Consider $A,$ the set of functions $u : [0,2] \to \mathbb{R}$ of the form $$ u(x) = \begin{cases} v(x) & x \in [0,1) \\ w(x) & x \in (1,2] \end{cases}$$ and $v,w$ both $C^2$ functions on their respective domains. Furthermore, assume that $u(0) = u(2) = 0$ and that $[u] := w(1) - v(1) = a,$ a known constant. Define $$E(u) = \frac{1}{2} \int_0^1 (u_x)^2 dx + \int_1^2 (u_x)^2 dx + \frac{w(1) + v(1)}{2}b,$$ where $b$ is another known constant. Show there exists a minimizer of $E$ over all $u \in A,$ and find the minimizing function.
This question is from a qualifying exam I've been studying. Many of the other questions on the qualifying exam have come from Evans PDE, which makes me believe that chapter 8 on variational methods might be the relevant material to solve this problem. However, the problem isn't of the form $E[u] = \int L(x,u,Du) \, dx,$ which causes much difficulty for using this material.
Normally, one would try to show $E$ is coercive and convex in the $u_x$ argument, and then since $A$ is nonempty, there would exist a minimizer. Does this still seem to be the approach here? Or if not, does anyone know of other common methods for showing the existence of a minimizer?
Any hints, suggestions, proofs, etc. on either existence or actually finding the minimizing function would be greatly appreciated! Thanks in advance.
The two integral terms are quadratic: taken in isolation, they admit a minimizer, a linear function. Let us assume the mimimiser $u$ were not piecewise linear: then, without altering $u(1)$ and $w(1)$ (i.e. the value of the third term), the energy could be lowered by reducing the second derivative of $v$ and $v$ (by "straightening" them), yielding a contraddiction.
Let us then consider two such linear functions, $$ \hat{v} = \alpha x$$ and $$ \hat{w} = \alpha + a -(\alpha + a)(x-1) $$ satisying the discontinuity jump constraint. This reduces the minimisation to a one-dimensional problem, as the only variable is $\alpha$, the slope of $v$: the fixed discontinuity jump and the right-boundary conditions determine the slope of $w$. One reformulates the energy as a function of $\alpha$ $$\mathcal{E}(\alpha) = \alpha^2 + (\alpha + a)^2 + \frac{(a + 2\alpha)b}{2}$$ and $$\frac{\mathrm{d}\mathcal{E} }{\mathrm{d}\alpha} = 0$$ will yield the candidate minimum.