Partial Differential Equations/Sobolev spaces

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Partial Differential Equations
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There are some partial differential equations which have no solution. However, some of them have something like ‘almost a solution’, which we call a weak solution. Among these there are partial differential equations whose weak solutions model processes in nature, just like solutions of partial differential equations which have a solution.

These weak solutions will be elements of the so-called Sobolev spaces. By proving properties which elements of Sobolev spaces in general have, we will thus obtain properties of weak solutions to partial differential equations, which therefore are properties of some processes in nature.

In this chapter we do show some properties of elements of Sobolev spaces. Furthermore, we will show that Sobolev spaces are Banach spaces (this will help us in the next section, where we investigate existence and uniqueness of weak solutions).

Repetition: Lp spaces[edit]

Lemma 12.2:

Let S \subseteq \mathbb R^d and let f, g: S \to \mathbb R be functions such that f, g \in L^1_{\text{loc}}(S) and \mathcal T_f = \mathcal T_g. Then f = g almost everywhere.


We define

h: \mathbb R^d \to \mathbb R, h(x) := \begin{cases}
f(x) - g(x) & x \in S \\
0 & x \notin S

Weak derivatives[edit]

Definition 12.1:

Let S \subseteq \R^d be a set, p \in [1, \infty] and f \in L^p(S). If \alpha \in \mathbb N_0^d is a d-dimensional multiindex and g \in L^p(S) such that

\partial_\alpha \mathcal T_f = \mathcal T_g

, we call g an \alphath-weak derivative of f.

Remarks 12.2: If f \in L^p(S) is a function and \alpha \in \mathbb N_0^d is a d-dimensional multiindex, any two \alphath-weak derivatives of f are equal except on a null set. Furthermore, if \partial_\alpha f exists, it also is an \alphath-weak derivative of f.


1. We prove that any two \alphath-weak derivatives are equal except on a nullset.

Let g, h \in L^p(S) be two \alphath-weak derivatives of f. Then we have

\mathcal T_g = \partial_\alpha \mathcal T_f = \mathcal T_h

Notation 12.3 If it exists, we denote the \alphath-weak derivative of f by \partial_\alpha f, which is of course the same symbol as for the ordinary derivative.

Theorem 12.4:

Let O \subseteq \mathbb R^d be open, p \in [1, \infty], f, g \in L^p(O) and \alpha \in \mathbb N_0^d. Assume that f, g have \alpha-weak derivatives, which we - consistent with notation 12.3 - denote by \partial_\alpha f and \partial_\alpha g. Then for all b, c \in \mathbb R:

\partial_\alpha (b f + c g) = b \partial_\alpha f + c \partial_\alpha g


Definition and first properties of Sobolev spaces[edit]

Before proceeding to the definition and first properties of the Sobolev spaces, we recall the following theorem from integration theory:

Lemma 12.5:

For an open set O \subseteq \mathbb R^d and an arbitrary p \in [1, \infty], the space L^p(O) with the norm

\|f\|_{L^p(O)} := \int_O |f(x)|^p dx

is a Banach space.

We omit the proof.

Now we are ready to define the Sobolev spaces:

Definition and theorem 12.6:

Let O \subseteq \mathbb R^d be open, p \in [1, \infty], f, g \in L^p(O) and n \in \N_0. The Sobolev space \mathcal W^{n, p}(O) is defined as follows:

\mathcal W^{n, p}(O) := \{f \in L^p(O) : \forall \alpha \in \mathbb N_0^d \text{ such that } |\alpha| \le n : \partial_\alpha f \text{ exists}\}

A norm on \mathcal W^{n, p}(O) is defined as follows:

\|f\|_{\mathcal W^{n, p}(O)} := \sum_{|\alpha| \le n} \left\| \partial_\alpha f \right\|_{L^p(O)}

With respect to this norm, \mathcal W^{n, p}(O) is a Banach space.

In the above definition, \partial_\alpha f denotes the \alphath-weak derivative of f.



We show that

\| f \|_{\mathcal W^{n, p}(O)} = \sum_{|\alpha| \le n} \left\| \partial_\alpha f \right\|_{L^p(O)}

is a norm.

We have to check the three defining properties for a norm:

  • \|f\|_{\mathcal W^{n, p}(O)} = 0 \Leftrightarrow f = 0 (definiteness)
  • \|c f\|_{\mathcal W^{n, p}(O)} = |c|\|f\|_{\mathcal W^{n, p}(O)} for every c \in \mathbb R (absolute homogeneity)
  • \|f + g\|_{\mathcal W^{n, p}(O)} \le \|f\|_{\mathcal W^{n, p}(O)} + \|g\|_{\mathcal W^{n, p}(O)} (triangle inequality)

We start with definiteness: If f = 0, then \|f\|_{\mathcal W^{n, p}(O)} = 0, since all the directional derivatives of the constant zero function are again the zero function. Furthermore, if \|f\|_{\mathcal W^{n, p}(O)} = 0, then it follows that \|f\|_{L^p(O)} = 0 implying that f=0 as \|f\|_{L^p(O)} is a norm.

We proceed to absolute homogeneity. Let c \in \mathbb R.

\|cf\|_{\mathcal W^{n, p}(O)} & := \sum_{|\alpha| \le n} \left\| \partial_\alpha c f \right\|_{L^p(O)} & \\
& = \sum_{|\alpha| \le n} \left\| c \partial_\alpha f \right\|_{L^p(O)} & \text{ theorem 12.4} \\
& = \sum_{|\alpha| \le n} |c| \left\| \partial_\alpha f \right\|_{L^p(O)} & \text{ by absolute homogeneity of } \| \cdot \|_{L^p(O)} \\
& = |c| \sum_{|\alpha| \le n} \left\| \partial_\alpha f \right\|_{L^p(O)} & \\
& =: |c| \|f\|_{\mathcal W^{n, p}(O)}

And the triangle inequality has to be shown:

\|f + g\|_{\mathcal W^{n, p}(O)} & := \sum_{|\alpha| \le n} \left\| \partial_\alpha (f + g) \right\|_{L^p(O)} & \\
& = \sum_{|\alpha| \le n} \left\| \partial_\alpha f + \partial_\alpha g \right\|_{L^p(O)} & \text{ theorem 12.4} \\
& \le \sum_{|\alpha| \le n} \left( \left\| \partial_\alpha f \right\|_{L^p(O)} + \left\| \partial_\alpha g \right\|_{L^p(O)} \right) & \text{ by triangle inequality of } \| \cdot \|_{L^p(O)} \\
& = \|f\|_{\mathcal W^{n, p}(O)} + \|g\|_{\mathcal W^{n, p}(O)}


We prove that \mathcal W^{n, p}(O) is a Banach space.

Let (f_l)_{l \in \mathbb N} be a Cauchy sequence in \mathcal W^{n, p}(O). Since for all d-dimensional multiindices \alpha \in \mathbb N_0^d with |\alpha| \le n and m, l \in \mathbb N

\|\partial_\alpha f_l - \partial_\alpha f_m)\|_{L^p(O)} = \|\partial_\alpha(f_l - f_m)\|_{L^p(O)} \le \sum_{|\alpha| \le n} \left\| \partial_\alpha (f_l - f_m) \right\|_{L^p(O)}

since we only added non-negative terms, we obtain that for all d-dimensional multiindices \alpha \in \mathbb N_0^d with |\alpha| \le n, (\partial_\alpha f_l)_{l \in \mathbb N} is a Cauchy sequence in L^p(O). Since L^p(O) is a Banach space, this sequence converges to a limit in L^p(O), which we shall denote by f_\alpha.

We show now that f := f_{(0, \ldots, 0)} \in \mathcal W^{n, p}(O) and f_l \to f, l \to \infty with respect to the norm \| \cdot \|_{\mathcal W^{n, p}(O)}, thereby showing that \mathcal W^{n, p}(O) is a Banach space.

To do so, we show that for all d-dimensional multiindices \alpha \in \mathbb N_0^d with |\alpha| \le n the \alphath-weak derivative of f is given by f_\alpha. Convergence then automatically follows, as

f_l \to f, l \to \infty & \Leftrightarrow \| f_l - f \|_{\mathcal W^{n, p}(O)} \to 0, l \to \infty & \\
& \Leftrightarrow \sum_{|\alpha| \le n} \left\| \partial_\alpha (f_l - f) \right\|_{L^p(O)} \to 0, l \to \infty &  \\
& \Leftrightarrow \sum_{|\alpha| \le n} \left\| \partial_\alpha f_l - \partial_\alpha f \right\|_{L^p(O)} \to 0, l \to \infty & \text{by theorem 12.4} \\

where in the last line all the summands converge to zero provided that \partial_\alpha f = f_\alpha for all d-dimensional multiindices \alpha \in \mathbb N_0^d with |\alpha| \le n.

Let \varphi \in \mathcal D(O). Since \partial_\alpha f_l \to f_\alpha and by the second triangle inequality

\|\partial_\alpha f - f_\alpha\| \ge |\|\partial_\alpha f\| - \|f_\alpha\||

, the sequence (\varphi \partial_\alpha f_l)_{l \in \mathbb N} is, for large enough l, dominated by the function 2 \|\varphi\|_\infty f_\alpha, and the sequence (\partial_\alpha \varphi f_l)_{l \in \mathbb N} is dominated by the function 2 \|\partial_\alpha \varphi\|_\infty f.

incomplete: Why are the dominating functions L1?


\int_{\R^d} \partial_\alpha \varphi(x) f(x) dx =& \lim_{l \to \infty} \int_{\R^d} \partial_\alpha \varphi(x) f_l(x) dx & \text{ dominated convergence} \\
& = \lim_{l \to \infty} (-1)^{|\alpha|} \int_{\R^d} \varphi(x) \partial_\alpha f_l(x) dx & \\
&= (-1)^{|\alpha|} \int_{\R^d} \varphi(x) f_\alpha(x) dx & \text{ dominated convergence}

, which is why f_\alpha is the \alphath-weak derivative of f for all d-dimensional multiindices \alpha \in \mathbb N_0^d with |\alpha| \le n.


Approximation by smooth functions[edit]

Multi-dimensional mollifiers[edit]

In the chapter about distributions, an example for a bump function was the standard mollifier, given by

\eta(x) = \frac{1}{z}\begin{cases} e^{-\frac{1}{1-\|x\|^2}}& \text{ if } \|x\| < 1\\
                 0& \text{ if } \|x\|\geq 1

, where z := \int_{B_1(0)} e^{-\frac{1}{1-\|x\|^2}} dx.

We can also define mollifiers with different support sizes as follows:

\eta_\epsilon(x) = \frac{1}{\epsilon^n} \eta\left( \frac{x}{\epsilon} \right)

With transformation of variables, we have that

\int_{B_\epsilon(0)} \eta_\epsilon(x) dx = \int_{B_1(0)} \eta(x) dx = 1

Approximation of continuous functions[edit]

Recall the definition of the multi-dimensional mollifiers. Let \Omega \subseteq \R^d be a domain and u \in C_0(\Omega), i. e. u is a continuous function with compact support.

Then u * \eta_\epsilon \in C^\infty(\Omega) and u * \eta_\epsilon \to u uniformly as \epsilon \to 0.

Proof: First, that u * \eta_\epsilon \in C^\infty(\Omega) follows by the definition of the convolution and the Leibniz integral rule (depending on how you define the convolution, you also need to prove that u * \eta_\epsilon \in C^\infty(\Omega) = \eta_\epsilon \in C^\infty(\Omega) * u, so that you apply the derivative to \eta_\epsilon). This shows the first part.

We further know that u is a continuous function. Since u has compact support, it is also uniformly continuous. Furthermore:

(u * \eta_\epsilon) (y) = \int_{\R^d} u(x) \eta_\epsilon(y - x) dx

Since u is uniformly continuous, we may choose \epsilon > 0 such that \|x - y\| < 2 \epsilon \Rightarrow |u(x) - u(y)| < \delta. Therefore, we find, since \int_{\R^d} \eta_\epsilon(x) dx = 1, the following:

\left| \int_{\R^d} u(x) \eta_\epsilon(y - x) dx - u(y) \right| \le \int_{\R^d} |u(x) - u(y)| \eta_\epsilon(y - x) dx < \delta

Since the smallest possible \delta decreases with decreasing \epsilon, and is arbitrary, the claim is proven.

Approximation of Lp-functions[edit]

Let \Omega \subseteq \R^d be a domain, 1 \le p < \infty and u \in L^p(\Omega), and let furthermore \text{supp } u + B_\epsilon(0) \subseteq \Omega.

Then u * \eta_\epsilon \in C^\infty(\Omega) and u * \eta_\epsilon \to u uniformly as \epsilon \to 0.

Proof: The first claim, namely u * \eta_\epsilon \in C^\infty(\Omega), follows exactly the same way as in the subsubchapter about mollifying continuous functions, by the Leibnis rule for integrating under the integral sign.

We will use that C_0(\Omega) is a dense subset of L^p(\Omega). This means by definition that

\forall f \in L^p(\Omega): \forall \epsilon > 0: \exists g \in C_0(\Omega) : \|f - g\|_{L^p} < \epsilon

i. e. one can approximate functions in L^p(\Omega) arbitrarily well with functions in C_0(\Omega).

This is proven by approximating the elementary functions, which are used in the definition of the Lebesgue integral, with continuous functions. It is OK that they have only compact support, because all L^p-functions h have the property that for arbitrary \epsilon > 0 there exists a bounded set M such that

\left| \int_{\R^d \setminus M} h(x) dx \right| < \epsilon


Let' choose an arbitrary \epsilon > 0. We choose now first g \in C_0(\Omega) such that \|g - f\|_{L^p} < \frac{\epsilon}{4}.

Then we choose \delta_1 > 0 such that \|g - g * \eta_{\delta_1} \| < \frac{\epsilon}{4}.

Third, we notice that f * \eta_\delta - g * \eta_\delta = (f - g) * \eta_\delta. With the triangle inequality and the reversed triangle inequality, we furthermore obtain:

|\|(f - g) * \eta_\delta\|_{L^p} - \|f - g\|_{L^p}| \le \|((f - g) * \eta_\delta) - (f - g)\|_{L^p}

If we choose now \delta_2 > 0 such that \|((f - g) * \eta_{\delta_2}) - (f - g)\| < \frac{\epsilon}{4}, we obtain, together with the fact that \|f - g\|_{L^p} < \frac{\epsilon}{4}, that

\|f * \eta_{\delta_2} - g * \eta_{\delta_2}\|_{L^p} < \frac{\epsilon}{2}

We now choose \delta = \min\{\delta_1, \delta_2\} and obtain with the triangle inequality:

\|f - f * \eta_\delta\|_{L^p} \le \|f - g\|_{L^p} + \|g - g * \eta_\delta\|_{L^p} + \|f * \eta_\delta - g * \eta_\delta\|_{L^p} < \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{2} = \epsilon

Since \epsilon > 0 was arbitrary, this finishes the proof.

Lemma 7.3[edit]

Let \Omega \subset \R^d be a domain, let r > 0, and U \subset \Omega, such that U + B_r(0) \subseteq \Omega. Let furthermore u \in \mathcal W^{m, p}(U). Then \mu_\epsilon * f is in C^\infty(U) for \epsilon < r and \lim_{\epsilon \to 0} \|\mu_\epsilon * f - f\|_{W^{m, p}(U)} = 0.

Proof: The first claim, that \mu_\epsilon * f \in C^\infty(U), follows from the fact that if we choose

\tilde f(x) = \begin{cases}
f(x) & x \in U \\
0 & x \notin U 

Then, due to the above section about mollifying L^p-functions, we know that the first claim is true.

The second claim follows from the following calculation, using the one-dimensional chain rule:

\frac{\partial^\alpha}{\partial x^\alpha} (\mu_\epsilon * f) (y)= \int_{\R^d} \frac{\partial^\alpha}{\partial x^\alpha}\mu_\epsilon(y -x) f(x) dx = (-1)^{|\alpha|} \int_{\R^d} \frac{\partial^\alpha}{\partial y^\alpha}\mu_\epsilon(y -x) f(x) dx
=\int_{\R^d} \mu_\epsilon(y -x) \frac{\partial^\alpha}{\partial y^\alpha} f(x) dx = (\mu_\epsilon * \frac{\partial^\alpha}{\partial y^\alpha}f) (y)

Due to the above secion about mollifying L^p-functions, we immediately know that \lim_{\epsilon \to 0} \|\mu_\epsilon * \frac{\partial^\alpha}{\partial y^\alpha}f - f\| = 0, and the second statement therefore follows from the definition of the W^{m, p}(U)-norm.

Partition of unity[edit]

Let \Omega \subseteq \R^d be an open set, and \bigcup_{\alpha \in A} M_\alpha be an open cover of \Omega. Then there exists a sequence of functions \eta_i such that the following conditions are satisfied:

  1. \forall i \in \N : \forall x \in \Omega : 0 \le \eta_i(x) \le 1
  2. \forall x \in \Omega : \exists \text{ only finitely many } i \in \N : \eta_i(x) \neq 0
  3. \forall i \in \N : \exists \alpha \in A : \text{supp } \eta_i \subseteq M_\alpha
  4. \forall x \in \Omega : \sum_{i=0}^\infty \eta_i(x) = 1

Proof: We will proof this by explicitly constructing such a sequence of functions. First, we construct a sequence of closed balls B_i with the properties

  1. \forall i \in \N : \exists \alpha \in A : B_i \subseteq M_\alpha
  2. \forall x \in \Omega : \exists \text{ only finitely many } i \in \N : x \in B_i
  3. \bigcup_{i \in \N} \text{interior}( B_i) = \Omega.

In order to do this, we first start with the definition of a sequence compact sets:

K_i := \{x \in \Omega : \text{dist}(\partial \Omega, x) \ge \frac{1}{i}, \|x\| \le i\}

This sequence obviously has the properties

  • \bigcup_{j \in \N} K_j = \Omega
  • K_j \subset \text{interior}(K_{j+1})

We now construct our sequence of closed balls such that

K_1 \subset \bigcup_{1 \le i \le k_1} \text{interior}(B_i) \subseteq \text{interior}(K_2)
and K_{j+1} \setminus \text{interior}(K_j) \subset \bigcup_{k_j < i \le k_{j+1}} B_i \subseteq \text{interior}(K_{j+2}) \setminus K_{j-1}

We do this in the following way: To meet the first condition, we first cover K_1 with balls by choosing for every x \in K_1 a ball B_x such that B_x \subseteq ( M_\alpha \cap \text{interior}(K_2)). Since these balls cover K_1, and K_1 is compact, we may choose a finite subcover B_1, \ldots B_{k_1}.

To meet the second condition, we do just the same thing, noticing that K_{j+1} \setminus \text{interior}(K_j) is compact and \text{interior}(K_{j+2}) \setminus K_{j-1} is open.

This sequence of open balls fulfills the conditions which we wanted.

We recall now the Definition of the multi-dimensional mollifier and note that we can recenter it to an arbitrary x_0 \in \R^d by using the formula

\eta_{\epsilon, x_0} (x) := \eta_\epsilon (x - x_0)

What we do now, is that for each B_i, we choose \tilde \eta_i as the mollifier which is adjusted to the ball, i. e. the ball is it's support. Then we choose

\eta(x) := \sum_{i=0}^\infty \tilde \eta_i(x)


\eta_i := \frac{\tilde \eta_i}{\eta},

and we have found a sequence of functions which satisfies the conditions we wanted.

Approximation of Wm, p-functions with smooth functions on open sets[edit]

Let \Omega \subseteq \R^d be an open set. Then for all functions v \in W^{m, p}(\Omega), there exists a sequence of functions in C^\infty(\Omega) \cap W^{m, p}(\Omega) approximating it.


Let's choose

U_i := \{x \in \Omega : \text{dist}(\partial \Omega, x) > \frac{1}{i} \wedge \|x\| < i\}


V_i =\begin{cases}
U_3 & i = 0 \\
U_{i+3} \setminus \overline{U_{i+1}} & i > 0

One sees that the V_i are an open cover of \Omega. Therefore, we can choose a sequence of functions (\tilde \eta_i)_{i \in \N} (partition of the unity) such that

  1. \forall i \in \N : \forall x \in \Omega : 0 \le \tilde \eta_i(x) \le 1
  2. \forall x \in \Omega : \exists \text{ only finitely many } i \in \N : \tilde \eta_i(x) \neq 0
  3. \forall i \in \N : \exists j \in \N : \text{supp } \tilde \eta_i \subseteq V_j
  4. \forall x \in \Omega : \sum_{i=0}^\infty \tilde \eta_i(x) = 1

By defining \Eta_i := \{\tilde \eta_j \in \{\tilde \eta_m\}_{m \in \N} : \text{supp } \tilde \eta_j \subseteq V_i\} and

\eta_i(x) := \sum_{\eta \in \Eta_i} \eta(x), we even obtain the properties
  1. \forall i \in \N : \forall x \in \Omega : 0 \le \eta_i(x) \le 1
  2. \forall x \in \Omega : \exists \text{ only finitely many } i \in \N : \eta_i(x) \neq 0
  3. \forall i \in \N: \text{supp } \eta_i \subseteq V_i
  4. \forall x \in \Omega : \sum_{i=0}^\infty \tilde \eta_i(x) = 1

where the properties are the same as before except the third property, which changed. Let |\alpha| = 1, \varphi be a bump function and (v_j)_{j \in \N} be a sequence which approximates v in the L^p(\Omega)-norm. The calculation

\int_\Omega \eta_i(x) v_j(x) \frac{\partial^\alpha}{\partial x^\alpha} \varphi(x) dx = - \int_\Omega \left(\frac{\partial^\alpha}{\partial x^\alpha}\eta_i(x) v_j(x) + \eta_i(x) \frac{\partial^\alpha}{\partial x^\alpha} v_j(x)\right) \varphi(x) dx

reveals that, by taking the limit j \to \infty on both sides, v \in W^{m, p}(\Omega) implies \eta_i v \in W^{m, p}(\Omega), since the limit of \eta_i(x) \frac{\partial^\alpha}{\partial x^\alpha} v_j(x) must be in L^p(\Omega) since we may choose a sequence of bump functions \varphi_k converging to 1.

Let's choose now

W_i = \begin{cases}
U_{i+4} \setminus \overline{U_i} & i \ge 1 \\
U_4 & i = 0

We may choose now an arbitrary \delta > 0 and \epsilon_i so small, that

  1. \|\eta_{\epsilon_i} * (\eta_i v) - \eta_i v\|_{W^{m, p}(\Omega)} < \delta \cdot 2^{-(j+1)}
  2. \text{supp } (\eta_{\epsilon_i} * (\eta_i v)) \subset W_i

Let's now define

w(x) := \sum_{i=0}^\infty \eta_{\epsilon_i} * (\eta_i v)(x)

This function is infinitely often differentiable, since by construction there are only finitely many elements of the sum which do not vanish on each W_i, and also since the elements of the sum are infinitely differentiable due to the Leibniz rule of differentiation under the integral sign. But we also have:

\|w - v\|_{W^{m, p}(\Omega)} = \left\|\sum_{i=0}^\infty \eta_{\epsilon_i} * (\eta_i v) -\sum_{i=0}^\infty (\eta_i v)\right\|_{W^{m, p}(\Omega)} \le \sum_{i=0}^\infty \|\eta_{\epsilon_i} * (\eta_i v) - \eta_i v\|_{W^{m, p}(\Omega)} < \delta \sum_{i=0}^\infty 2^{-(j+1)} = \delta

Since \delta was arbitrary, this finishes the proof.

Approximation of Wm, p-functions with smooth functions on compact sets[edit]

Let \Omega be a bounded domain, and let \partial \Omega have the property, that for every point x \in \partial \Omega, there is a neighbourhood \mathcal U_x such that

\Omega \cap \mathcal U_x = \{(x_1, \ldots, x_d) \in \R^d : x_i < f(x_1, \ldots, x_{i-1}, x_{i+1}, \ldots, x_{d-1}) \}

for a continuous function f. Then every function in W^{m, p}(\Omega) can be approximated by C^\infty(\overline{\Omega})-functions in the W^{m, p}(\Omega)-norm.


to follow

Continuous representatives[edit]

Sobolev inequalities[edit]



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