Fixed Points, Monotonicity and Sub- and Supersolutions

Proof. In the language of differential forms (8A.8 Differential Forms and Stokes’ Theorem), $\det Du\,dx^1\wedge\cdots\wedge dx^n = du^1\wedge du^2\wedge\cdots\wedge du^n$, the pullback of the volume form by $u$. Since $d(du^i) = 0$,

By Stokes' theorem, $\int_U\det Du\,dx = \int_{\partial U}u^1\,du^2\wedge\cdots\wedge du^n$. The form on the right is restricted to $\partial U$, and the restriction of $du^i$ to the boundary is $d(u^i|_{\partial U})$: it depends only on the values of $u$ on $\partial U$. So the integral is the same for $v$.

Proof. Suppose there were. Apply Lemma 5.2 to $w$ and the identity map, which agree on $\partial B$: $\int_B\det Dw\,dx = \int_B\det I\,dx = |B| > 0$. But $|w|^2 = 1$ everywhere, and differentiating, $(Dw)^Tw = 0$: the nonzero vector $w(x)$ is in the kernel of $Dw(x)^T$, so $\det Dw(x) = 0$ at every point. Contradiction.

Proof (Brouwer's theorem). Suppose $f : \overline B \to \overline B$ is continuous with no fixed point. Then $|f(x) - x| \geq 3\varepsilon > 0$ on the compact ball, for some $\varepsilon$. Extend $f$ continuously to $\mathbb{R}^n$ by $f(x) = f(x/|x|)$ for $|x| > 1$ and mollify it: since $f$ is uniformly continuous near $\overline B$, convolution with a narrow mollifier gives a smooth map $g_0$ with $|g_0 - f| < \varepsilon$ on $\overline B$; then $g = g_0/(1 + \varepsilon)$ maps $\overline B$ into $\overline B$, is smooth, and $|g - f| \leq |g_0 - f| + \varepsilon|g_0|/(1 + \varepsilon) < 2\varepsilon$, so $|g(x) - x| > \varepsilon$: $g$ has no fixed point either.

Now define $w(x)$ as the point where the ray from $g(x)$ through $x$ leaves the ball: $w(x) = x + t(x)\big(x - g(x)\big)$ with $t(x) \geq 0$ the larger root of $|x + t(x - g(x))|^2 = 1$. Writing $e = x - g(x) \neq 0$, this is the quadratic $|e|^2t^2 + 2(x\cdot e)t + |x|^2 - 1 = 0$, whose discriminant $4(x\cdot e)^2 + 4|e|^2(1 - |x|^2)$ is positive except where $|x| = 1$ and $x\cdot e = 0$. At such a point $x\cdot g(x) = |x|^2 = 1$, which for $|g(x)| \leq 1 = |x|$ forces $g(x) = x$, excluded. So the discriminant is positive on $\overline B$, $t(x)$ is a smooth function of $x$, and $w$ is smooth with values in $S$. For $|x| = 1$, $x\cdot e = 1 - x\cdot g(x) \geq 0$, so the larger root is $t = 0$ and $w(x) = x$. This is a smooth retraction, contradicting Lemma 5.3.

For a compact convex $K$: $K$ is a retract of a large closed ball $\overline{B_R} \supseteq K$ by the nearest-point projection $P_K$, which is continuous (4A.5 Hilbert Spaces and Lax–Milgram; it is $1$-Lipschitz). If $f : K \to K$, then $f\circ P_K : \overline{B_R} \to K \subseteq \overline{B_R}$ has a fixed point $x$, which lies in $K$, so $P_Kx = x$ and $f(x) = x$.

Proof. Fix $\varepsilon > 0$. Since $K$ is compact, it has a finite $\varepsilon$-net $u_1, \ldots, u_N \in K$: every $u \in K$ is within $\varepsilon$ of some $u_i$. The Schauder projection is

The denominator is positive (some $\lambda_i(u) > 0$), so $P_\varepsilon$ is continuous, with values in $K_\varepsilon$, the convex hull of $u_1, \ldots, u_N$, which lies in $K$ (convexity) and is a compact convex subset of a finite-dimensional space. Since $P_\varepsilon(u)$ is an average of points $u_i$ with $\|u - u_i\| < \varepsilon$, $\|P_\varepsilon(u) - u\| < \varepsilon$.

The map $P_\varepsilon\circ A$ sends $K_\varepsilon$ into $K_\varepsilon$ continuously, so by Brouwer's theorem for compact convex sets (in the finite-dimensional span of the $u_i$) it has a fixed point $v_\varepsilon = P_\varepsilon(Av_\varepsilon)$. Then $\|Av_\varepsilon - v_\varepsilon\| = \|Av_\varepsilon - P_\varepsilon(Av_\varepsilon)\| < \varepsilon$. Take $\varepsilon = \frac1k$; by compactness of $K$ a subsequence $v_{1/k_j} \to v \in K$, and by continuity $\|Av - v\| = \lim\|Av_{1/k_j} - v_{1/k_j}\| = 0$.

Proof. Choose $M$ with $\|u\| < M$ for every $u$ in the set. Define

continuous, mapping the closed ball $\overline{B_M}$ into itself, with $\tilde A(\overline{B_M})$ precompact. Let $K$ be the closed convex hull of $\tilde A(\overline{B_M})$: it is compact (the closed convex hull of a compact set in a Banach space is compact, Mazur's theorem) and contained in $\overline{B_M}$, so $\tilde A$ maps $K$ into $K$. By Theorem 5.5 there is $u \in K$ with $\tilde A(u) = u$. If $\|A(u)\| > M$, then $u = \lambda A(u)$ with $\lambda = M/\|A(u)\| < 1$ and $\|u\| = M$, contradicting the choice of $M$. So $\|A(u)\| \leq M$ and $u = A(u)$.

Proof. The map. For $u \in H^1_0(U)$, let $A(u) = w$ be the solution of the linear problem $-\Delta w + \mu w = f - b(\nabla u)$, $w \in H^1_0$. The right side is in $L^2$, with $\|f - b(\nabla u)\|_{L^2} \leq \|f\|_{L^2} + C(\|\nabla u\|_{L^2} + |U|^{1/2})$, so by Lax–Milgram and boundary regularity (6A.6 Weak Solutions and Elliptic Regularity) $w \in H^2(U)$ with $\|w\|_{H^2} \leq C'(\|f\|_{L^2} + \|u\|_{H^1} + 1)$.

$A$ is compact. Bounded sets of $H^1_0$ go to bounded sets of $H^2$, which are precompact in $H^1_0$ by Rellich. And $A$ is continuous: $\|A(u) - A(v)\|_{H^2} \leq C'\|b(\nabla u) - b(\nabla v)\|_{L^2} \leq C'\operatorname{Lip}(b)\|u - v\|_{H^1}$.

The a priori estimate. Suppose $u = \lambda A(u)$, $0 \leq \lambda \leq 1$: then $-\Delta u + \mu u = \lambda\big(f - b(\nabla u)\big)$. Multiply by $u$ and integrate:

By Young's inequality, $C\int|\nabla u||u| \leq \frac12\int|\nabla u|^2 + \frac{C^2}{2}\int u^2$ and $\int(|f| + C)|u| \leq \frac12\int u^2 + \frac12\int(|f| + C)^2$. So

a bound on $\|u\|_{H^1}$ independent of $\lambda$. By Theorem 5.6, $A$ has a fixed point, which is the solution.

Proof. If not, $w(x) = -r\frac{v(x)}{|v(x)|}$ is a continuous map of the closed ball $\overline{B_r}$ into itself, with a fixed point $z$ by Brouwer. Then $|z| = r$, and $r^2 = z\cdot z = -r\frac{v(z)\cdot z}{|v(z)|} \leq 0$, a contradiction.

Proof. Galerkin approximations. Let $w_1, w_2, \ldots$ be the Dirichlet eigenfunctions of $-\Delta$ (6A.6 Weak Solutions and Elliptic Regularity), an orthogonal basis of $H^1_0$. We look for $u_m = \sum_{k=1}^md_m^kw_k$ with

Define $v : \mathbb{R}^m \to \mathbb{R}^m$ by $v(d)^k = \int a\big(\nabla\sum_jd^jw_j\big)\cdot\nabla w_k - \int fw_k$, continuous by 2 and dominated convergence. With $u = \sum d^jw_j$, by 3,

The norms $\|\nabla u\|_{L^2}$ and $|d|$ are equivalent on the span of $w_1, \ldots, w_m$, so this is positive for $|d| = r$ large, and Lemma 5.8 gives a solution of (G).

Bounds. Multiplying (G) by $d_m^k$ and summing, $\int a(\nabla u_m)\cdot\nabla u_m = \int fu_m$, so by 3 and Poincaré, $\alpha\|\nabla u_m\|^2 \leq \beta|U| + C_P\|f\|\|\nabla u_m\|$: the $u_m$ are bounded in $H^1_0$, and by 2 the $a(\nabla u_m)$ are bounded in $L^2(U; \mathbb{R}^n)$. Pass to a subsequence: $u_m \rightharpoonup u$ in $H^1_0$, $u_m \to u$ in $L^2$ (Rellich), and $a(\nabla u_m) \rightharpoonup \xi$ in $L^2$. Letting $m \to \infty$ in (G) with $k$ fixed, $\int\xi\cdot\nabla w_k = \int fw_k$ for every $k$, hence

Minty's trick: $\xi = a(\nabla u)$. For any $w \in H^1_0$, monotonicity gives

Expand. The term $\int a(\nabla u_m)\cdot\nabla u_m = \int fu_m \to \int fu = \int\xi\cdot\nabla u$, by the energy identity, strong $L^2$ convergence and (L) with $v = u$. The other terms are pairings of a weakly convergent sequence with a fixed function. So in the limit

Take $w = u - \lambda v$ with $\lambda > 0$ and $v \in H^1_0$, and divide by $\lambda$: $\int\big(\xi - a(\nabla u - \lambda\nabla v)\big)\cdot\nabla v \geq 0$. As $\lambda \to 0$, $a(\nabla u - \lambda\nabla v) \to a(\nabla u)$ pointwise, dominated by $C(1 + |\nabla u| + |\nabla v|)$, so $\int\big(\xi - a(\nabla u)\big)\cdot\nabla v \geq 0$ for every $v$; replacing $v$ by $-v$ gives equality. With (L), $\int a(\nabla u)\cdot\nabla v = \int fv$: $u$ is a weak solution.

Uniqueness. If $u_1, u_2$ are solutions, subtract the equations and test with $u_1 - u_2$: $\theta\|\nabla(u_1 - u_2)\|^2 \leq \int\big(a(\nabla u_1) - a(\nabla u_2)\big)\cdot\nabla(u_1 - u_2) = 0$.

Proof. Test with $v = w^+$, using $\nabla w^+ = \nabla w$ on $\{w > 0\}$ and $0$ elsewhere (4A.11 Sobolev Embeddings and Critical Exponents): $\int|\nabla w^+|^2 + \lambda\int(w^+)^2 \leq 0$. So $\nabla w^+ = 0$, and since $w^+ \in H^1_0$, Poincaré's inequality gives $w^+ = 0$.

Proof. Let $m \leq \underline u \leq \overline u \leq M$, and choose $\lambda \geq 0$ so that $z \mapsto f(x, z) + \lambda z$ is non-decreasing on $[m, M]$ (any $\lambda$ at least the Lipschitz constant of $f$ there will do). Set $u_0 = \underline u$ and define $u_{k+1} \in H^1_0$ as the weak solution of the linear problem

which exists by Lax–Milgram whenever $u_k$ is bounded.

Claim: $\underline u = u_0 \leq u_1 \leq u_2 \leq \cdots \leq \overline u$. First, $w = u_0 - u_1$ satisfies $-\Delta w + \lambda w \leq \big(f(x, u_0) + \lambda u_0\big) - \big(f(x, u_0) + \lambda u_0\big) = 0$ weakly, and $w^+ \in H^1_0$ because $u_0 \leq 0$ on $\partial U$ and $u_1 \in H^1_0$; by Lemma 5.10, $u_0 \leq u_1$. Next, if $u_{k-1} \leq u_k$ (both in $[m, M]$), then $w = u_k - u_{k+1}$ satisfies

by monotonicity of $z \mapsto f + \lambda z$, so $u_k \leq u_{k+1}$. Finally, if $u_k \leq \overline u$, then $w = u_{k+1} - \overline u$ satisfies $-\Delta w + \lambda w \leq \big(f(x, u_k) + \lambda u_k\big) - \big(f(x, \overline u) + \lambda\overline u\big) \leq 0$, with $w^+ \in H^1_0$ since $\overline u \geq 0$ on $\partial U$, so $u_{k+1} \leq \overline u$. By induction the claim holds, and all $u_k$ take values in $[m, M]$.

Convergence. The sequence increases pointwise and is bounded, so $u_k \to u$ pointwise and in $L^2$ (dominated convergence), and $f(x, u_k) \to f(x, u)$ in $L^2$ (Lipschitz). Testing the equation for $u_{k+1}$ with $u_{k+1}$ itself shows $\|\nabla u_{k+1}\|^2 \leq \|f(x, u_k) + \lambda u_k\|\,\|u_{k+1}\|$, bounded, so a subsequence converges weakly in $H^1_0$, to $u$. Passing to the limit in $\int\nabla u_{k+1}\cdot\nabla v + \lambda\int u_{k+1}v = \int\big(f(x, u_k) + \lambda u_k\big)v$ gives $\int\nabla u\cdot\nabla v = \int f(x, u)v$.