20201029-gamma-convergence

We define the following problem:

\left\{\begin{array}{l} -\vartheta^{\prime \prime}-\vec{h} \cdot D \vec{m}(\alpha+\vartheta)=0 \quad \text { in }(0,1) \\ \vartheta(0)=0 \\ \vartheta^{\prime}(1)=0 \end{array}\right.\tag{$P_{\vec h,\alpha}$}

Admissible space for the $\Gamma$-limit:

\mathcal H^*=\{(\vec h,\alpha,\vartheta)\in\mathbb R^2\times C^2(]0,1[)\cap C^1_{0L}([0,1])\times C^0_{0L}([0,1]),\alpha(0)=0\}

Admissible set for the $\Gamma$-limit

\mathcal A^*=\{(\vec h,\alpha,\vartheta)\in \mathcal H^*: \vartheta \text{ solves }(P_{\vec h,\alpha})\}

Note that $(0,0,0)\in \mathcal A^*$, thus the admissible set is not empty.

Let $\overline\vartheta\in C^2(]0,1[)\cap C^1([0,1])$ be such that $\overline\vartheta(0)=0$ and $\overline\vartheta’(1)=0$. Set

\mathcal C^*(\vartheta,\alpha,\vec h)= \left\{ \begin{aligned} &\frac 12 \int |\vartheta-\overline\vartheta|^2&& \text{if }(\vartheta,\alpha,\vec h)\in\mathcal A^*\\ &+\infty &&\text{otherwise} \end{aligned}\right.

Note that the minimum of $\mathcal C^*$ is 0 and that all minimizers have the form $(\overline{\vec h},\overline\alpha,\overline{\vartheta})$, where $\overline{\vec h}=(H\cos\phi,H\sin\phi)$ with $0<H\le{\sup\theta}' '$ where

\phi=\overline\theta(0)+\arcsin\left(\frac{\vartheta' '(0)}{H}\right),

and

\overline\alpha=\phi-\overline\theta-\arcsin\left(\frac{\overline\theta' '}{H}\right).

Consider a sequence $\varepsilon(n)=\gamma(n)=\frac 1 n\to 0$. We can define the cost functionals

C^*_n\left(\vec h,\alpha,\vartheta\right)= \left\{ \begin{aligned} &\frac{1}{2} \int|\vartheta-\overline{\vartheta}|^{2}+\frac{1}{2n} \int {\alpha'}^{2}+\frac{1}{2n}|\vec h|^{2}&& \text{if }(\vec h,\alpha,\vartheta)\in\mathcal A\\ &+\infty &&\text{otherwise} \end{aligned}\right.

where

\mathcal{A}=\left\{(\overrightarrow{\boldsymbol{h}}, \alpha, \boldsymbol{\vartheta}) \in \mathcal{H}: \vartheta \text { solves }\left(P_{\vartheta}\right)\right\}

with $\mathcal H=\mathbb R^2\times H^1_{0L}\times H^1_{0L}$.

Let $(\vartheta_n,\alpha_n,\vec h_n)$ be a sequence of minimizers. Then choosing $(\overline\vartheta,\overline\alpha,\overline{\vec h})$ as competitor we find

|\vec h_n|^2+\int {\alpha_n'}^2\le |\overline{\vec h}|^2+\int {\overline\alpha}'^2=C.

Hence

\alpha_n\stackrel{H^1_{0L}}\rightharpoonup\alpha,\qquad \vec h_n\to\vec h.

\begin{aligned} \int \vartheta_n'^2&=\int \vec{h}_{n} \cdot D\vec m\left(\alpha_{n}+\vartheta_{n}\right) \vartheta_{n}\le \frac 1 {2c_p^2}\int\vartheta_n^2+2c_p^{2}\int\vec h_n\cdot D\vec m(\alpha_n+\vartheta_n)\\ &\le \frac 1 {2c_p^2}\int\vartheta_n^2+C\le \frac 12 \int\vartheta_n'^2+C \end{aligned}

which gives

\vartheta_n\stackrel{H^1_{0L}}\rightharpoonup\vartheta.

Note that $\alpha\in C^0([0,1])$ by standard embedding. Moreover, if we pass to the limit in the weak formulation of $(P_{\vec h_n,\alpha_n})$ we see that $\vartheta$ is a weak solution of problem $(P_{\vec h,\alpha})$. This in turn implies by regularity theory for ODEs that $\vartheta\in C^2(]0,1[)\times C^1([0,1])$ and that it is a strong solution. As a consequence,

(\vec h,\alpha,\vartheta)\in\mathcal A^*.

Note also that

\mathcal C_n(\vec h_n,\alpha_n,\vartheta_n)\to \mathcal C^*(\vec h,\alpha,\vartheta).

It remains to show that $(\vec h,\alpha,\vartheta)$ is a minimizer of $C^*$. To this aim, we construct a recovery sequence $(\overline{\vec h}_n,\overline\alpha_n,\overline\vartheta_n)$ for the minimum $(\overline{\vec h},\overline\alpha,\overline\vartheta)$. We define

\overline{\vec h}_n=\overline{\vec h}.

Moreover, we take $\overline\alpha_n$ to be any sequence in $H^1_{0L}$ such that

\color{red}\overline\alpha_n\stackrel{C^0([0,1])}\to\overline\alpha,\qquad \int\overline\alpha_n'^2\le \sqrt n.

and we define $\overline\vartheta_n$ to be the solution of $(P_{\overline{\vec h}_n,\overline\alpha_n})$. Then $\overline\vartheta_n\stackrel{C([0,1])}\to\overline\vartheta$, and hence

\mathcal C_n(\vec h_n,\alpha_n,\vartheta_n)\le \mathcal C_n(\overline{\vec h}_n,\overline\alpha_n,\overline\vartheta_n)\to 0,

which implies that $C^*(\vec h,\alpha,\vartheta)=0$. Thus $(\vec h,\alpha,\vartheta)$ is a minimizer.