10. Coercive variational problems and their approximation

In this chapter we discuss variational problems posed in Hilbert spaces. Let \(V\) be a Hilbert space, and let \(A(\cdot,\cdot) : V \times V \rightarrow {\mathbb R}\) be a bilinear form which is

• coercive (also known as elliptic)

\[ A(u,u) \geq \alpha_1 \| u \|_V^2 \qquad \forall \, u \in V, \]

• and continuous

\[ A(u,v) \leq \alpha_2 \| u \|_V \, \| v \|_V \qquad \forall \, u , v \in V, \]

with bounds \(\alpha_1\) and \(\alpha_2\) in \({\mathbb R}^+\). It is not necessarily symmetric. Let \(f(.) : V \rightarrow {\mathbb R}\) be a continuous linear form on \(V\), i.e.,

\[ f(v) \leq \| f \|_{V^\ast} \| v \|_V. \]

We are posing the variational problem: find \(u \in V\) such that

\[ A(u,v) = f(v) \qquad \forall \, v \in V. \]

Example: Diffusion-reaction equation:

Consider the PDE

\[ -\operatorname{div} (a(x) \nabla u) + c(x) u = f \qquad \mbox{in } \Omega, \]

with Neumann boundary conditions. Let \(V\) be the Hilbert space generated by the inner product \((u,v)_V := (u,v)_{L_2}+ (\nabla u , \nabla v)_{L_2}\). The variational formulation of the PDE involves the bilinear form

\[ A(u,v) = \int_{\Omega} (a(x) \nabla u) \cdot \nabla v \, dx + \int_\Omega c(x) u v \, dx. \]

Assume that the coefficients \(a(x)\) and \(c(x)\) fulfill \(a(x) \in {\mathbb R}^{d \times d}\), \(a(x)\) symmetric and \(0 < \lambda_1 \leq \lambda_{\min} (a(x)) \leq \lambda_{\max} (a(x)) \leq \lambda_2\), and \(c(x)\) such that \(0 < \gamma_1 \leq c(x) \leq \gamma_2\) almost everywhere. Then \(A(\cdot,\cdot)\) is coercive with constant \(\alpha_1 = \min \{ \lambda_1, \gamma_1 \}\) and \(\alpha_2 = \max \{ \lambda_2, \gamma_2 \}\).

Example: Diffusion-convection-reaction equation:

The partial differential equation

\[ -\Delta u + b \cdot \nabla u + u = f \qquad \mbox{in} \; \Omega \]

with Dirichlet boundary conditions \(u = 0\) on \(\partial \Omega\) leads to the bilinear form

\[ A(u,v) = \int \nabla u \, \nabla v \, dx + \int b \cdot \nabla u \, v \, dx + \int u v \, dx. \]

If \(\operatorname{div} \, b \leq 0\), what is an important case arising from incompressible flow fields (\(\operatorname{div} \, b = 0\)), then \(A(\cdot,\cdot)\) is coercive and continuous w.r.t. the same norm as above (Exercise!).

Instead of the linear form \(f(\cdot)\), we will often write \(f \in V^\ast\). The evaluation is written as the duality product

\[ \left< f , v \right>_{V^\ast \times V} = f(v). \]

Lemma: A continuous bilinear form \(A(\cdot,\cdot) : V \times V \rightarrow {\mathbb R}\) induces a continuous linear operator \(A : V \rightarrow V^\ast\) via

\[ \left< A u, v \right> = A(u,v) \qquad \forall \, u,v \in V. \]

The operator norm \(\| A \|_{V \rightarrow V^\ast}\) is bounded by the continuity bound \(\alpha_2\) of \(A(\cdot,\cdot)\).

Proof: For every \(u \in V\), \(A(u,\cdot)\) is a bounded linear form on \(V\) with norm

\[ \| A(u,\cdot) \|_{V^\ast} = \sup_{v \in V} \frac{A(u,v)}{ \| v \|_V} \leq \sup_{v \in V} \frac{\alpha_2 \| u \|_V \, \| v \|_V }{ \| v \|_V} = \alpha_2 \| u \|_V \]

Thus, we can define the operator \(A : u \in V \rightarrow A(u,\cdot) \in V^\ast\). It is linear, and its operator norm is bounded by

\[\begin{align*} \| A \|_{V \rightarrow V^\ast} &= \sup_{u\in V} \frac{ \| A u \|_{V^\ast}}{\| u \|_V} = \sup_{u\in V} \sup_{v \in V} \frac{ \left< A u, v\right>_{V^\ast \times V} } {\| u \|_V \, \| v \|_V} \\ &= \sup_{u\in V} \sup_{v \in V} \frac{ A (u,v) } {\| u \|_V \, \| v \|_V} \leq \sup_{u\in V} \sup_{v\in V} \frac{ \alpha_2 \| u \|_V \| v \|_V } {\| u \|_V \, \| v \|_V} = \alpha_2. \end{align*}\]

\(\Box\)

Using this notation, we can write the variational problem as operator equation: find \(u \in V\) such that

\[ A u = f \qquad (\mbox{in } V^\ast). \]

Banach’s contraction mapping theorem:

Given a Banach space \(V\) and a mapping \(T : V \rightarrow V\), satisfying the Lipschitz condition

\[ \| T(v_1) - T(v_2) \| \leq L \, \| v_1 - v_2 \| \qquad \forall \, v_1, v_2 \in V \]

for a fixed \(L \in [0,1)\). Then there exists a unique \(u \in V\) such that

\[ u = T(u), \]

i.e. the mapping \(T\) has a unique fixed point \(u\). The iteration \(u^1 \in V\) given, compute

\[ u^{k+1} := T(u^k) \]

converges to \(u\) with convergence rate \(L\):

\[ \| u - u^{k+1} \| \leq L \| u - u^k \| \]

Theorem by Lax and Milgram:

Given a Hilbert space \(V\), a coercive and continuous bilinear form \(A(\cdot,\cdot)\), and a continuous linear form \(f(.)\). Then there exists a unique \(u \in V\) solving

\[ A(u,v) = f(v) \qquad \forall \, v \in V. \]

There holds

\[ \| u \|_V \leq \alpha_1^{-1} \| f \|_{V^\ast} \]

Proof: Start from the operator equation \(A u = f\). Let \(J_V : V^\ast \rightarrow V\) be the Riesz isomorphism defined by

\[ (J_V g, v)_V = g(v) \qquad \forall \, v \in V, \; \forall \, g \in V^\ast. \]

Then the operator equation is equivalent to

\[ J_V A u = J_V f \qquad (\mbox{in } V), \]

and to the fixed point equation (with some \(0 \neq \tau \in {\mathbb R}\) chosen below)

\[\begin{align*} u = u - \tau J_V (A u - f). \end{align*}\]

We will verify that

\[ T (v) := v - \tau J_V (A v - f) \]

is a contractive mapping, i.e., \(\| T (v_1) - T (v_2) \|_V \leq L \| v_1 - v_2 \|_V\) with some Lipschitz constant \(L \in [0,1)\). Let \(v_1, v_2 \in V\), and set \(v = v_1 - v_2\). Then

\[\begin{align*} \| T (v_1) - T (v_2) \|_V^2 &= \| \{ v_1 - \tau J_V (A v_1 - f) \} - \{ v_2 - \tau J_V (A v_2 -f) \} \|_V^2 \\ &= \| v - \tau J_V A v \|_V^2 \\ &= \| v \|_V^2 - 2 \tau (J_V A v, v)_V + \tau^2 \| J_V A v \|_V^2 \\ &= \| v \|_V^2 - 2 \tau \left< A v , v \right> + \tau^2 \| A v \|_{V^\ast}^2 \\ &= \| v \|_V^2 - 2 \tau A(v,v) + \tau^2 \| A v \|_{V^\ast}^2 \\ & \leq \| v \|_V^2 - 2 \tau \alpha_1 \| v \|_V^2 + \tau^2 \alpha_2^2 \| v \|_V^2 \\ &= (1 - 2 \tau \alpha_1 + \tau^2 \alpha_2^2) \| v_1 - v_2 \|_V^2 \end{align*}\]

Now, we choose \(\tau = \alpha_1 / \alpha_2^2\), and obtain a Lipschitz constant

\[ L^2 = 1 - \alpha_1^2 / \alpha_2^2 \in [0,1). \]

Banach’s contraction mapping theorem state that \(T(\cdot)\) has a unique fixed point \(u\), which satisfies \(u = u - \tau J_V (A u - f)\), and thus \(A u = f\). Finally, we obtain the claimed bound from

\[ \| u \|_V^2 \leq \alpha_1^{-1} A(u,u) = \alpha_1^{-1} \, f(u) \leq \alpha_1^{-1} \| f\|_{V^\ast} \| u \|_V, \]

and dividing by one factor \(\|u\|\). \(\Box\)

10.1. Approximation of coercive variational problems

Now, let \(V_h\) be a closed subspace of \(V\). We compute the approximation \(u_h \in V_h\) by the Galerkin method

\[ A(u_h, v_h) = f(v_h) \qquad \forall \, v_h \in V_h. \]

This variational problem is uniquely solvable by Lax-Milgram, since, \((V_h,\|.\|_V)\) is a Hilbert space, and continuity and coercivity on \(V_h\) are inherited from the original problem on \(V\).

The next theorem says that the solution defined by the Galerkin method is, up to a constant factor, as good as the best possible approximation in the finite dimensional space.

Céa’s Lemma:

The approximation error of Galerkin’s method is quasi optimal:

\[ \| u - u_h \|_V \leq \frac{\alpha_2}{\alpha_1} \inf_{v \in V_h} \| u - v_h \|_V \]

Proof: A fundamental property is the Galerkin orthogonality

\[ A(u-u_h, w_h) = A(u,w_h) - A(u_h, w_h) = f(w_h) - f(w_h) = 0 \qquad \forall \, w_h \in V_h. \]

Now, pick an arbitrary \(v_h \in V_h\), and bound

\[\begin{align*} \| u - u_h \|_V^2 & \leq \alpha_1^{-1} A(u-u_h, u-u_h) \\ & = \alpha_1^{-1} A(u-u_h, u-v_h) + \alpha_1^{-1} A(u-u_h, \underbrace{v_h-u_h}_{\in V_h}) \\ & \leq \alpha_2 / \alpha_1 \, \| u - u_h \|_V \| u - v_h \|_V. \end{align*}\]

Divide one factor \(\|u - u_h\|\). Since \(v_h \in V_h\) was arbitrary, the estimation holds true also for the infimum in \(V_h\). \(\Box\)

If \(A(\cdot,\cdot)\) is additionally symmetric, then it is an inner product. In this case, the coercivity and continuity properties are equivalent to

\[ \alpha_1 \| u \|_V^2 \leq A(u,u) \leq \alpha_2 \, \| u \|_V^2 \qquad \forall \, u \in V. \]

The generated norm \(\|.\|_A\) is an equivalent norm to \(\|.\|_V\). In the symmetric case, we can use the orthogonal projection with respect to \((.,.)_A\) to improve the bounds to

\[ \| u - u_h \|_V^2 \leq \alpha_1^{-1} \| u - u_h \|_A^2 \leq \alpha_1^{-1} \inf_{v_h \in V_h} \| u - v_h \|_A^2 \leq \alpha_2 / \alpha_1 \| u - v_h \|_V^2. \]

The factor in the quasi-optimality estimate is now the square root of the general, non-symmetric case.