A C1 planar gradient at a nondegenerate saddle has local stable and unstable curves

Translating the source point to the origin and subtracting the constant $u(p)$, assume $p=0$, $u(0)=0$ and $\mathrm{\nabla }u(0)=0$. Put $H:=D^{2}u(0)$ and $A:=-H$. By [F8] the symmetric matrix $A$ has an orthonormal basis of eigenvectors; its eigenvalues are $-a$ and $b$ for some $a,b>0$, because $H$ has one negative and one positive eigenvalue. Let $E^s$ be the eigenline of $-a$ and $E^u$ the eigenline of $b$, and let $P_s,P_u$ be the orthogonal projections onto them, so $P_s+P_u=I$, $e^{tA}=e^{-at}{P}_s+e^{bt}{P}_u$, and for $s\ge t\ge 0$ one has $\mathrm{\mid }{e}^{(t-s)A}{P}_u\mathrm{\mid }\le e^{-b(s-t)}$ while $\mathrm{\mid }{e}^{tA}{P}_s\mathrm{\mid }\le e^{-at}$. Fix $\beta$ with $0<\beta <\min (a,b)$. Since $\mathrm{\nabla }u$ is $C^1$ with derivative $H$ at the origin, the field $X:=-\mathrm{\nabla }u$ has the form $X(x)=Ax+R(x)$ with $R$ of class $C^1$, $R(0)=0$ and $DR(0)=0$.

A compactly supported perturbation of $R$ with small derivative. Let $\epsilon >0$ (to be fixed below). Choose $\rho >0$ with $\mathrm{\mid }DR(x)\mathrm{\mid }\le \epsilon$ for $\mathrm{\mid }x\mathrm{\mid }\le 2\rho$; this is possible because $DR(0)=0$ and $DR$ is continuous. Choose a smooth cutoff $\chi :{\mathbb{R}}^2\to [0,1]$ with $\chi =1$ on the ball $B_{\rho }$ and $\chi =0$ outside $B_{2\rho }$, and put $\tilde{R}:=\chi R$. On $B_{\rho }$ one has $\tilde{R}=R$, and $\tilde{R}(0)=0$ and $D\tilde{R}=0$ there at the origin. On the support of $D\chi$, which is a compact subset of $B_{2\rho }\setminus B_{\rho }$, the mean value theorem [F6] gives $\mathrm{\mid }R(x)\mathrm{\mid }\le \epsilon \mathrm{\mid }x\mathrm{\mid }\le 2\epsilon \rho$, while $\mathrm{\mid }D\chi \mathrm{\mid }\le C\mathrm{/}\rho$ for a constant $C$ depending only on the fixed cutoff profile; hence $\mathrm{\mid }D\tilde{R}(x)\mathrm{\mid }\le \epsilon +(2C)\epsilon =(1+2C)\epsilon$ for every $x$, using $D\tilde{R}=\chi DR+RD\chi$. Since $\epsilon$ is arbitrary, ${\epsilon }^{\mathrm{\prime }}:={\sup }_x\mathrm{\mid }D\tilde{R}(x)\mathrm{\mid }$ can be made as small as we please. Moreover $D\tilde{R}$ is continuous with compact support, hence uniformly continuous on ${\mathbb{R}}^2$ by [F7]. Set $\tilde{X}(x):=Ax+\tilde{R}(x)$; then $\tilde{X}=X$ on $B_{\rho }$.

The weighted space of paths. Let $\mathcal{B}$ be the set of continuous $z:[0,\mathrm{\infty })\to {\mathbb{R}}^2$ with $\parallel z{\parallel }_{\beta }:={\sup }_{t\ge 0}{e}^{\beta t}\mathrm{\mid }z(t)\mathrm{\mid }<\mathrm{\infty }$. This is a normed vector space, and it is complete: if $(z_n)$ is $\parallel \cdot {\parallel }_{\beta }$-Cauchy then each sequence $(z_n(t))$ is Cauchy in ${\mathbb{R}}^2$, which is complete by [F5], so $z(t):={\lim }_n{z}_n(t)$ exists; the Cauchy bound $e^{\beta t}\mathrm{\mid }{z}_n(t)\mathrm{\mid }\le M$ passes to the limit, so $z\in \mathcal{B}$, and on each $[0,T]$ the convergence is uniform, so every component of $z$ is continuous by [F4]; finally $\parallel z_n-z{\parallel }_{\beta }\to 0$. Also $\mathrm{\mid }z(t)\mathrm{\mid }\le e^{-\beta t}\parallel z{\parallel }_{\beta }$ for every $z\in \mathcal{B}$ and $t\ge 0$.

The integral operator and its contraction constant. For $z\in \mathcal{B}$ define $N(z)(t):={\int }_0^t{e}^{(t-s)A}{P}_s\tilde{R}(z(s))ds-{\int }_t^{\mathrm{\infty }}{e}^{(t-s)A}{P}_u\tilde{R}(z(s))ds\mathrm{.}$ Both integrals converge absolutely, because $\mathrm{\mid }\tilde{R}(p)\mathrm{\mid }\le {\epsilon }^{\mathrm{\prime }}\mathrm{\mid }p\mathrm{\mid }$ by [F6] and the kernel bounds of step 1.1 give $\mathrm{\mid }{e}^{(t-s)A}{P}_u\tilde{R}(z(s))\mathrm{\mid }\le {\epsilon }^{\mathrm{\prime }}{e}^{-b(s-t)}{e}^{-\beta s}\parallel z{\parallel }_{\beta }$ whose $s$-integral over $[t,\mathrm{\infty })$ is ${\epsilon }^{\mathrm{\prime }}\parallel z{\parallel }_{\beta }{e}^{-\beta t}\mathrm{/}(b+\beta )$. The same kernel bounds give, after multiplying by $e^{\beta t}$, $\parallel N(z){\parallel }_{\beta }\le {\epsilon }^{\mathrm{\prime }}(\frac{1}{a-\beta }+\frac{1}{b+\beta })\parallel z{\parallel }_{\beta },\parallel N(z)-N(w){\parallel }_{\beta }\le {\epsilon }^{\mathrm{\prime }}(\frac{1}{a-\beta }+\frac{1}{b+\beta })\parallel z-w{\parallel }_{\beta },$ because $\mathrm{\mid }\tilde{R}(p)-\tilde{R}(q)\mathrm{\mid }\le {\epsilon }^{\mathrm{\prime }}\mathrm{\mid }p-q\mathrm{\mid }$ by the componentwise mean value theorem [F6]; recall $a>\beta$. Also $N(z)$ is continuous, being the difference of two continuous functions of $t$. Fix $\epsilon$ in step 2.1 so small that $k:={\epsilon }^{\mathrm{\prime }}(1\mathrm{/}(a-\beta )+1\mathrm{/}(b+\beta ))\le \frac{1}{2}$ and $k_0:={\epsilon }^{\mathrm{\prime }}(1\mathrm{/}a+1\mathrm{/}b)\le \frac{1}{2}$.

$\mathcal{B}$-Fréchet differentiability of $N$. For $z,w\in \mathcal{B}$ let $DN(z)w$ be given by the same formula with $\tilde{R}(z(s))$ replaced by $D\tilde{R}(z(s))w(s)$. The estimates of step 3.1 show that $DN(z)$ is linear and bounded with $\parallel DN(z)w{\parallel }_{\beta }\le k\parallel w{\parallel }_{\beta }$. By the componentwise mean value theorem [F6] applied on the segment from $z(s)$ to $z(s)+w(s)$, $\mathrm{\mid }\tilde{R}(z(s)+w(s))-\tilde{R}(z(s))-D\tilde{R}(z(s))w(s)\mathrm{\mid }\le \omega (\mathrm{\mid }w(s)\mathrm{\mid })\mathrm{\mid }w(s)\mathrm{\mid },$ where $\omega (\delta ):=\sup \{\mathrm{\mid }D\tilde{R}(p^{\mathrm{\prime }})-D\tilde{R}(p)\mathrm{\mid }:\mathrm{\mid }{p}^{\mathrm{\prime }}-p\mathrm{\mid }\le \delta \}$ satisfies $\omega (\delta )\to 0$ as $\delta \to 0$ by the uniform continuity of $D\tilde{R}$ established in step 2.1. Since $\mathrm{\mid }w(s)\mathrm{\mid }\le \parallel w{\parallel }_{\beta }$, the weighted kernel estimates give $\parallel N(z+w)-N(z)-DN(z)w{\parallel }_{\beta }\le C_{\beta }\omega (\parallel w{\parallel }_{\beta })\parallel w{\parallel }_{\beta }$, where $C_{\beta }=1\mathrm{/}(a-\beta )+1\mathrm{/}(b+\beta )$; this is $o(\parallel w{\parallel }_{\beta })$: thus $DN(z)$ is the Fréchet derivative of $N$ at $z$, and $z\mapsto DN(z)$ is continuous in operator norm because $\omega$ is a modulus of continuity.

Fixed points of the contractions $T_{\xi }$. For $\xi \in E^s$ put $y_{\xi }(t):=e^{tA}\xi$, so $y_{\xi }\in \mathcal{B}$ and $\parallel y_{\xi }{\parallel }_{\beta }\le \mathrm{\mid }\xi \mathrm{\mid }$ by step 1.1, and define $T_{\xi }(z):=y_{\xi }+N(z)$. By step 3.1, $T_{\xi }$ is a $k$-contraction of the nonempty complete space $\mathcal{B}$ with $k\le \frac{1}{2}$; by the Banach fixed point theorem [F2] it has a unique fixed point $z_{\xi }\in \mathcal{B}$. Since $N(0)=0$, the estimates of step 3.1 give $\parallel z_{\xi }{\parallel }_{\beta }\le \parallel y_{\xi }{\parallel }_{\beta }+k\parallel z_{\xi }{\parallel }_{\beta }$, hence $\parallel z_{\xi }{\parallel }_{\beta }\le 2\mathrm{\mid }\xi \mathrm{\mid }$ and $\mathrm{\mid }{z}_{\xi }(t)\mathrm{\mid }\le 2\mathrm{\mid }\xi \mathrm{\mid }{e}^{-\beta t}(t\ge 0)\mathrm{.}$

$z_{\xi }$ is a trajectory of $X$ with exponential decay. Write the first integral of $N(z_{\xi })(t)$ as $e^{tA}{P}_s{\int }_0^t{e}^{-sA}{P}_s\tilde{R}(z_{\xi }(s))ds$ and the second as $e^{tA}{P}_u(I_{\mathrm{\infty }}-{\int }_0^t{e}^{-sA}{P}_u\tilde{R}(z_{\xi }(s))ds)$ with $I_{\mathrm{\infty }}:={\int }_0^{\mathrm{\infty }}{e}^{-sA}{P}_u\tilde{R}(z_{\xi }(s))ds$; the integrands are continuous and the integrals converge absolutely as in step 3.1. Differentiating with the product rule and the primitive of a continuous function [F9] gives $z_{\xi }^{\mathrm{\prime }}(t)=A{z}_{\xi }(t)+P_s\tilde{R}(z_{\xi }(t))+P_u\tilde{R}(z_{\xi }(t))=A{z}_{\xi }(t)+\tilde{R}(z_{\xi }(t))=\tilde{X}(z_{\xi }(t))$ for every $t\ge 0$, so $z_{\xi }$ solves the ODE of $\tilde{X}$; by [F1] it is the trajectory ${\mathrm{\Phi }}_t(z_{\xi }(0))$. If $\mathrm{\mid }\xi \mathrm{\mid }\le \rho \mathrm{/}2$ then $\mathrm{\mid }{z}_{\xi }(t)\mathrm{\mid }\le \rho$ for all $t\ge 0$ by the exponential bound, so the trajectory stays in the ball where $\tilde{X}=X$; hence it is a trajectory of $X$ and $z_{\xi }(0)=\xi +h(\xi ),h(\xi ):=-{\int }_0^{\mathrm{\infty }}{e}^{-sA}{P}_u\tilde{R}(z_{\xi }(s))ds\in E^u,$ is the initial point of a trajectory of $X$ converging exponentially to the origin with rate $\beta$.

The initial points depend $C^1$ on $\xi$. Let $\mathrm{\Psi }(y)$ be the unique fixed point of $z\mapsto y+N(z)$; it exists by the same contraction argument as in step 4.2 and $z_{\xi }=\mathrm{\Psi }(y_{\xi })$, where $\xi \mapsto y_{\xi }=e^{\cdot A}\xi$ is linear and bounded into $\mathcal{B}$. We show that $\mathrm{\Psi }$ is Fréchet differentiable with $D\mathrm{\Psi }(y)=(I-DN(\mathrm{\Psi }(y)){)}^{-1}$. Let $u:=\mathrm{\Psi }(y)$, $v:=\mathrm{\Psi }(y+\eta )$ and $r:=N(v)-N(u)-DN(u)(v-u)$; step 4.1 gives $\parallel r\parallel \le C_{\beta }\omega (\parallel v-u\parallel )\parallel v-u\parallel$, and $(I-DN(u))(v-u)=\eta +r$. Since $\parallel DN(u)\parallel \le k<1$, the operator $I-DN(u)$ is invertible with inverse of norm at most $(1-k{)}^{-1}$ by the Neumann series [F3], so $\parallel v-u\parallel \le (1-k{)}^{-1}(\parallel \eta \parallel +\parallel r\parallel )$, while the contraction inequality directly gives $\parallel v-u\parallel \le (1-k{)}^{-1}\parallel \eta \parallel$. Therefore $v-u-(I-DN(u){)}^{-1}\eta =(I-DN(u){)}^{-1}r=o(\parallel \eta \parallel )$, so $D\mathrm{\Psi }(y)$ exists and has the stated form; it is a bounded linear map. Continuity of $y\mapsto D\mathrm{\Psi }(y)$ follows from the Lipschitz continuity of $\mathrm{\Psi }$, the continuity of $DN$ (step 4.1) and the continuity of the inversion map $S\mapsto (I-S{)}^{-1}$ [F3]. Consequently $\xi \mapsto z_{\xi }=\mathrm{\Psi }(e^{\cdot A}\xi )$ is $C^1$, and since the evaluation $z\mapsto z(0)$ is linear and bounded and $DN(0)=0$, differentiating $z_{\xi }$ at $\xi =0$ in a direction $\delta \in E^s$ gives $e^{\cdot A}\delta$, so $DG(0)=\mathrm{i}\mathrm{d}$ for $G(\xi ):=z_{\xi }(0)=\xi +h(\xi ),Dh(0)=P_{u}DG(0){\mathrm{\mid }}_{E^s}=0.$

Bounded trajectories. A bounded trajectory $y$ in $B_{\rho }$ satisfies variation of constants. In its unstable component, $e^{-bt}{P}_{u}y(t)\to 0$ and the integral of $e^{-bs}{P}_u\tilde{R}(y(s))$ converges, since its integrand is bounded by $\epsilon \mathrm{\prime }\rho e^{-bs}$. Therefore $y=T_{\xi }y$ with $\xi =P_{s}y(0)$. This does not yet put $y$ in the weighted space. Instead use the complete space $C_b([0,\mathrm{\infty }),{\mathbb{R}}^2)$ with the supremum norm; its completeness follows by the pointwise-limit and uniform-continuity argument of step 2.2. The same operator has contraction constant $k_0=\epsilon \mathrm{\prime }(1\mathrm{/}a+1\mathrm{/}b)\le \frac{1}{2}<1$ there. The weighted fixed point $z_{\xi }$ is bounded and solves the same equation, so uniqueness in this larger space gives $y=z_{\xi }$, and hence exponential decay.

The stable curve, the unstable curve and the half-trajectories. Choose ${\delta }_0\in (0,\rho \mathrm{/}2)$ with $\mathrm{\mid }h(\xi )\mathrm{\mid }\le \mathrm{\mid }\xi \mathrm{\mid }$ for $\mathrm{\mid }\xi \mathrm{\mid }<{\delta }_0$; this holds by step 5.1 because $Dh(0)=0$. Then $G=(\mathrm{i}\mathrm{d},h)$ maps $(-{\delta }_0,{\delta }_0)\subset E^s$ $C^1$-injectively onto a $C^1$ embedded curve $S$, because the linear projection $P_s$ restricts to the inverse of $G$ on $S$ and $S$ is the graph of the $C^1$ function $h$; and $T_{0}S=E^s$ because $DG(0)=\mathrm{i}\mathrm{d}$. For $x=G(\xi )$, uniqueness of the fixed point after shifting time gives $z_{\xi }(t)=G(P_s{z}_{\xi }(t))$. On this graph, $\mathrm{\mid }h(\xi )\mathrm{\mid }\le \mathrm{\mid }\xi \mathrm{\mid }$ and the small derivative bound give $\dot{\xi }=-a\xi +P_s\tilde{R}(G(\xi ))$, with $\mathrm{\mid }{P}_s\tilde{R}(G(\xi ))\mathrm{\mid }\le 2{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }\xi \mathrm{\mid }<a\mathrm{\mid }\xi \mathrm{\mid }\mathrm{/}2$ after decreasing $\epsilon$ initially. Thus $\mathrm{\mid }\xi \mathrm{\mid }$ strictly decreases toward zero and each local half-graph is invariant. By step 4.3 every point of $S$ has its forward trajectory in $S$, converging to $0$ at rate $\beta$; by step 5.2 every trajectory of $\tilde{X}$, hence of $X$, that stays in a sufficiently small ball $V\subset B_{\rho }$ for all $t\ge 0$ equals some $z_{\xi }$, with $\mathrm{\mid }\xi \mathrm{\mid }<{\delta }_0$, and therefore starts on $S$. If $x=G(\xi )\mathrm{\ne }0$, the forward trajectory of $x$ is a connected subset of $S\setminus \{0\}$ whose parameter values $P_s{z}_{\xi }(t)$ tend to $0$ and contain $\xi$, so by the intermediate value property it meets both components of $S\setminus \{0\}$ only according to the sign of $\xi$: the two components $\{G(\xi ):\xi >0\}$ and $\{G(\xi ):\xi <0\}$ are each a single half-trajectory. Applying the same construction to the function $\tilde{u}:=-u$, whose Hessian $-H$ is again a nondegenerate saddle and whose gradient field is $-X$, produces the unstable curve $U$ of $X$, tangent to the negative eigenline of $H$ and swept out by the two backward half-trajectories with exponential backward convergence; the neighbourhood $V$ is the intersection of the two trapping balls. This proves all the assertions; every step used only the displayed estimates, the Banach fixed point theorem, the Neumann series and the primitive of continuous functions, none of which needs a choice principle.