35. Abstract Theory

\(\DeclareMathOperator{\opdiv}{div}\) We have seen a couple of examples leading to variational problems of the form:

Mixed variational problem: Find \(u \in V\) and \(p \in Q\) such that

\[\begin{split} \begin{array}{ccccll} a(u,v) & + & b(v,p) & = & f(v) & \forall \, v \in V \\ b(u,q) & & & = & g(q) & \forall \, q \in Q \end{array} \end{split}\]

We introduce operators

\[\begin{split} \begin{array}{rll} A : & V \rightarrow V^\ast : & u \mapsto a(u,\cdot), \\ B : & V \rightarrow Q^\ast : & u \mapsto b(u, \cdot), \\ B^* : & Q \rightarrow V^\ast : & p \mapsto b(\cdot ,p) \end{array} \end{split}\]

to rewrite the variational problem as operator equation

\[\begin{split} \begin{array}{ccccl} A u & + & B^\ast p & = & f \\ B u & & & = & g \end{array} \end{split}\]

To be solvable, the \(B\) operator must be onto (surjective). If we think of matrices, it cannot be a high rectangular matrix, i.e. the space \(Q\) must not be too rich.

Surjectivity of \(B\) can be rephrased by the LBB-condition:

\[ \sup_{u \in V} \frac{b(u,p)}{\| u \|_V} \geq \beta \| p \|_Q \]

Typically, the \(B\) operator has a non-trivial null-space:

\[ V_0 = \{ v \in V : B v = 0 \} \]

We split the space \(V\) orthogonally as \(V = V_0 \times V_0^\bot\), and decompose vectors and operators accordingly:

\[\begin{split} \left( \begin{array}{ccc} A_{\bot \bot} & A_{\bot 0}  & B^*_\bot \\ A_{0 \bot} & A_{0 0}  & 0 \\ B_{\bot} & 0 & 0 \end{array} \right) \left( \begin{array}{c} u_\bot \\ u_0 \\ p \end{array} \right) = \left( \begin{array}{c} f_\bot \\ f_0 \\ g \end{array} \right) \end{split}\]

This is a block-triangular system. First we use the third equation to solve for \(u_\bot\), then we use the second equation to solve for \(u_0\), and finally we use the first equation to solve for \(p\). For this construction, we need regularity of \(B_\bot\) (and thus also \(B_\bot^*\)), and \(A_{00}\). Regularity of \(B_\bot\) follow from the LBB condition. We require that the bilinear-form \(a(.,.)\) is coercive on the null-space \(V_0\) to guarantee regularity of \(A_{00}\).

By forming the product space \(X = V \times Q\), we can consider one problem: find \((u,p) \in X\) such that

\[ B( (u,p), (v,q) ) = F( (v,q) ) \qquad \forall \, (v,q) \in X \]

with the big bilinear-form

\[ B( (u,p), (v,q) ) = a(u,v) + b(u,q) + b(v,p) \]

and the big linear-from

\[ F ( (v,q) ) = f(v) + g(q) \]

We collect the finding to the following

Brezzi’s theorem:

• V and Q Hilbert spaces,

• \(a(.,.)\) and \(b(.,.)\) continuous bilinear-forms,

• \(f(.)\) and \(g(.)\) continuous linear-forms.

Assume there holds the LBB condition

\[ \sup_{u \in V} \frac{b(u,p)}{\| u \|_V} \geq \beta \| p \|_Q \]

and the kernel coercivity

\[ A(v,v) \geq \alpha \| v \|_V^2 \quad \forall \, v \in V_0. \]

Then the mixed variational problem is uniquely solvable with

\[ \| u \|_V + \| p \|_Q \leq c \, ( \| f \|_{V^\ast} + \| g \|_{Q^\ast}) \]

Proof: The proof follows solving the triangular system above.

The big bilinear-form \(B(.,.)\) is continuous

\[ B((u,p),(v,q)) \preceq ( \| u \| + \| p \| ) \, ( \| v \| + \| q \| ). \]

We prove that it fulfills the \(\inf-\sup\) condition

\[ \inf_{v,q} \sup_{u,p} \frac{B((u,p),(v,q))}{ ( \|v\|_V+\|q\|_Q) (\|u\|_V+\|p\|_Q)} \succeq \beta. \]

Then, we use Theorem by Babuska-Aziz to conclude continuous solvability.

To prove the \(\inf-\sup\)-condition, we choose arbitrary \(v \in V\) and \(q \in Q\). We will construct \(u \in V\) and \(p \in Q\) such that

\[ \| u \|_V + \| p \|_Q \preceq \| v \|_V + \| q \|_Q \]

and

\[ B((u,p),(v,q)) = \| v \|_V^2 + \| q \|_Q^2. \]

First, we use the LBB-condition to choose \(u_1 \in V\) such that

\[ b(u_1,q) = \| q \|_Q^2 \qquad \mbox{and} \qquad \| u_1 \|_V \leq 2 \beta_1^{-1} \, \| q \|_Q. \]

Next, we solve a problem on the kernel:

\[ \mbox{Find } u_0 \in V_0 : \quad a(u_0, w_0) = (v,w_0)_V - a(u_1, w_0) \qquad \forall \, w_0 \in V_0 \]

Due to kernel ellipticity, the left hand side is a coercive bilinear-form on \(V_0\). The right hand side is a continuous linear-form. By Lax-Milgram, the problem has a unique solution fulfilling

\[ \| u_0 \|_V \preceq \| v \|_V + \| u_1 \|_V \]

We set

\[ u = u_0 + u_1. \]

By the Riesz-isomorphism, we define a \(z \in V\) such that

\[ (z,w)_V = (v,w)_V - a(u, w) \qquad \forall \, w \in V \]

By construction, it fulfills \(z \bot_V V_0\). The LBB condition implies

\[ \| p \|_Q \leq \beta_1^{-1} \sup_v \frac{b(v,p)}{\| v\|_V} = \beta_1^{-1} \sup_v \frac{(v, B^\ast p)_V}{\| v\|_V} = \beta_1^{-1} \| B^\ast p \|_V, \]

and thus \(B^\ast Q\) is closed, and \(z \in B^\ast Q\). Take the \(p \in Q\) such that

\[ z = B^\ast p. \]

It fulfills

\[ \| p \|_Q \leq \beta_1^{-1} \| z \|_V \preceq \| v \|_V + \| q \|_Q \]

Concluding, we have constructed \(u\) and \(p\) such that

\[ \| u \|_V + \| p \|_Q \preceq \| v \|_V + \| q \|_Q, \]

and

\[\begin{align*} B((u,p),(v,q)) & = a(u,v) + b(v,p) + b(u,q) \\ & = a(u,v) + (z,v)_V + b(u,q) \\ & = a(u,v) + (v,v)_V - a(u,v) + b(u,q) \\ & = \| v \|_V^2 + b(u_1, q) \\ & = \| v \|_V^2 + \| q \|_Q^2. \end{align*}\]

\(\Box\)

35.1. Constrained minimization problem

Now assume that \(a(.,.)\) is symmetric and coercive. Consider the constrained minimization problem

\[ \min_{B v = g} \tfrac{1}{2} a(v,v) - f(v) \]

The Lagrangian is

\[\begin{eqnarray*} L(v,q) & = & \tfrac{1}{2} a(v,v) - f(v) + \left<Bv-g,q\right> \\ & = & \tfrac{1}{2} a(v,v) - f(v) + b(v,q) - g(q) \end{eqnarray*}\]

Zeroing the directional derivatives with respect to the \(V\)-variable we get

\[ \partial_v L(u,p) = a(u,v) + b(v,p) - f(v) = 0 \]

and for the \(Q\)-variable we obtain

\[ \partial_q L(u,p) = b(u,q) - g(q) = 0 \]

The mixed variational problem is recovered from the Karush Kuhn Tucker (KKT) conditions.

35.2. Stokes equation within the abstract theory

The Hilbert spaces are \(V := [H_0^1(\Omega)]^d\) and \(Q = L_2^0\), and the forms

\[\begin{eqnarray*} a(u,v) & = & \int \nabla u \nabla v \\ b(u,q) & = & \int \operatorname{div} u \, q \\ f(v) & = & \int f v \\ g(q) & = & 0 \end{eqnarray*}\]

are continuous. The LBB condition

\[ \sup_{u \in V} \frac{ \int_\Omega \operatorname{div} u \, q } { \| u \|_{H^1} } \geq \beta \, \| q \|_{L_2} \]

for \(q \in L_2\) with \(\int q = 0\) is true, but non-trivial to prove.

The kernel space is

\[ V_0 = \{ v \in [H_0^1]^d : \operatorname{div} v = 0 \}, \]

the bilinear-form \(a(.,.)\) is coercive on the whole space \(V\), and so also on the sub-space \(V_0\).

35.3. Dirichlet boundary conditions as mixed system

\(V = H^1(\Omega)\), \(Q = H^{-1/2}(\partial \Omega)\)

\[\begin{eqnarray*} a(u,v) &= & \int \nabla u \nabla v \\ b(u,\mu) & = & \left< u_{|\partial \Omega}, \mu \right>_{H^{1/2} \times H^{-1/2} } \\ f(v) & = & \int f v \\ g(\mu) & = & \left< u_D, \mu \right>_{H^{1/2} \times H^{-1/2} } \end{eqnarray*}\]

Continuity of \(a(.,.)\) and \(f(.)\) are clear. Continuity of \(b(.,.)\) and \(g(.)\) require the trace theorem, and \(\left< v, \mu \right>_{H^{1/2} \times H^{-1/2} } \leq \| v \|_{H^{1/2}} \| \mu \|_{H^{-1/2}}\).

To show the LBB-condition, we use that we can continuously extend a boundary function \(v \in H^{1/2}\) to a domain function \(w \in H^1\):

\[ \| \mu \|_{H^{-1/2}} = \sup_{v \in H^{1/2}(\partial \Omega)} \frac{ <v,\mu>_{\partial \Omega} } { \| v \|_{H^{1/2}}} \approx \sup_{w \in H^1(\Omega)}\frac{ <w,\mu>_{\partial \Omega} } { \| w \|_{H^1(\Omega)}} \]

The kernel space is

\[ V_0 = \{ v \in H^1(\Omega) : v_{|\partial \Omega} = 0 \} = H_0^1 \]

On \(H_0^1\), the bilinear-form \(a(.,.)\) is continuous by Friedrichs’ inequality.

35.4. Mixed method for second order equation

Our Hilbert-spaces are \(\Sigma \times V = H(\operatorname{div}) \times L_2\), the forms are

\[\begin{eqnarray*} a(\sigma,\tau) &= & \int \sigma \cdot \tau \\ b(\sigma, v) & = & \int \operatorname{div} \sigma v \\ f(\tau) & = & 0\\ g(v) & = & -\int f v \end{eqnarray*}\]

On \(H(\operatorname{div})\) we have the norm

\[ \| \sigma \|_{H(\operatorname{div}}^2 = \| \sigma \|_{L_2}^2 + \| \operatorname{div} \sigma \|_{L_2}^2 \]

Thus, all forms are continuous. The LBB condition

\[ \sup_{\sigma \in \Sigma} \frac{\int \operatorname{div} \sigma \, v} { \| \sigma \|_{H(\operatorname{div})}} \succ \| v \|_{L_2} \]

is shown as follows: Given a \(v \in L_2\), we are going to find a candidate \(\sigma\). We solve the Poisson equation

\[ \Delta w = v \qquad w = 0 \text{ on } \partial \Omega \]

By Friedrichs’ inequality we get \(\| \nabla w \|_{L_2} \prec \| v \|_{L_2}\). We set \(\sigma = \nabla w\), and see that \(\operatorname{div} \sigma = v\). Thus

\[ \frac{\int \operatorname{div} \sigma \, v} { \| \sigma \|_{H(\operatorname{div})}} = \frac{\| v \|_{L_2}^2} { (\| \sigma \|_{L_2}^2 + \| v \|_{L_2}^2)^{1/2} } \succ \| v \|_{L_2} \]

The kernel space is

\[ V_0 = \{ \tau \in H(\opdiv) : \opdiv \tau = 0 \}  \]

The bilinear-form \(a(.,.)\) is not elliptic in general, but on the kernel:

\[ a(\tau, \tau) = \| \tau \|^2_{L_2} = \| \tau \|^2_{L_2} + \| \opdiv \tau \|_{L_2}^2 = \| \tau \|_{H(\opdiv)}^2 \quad \forall \tau \in V_0 \]