:

  1. T is linear
  2. T is continuous; i. e. if \phi_i \to \phi in the notion of convergence of the function space, then it must follow that T \phi_i \to T \phi in the ordinary notion of convergence in the real numbers known from first semester Analysis (i. e. |T \phi_i - T \phi| \to 0, i \to \infty)

If \mathcal A is the space of the bump functions, we call a distribution T: \mathcal D \to \R a distribution (because usually distributions are distributions with the bump functions as function space). If however \mathcal A is the space of Schwartz distributions, then we call a distribution T: \mathcal S \to \R a tempered distribution.

Examples[edit]

An example for a distribution is the dirac delta distribution for an a \in \R^d, which is defined by

\delta_a(\phi) := \phi(a)

for functions \phi: \R^d \to \R.

Regular distributions[edit]

Let f be a function and \mathcal A \subseteq L^\infty be a function space, where L^\infty denotes the set of the essentially bounded functions (i. e. the functions which are below a certain constant exept for a Lebesgue nullset). Then we can define a mapping \mathcal A \to \R as follows:

T_f (\varphi) := \int_{\R^n} \varphi(x) f(x) dx

We call a distribution T a regular distribution, if and only if there is a function f such that T = T_f.

Theorem 1.3[edit]

The following three claims are true:

  1. If f is an integrable function and \mathcal A \subseteq L^\infty(\R^d), where the inverse of the embedding Id: \mathcal A \to L^\infty is continuous, then T_f as defined above is a distribution.
  2. If f is a locally integrable function, \Omega \subseteq \R^d is a domain and \mathcal A = \mathcal D(\Omega), then T_f as defined above is a distribution.
  3. If f \in L^2(\R^d) and \mathcal A = \mathcal S(\R^d), then T_f as defined above is a distribution.

Proof:

1) The linearity is due to the linearity of the integral. Well-definedness follows from the calculation

\int_{\R^d} |\varphi(x) f(x)| dx \le \|\varphi\|_{L^\infty} \|f\|_{L^1}

Since the inverse of the embedding is continuous, we have

\|\varphi_i - \varphi\|_{\mathcal A} \to 0 \Rightarrow \|\varphi_i - \varphi\|_{L^\infty} \to 0

Therefore, continuity follows from

|T_f \varphi_i - T_f \varphi| = \left| \int_\Omega (\varphi_i - \varphi)(x) f(x) dx \right| \le \|\varphi_i - \varphi\|_{L^\infty} \underbrace{\int_\Omega f(x) dx}_{\text{constant}} \to 0, i \to \infty

2) The proof follows by observing that f \in L^1(\text{supp } \varphi), since \text{supp } \varphi is bounded, and that the notion of convergence in \mathcal D(\Omega) requires that if \phi_i \to \phi, then there exists a compact set K \subset \R^d such that \forall i \in \N : \text{ supp} \phi_i \subseteq K, and then performing almost the same calculations as above.

3) Due to the triangle inequality for integrals and Hölder's inequality, we have

|T_f(\phi_i) - T_f(\phi)| \le \int_{\R^d} |(\phi_i - \phi)(x)| |f(x)| dx \le \|\phi_i - \phi\|_{L^2} \|f\|_{L^2}

But we furthermore have

\begin{align}
\|\phi_i - \phi\|_{L^2}^2 & \le \|\phi_i - \phi\|_{L^\infty} \int_{\R^d} |(\phi_i - \phi)(x)| dx \\
& = \|\phi_i - \phi\|_{L^\infty} \int_{\R^d} \prod_{j=1}^d (1 + x_j^2) |(\phi_i - \phi)(x)| \frac{1}{\prod_{j=1}^d (1 + x_j^2)} dx \\
& \le \|\phi_i - \phi\|_{L^\infty} \|\prod_{j=1}^d (1 + x_j^2) (\phi_i - \phi)\|_{L^\infty} \underbrace{\int_{\R^d} \frac{1}{\prod_{j=1}^d (1 + x_j^2)} dx}_{= \pi^d}
\end{align}

If \phi_i \to \phi in the notion of convergence of the Schwartz function space, then this expression goes to zero. Therefore, continuity is verified. Linearity again follows by the properties of the integral. Well-definedness follows from

\int_{\R^d} |\phi(x)| |f(x)| dx \le \|\phi\|_{L^2} \|f\|_{L^2} < \infty

Distribution spaces[edit]

If \mathcal A (\Omega) is a function space of functions defined on \Omega with a notion of convergence, then the set of all distributions on this space is usually denoted with \mathcal A ' (\Omega). This set is also called a "distribution space". It is the dual space of \mathcal A (\Omega).

Theorem 1.4[edit]

\forall \text{ domain } \Omega \subseteq \R^d : \mathcal D' (\Omega) \supset \mathcal S' (\R^d)

Proof: Let T \in \mathcal S' (\R^d), let \phi_i \to \phi be a convergent sequence of bump functions with their limit, and let \varphi, \psi be two bump functions.

Theorem 1.1 gives us that \phi_i are Schwartz functions.

Theorem 1.2 gives us that \phi_i \to \phi in the sense of Schwartz functions.

From these two statements we can conclude due to T \in \mathcal S' (\R^d), that T \phi_i \to T\phi.

Theorem 1.1 tells us furthermore that \varphi, \psi are Schwartz functions. From this we can conclude due to T \in \mathcal S' (\R^d) that T(\alpha \varphi + \beta \psi) = \alpha T \varphi + \beta T \psi.

This completes the proof.

Operations on Distributions[edit]

Lemma 1.5[edit]

Let \mathcal A(\Omega_1), \mathcal A(\Omega_2) \subseteq L^\infty(\R^d) be function spaces, and L: \mathcal A (\Omega_1) \to L^1_\text{loc} (\Omega_2) be a linear function.

If there exists a linear operator L^*: \mathcal A (\Omega_2) \to \mathcal A (\Omega_1), which is sequentially continuous[1], and it holds that:

\int\limits_{\Omega_1} \varphi(x) (L^*\psi)(x) dx = \int\limits_{\Omega_2} (L \varphi)(x) \psi(x) dx

Then, under these conditions, we may define the operator

\tilde L: \mathcal A' (\Omega_1) \to \mathcal A' (\Omega_2), (\tilde L T) (\varphi) = T(L^* \varphi)

, which really maps to \mathcal A' (\Omega_2), and for regular distributions and f \in \mathcal A(\Omega_1) it will have the property

\tilde L T_f = T_{Lf}

Proof: Well-definedness follows from the fact that L^* \varphi is a function of \mathcal A (\Omega_1) due to the first requirement on L^*. Linearity follows from the linearity of T and linearity of L^*:

(\tilde L T)(\varphi + \psi) := T(L^*(\varphi+\psi)) = T(L^* \varphi + L^* \psi) = T(L^* \varphi) + T(L^* \psi) =: (\tilde L T)(\varphi) + (\tilde L T)(\psi)

Continuity follows just the same way from continuity of T and L^*: Let \phi_i \to \phi w.r.t. the notion of conv. of \mathcal A (\Omega_2). Then

\phi_i \to \phi \Rightarrow L^* \phi_i \to L^* \phi \Rightarrow T (L^* \phi_i) \to T (L^* \phi) \Leftrightarrow: (\tilde L T) (\phi_i) \to (\tilde L T) (\phi)

The property

\tilde L T_f = T_{Lf}

follows directly from the equation

\int\limits_{\Omega_1} \varphi(x) (L^*\psi)(x) dx = \int\limits_{\Omega_2} (L \varphi)(x) \psi(x) dx.

Multiplication by a smooth function[edit]

Let \psi be a smooth function ("smooth" means it is \infty often differentiable). Then, by defining L(\phi(x)) = \psi(x) \cdot \phi(x) and L^*(\varphi(x)) = \psi(x) \cdot \varphi(x), we meet the requirements of the above lemma and may define multiplication of distributions by smooth functions as follows:

Let T \in \mathcal A ', then \psi \cdot T(\varphi) := T(\psi \cdot \varphi)

Differentiation[edit]

For the bump functions and the Schwartz functions, we also may define the differentiation of distributions. Let k \in \N and L = \sum_{|\alpha| \le k} a_\alpha (x) \frac{\partial^\alpha}{\partial x^\alpha}. Let's now define

L^*(\phi) := \sum_{|\alpha| \le k} (-1)^{|\alpha|}\frac{\partial^\alpha}{\partial x^\alpha} (a_\alpha (x) \phi (x)).

Then, for the spaces \mathcal A (\Omega_1) = \mathcal A (\Omega_2) = \mathcal D(\Omega) or \mathcal S(\R^d), the requirements for the above lemma 1.4 are met and we may define the differentiation of distribution in the following way:

L T(\varphi) := T(L^* \varphi)

This definition also satisfies LT_f = T_{Lf}.


Proof: By integration by parts, we obtain:

\int_\Omega \phi(x) \alpha(x) \frac{\partial}{\partial x_i} \psi(x) dx = -\int_\Omega \frac{\partial}{\partial x_i} (\phi(x) \alpha(x)) \psi(x) dx + \int_{\partial \Omega} \alpha(x) \phi(x) \psi(x) \nu_i(x) dx

, where \nu_i is the i-th component of the outward normal vector and \partial \Omega is the boundary of \Omega. For bump functions, the boundary integral \int_{\partial \Omega} \alpha(x) \phi(x) \psi(x) \nu_i(x) dx vanishes anyway, because the functions in \mathcal D (\Omega) are zero there. For Schwartz functions, we may use the identity

\int_{\R^d} \phi(x) \alpha(x) \frac{\partial}{\partial x_i} \psi(x) dx = \lim_{r \to \infty} \int_{B_r(0)} \phi(x) \alpha(x) \frac{\partial}{\partial x_i} \psi(x) dx

and the decreasing property of the Schwartz functions to see that the boundary integral goes to zero and therefore

\int_{\R^d} \phi(x) \alpha(x) \frac{\partial}{\partial x_i} \psi(x) dx = -\int_{\R^d} \frac{\partial}{\partial x_i} (\phi(x) \alpha(x)) \psi(x) dx

To derive the equation

\int\limits_{\Omega} \varphi(x) (L^*\psi)(x) dx = \int\limits_{\Omega} (L \varphi)(x) \psi(x) dx

, we may apply the formula from above several times. This finishes the proof, because this equation was the only non-trivial property of L^*, which we need for applying lemma 1.5.

Push-Forward[edit]

Let \Omega_1 \subseteq \R^d, \Omega_2 \subseteq \R^d be domains, and let \Theta: \Omega_1 \to \Omega_2 be a smooth function from \Omega_1 to \Omega_2, such that for all compact subsets c_2 \subset \Omega_2, \Theta^{-1}(c_2) \subset \Omega_1 is compact. Then we call the function

\Theta^*: \mathcal D(\Omega_2) \to \mathcal D(\Omega_1), \Theta^* (\varphi) = \varphi \circ \Theta

the pull-back of bump functions.

If we choose \Omega_1 = \Omega_2 = \R^d, i. e. \Theta: \R^d \to \R^d is a smooth function from \R^d to \R^d such that for all compact sets c_2 \subset \R^d, \Theta^{-1}(c_1) \subset \R^d is compact, then we also define the pull-back of Schwartz functions just exactly the same way:

\Theta^*: \mathcal S(\R^d) \to \mathcal S(\R^d), \Theta^* (\varphi) = \varphi \circ \Theta

For bump functions and Schwartz functions, we may define the push-forward:

For the bump functions

\Theta_*: \mathcal D'(\Omega_1) \to \mathcal D'(\Omega_2), \Theta_*(T(\phi)) = T(\Theta^*(\phi))

or, for Schwartz functions:

\Theta_*: \mathcal S'(\R^d) \to \mathcal S'(\R^d), \Theta_*(T(\phi)) = T(\Theta^*(\phi))

Convolution[edit]

Let \vartheta \in \mathcal D(B_r(0), and let \Omega_1 \supseteq \Omega_2 + B_r(0). Let's define

L: \mathcal D(\Omega_1) \to C^\infty(\Omega_2), (L \varphi)(y) = (\varphi * \vartheta)(y) := \int_{\Omega} \varphi(x) \vartheta(y - x) dx.

This function (L) is linear, because the integral is linear. It is called the convolution of \vartheta and \varphi.

We can also define: \tilde \vartheta(x) = \vartheta(-x), and:

L^* \varphi := \tilde \vartheta * \varphi

By the theorem of Fubini, we can calculate as follows:

\int_{\Omega_2} (L \varphi)(x) \psi(x) dx = \int_{\Omega_2} \int_{\Omega_1} \vartheta(x - y) \varphi(y) \psi(x) dy dx
= \int_{\Omega_1} \int_{\Omega_2} \vartheta(x - y) \varphi(y) \psi(x) dx dy = \int_{\Omega_1} \varphi(y) (L^*\psi)(y) dy

Therefore, the first assumption for Lemma 1.5 holds.

Due to the Leibniz integral rule, we obtain that for f \in L^1 (i. e. f is integrable) and g \in C^k (\R^d) (i. e. the partial derivatives of g exist up to order k and are also continuous):

\frac{\partial^\alpha}{\partial x^\alpha} (f * g) = f * \left( \frac{\partial^\alpha}{\partial x^\alpha} g \right), |\alpha| \le k

With this formula, we can see (due to the monotony of the integral) that

\sup_{x \in \R^d} \left|\frac{\partial^\alpha}{\partial x^\alpha} (f * g)(x)\right| = \sup_{x \in \R^d} \left| \int_{\R^d} f(y) \frac{\partial^\alpha}{\partial x^\alpha} g(x-y)dy \right| \le \overbrace{\sup_{x \in \R^d} \left| \int_{\R^d} f(y) dy \right|}^{\text{constant}} \cdot \sup_{x \in \R^d} \left| \frac{\partial^\alpha}{\partial x^\alpha} g(x) \right|

From this follows sequential continuity for Schwartz and bump functions by defining f = \vartheta and g = \phi_i - \phi. Thus, with the help of lemma 1.5, we can define the convolution with a distribution of \mathcal D'(\Omega) or \mathcal S'(\R^d) as follows:

(\vartheta * T)(\varphi) := T(\tilde \vartheta * \varphi)

Notes[edit]

  1. This means in this case that if \phi_i \to \phi with respect to the notion of convergence of \mathcal A (\Omega_2), then must also L^* \phi_i \to L^* \phi w.r.t. (="with respecct to") the notion of convergence of \mathcal A (\Omega_1)

Exercises[edit]

  1. Show that R^d endowed with the usual topology is a topological vector space.

Sources[edit]