1
$\begingroup$

I have a functional $$ J(\theta) = \frac{1}{2}\int_0^t \left(k(\tau)-\theta^{T}(t)l(\tau)\right)^2 + (\theta(t)-\theta(\tau))^T W(\theta(t)-\theta(\tau)) d\tau $$ which I would like to minimize wrt. $\theta(t)$. I see it is convex, if W is positive semidefinite, which is assumed to be true.

Let $\Theta(\epsilon,t)$ be a family of test-functions and $\epsilon$ the variation parameter, s.t. $\Theta(\epsilon_0,t)=\theta(t)$. Now I get the first variation $$ \delta J(\theta) = \int_0^t -l(\tau)(k(\tau)-l^T(\tau)\Theta(\epsilon,t)) \; \delta \theta(t) + W(\Theta(\epsilon,t)-\Theta(\epsilon,\tau)) \; (\delta\theta(t)-\delta\theta(\tau)) d\tau $$ using the condition for a local stationary point of $J(\epsilon)$ $$\delta J |_{\epsilon = \epsilon_0} = 0 \,, \qquad \forall \delta \theta$$ I end up with the two Euler-Lagrange equations $$ -l(\tau)(k(\tau)-l^T(\tau)\theta(t)) + W(\theta(t)-\theta(\tau)) =0 \\ - W(\theta(t)-\theta(\tau)) =0 \;. $$

Comparing to the classical least squares solution $$ \int_0^t -l(\tau)\left(k(\tau)-l^T(\tau)\theta\right) d\tau = 0 \,, $$ I cannot see, why the integral disappears in the variational formulation and what exactly those Euler-Lagrange equations (ELE) as a function of $\theta$ and $\tau$ mean (if the above is correct). Can anyone explain to me the connection between those and LS (I know, that there $\theta$ is constant by assumption). In other words: why can I not recover the LS solution with the variational formulation? And what does this second ELE mean?

$\endgroup$
0
$\begingroup$

I overlooked, that since $\delta\theta(t)$ does not depend on $\tau$, I can facter it out of the integral. Then, one doesn't even need to apply the fundamental lemma of the calculus of variations, since there is the variation times the integral equals zero. It's easy to see, that then the integral has to be zero, since the equation has to hold for all variations.

| 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.