Structure of Saddle-point Problems

73. Structure of Saddle-point Problems#

73.1. Constrained minimization problems#

Let $A \in {\mathbf R}^{n\times n}$ be spd, and $f \in {\mathbf R}^n$. Then the minimization problem

\[ \min_u \frac{1}{2} \, u^T A u - f^T u \]

is equivalent to the linear system

\[ A u = f. \]

Since $A$ is spd, the function to be minimized is convex.

Now let $B \in {\mathbf R}^{m\times n}$ of full rank, with $m \leq n$ and, and $g \in {\mathbf R}^m$. We consider the constrained minimization problem

\[ \min_{u \atop B u = g} \frac{1}{2} \, u^T A u - f^T u. \]

The minimizers are found among the critical points of the Lagrange function

\[ L(u, p) = \frac{1}{2} \, u^T A u - f^T u + p^T (B u - g), \]

where $p \in {\mathbb R}^m$ is the Lagrange parameter. Differentiation with respect to $u$ and $p$, and zeroing leads to the Karush-Kuhn-Tucker (KKT) equations

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

The solution is not a minimizer, but a saddle-point of the Lagrange function $L(.,.)$.

Theorem: The Hessian matrix of $L(.,.)$,

\[\begin{split} \left( \begin{array}{cc} A & B \\ B & \end{array} \right) \in {\mathbf R}^{(m+n) \times (m+n)} \end{split}\]

has $n$ positive and $m$ negative eigenvalues.

Proof: We define the sub-spaces

\[\begin{eqnarray*} V_1 & := & \{ (u, 0) : u \in {\mathbf R}^n \}, \\ V_2 & := & \{ (-A^{-1} B^T p, p) : p \in {\mathbf R}^m \}, \end{eqnarray*}\]

and evaluate for $u \neq 0$

\[\begin{split} \left( \begin{array}{c} u \\ 0 \end{array} \right)^T \left( \begin{array}{cc} A & B \\ B & \end{array} \right) \left( \begin{array}{c} u \\ 0 \end{array} \right) = u^T A u > 0 \end{split}\]

and for $p \neq 0$

\[\begin{split} \left( \begin{array}{c} -A^{-1}B^T p \\ p \end{array} \right)^T \left( \begin{array}{cc} A & B \\ B & \end{array} \right) \left( \begin{array}{c} -A^{-1}B^T p \\ p \end{array} \right) = - p^T B A^{-1} B^T p < 0. \end{split}\]

The matrix $S := B A^{-1} B^T$ is called the (negative) Schur complement. For an spd matrix $A$ and for the rank-$m$ matrix $B$ it is positive definit.

Since we found an $n$-dimensional sub-space where the quadratic form is positive, and an $m$-dimensional sub-space where it is negative, the $\min-\max$-theorem proves that we have $n$ positive and $m$ negative eigenvalues.

73.2. Examples#

Stokes Equation

The vector-field $u \in [H^1_{0,D}]^2$ is the velocity of an incompressible flow. It minimizes the $H^1$-norm, under the incompressibility constraint $\operatorname{div} u = 0$:

\[ \min_{u \atop \operatorname{div} u} \frac{1}{2} \int | \nabla u |^2 - \int f u \]

The variational formulation of the KKT system is

\[\begin{split} \begin{array}{ccccll} \int \nabla u : \nabla v & + & \int \operatorname{div} v \, p & = & \int f v & \qquad \forall \, v \\ \int \operatorname{div} u \, q & & & = & 0 & \qquad \forall \, q \end{array} \end{split}\]

Discretization leads to the matrix equation

\[\begin{split} \begin{array}{ccccl} A u & + & B ^T p & = & f, \\ B u & & & = & 0, \end{array} \end{split}\]

where $A$ is the discretization of the $-\Delta$-operator for both components of the velocity, and $B$ is the discretization of the $\operatorname{div}$-operator. It’s transposed matrix $B^T$ is the discretization of the negative $\nabla$-operator. The physical meaning of the Lagrange parameters $p$ is pressure.

Dirichlet boundary conditions

We can interpret a Dirichlet problem as a constrained minimization problem

\[ \min_{u \in H^1 \atop u = u_D \text{ on } \Gamma_D} \frac{1}{2} \int |\nabla u |^2 - \int f u \]

The KKT system is to find $u \in H^1(\Omega)$ and $\lambda$ on the boundary such that

\[\begin{split} \begin{array}{ccccll} \int_\Omega \nabla u \nabla v & + & \int_{\Gamma_D} v \lambda & = & \int f v & \qquad \forall \, v \\ \int_{\Gamma_D} u \, \mu & & & = & \int_{\Gamma_D} u_D \, \mu & \qquad \forall \, \mu \end{array} \end{split}\]

By integration by parts in the first equation, we obtain

\[ -\int_\Omega \Delta u v + \int_{\partial \Omega} \frac{\partial u}{\partial n} v + \int_{\Gamma_D} \lambda v = \int_\Omega f v. \]

We see that the meaning of the Lagrange parameter $\lambda$ is the negative normal derivative $- \tfrac{\partial u}{\partial n}$

73.3. Schur complement iteration#

We solve the block-system

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

Since $A$ is spd and thus regular, we can solve for $u$ in the first block-row

\[ u = A^{-1} (f - B^T p). \]

Inserting this $u$ into the second block-row gives

\[ B A^{-1} (f - B^T p) = g, \]

or equivalently

\[ B A^{-1} B^T p = B A^{-1} f - g. \]

The Schur complement matrix $S = B A^{-1} B^T$ is positive definite. The Uzawa algorithm (and there are many variants of it) is now to use a Richardson iteration, or conjugate gradients, to solve the Schur complement equation

\[ S p = \tilde g \]

with $\tilde g = B A^{-1} f - g$. After computing the Lagrange parameter $p$, the first component $u$ is found from $u = A^{-1} (f - B^T p)$.

Of course, the drawback of this method is that it requires the inverse $A^{-1}$, which is usually not cheaply available. If the Uzawa iteration does not take many iteration steps, one can also use an (inner) iterative solver to apply $A^{-1}$ in every step.

To design preconditioners for the Uzawa iteration, it is important to understand the norm induced by the Schur-complement $S$:

\[ \| p \|_S^2 = \| B^T p \|_{A^{-1}}^2 = \left( \sup_u \frac{ u^T B^T p}{ \| u \|_{A}} \right)^2 \]

If $A$ and $B$ come from finite-element discretization of bilinear-forms $a(.,.)$ and $b(.,.)$, then the last term can be recast as

\[ \sup_{u \in V_h} \frac{b(u,p)^2}{a(u,u)} \]

If we can prove continuity of $b(.,.)$

\[ b(u,p) \leq \overline \beta \, \| u \|_A \| p \|_Q, \]

and the famous LBB-condition

\[ \sup_u \frac{b(u,p)}{\| u \|_A} \geq \underline \beta \, \| p \|_Q, \]

we immediately get

\[ \underline \beta^2 \, \| p \|_Q^2 \leq \| p \|_S^2 \leq \overline \beta^2 \, \| p \|_Q^2 \]

If we have a preconditioner for the norm-matrix of the $Q$-norm, we have a preconditioner for the Schur complement $S$.

from ngsolve import *
from ngsolve.webgui import Draw
mesh = Mesh(unit_square.GenerateMesh(maxh=0.1))

fesu = H1(mesh, order=1)
feslam = H1(mesh, order=1, definedon=mesh.Boundaries(".*"))
u,v = fesu.TnT()
lam,mu = feslam.TnT()

a = BilinearForm(grad(u)*grad(v)*dx + u*v*dx).Assemble()
b = BilinearForm(u*mu*ds).Assemble()
f = LinearForm(10*v*dx).Assemble()
g = LinearForm((x+y)*mu*ds).Assemble()

ainv = a.mat.Inverse()
gtilde = (b.mat @ ainv * f.vec - g.vec).Evaluate()

S = b.mat @ ainv @ b.mat.T
preS = Projector(feslam.FreeDofs(), True)

from ngsolve.krylovspace import CGSolver
Sinv = CGSolver(S, preS, printrates=True, maxiter=200)

gflam = GridFunction(feslam)
gfu = GridFunction(fesu)

gflam.vec.data = Sinv * gtilde 
gfu.vec.data = ainv * (f.vec - b.mat.T * gflam.vec)

Draw (gfu);

CG iteration 1, residual = 5.703743410155175     
CG iteration 2, residual = 0.32250970878595636     
CG iteration 3, residual = 0.08335855188353485     
CG iteration 4, residual = 0.011192460122905604     
CG iteration 5, residual = 0.004812013162339133     
CG iteration 6, residual = 0.004204370475404719     
CG iteration 7, residual = 0.0035184047718742643     
CG iteration 8, residual = 0.004623704351628802     
CG iteration 9, residual = 0.0027644465369766743     
CG iteration 10, residual = 0.001095452880817425     
CG iteration 11, residual = 0.00720533141359291     
CG iteration 12, residual = 0.0009022414011419461     
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CG iteration 40, residual = 6.181384164542866e-09     
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CG iteration 42, residual = 9.600238226564242e-10     
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CG iteration 44, residual = 2.5856683186987715e-09     
CG iteration 45, residual = 5.619150701221642e-10     
CG iteration 46, residual = 1.036006096856494e-09     
CG iteration 47, residual = 3.553474254749655e-10     
CG iteration 48, residual = 1.0401499370989996e-10     
CG iteration 49, residual = 1.8531733433027572e-11     
CG iteration 50, residual = 1.107279980834582e-11     
CG iteration 51, residual = 9.528784983278807e-12     
CG iteration 52, residual = 2.996273588831822e-12     

If we experiment with the mesh-size, we observe that the iteration number doubles, if we refine the mesh by a factor of 4, which numerically indicate that

\[ \kappa \{ S \} = O(h^{-1}) \]

We will come back to this, later.

73.4. Block-preconditioning#

Now we want to solve directly the saddle-point system, without forming the Schur complement explicitly. We want to use preconditioners for both fields separately.

First, we show that the spectrum of

\[\begin{split} \left( \begin{array} {cc} A & 0 \\ 0 & S \end{array} \right)^{-1} \left( \begin{array} {cc} A & B^T \\ B & 0 \end{array} \right) \end{split}\]

consists just of a few, small eigenvalues, and thus this matrix is well conditioned.

Theorem The spectrum $\sigma$ of the generalized eigenvalue problem

\[\begin{split} \left( \begin{array} {cc} A & B^T \\ B & 0 \end{array} \right) \left( \begin{array}{cc} u \\ p \end{array} \right) = \lambda \left( \begin{array} {cc} A & 0 \\ 0 & S \end{array} \right) \left( \begin{array}{cc} u \\ p \end{array} \right) \end{split}\]

satisfies $\sigma \subset \{ 1, (1+\sqrt{5})/2, (1-\sqrt{5})/2 \}$.

Proof: The eigenvalue problem reads as

\[\begin{split} \left( \begin{array} {ccc} A u & + & B^T p \\ B u & & \end{array} \right) = \lambda \left( \begin{array} {c} A u \\ S p \end{array} \right), \end{split}\]

with $(u,p) \neq 0$. The first row reads as

\[ A u + B^T p = \lambda A u \]

or $$ B^T p = (\lambda-1) A u. $$ This implies $$ B A^{-1} B^T p = (\lambda-1) B u $$ or $$ S p = (\lambda-1) B u $$

Inserting the second equation $Bu = \lambda S p$, we obtain

\[ S p = (\lambda-1) \lambda S p \]

If $p$ is non-zero, the eigenvalue $\lambda$ must satisfy the quadratic equation

\[ 1 = (\lambda - 1) \lambda, \]

i.e. $\lambda = \frac{1}{2} (1 \pm \sqrt{5})$. If $p = 0$, but $u \neq 0$, the equation $B^T p = (\lambda-1) A u$ can be satisfied only if $\lambda = 1$.

Since

\[\begin{split} \left(\begin{array}{cc} A&B^T \\ B&0\end{array}\right) \end{split}\]

is symmetric, and

\[\begin{split} \left(\begin{array}{cc} A&0 \\ 0&S\end{array}\right) \end{split}\]

provides a norm, the fraction $$ \frac{\max | \sigma |}{\min | \sigma |} = \frac{1+\sqrt{5}}{| 1-\sqrt{5} |} = 2.618... $$ bounds the norm of the preconditioned matrix. It also bounds the number of iterations of a Krylov-space method. However, since eigenvalues are positive and negative, we cannot use conjugate gradients. One can apply MinRes, or GMRes.

from ngsolve import *
from ngsolve.webgui import Draw

mesh = Mesh(unit_square.GenerateMesh(maxh=0.1))

fesu = H1(mesh, order=1, autoupdate=True)
feslam = H1(mesh, order=1, definedon=mesh.Boundaries(".*"), autoupdate=True)
u,v = fesu.TnT()
lam,mu = feslam.TnT()

a = BilinearForm(grad(u)*grad(v)*dx + u*v*dx)
b = BilinearForm(u*mu*ds)
pre = Preconditioner(a, type="multigrid")
a.Assemble()
b.Assemble()

for i in range(2):
    mesh.Refine()
    a.Assemble()
    b.Assemble()

f = LinearForm(10*v*dx).Assemble()
g = LinearForm((x+y)*mu*ds).Assemble()

K = BlockMatrix( [[a.mat, b.mat.T], [b.mat, None]] )
F = BlockVector( [f.vec, g.vec])
U = K.CreateRowVector()
prelam = Projector(feslam.FreeDofs(), True)
PRE = BlockMatrix( [[pre.mat, None], [None, prelam]] )

from ngsolve.krylovspace import MinRes

U.data = MinRes(K, pre=PRE, rhs=F, maxsteps=400)

LinearSolver iteration 1, residual = 10.006052884938201     
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gfu = GridFunction(fesu)
gfu.vec.data = U[0]
Draw (gfu);