# 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 Helmholtz' equation), we will now take a look at distributions. Distributions are continuous and linear functions which map a test function to a real number.

## Definition and characterisationEdit

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

Definition 4.1:

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

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

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

Having defined the dual space of a topological vector space, we can define distributions as follows:

Definition 4.2:

Let $O \subseteq \mathbb R^d$ be open. We call the elements of $\mathcal D(O)^*$ distributions.

Furthermore, we also give a name to the elements of the dual space of the Schwartz space, but we give a slightly different name:

Definition 4.3:

We call the elements of $\mathcal S(\mathbb R^d)^*$ tempered distributions.

We will now show that distributions are exactly the linear and sequentially continuous functions from $\mathcal D(\Omega)$ to $\mathbb R$ for a domain $\Omega \subseteq \mathbb R^d$.

Theorem 4.4:

Let $O \subseteq \mathbb R^d$ be open and let $\mathcal T: \mathcal D(O) \to \mathbb R$ be a function. Then the following two properties are equivalent:

1. $\mathcal T$ is continuous
2. $\mathcal T$ is sequentially continuous ($\varphi_l \to \varphi \Rightarrow \mathcal T(\varphi_l) \to \mathcal T(\varphi)$)

The same holds true for tempered distributions:

Theorem 4.5:

Let $\mathcal T: \mathcal S(\mathbb R^d) \to \mathbb R$ be a function. Then the following two properties are equivalent:

1. $\mathcal T$ is continuous
2. $\mathcal T$ is sequentially continuous ($\varphi_l \to \varphi \Rightarrow \mathcal T(\varphi_l) \to \mathcal T(\varphi)$)

Proof of theorems 4.4 and 4.5:

1. $\Rightarrow$:

Let $\mathcal T$ be continuous, and let $(\varphi_l)_{l \in \mathbb N}$ be a sequence such that $\varphi_l \to \varphi, l \to \infty$. We choose

$U = (\mathcal T(\varphi) - \epsilon, \mathcal T(\varphi + \epsilon)) \in \mathbb R$

for an arbitrary $\epsilon \in \mathbb R_{>0}$. As $\mathcal T$ is continuous, $\mathcal T^{-1}(U)$ is open. Therefore, $\mathcal X \setminus \mathcal T^{-1}(U)$ is closed. Assume by contradiction that infinitely many elements of the sequence $(\varphi_l)_{l \in \mathbb N}$ are not contained in $\mathcal T^{-1}(U)$. We choose $(\varphi_{l_m})_{m \in \mathbb N}$ as the subsequence of $(\varphi_l)_{l \in \mathbb N}$ not contained in $\mathcal T^{-1}(U)$. Then by the second property of the notion of convergence, $\varphi_{l_m} \to \varphi, m \to \infty$, and since $\mathcal X \setminus \mathcal T^{-1}(U)$ is closed, $\varphi \notin \mathcal T^{-1}(U)$ which contradicts $\mathcal T(\varphi) \in U$. Thus, for large enough $l$, we have $\varphi_l \in \mathcal T^{-1}(U)$ and therefore $\mathcal T(\varphi_l) \in U$. Since $\epsilon \in \mathbb R_{>0}$ was arbitrary, this shows that $\mathcal T(\varphi_l) \to \mathcal T(\varphi)$ and thus $\mathcal T$ is sequentially continuous.

2. $\Leftarrow$:

Let $\mathcal T$ be sequentially continuous. Let $F \subseteq \mathbb R$ be a closed set. We show that $\mathcal T^{-1}(F)$ is closed, thereby showing continuity (due to Lemma 3.13, which showed that in the definition of continuity, we may replace open with closed sets). Let $(\varphi_l)_{l \in \mathbb N}$ be a sequence such that $\forall l \in \mathbb N : \varphi_l \in \mathcal T^{-1}(F)$ and $\varphi_l \to \varphi, l \to \infty$. We have

$\forall l \in \mathbb N : \mathcal T(\varphi_l) \in F$

, because else $\varphi_l$ would by definition of the preimage not be in $\mathcal T^{-1}(F)$. Since $\mathcal T$ is sequentially continuous, we have

$\lim_{l \to \infty} \mathcal T(\varphi_l) = \mathcal T(\varphi)$

Due to the closedness of $F$ it further follows (see exercise 1) that

$\mathcal T(\varphi) \in F$

and thus, again by definition of the preimage,

$\varphi \in \mathcal T^{-1}(F)$

By the definitions of closed sets in $\mathcal D(\Omega)$ and $\mathcal S(\mathbb R^d)$, which were given in the last chapter to be exactly the sets which are closed with respect to convergence, $\mathcal T^{-1}(F)$ is closed.

$////$

Next, we will show that every tempered distribution is a distribution. But to do so, we need the following lemma:

Lemma 4.6:

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

Proof:

Let $(\varphi_l)_{l \in \mathbb N}$ be a sequence in $\mathcal D(\Omega)$ such that $\varphi_l \to \varphi \in \mathcal D(\Omega)$ with respect to the notion of convergence of $\mathcal D(\Omega)$. Let thus $K \subset \mathbb R^d$ be the compact set in which all the $\text{supp } \varphi_l$ are contained. From this also follows that $\text{supp } \varphi \subseteq K$, since otherwise $\|\varphi_l - \varphi\|_\infty \ge |c|$, where $c \in \mathbb R$ is any nonzero value $\varphi$ takes outside $K$; this would contradict $\varphi_l \to \varphi$ with respect to our notion of convergence.

In $\mathbb R^d$, ‘compact’ is equivalent to ‘bounded and closed’. Therefore, $K \subset B_R(0)$ for an $R > 0$. Therefore, we have for all multiindices $\alpha, \beta \in \N_0^d$:

\begin{align} \|x^\alpha \partial_\beta \varphi_l - x^\alpha \partial_\beta \varphi\|_\infty &= \sup_{x \in \mathbb R^d} \left| x^\alpha \partial_\beta \varphi_l(x) - x^\alpha \partial_\beta \varphi(x) \right| & \text{ definition of the supremum norm} \\ &= \sup_{x \in B_R(0)} \left| x^\alpha \partial_\beta \varphi_l(x) - x^\alpha \partial_\beta \varphi(x) \right| & \text{ as } \text{supp } \varphi_l, \text{supp } \varphi \subseteq K \subset B_R(0) \\ &\le R^{|\alpha|} \sup_{x \in B_R(0)} \left| \partial_\beta \varphi_l(x) - \partial_\beta \varphi(x) \right| & \forall i \in \{1, \ldots, d\}, (x_1, \ldots, x_d) \in \overline{B_R(0)} : |x_i| \le R \\ &= R^{|\alpha|} \sup_{x \in \mathbb R^d} \left| \partial_\beta \varphi_l(x) - \partial_\beta \varphi(x) \right| & \text{ as } \text{supp } \varphi_l, \text{supp } \varphi \subseteq K \subset B_R(0) \\ &= R^{|\alpha|} \left\| \partial_\beta \varphi_l(x) - \partial_\beta \varphi(x) \right\|_\infty & \text{ definition of the supremum norm} \\ & \to 0, l \to \infty & \text{ since } \varphi_l \to \varphi \text{ in } \mathcal D(\Omega) \end{align}

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

$////$

Corollary 4.7:

Every tempered distribution is contained in every $\mathcal D(\Omega)^*$, where $\Omega \subseteq \mathbb R^d$ is an arbitrary domain.

Or to be more precise, the restriction of every tempered distribution to $\mathcal D(\Omega)$ is contained in $\mathcal D(\Omega)^*$ for every domain $\Omega$.

Proof:

Let $\mathcal T$ be a tempered distribution, and let $\Omega \subseteq \mathbb R^d$ be an arbitrary domain.

1.

We show that $\mathcal T(\varphi)$ has a well-defined value for $\varphi \in \mathcal D(\Omega)$.

Due to theorem 3.9, every bump function is a Schwartz function, which is why the expression

$\mathcal T (\varphi)$

makes sense for every $\varphi \in \mathcal D(\Omega)$.

2.

We show that the restriction is linear.

Let $a, b \in \mathbb R$ and $\varphi, \vartheta \in \mathcal D(\Omega)$. Since due to theorem 3.9 $\varphi$ and $\vartheta$ are Schwartz functions as well, we have

$\forall a, b \in \mathbb R, \varphi, \vartheta \in \mathcal D(\Omega) : \mathcal T (a \varphi + b \vartheta) = a \mathcal T (\varphi) + b \mathcal T (\vartheta)$

due to the linearity of $\mathcal T$ for all Schwartz functions. Thus $\mathcal T$ is also linear for bump functions.

3.

We show that the restriction of $\mathcal T$ to $\mathcal D(\Omega)$ is sequentially continuous.

Let $\varphi_l \to \varphi, l \to \infty$ with respect to the notion convergence of $\mathcal D(\Omega)$. Due to lemma 4.6, we have that $\varphi_l \to \varphi, l \to \infty$ with respect to the notion convergence of $\mathcal S(\mathbb R^d)$ and therefore, since $\mathcal T$ is sequentially continuous with respect to the notion convergence of $\mathcal S(\mathbb R^d)$ (by theorem 4.5), $\mathcal T(\varphi_l) \to \mathcal T(\varphi)$. This shows sequential continuity and thus continuity (by theorem 4.4).

$////$

## Regular distributionsEdit

In this section, we shortly study a class of distributions which we call regular distributions. In particular, we will see that for certain kinds of functions there exist corresponding regular distributions. But let's first define what regular distributions are.

Definition 4.8:

Let $\Omega \subseteq \mathbb R^d$ be a domain and let $\mathcal T \in \mathcal D(\Omega)^*$. If for all $\varphi \in \mathcal D(\Omega)$ $\mathcal T(\varphi)$ can be written as

$\mathcal T(\varphi) = \int_\Omega f(x) \varphi(x) dx$

for a function $f: \mathbb R^d \to \mathbb R$ which is independent of $\varphi$, then we call $\mathcal T$ a regular distribution.

Definition 4.9:

Let $\mathcal T \in \mathcal S(\mathbb R^d)^*$. If for all $\phi \in \mathcal S(\mathbb R^d)$ $\mathcal T(\phi)$ can be written as

$\mathcal T(\phi) = \int_{\mathbb R^d} f(x) \phi(x) dx$

for a function $f: \mathbb R^d \to \mathbb R$ which is independent of $\phi$, then we call $\mathcal T$ a regular tempered distribution.

Two questions related to this definition could be asked: Given a function $f: \mathbb R^d \to \mathbb R$, is $\mathcal T_f$ given by

$\mathcal T_f(\varphi) := \int_\Omega f(x) \varphi(x) dx$

a distribution? Or is $\mathcal T_f$ given by

$\mathcal T_f(\phi) := \int_{\mathbb R^d} f(x) \phi(x) dx$

a tempered distribution? In general, the answer to these two questions is no, but both questions can be answered with yes if the respective function $f$ has the respectively right properties, as the following two theorems show. But before we state the first theorem, we have to define what local integrability means, because in the case of bump functions, local integrability will be exactly the property which $f$ needs in order to define a corresponding regular distribution:

Definition 4.10:

Let $f: \mathbb R^d \to \mathbb R$ be a function. We say that $f$ is locally integrable iff for all compact subsets $K$ of $\mathbb R^d$

$-\infty < \int_K f(x) dx < \infty$

Now we are ready to give some sufficient conditions on $f$ to define a corresponding regular distribution or regular tempered distribution by the way of

$\mathcal T_f(\varphi) := \int_{\Omega} f(x) \varphi(x) dx$

or

$\mathcal T_f(\phi) := \int_{\mathbb R^d} f(x) \phi(x) dx$

Theorem 4.11:

Let $f: \mathbb R^d \to \mathbb R$ be a locally integrable function, and let $\Omega \subseteq \mathbb R^d$ be a domain. Then

$\mathcal T_f : \mathcal D(\Omega) \to \mathbb R, \mathcal T_f(\varphi) := \int_{\Omega} f(x) \varphi(x) dx$

is a regular distribution.

Proof:

Well-definedness follows from the monotony of the integral:

$\int_{\Omega} |\varphi(x) f(x)| dx \le \int_{\Omega} \|\varphi\|_\infty |f(x)| dx = \|\varphi\|_\infty \int_{\Omega} |f(x)| dx < \infty$

In order to be less or equal than something, the first integral must have a well-defined value in the first place. Since we excluded infinity among the possible values, $\mathcal T_f$ really maps to $\mathbb R$ and well-definedness is proven.

Continuity follows similarly due to

$|T_f \varphi_l - T_f \varphi| = \left| \int_\Omega (\varphi_l - \varphi)(x) f(x) dx \right| \le \|\varphi_l - \varphi\|_\infty \underbrace{\int_\Omega |f(x)| dx}_{\text{independent of } l} \to 0, l \to \infty$

Linearity follows due to the linearity of the integral.

$////$

Theorem 4.12:

Let $f \in L^2(\mathbb R^d)$, i. e.

$\int_{\mathbb R^d} |f(x)|^2 dx < \infty$

Then

$\mathcal T_f : \mathcal S(\mathbb R^d) \to \mathbb R, \mathcal T_f(\phi) := \int_{\mathbb R^d} f(x) \phi(x) dx$

is a regular tempered distribution

Proof:

Well-definedness follows from Hölder's inequality:

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

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

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

Furthermore

\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 follows from the linearity of the integral.

## Operations on DistributionsEdit

For $\varphi, \vartheta \in \mathcal D(\mathbb R^d)$ there are operations such as the differentiation of $\varphi$, the convolution of $\varphi$ and $\vartheta$ and the multiplication of $\varphi$ and $\vartheta$. In the following section, we want to define these three operations (differentiation, convolution with $\vartheta$ and multiplication with $\vartheta$) for a distribution $\mathcal T$ instead of $\varphi$.

Lemma 4.13:

Let $O, U \subseteq \mathbb R^d$ be open sets and let $L : \mathcal D(O) \to \mathcal D(U)$ be a linear function. If there is a linear and continuous function $\mathcal L : \mathcal D(U) \to \mathcal D(O)$ such that

$\forall \varphi \in \mathcal D(O), \vartheta \in \mathcal D(U) : \int_O \varphi(x) \mathcal L(\vartheta)(x) dx = \int_U L(\varphi)(x) \vartheta(x) dx$

, then for every distribution $\mathcal T \in \mathcal D(O)^*$, the function $\varphi \mapsto \mathcal T(\mathcal L(\varphi))$ is a distribution. Therefore, we may define a function

$\Lambda : \mathcal D(O)^* \to \mathcal D(U)^*, \Lambda(\mathcal T) := \mathcal T \circ \mathcal L$

This function has the property

$\forall \varphi \in \mathcal D(O) : \Lambda(\mathcal T_\varphi) = \mathcal T_{L \varphi}$

Noticing that differentiation, convolution and multiplication are linear, we will define these operations for distributions by taking $L$ as the respective of these three operations.

Proof:

We have to prove two claims: First, that the function $\varphi \mapsto \mathcal T(\mathcal L(\varphi))$ is a distribution, and second that $\Lambda$ as defined above has the property

$\forall \varphi \in \mathcal D(O) : \Lambda(\mathcal T_\varphi) = \mathcal T_{L \varphi}$

1.

We show that the function $\varphi \mapsto \mathcal T(\mathcal L(\varphi))$ is a distribution.

$\mathcal T(\mathcal L(\varphi))$ has a well-defined value in $\mathbb R$ as $\mathcal L$ maps to $\mathcal D(O)$, which is exactly the preimage of $\mathcal T$. The function $\varphi \mapsto \mathcal T(\mathcal L(\varphi))$ is continuous since it is the composition of two continuous functions, and it is linear for the same reason (see exercise 2).

2.

We show that $\Lambda$ has the property

$\forall \varphi \in \mathcal D(O) : \Lambda(\mathcal T_\varphi) = \mathcal T_{L \varphi}$

For every $\vartheta \in \mathcal D(U)$, we have

$\Lambda(\mathcal T_\varphi)(\vartheta) := (\mathcal T_\varphi \circ \mathcal L)(\vartheta) := \int_O \varphi(x) \mathcal L(\vartheta)(x) dx \overset{\text{by assumption}}{=} \int_U L(\varphi)(x) \vartheta(x) dx =: \mathcal T_{L \varphi}(\vartheta)$

Since equality of two functions is equivalent to equality of these two functions evaluated at every point, this shows the desired property.

$////$

Using the last lemma 4.13, we define operations like multiplication with a smooth function, differentiation and convolution with a smooth function for distributions.

Definition 4.14:

### 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.

### 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:

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

## ExercisesEdit

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