A brief introduction to Gamma-convergence

1. A motivating example

Consider the p-Laplace equation

\text{div}(|Du|^{p-2}Du) = 0 in U,

where U is a nonempty bounded open subset of \mathbb{R}, and u=g on \partial U. The energy functional associated with the PDE is

I[w] = \int_U |Dw|^p dx.

For p>1, because of convexity, I[\cdot] has a unique minimizer u^{(p)}\in W^{1,p}(U) which solves the p-Laplace equation.

Now, suppose that u^{(p)}\rightarrow u^{(\infty)} for some function u^{(\infty)} as p\rightarrow\infty (in certain sense). Does u^{(\infty)} solve any PDE?

First note that by some computations, for p>1,

\displaystyle\text{div}(|Du|^{p-2}Du) = |Du|^{p-4}\left[|Du|^2\Delta{u} + (p-2) \sum_{i,j=1}^n u_{x_ix_j}u_{x_i}u_{x_j}\right] = 0.

Formally, u^{(p)} solves |Du|^2\Delta{u} + (p-2) \sum_{i,j=1}^n u_{x_ix_j}u_{x_i}u_{x_j} = 0. Dividing both sides by p-2, the equation becomes

\displaystyle \frac{1}{p-2}|Du|^2\Delta{u} + \sum_{i,j=1}^n u_{x_ix_j}u_{x_i}u_{x_j} = 0.

Letting p\rightarrow\infty, u^{(\infty)} should solve \Delta_{\infty}u := \sum_{i,j=1}^n u_{x_ix_j}u_{x_i}u_{x_j} = 0.

Moreover, if we define I_p[w] = \frac{1}{|U|} \int_U |Dw|^p (which is essentially the same as I above), then (I_p[u])^{1/p}\rightarrow \|Du\|_{L^\infty(U)}=:I_\infty[u] as p\rightarrow\infty. So formally u^{(p)} converges to a minimizer of I_\infty[\cdot]. But does the above discussion really make sense? For example, (I_p)^{1/p}\rightarrow I_\infty might make no sense because the domains of I_p‘s are different!

To rigor above discussion, our goal is to develop a theory of functional convergence such that

  1. u^{(p)} converges to u^{(\infty)} which is a minimizer of I_\infty; and even better (if possible)
  2. u^{(\infty)} solves a “limiting” equation.

We will do 1 only. For 2, one may refer to the paper “Extension of functions satisfying Lipschitz conditions” by Gunnar Aronssov.

2. \Gamma-convergence

Definition (\Gamma-convergence). Let M_\varepsilon and \mathcal{N} be a family of topological spaces. A family of functionals E_\varepsilon (defined on M_\varepsilon) \Gamma-converges to a functional F (defined on \mathcal{N}) if the following hold:

  1. (Compactness) If E_\varepsilon(u_\varepsilon)\leq C, where u_\varepsilon\in M_\varepsilon, then up to extraction, u_\varepsilon\rightarrow u for some u\in\mathcal{N}.
  2. (Lower bound) If v_\varepsilon\rightarrow v for some v\in\mathcal{N} (v_\varepsilon\in M_\varepsilon), then \displaystyle \liminf_{\varepsilon\rightarrow 0} E_\varepsilon(v_\varepsilon) \geq F(v).
  3. (Recovery sequence) For any w\in\mathcal{N}, there exist w_\varepsilon\in M_\varepsilon such that w_\varepsilon\rightarrow w and \displaystyle \limsup_{\varepsilon\rightarrow 0} E_\varepsilon(w_\varepsilon) \leq F(w).

Note that in the above definition, we do not specify what is meant by \rightarrow. It has to be specified in each application. Also, we may replace the \leq in 3 by = because of 2.

Now we prove the fundamental theorem, and probably the only theorem of \Gamma-convergence:

Theorem 1. If E_\varepsilon \Gamma-converges to F and if u_\varepsilon is a minimizer of E_\varepsilon with E_\varepsilon(u_\varepsilon)\leq C, then up to extraction, u_\varepsilon\rightarrow u\in\mathcal{N} and u is a minimizer of F.

Proof. By compactness, u_\varepsilon\rightarrow u\in\mathcal{N} up to extraction. By lower bound,

\displaystyle \liminf_{\varepsilon\rightarrow 0}E_\varepsilon(u_\varepsilon) \geq F(u).

We claim that u is a minimizer of F. Let w\in\mathcal{N}. Then there exists a recovery sequence w_\varepsilon\in M_\varepsilon such that w_\varepsilon\rightarrow w and \displaystyle\limsup_{\varepsilon\rightarrow 0} E_\varepsilon(w_\varepsilon)\leq F(w). Since u_\varepsilon minimizes E_\varepsilon, E_\varepsilon(u_\varepsilon)\leq E_\varepsilon(w_\varepsilon) and so

\displaystyle F(u)\leq \liminf_{\varepsilon\rightarrow 0} E_\varepsilon(u_\varepsilon) \leq \limsup_{\varepsilon\rightarrow 0} E_\varepsilon(w_\varepsilon) \leq F(w).

Therefore u minimizes F. \Box

3. Returning to our example

Now we go back to our example in the first section and see how \Gamma-convergence applies.

Let g be a Lipschitz function on \partial U. We define the following:

  • E_p = I_p^{1/p} defined on M_p = \{u\in W^{1,p}(U): u=g\text{ on }\partial U\}.
  • F=I_\infty defined on \mathcal{N}=\{u\in W^{1,\infty}(U): u=g\text{ on }\partial U\}.
  • \rightarrow means the weak convergence in W^{1,q}(U) for all 1<q<\infty.

With the above definitions, we obtain

Theorem 2. E_p \Gamma-converges to F as p\rightarrow\infty.

Therefore, by Theorem 1, u^{(p)} converges to a minimizer u^{(\infty)} of I_\infty.

Proof. The recovery sequence is very easy to construct. If w\in\mathcal{N}, then w\in M_p for all p>1 (recall that the domain U is bounded). Therefore, we can take the recovery sequence to be just the constant sequence w.

We now prove the compactness. Suppose u_p\in M_p with E_p(u_p)\leq C. Fix 1<q<\infty. By Hölder’s inequality, if p\geq q,

\displaystyle\|Du_p\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \|Du_p\|_{L^p(U)}\frac{1}{|U|^{1/p}} = E_p(u_p)\leq C.

Therefore, \{u_p\}_{p\geq q} is bounded in W^{1,q}(U) (you might need to think carefully why). Thus, we can extract a subsequence, still called \{u_p\} for simplicity, converging weakly to u\in W^{1,q}(U) with u=g on \partial U. We can repeat this argument for q+1, q+2, \ldots so that in each step, we select the weakly convergent sequence as a subsequence of the previous sequence. Taking the diagonal sequence, we conclude that there exists a subsequence \{u_{p_j}\} such that u_{p_j}\rightharpoonup u in W^{1,q}(U) for all 1<q<\infty. This implies u\in W^{1,q}(U) for all 1<q<\infty. Now, by lower semicontinuity,

\displaystyle\|Du\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \liminf_{p\rightarrow\infty}\|Du_p\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \liminf_{p\rightarrow\infty} E_p(u_p)\leq C.

Taking q\rightarrow\infty, \|Du\|_{L^\infty(U)}\leq C. That is, u\in\mathcal{N}.

It remains to show the lower bound condition holds. Suppose v_p\in M_p such that v_p\rightarrow v for some v\in\mathcal{N}. That is, v_p\rightharpoonup v in W^{1,q}(U) for all 1<q<\infty. Then same argument as before shows

\displaystyle\|Dv\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \liminf_{p\rightarrow\infty}\|Dv_p\|_{L^q(U)}\frac{1}{|U|^{1/q}} \leq \liminf_{p\rightarrow\infty} E_p(v_p).

Let q\rightarrow\infty, we get \displaystyle F(v)\leq \liminf_{p\rightarrow\infty} E_p(v_p). \Box

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