Energy minimization

math

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Energy Minimization: the Method

Consider $E(u) = \frac{1}{2}\|u\|_A^2 + \int_{\Omega} F(u) = \frac{1}{2}\|u\|_A^2 + \int_{\Omega} \frac{1}{2} p^2(u),$
where

• $$p(u) = \frac{1}{2}(u^2-1), \quad p'(u) = u, \quad p''=1.$$
• $$F(u) = p^2(u) = \frac{1}{4}(u^2-1)^2$$
• $$f(u) = F'(u) = 2p(u)p'(u) = u(u^2-1)$$
• $$f'(u) = F''(u) = 3u^2 - 1$$ .

Now we replace $p$ by $p_L(u; u_k)$ as the first order approximation of the quadratic function, i.e.,

$$p_L(u; u_k) = p(u_k) + p'(u_k)(u-u_k).$$ Note that $p_L(u_k;u_k) = p(u_k), p'_L = const = p'(u_k).$
We then obtain a quadratic form
$$E_L(u) = \frac{1}{2}\|u\|_A^2 + \int_{\Omega} \frac{1}{2} p_L^2(u).$$

Obviously $E_L(u)$ is convex as a quadratic form. It is a good approximaiton of $E(u)$ as a quadratic function is approximated by its linear Taylor series. We then let

$$u_{k+1} ={\rm argmin} E_L(u; u_k).$$

Goal

• energy stable
• convergence
• order of convergence

Energy Stable

$$E(u_{k+1}) ?\leq ? E_L(u_{k+1}) \leq E_L(u_k) = E(u_k).$$
We write as
$$E(u_{k+1}) - E(u_k) = E(u_{k+1}) - E_L(u_{k+1}) + E_L(u_{k+1}) - E(u_k).$$
As a quadratic energy minimization, the energy decreased in $E_L$ is $$E_L(u_k)-E_L(u_{k+1}) = \int_{\Omega}[p'(u_k)]^2(u_{k+1} - u_k)^2 + \|u_{k+1} - u_k\|_A^2,$$ which can be proved by the Taylor series of $E_L(u)$ at the minimum point $u_{k+1}$.
The first part is
$$E(u_{k+1}) - E_L(u_{k+1}) = \int_{\Omega} p^2(u_{k+1}) - p_L^2(u_{k+1}),$$
$$p^2 - p_L^2 = (p-p_L)(p+p_L)=\frac{1}{2}(u_{k+1} - u_k)^2(p+p_L)(u_{k+1})$$
The difference is
$$\int_{\Omega} \left [(p+p_L)_{k+1}/2 - (p'_k)^2\right ] (u_{k+1} - u_k)^2 \leq ? \; 0$$

We skip the $A$-norm part, as we skip the parameter $1/\epsilon^2$ in the definition of $F(u)$ which will dominate the energy.

Write out the quadratic form $\left [(p+p_L)_{k+1}/2 - (p'_k)^2\right ]$ as a function of $u_{k+1}$ and check when it is positive. My computation shows that when it is positive, we have $|u_{k+1} - u_k| > \sqrt{6} - 2 \approx 0.45.$

So consider the time dependent problem and introduced a time discretization $(u_{k+1} - u_k)/\delta t$. Namely consider the evolution problem

$$u_t = - \delta_u E_L(u; u_k).$$ When $\delta t$ is small, then $u_{k+1}$ should be close to $u_k$ and thus can control the derivation. How small is the $\delta t$? Write out the Euler equation for $u_{k+1} - u_k$

Not quite right. If we add a quadratic term $1/\delta t \|u-u_k\|^2$ into $E_L(u)$, then the energy decreated is more. The coefficient will be like

$$\frac{1}{2}(p+p_L)_{k+1} - (p'_k)^2 - \frac{2}{\delta t}.$$

Discrete Maximum Principle

If $|u_{k}|\leq 1, |u_{k+1}|\leq 1$, i.e., the discrete maximum principle holds, then from the figure $(p+p_L)_{k+1}/2 < 0.$ But unfortunately, we can't prove the discrete maximum principle as a lineared $f(u)$ is used.

We can modify our minimization problem to a constraint minimization problem

$$\min_{u, |u|\leq 1} E_L(u).$$
The energy is quadratic and the constraint is a convex set. So the minimizer exists and unique. This known as obstacle problem and fast (multigrid) methods exists. If we let $$u_{k+1} ={\rm argmin}_{|u|\leq 1} E_L(u; u_k).$$
We still have $E_L(u_{k+1})\leq E_L(u_k)=E(u_k)$. Now since $|u_{k+1}|\leq 1$, we have $(p+p_L)(u_{k+1})\leq 0$ and thus $E(u_{k+1}) - E_L(u_{k+1}) = \int_{\Omega} \frac{1}{2}(u_{k+1} - u_k)^2(p+p_L)(u_{k+1}) < 0$.