0
$\begingroup$

Given two Hilbert spaces $X$ and $Y$ and $a \colon X \times X \to \mathbb{R}$ as well as $b \colon X \times M \to \mathbb{R}$, both being continuous bilinear forms with $a(\cdot, \cdot)$ being symmetric and coercive for all $v \in X$. $f \in X'$ and $g \in M'$.

I am worried about the proper connection between the minimum formulation and the saddle point formulation.

Find the minimum $u \in X(g) = \{ u \in X \mid b(u,q) = g(q) \, \forall q \in M \}$ of the quadratic functional $J(u) := \frac12 \, a(u,u) - l(u)$. With other words:

$$\min_{u \in X(g)} \, J(u).$$

Now the Lagrangian Functional $L \colon X \times M \to \mathbb{R}$ is introduced, given as

$$ L(u, \lambda) := J(u) + (b(u,\lambda) - g(\lambda)).$$

Now, for all $\lambda$ we do the following:

$$ X \times M \to X \times {\lambda} \cong X \to \mathbb{R}$$

with finding $u_\lambda \in X$ for all $\lambda$.

Then: We are trying to find out, which $\lambda \in M$ do fulfill the condition that $u_\lambda \in X$. This must be a solution of the minimising problem.

Then there is a conversion to a system of equations of the approach given above, namely:

Find $(u, \lambda) \in X \times M$ such that $$ a(u,v) + b(v, \lambda) = l(v), \quad \forall v \in X \\ b(u,\mu) = g(\mu), \quad \forall \mu \in M$$

I cannot see how to transfer the given procedure to the equations given above.

My thoughts so far: As we are dealing with a Lagrangian, the best seems to be to take the derivatives, both of $u$ and $\lambda$. What puzzles me most is what happens to the factor $\frac12$ in the defintion of $a(\cdot, \cdot)$. Is this gone due to the symmetry? How do the $v$ and $\mu$ come into place when we take a look at the derivative of the Lagrangian? Especially: how do we know which argument of $a$ can be replaced by $v$?

$\endgroup$
  • 2
    $\begingroup$ are you familiar with the first order KKT (optimality) conditions for a (local) minimum (see [chapter 12 of this book)? Any local minimum $(u^*,\lambda^*)$ will have to verify your two linear equations. The first equation in the two-equation system is the so-called adjoint equation $L_u (u^*,\lambda^*)(v) = 0$, $\forall v \in X$. The second equation (nothing other than your "primal model" constraint) comes from $L_\lambda (u^*,\lambda^*)(\mu) = 0$, $\forall \mu \in M$. $\endgroup$ – GoHokies Jan 16 '18 at 17:31
  • $\begingroup$ and yes, the 1/2 factor vanishes because of the symmetry and linearity of the bilinear form $a$. $\endgroup$ – GoHokies Jan 16 '18 at 17:31
  • $\begingroup$ @GoHokies No, I am not. But thank you, I will get that book tomorrow. Helpful to have a point where to start now! $\endgroup$ – mdot Jan 16 '18 at 17:49
3
$\begingroup$

To expand on @GoHokies comment: The answer to your all your questions is basically "because $a$ is bilinear and symmetric" (necessarily so; without symmetry this approach wouldn't work). Specifically, don't think of gradients but of directional derivatives: You have a saddle point if the derivative in every direction vanishes, i.e., if the Fréchet derivative (you should look this up; understanding how derivatives of functionals are defined is important) of $L$ at $u\in X$ applied to every direction $\nu\in X$ gives zero: $$L'(u,\lambda)\nu = 0 \qquad\text{for all } \nu\in X.\label{cc:1}\tag{1}$$ Now $l$ and $b$ are linear in $u$, hence $l'(u)\nu = l(\nu)$ and similarly for $b$. The bilinear form is quadratic in $u$ and so by the product rule, $$\frac{d}{du} a(u,u)\nu = \frac{d}{du} a_1(u,u)\nu + \frac{d}{du} a_2(u,u)\nu = a(\nu,u) + a(u,\nu) = 2 a(u,\nu),$$ where $a_1$ is $a$ with the first argument frozen (i.e., you only take the derivative with respect to the first argument) and similarly for $a_2$, and I have used that $a$ is bilinear in each argument separately. The last step is of course due to the symmetry of $a$ (which also means that it doesn't matter which argument of $a$ is replaced, answering your third question).

Hence, \eqref{cc:1} reduces to $$a(u,\nu) + b(\nu,\lambda) = l(\nu)\qquad\text{for all }\nu \in X.$$ The second equation works in the same way by looking at the derivative with respect to $\lambda$ in all directions $\mu$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.