the symmetric condition of the minimizer

Question

Let $I:=(0,1)$ and $u_0\in H^2(I)$ be given. Define $$ u_1:=\operatorname{argmin}_{u\in H^1(0,1/2)}\int_0^{1/2}|u'|^2 + \int_0^{1/2}|u-u_0|^2dx $$ and $$ u_2:=\operatorname{argmin}_{u\in H^1(1/2,1)}\int_{1/2}^1|u'|^2 + \int_{1/2}^1|u-u_0|^2dx. $$ My question: for what condition should I put on $u_0$ such that $$ u_1(1/2)\neq u_2(1/2) $$ hold?

Answer 1

Let me expand on my comment. First, write down the Euler-Lagrange equation for

$$\min_u \int_a^b (u')^2 + (u-u_0)^2 \, dx,$$

which is

$$-u''(x) + u(x) = u_0(x) \ \ \ x\in (a,b),$$

with boundary conditions $u'(a)=0=u'(b)$. Consider now a special case of $u_0(x) = x$. Then the solution is

$$u(x) = Ae^x + Be^{-x} + x$$

where

$$A = \frac{e^a - e^b}{e^{2b}-e^{2a}} \ \ \text{ and } \ \ B = \frac{e^{-a} - e^{-b}}{e^{-2a} - e^{-2b}}.$$

Specializing to your two cases of $a=0,b=1/2$, and $a=1/2,b=1$, we find (after some tedious calculations) that

$$u_1(1/2) = \frac{1}{2}-\gamma \ \ \text{ and } \ \ u_2(1/2) = \frac{1}{2} + \gamma,$$

where

$$\gamma = \frac{e^{1/2}-1}{e^{1/2}+1}.$$

So in this simple setting $u_1(1/2)\neq u_2(1/2)$.

It is reasonable to suspect that this is always the case, in some generic sense, but I'm not sure how to prove this. The two problems are disconnected in a sense, so it would be surprising for the solutions to match up nicely at the boundary. I suppose they would match up when $u_0$ is even with respect to $x=1/2$.

By the way, I suppose you know your problem is a special case of the Mumford-Shah functional used in computer vision and image processing. There is a lot of good mathematical work on the Mumford-Shah functional that might be worth looking at, in case you were not aware.