Theorems for existence and uniqueness of variational problem

Introduction

Among simulation engineers, it is well accepted that the solution of a PDE can be envisioned as the following three general steps (actually, this was also my premature understanding during the early era of my study on numerical simulation).

Expand the unknown function to be solved by a set of basis functions.
Multiply both sides of the equation by a set of test functions and integrate the product over the solution domain.
With the application of integration by parts, the space dimension of the integral is reduced by 1 and the appeared boundary integral can be used to apply predefined boundary conditions.

The formulation thus obtained, which discretizes the original continuous problem, is called weak form or variational problem, from which the weak solution results. At first glance, the above envisioned procedures could be applicable to any PDEs, at least, discretized system of equations can be constructed and system matrix can be filled. However, without a careful and crystal clear proof about the existence and uniqueness of the solution for the weak form or variational problem, the results can never be relied on - after all, any operation on the computer can produce something, usually huge amount of data, which is either truth or rubbish. What kind of meaning will be assigned to it and how much value we can extract from it depend on the wisdom, rationality and rigorousness of the human operator.

In this post, the proof for the existence and uniqueness of the solution of the following variational problem will be presented, which is the corner stone of numerical schemes such as the finite element method and boundary element method (BEM).

Let $H_1$ and $H_2$ be two Hilbert spaces, $a(\cdot, \cdot): H_1 \times H_2 \rightarrow \mathbb{K}$ be a sesquilinear form with $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$ (Note: when $\mathbb{K} = \mathbb{R}$, $a(\cdot, \cdot)$ is a bilinear form), $l(\cdot): H_2 \rightarrow \mathbb{K}$ be a bounded linear functional on $H_2$. The solution $u \in H_1$ of the equation below will be sought under an arbitrarily given $v \in H_2$:

\[
\begin{equation}
\label{eq:variational-problem}
a(u, v) = l(v) \quad (\forall v \in H_2)
\end{equation}
\]

In this post, we'll first show that the existence and uniqueness of $u$ along with a priori estimate of its norm can be obtained thanks to the inf-sup condition. Secondly, by introducing the famous Lax-Milgram Lemma, the inf-sup condition can be relaxed to H-ellipticity condition which still preserves the solvability of the variational problem.

inf-sup condition

Definition (inf-sup condition) The continuous sesquilinear form $a(\cdot, \cdot)$ satisfies the in-sup condition if there exists a constant $\gamma > 0$, such that

\[
\begin{align}
\label{eq:inf-sup-condition-a}
\inf_{u \in H_1 \backslash \{0\}} \sup_{v \in H_2 \backslash \{0\}} \frac{\abs{a(u, v)}}{\norm{u}_{H_1} \norm{v}_{H_2}} & \geq \gamma > 0, \tag{a} \\
\label{eq:inf-sup-condition-b}
\forall v \in H_2 \backslash \{0\}: \sup_{u \in H_1 \backslash \{0\}} \abs{a(u, v)} &> 0. \tag{b}
\end{align}
\]

Remark

For (a), first fix $u \in H_1 \backslash \{0\}$, then vary $v \in H_2 \backslash \{0\}$ and take the supremum of $\frac{\abs{a(u, v)}}{\norm{u}_{H_1} \norm{v}_{H_2}}$. Let $A \in L(H_1, H_2')$ be the associated operator of $a(\cdot, \cdot)$ satisfying $(Au, v) = a(u, v) \; (\forall u \in H_1, v \in H_2)$. Let $\phi_u := Au \in H_2'$. Because $u$ is already fixed so is its norm $\norm{u}_{H_1}$, the supremum can be considered as a measure of the norm $\norm{\phi_u}_{H_2'}$.
For (b), $v$ is firstly fixed and $u$ is then varied. Because the supremum of the absolute value of the sesquilinear form should be strictly larger than 0, $\phi_u$ cannot be a zero operator.

Theorem (Existence and uniqueness) The condition that the sesquilinear form $a(\cdot, \cdot)$ satisfies the inf-sup condition is equivalent to the following condition: for all $l \in H_2'$, the variational problem \eqref{eq:variational-problem} has a unique solution $u \in H_1$, which satisfies the priori estimate

\[
\begin{equation}
\label{eq:priori-estimate}
\norm{u}_{H_1} \leq \frac{1}{\gamma} \norm{l}_{H_2'}.
\end{equation}
\]

Proof: A. Given the inf-sup condition, we prove the existence and uniqueness of the solution and the priori estimate.

We'll show the associated operator $A: H_1 \rightarrow H_2'$ of $a(\cdot, \cdot)$ is continuous.

Because $a(\cdot, \cdot)$ is continuous, we have

\[
\abs{a(u, v)} \leq \norm{a} \norm{u}_{H_1} \norm{v}_{H_2} \quad (\forall u \in H_1, v \in H_2).
\]

Let $\phi_u := Au = a(u, \cdot)$, then

\[
\abs{\phi_u(v)} = \abs{a(u, v)} \leq \norm{a} \norm{u}_{H_1} \norm{v}_{H_2} \quad (\forall u \in H_1, v \in H_2).
\]

Because $u$ is given and fixed in $\phi_u$, we define the constant $C(a, u) := \norm{a} \norm{u}_{H_1}$, therefore $\phi_u$ is bounded:

\[
\abs{\phi_u(v)} \leq C(a, u) \norm{v}_{H_2} \quad (\forall v \in H_2).
\]

It should be noted that when $\phi_u$ is applied to $v \in H_2$, a complex conjugate operation must be applied first to $v$ due to the definition of $a(\cdot, \cdot)$ which is complex conjugate linear with respect to its second argument. Therefore, $\phi_u$ is a bounded complex conjugate linear operator from $H_1$ to $H_2^*$, where $H_2^*$ is the anti-dual space of $H_2$. Because the only difference between $H_2^*$ and $H_2'$ is a complex conjugate, the two spaces can be identified isometrically $H_2^* \cong H_2'$. In the following, we use $H_2'$ replacing $H_2^*$ and let $\phi_u$ inherently includes a complex conjugate operation, which makes $\phi_u$ a bounded linear operator in $H_2'$. According to this analysis, we know the operator $A$ really maps $u$ to an element in $H_2'$.

Then the norm of $\phi_u$ in $H_2'$ is

\[
\norm{\phi_u}_{H_2'} = \norm{Au}_{H_2'} = \sup_{v \in H_2 \backslash \{0\}} \frac{\abs{\phi_u(v)}}{\norm{v}_{H_2}} \leq \norm{a} \norm{u}_{H_1} < \infty \quad (\forall u \in H_1),
\]

based on which the operator $A: H_1 \rightarrow H_2'$ is continuous.
Prove the image of $H_1$ under $A$ is closed in $H_2'$. The basic idea is that the closeness of $A(H_1)$ in $H_2'$ can be proved by showing that any Cauchy sequence in $A(H_1)$ is convergent in $A(H_1)$.

According to (a) of the inf-sup condition

\[
\forall u \in H_1 \backslash \{0\}: \sup_{v \in H_2 \backslash \{0\}} \frac{\abs{a(u, v)}}{\norm{u}_{H_1} \norm{v}_{H_2}} \geq \gamma > 0,
\]

we have the equivalent

\[
\forall u \in H_1 \backslash \{0\}: \norm{Au}_{H_2'} = \sup_{v \in H_2 \backslash \{0\}} \frac{\abs{Au(v)}}{\norm{v}_{H_2}} \geq \norm{u}_{H_1} \gamma.
\]

When $u = 0$, the equality in the above holds. Therefore,

\[
\forall u \in H_1: \norm{Au}_{H_2'} = \sup_{v \in H_2 \backslash \{0\}} \frac{\abs{Au(v)}}{\norm{v}_{H_2}} \geq \norm{u}_{H_1} \gamma.
\]

Let $(y_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $A(H_1)$ and $(x_n)_{n \in \mathbb{N}}$ be the corresponding sequence in $H_1$ such that $A(x_n) = y_n$. For all $\varepsilon > 0$, there exists a larger enough $N_0 \in \mathbb{N}$ such that when $m, n > N_0$,

\[
\varepsilon > \norm{y_m - y_n}_{H_2'} = \norm{A(x_m - x_n)}_{H_2'} \geq \norm{x_m - x_n}_{H_1} \gamma.
\]

From this we know that $(x_n)_{n \in \mathbb{N}}$ is also a Cauchy sequence in $H_1$. Because $H_1$ is a Hilbert space, there exists an $x \in H_1$ such that

\[
\lim_{n \rightarrow \infty} \norm{x_n - x}_{H_1} = 0.
\]

According to step 1, $A$ is a continuous linear operator, so we have

\[
\lim_{n \rightarrow \infty} \norm{A(x_n) - A(x)}_{H_2'} \leq \lim_{n \rightarrow \infty} \norm{A}_{H_2' \leftarrow H_1} \norm{x_n - x}_{H_1} = 0.
\]

Because $A(x) \in A(H_1)$, any Cauchy sequence in $A(H_1)$ is also convergent in $A(H_1)$ and $A(H_1)$ is closed.
Prove $A(H_1) = H_2'$ and the solution for the variational problem \eqref{eq:variational-problem} exists.

Assume $A(H_1)$ is a proper subset of $H_2'$. Then there exists a non-zero $y_0 \in A(H_1)^{\perp}$. Due to Riesz representation theorem, there exists $y_0' \in (A(H_1)')$ such that

\[
\forall y \in A(H_1): \langle y_0', y \rangle_{A(H_1)' \times A(H_1)} = (y_0, y)_{H_2'} \; \text{and} \; \norm{y_0'}_{H_2''} = \norm{y_0}_{H_2'},
\]

where $\langle \cdot, \cdot \rangle_{A(H_1)' \times A(H_1)}$ is the dual pairing. It can be seen that $y_0'$ is a non-zero functional, which operates on $A(H_1)$ and evaluates to zero. In addition, because $A(H_1)$ is closed in $H_2'$ according to the proof in step 2, Hahn-Banach theorem can be used to extend the domain of $y_0'$ from $A(H_1)$ to the whole space $H_2'$, i.e. there exists a non-zero $\tilde{y}_0' \in H_2''$ such that $\tilde{y}_0'(y) = 0$ for all $y \in A(H_1)$.

Further because $H_2$ is a Hilbert space, it is reflexive: $H_2 \cong H_2''$, then $\tilde{y}_0' \in H_2$ and for all $y \in A(H_1)$
\[
\tilde{y}_0' (y) = y(\tilde{y}_0') = (Au) (\tilde{y}_0') = a(u, \tilde{y}_0') = 0 \quad (u \in H_1, Au = y).
\]
Because $y$ is arbitrarily selected in the image of $A$, $u$ can also vary arbitrarily in $H_1$. Hence we can conclude that there exists a non-zero $\tilde{y}_0' \in H_2$ such that
\[
\sup_{u \in H_1 \backslash \{0\}} \abs{a(u, \tilde{y}_0')} = 0,
\]
which contradicts (b) of the inf-sup condition. So we've proved $A(H_1) = H_2'$ and the solution of the variational problem \eqref{eq:variational-problem} exists.
Prove $A \in L(H_1, H_2')$ is injective and the variational problem \eqref{eq:variational-problem} has a unique solution for all $l \in H_2'$.
For all $l \in H_2'$, there exists a $u \in H_1$ such that $Au = l$ according to the proof in step 3. Assume there are two such solutions, namely, $u_1$ and $u_2$ being different, we have the following according to step 2
\[
\norm{Au_1 - Au_2}_{H_2'} = \norm{A(u_1 - u_2)}_{H_2'} \geq \norm{u_1 - u_2}_{H_1} \gamma \quad (u_1 - u_2 \in H_1 \backslash \{0\}),
\]
which contradicts $\norm{Au_1 - Au_2}_{H_2'} = 0$. Therefore, $A \in L(H_1, H_2')$ is injective and the variational problem \eqref{eq:variational-problem} has a unique solution for all $l \in H_2'$.
Prove the priori estimate.
For all $l \in H_2'$, there exists a unique $u \in H_1$ such that $Au = l$. According to step 2,
\[
\norm{l}_{H_2'} = \norm{Au}_{H_2'} \geq \norm{u}_{H_1} \gamma,
\]
which proves the priori estimate.

B. Given the existence and uniqueness of the solution and prove the inf-sup condition \eqref{eq:inf-sup-condition-a} and \eqref{eq:inf-sup-condition-b}.

Prove \eqref{eq:inf-sup-condition-b} of the inf-sup condition.

If the variational problem \eqref{eq:variational-problem} has a unique solution for all $l \in H_2$, associated operator $A \in L(H_1, H_2')$ of $a(\cdot, \cdot)$ is bijective.

If (b) of the inf-sup condition does not hold, there must exists $y_0 \in H_2 \backslash \{0\}$ such that
\[
\sup_{u \in H_1 \backslash \{0\}} = \abs{a(u, y_0)} = 0 \Leftrightarrow \forall u \in H_1, a(u, y_0) = (Au)(y_0) = 0.
\]
Because $H_2$ is reflexive, $y_0$ can be considered in $H_2''$:
\[
(Au)(y_0) = y_0(Au) = 0 \quad (\forall u \in H_1).
\]
Then from Riesz representation theorem, there exists $\tilde{y}_0 \in H_2'$ corresponding to $y_0 \in H_2''$ such that
\[
y_0(Au) = (\tilde{y}_0, Au)_{H_2'} = 0 \quad (\forall u \in H_1).
\]
Therefore, $A(H_1)^{\perp} \neq \{0\}$, which contradicts the fact that $A$ is bijective.
Prove \eqref{eq:inf-sup-condition-a} of the inf-sup condition.
\[
\begin{aligned}
\inf_{u \in H_1 \backslash \{0\}} \sup_{v \in H_2 \backslash \{0\}} \frac{\abs{a(u, v)}}{\norm{u}_{H_1} \norm{v}_{H_2}} &= \inf_{u \in H_1 \backslash \{0\}} \sup_{v \in H_2 \backslash \{0\}} \frac{\langle Au, v \rangle_{H_2' \times H_2}}{\norm{u}_{H_1} \norm{v}_{H_2}} \\
&= \inf_{w \in H_2' \backslash \{0\}} \sup_{v \in H_2 \backslash \{0\}} \frac{\langle w, v \rangle_{H_2' \times H_2}}{\norm{A^{-1} w}_{H_1} \norm{v}_{H_2}} \quad (w \in H_2', w = Au) \\
&\geq \inf_{w \in H_2' \backslash \{0\}} \sup_{v \in H_2 \backslash \{0\}} \frac{\langle w, v \rangle_{H_2' \times H_2}}{\norm{A^{-1}}_{H_2 \leftarrow H_2'} \norm{w}_{H_2'} \norm{v}_{H_2}}
\end{aligned}
\]
For the Hilbert space $H_2$, there exists an isometry $J_{H_2}: H_2 \rightarrow H_2'$. Let $\tilde{w} \in H_2$ and $J_{H_2} (\tilde{w}) = w$, we further have
\[
\begin{aligned}
\inf_{u \in H_1 \backslash \{0\}} \sup_{v \in H_2 \backslash \{0\}} \frac{\abs{a(u, v)}}{\norm{u}_{H_1} \norm{v}_{H_2}} &\geq \gamma \inf_{\tilde{w} \in H_2 \backslash \{0\}} \sup_{v \in H_2 \backslash \{0\}} \frac{\langle J_{H_2} \tilde{w}, v \rangle_{H_2' \times H_2}}{\norm{J_{H_2} \tilde{w}}_{H_2'} \norm{v}_{H_2}} \quad (\text{Let $\norm{A^{-1}}_{H_2 \leftarrow H_2'} = \gamma^{-1}$.}) \\
&= \gamma \inf_{\tilde{w} \in H_2 \backslash \{0\}} \frac{1}{\norm{\tilde{w}}_{H_2}} \sup_{v \in H_2 \backslash \{0\}} \frac{\langle J_{H_2} \tilde{w}, v \rangle_{H_2' \times H_2}}{\norm{v}_{H_2}} \quad (\because \norm{J_{H_2} \tilde{w}}_{H_2'} = \norm{\tilde{w}}_{H_2}) \\
&= \gamma \inf_{\tilde{w} \in H_2 \backslash \{0\}} \frac{\norm{J_{H_2} \tilde{w}}_{H_2'}}{\norm{\tilde{w}}_{H_2}} \\
&= \gamma
\end{aligned}.
\]

H-ellipticity condition and Lax-Milgram Lemma

Definition (H-ellipticity) Let $H_1 = H_2 = H$ is reflexive Banach space, $a: H \times H \rightarrow \mathbb{C}$ be a sesquilinear form. $a(\cdot, \cdot)$ is H-elliptic if there exits $\gamma > 0$ and $\sigma \in \mathbb{C}$ with $\abs{\sigma} = 1$, such that
\[
\forall u \in H: \Re(\sigma a(u, u)) \geq \gamma \norm{u}_H^2.
\]
Lemma (Lax-Milgram) Let $H$ be a Hilbert space. The sesquilinear form $a: H \times H \rightarrow \mathbb{C}$ is H-elliptic. Then the inf-sup condition holds.

Proof: From the H-ellipticity condition for $a(\cdot, \cdot)$, we have
\[
\gamma \norm{u}_H^2 \leq \Re(\sigma a(u, u)) \leq \abs{\Re(\sigma a(u, u))} \leq \abs{\sigma a(u, u)} = \abs{\sigma} \abs{a(u, u)} = \abs{a(u, u)}.
\]
Substitute this inequality into the LHS of \eqref{eq:inf-sup-condition-a} of the inf-sup condition while selecting $v$ to be equal to $u$,
\[
\inf_{u \in H \backslash \{0\}} \sup_{v \in H \backslash \{0\}} \frac{\abs{a(u, v)}}{\norm{u}_H \norm{v}_H} \geq \inf_{u \in H \backslash \{0\}} \frac{\gamma \norm{u}_H^2}{\norm{u}_H \norm{u}_H} = \gamma > 0.
\]
This proves \eqref{eq:inf-sup-condition-a} of the inf-sup condition.

To prove \eqref{eq:inf-sup-condition-b} of the inf-sup condition, given an arbitrary $v \in H \backslash \{0\}$ and let $u = v$, we have
\[
\sup_{u \in H \backslash \{0\}} \abs{a(u, v)} \geq \abs{a(v, v)} \geq \gamma \norm{v}_H > 0.
\]
According to Lax-Milgram Lemma, we can still have the existence, uniqueness and priori estimate for the solution of the variation problem from the H-elliptic condition on $a(\cdot, \cdot)$.

Summary

In this post, we present conditions and theorems along with their proofs, which ensures the existence and uniqueness for the solution of the general variational problem $a(u, v) = l(v) \; (\forall l \in H_2)$. The underlying condition is the inf-sup condition. During the proof, the application of Hahn-Banach theorem is a key step for proving that the associated operator $A \in L(H_1, H_2')$ of $a(\cdot, \cdot)$ is surjective. Because of Lax-Milgram Lemma, the inf-sup condition can be relaxed to H-ellipticity condition.