# Partial Differential Equations/Distributions

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.

## The dual spaceEdit

For each topological vector space, we can define a dual space.

Definition 3.1:

Let $\mathcal X$ be a topological vector space. The set

$\mathcal X^* := \{T: \mathcal X \to \mathbb R : T \text{ linear and continuous} \}$

is called the dual space of $\mathcal X$.

## DistributionsEdit

### Definition: DistributionsEdit

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.

### Theorem 1.2Edit

Let $(\phi_i)_{i \in \N}$ be an arbitrary sequence of bump functions. If $\phi_i \to \phi$ with respect to the notion of convergence for bump functions, then also $\phi_i \to \phi$ with respect to the notion of convergence for Schwartz functions.

Proof:

Let $K \subset R^d$ be the compact set in which all the $\text{supp } \phi_i$ are contained. In $\R^d$, ‘compact’ is the same as ‘bounded and closed’. Therefore, $K \subseteq B_M(0)$ for some $M > 0$. Then we have for every multiindices $\alpha, \beta \in \N_0^d$ that

$\sup_{x \in R^d} \left| x^\alpha \frac{\partial^\beta}{\partial x^\beta} (\phi_i(x) - \phi(x)) \right| = \sup_{x \in K} \left|x^\alpha \frac{\partial^\beta}{\partial x^\beta} (\phi_i(x) - \phi(x)) \right|$
$\le \sup_{x \in K} |x^\alpha| \left| \frac{\partial^\beta}{\partial x^\beta} (\phi_i(x) - \phi(x)) \right| \le M^{|\alpha|} \sup_{x \in K} \left| \frac{\partial^\beta}{\partial x^\beta} (\phi_i(x) - \phi(x)) \right| \to 0, i \to \infty$

due to the definition of convergence for bump functions. Therefore the sequence converges with respect to the notion of convergence for Schwartz functions.

### ExamplesEdit

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 distributionsEdit

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 1.3Edit

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 spacesEdit

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 1.4Edit

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

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 DistributionsEdit

### Lemma 1.5Edit

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 functionEdit

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 $\psi \cdot T(\varphi) := T(\psi \cdot \varphi)$

### DifferentiationEdit

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-ForwardEdit

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:

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

### ConvolutionEdit

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:

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

## NotesEdit

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 $\mathcal A (\Omega_1)$

## ExercisesEdit

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