Hamilton–Jacobi Equations and Viscosity Solutions

Proof. Straight lines are optimal. For a $C^1$ path $w$ from $y$ to $x$, Jensen's inequality for the convex function $L$ and the probability measure $ds/t$ on $[0, t]$ gives

with equality for the straight line $w(s) = y + \frac st(x - y)$. So the infimum over paths starting at $y$ is $tL(\frac{x - y}{t}) + g(y)$, and the infimum over all paths is the infimum over $y$ of this.

The minimum is attained. With $\operatorname{Lip}(g) = \Lambda$, $tL(\frac{x - y}{t}) + g(y) \geq tL(\frac{x - y}{t}) + g(x) - \Lambda|x - y|$. By superlinearity, $L(q) \geq (\Lambda + 1)|q| - C$, so the expression is at least $g(x) + |x - y| - Ct$, which tends to infinity as $|y| \to \infty$. A continuous function tending to infinity attains its minimum.

Proof. $\leq$. Fix $y$, and choose $z$ attaining the minimum in the formula for $u(y, s)$: $u(y, s) = sL(\frac{y - z}{s}) + g(z)$. Since

convexity gives $tL(\frac{x - z}{t}) \leq (t - s)L(\frac{x - y}{t - s}) + sL(\frac{y - z}{s})$, so $u(x, t) \leq tL(\frac{x - z}{t}) + g(z) \leq (t - s)L(\frac{x - y}{t - s}) + u(y, s)$.

$\geq$. Choose $w$ with $u(x, t) = tL(\frac{x - w}{t}) + g(w)$ and set $y = \frac stx + (1 - \frac st)w$, the point at time $s$ on the straight line from $w$ to $x$. Then $\frac{x - y}{t - s} = \frac{y - w}{s} = \frac{x - w}{t}$, and

Proof. In $x$. Choose $y$ attaining the minimum for $u(\hat x, t)$. Using $z = x - \hat x + y$ as a competitor for $u(x, t)$ (so that $x - z = \hat x - y$),

Exchange $x$ and $\hat x$. In $t$. By Exercise 9.14, $|u(x, t) - g(x)| \leq Ct$ for all $x$ and $t$; applying this to the semigroup started at time $\hat t$ (whose initial data $u(\cdot, \hat t)$ has the same Lipschitz constant) gives $|u(x, t) - u(x, \hat t)| \leq C|t - \hat t|$.

Proof. Let $u$ be differentiable at $(x, t)$.

$\leq 0$. For $q \in \mathbb{R}^n$ and $h > 0$, Lemma 9.2 at time $t + h$ with $s = t$ and the competitor $y = x$ gives $u(x + hq, t + h) \leq hL(q) + u(x, t)$. Subtract $u(x, t)$, divide by $h$ and let $h \to 0$: $q\cdot\nabla u + u_t \leq L(q)$. Hence $u_t + \sup_q\big(q\cdot\nabla u - L(q)\big) = u_t + H(\nabla u) \leq 0$.

$\geq 0$. Choose $z$ with $u(x, t) = tL(\frac{x - z}{t}) + g(z)$, put $q = \frac{x - z}{t}$, and for $0 < h < t$ let $s = t - h$ and $y = x - hq$ (the point at time $s$ on the line from $z$ to $x$). Then $\frac{y - z}{s} = q$, so $u(y, s) \leq sL(q) + g(z)$, and

Divide by $h$ and let $h \to 0$: $u_t + q\cdot\nabla u \geq L(q)$, so $u_t + H(\nabla u) \geq u_t + q\cdot\nabla u - L(q) \geq 0$.

Initial data. Taking $y = x$ gives $u(x, t) \leq tL(0) + g(x)$, and Exercise 9.14 gives the lower bound, so $u(x, t) \to g(x)$ as $t \to 0$.

Proof. (a) If $u - v$ has a local maximum or minimum at $(x_0, t_0)$ and both are $C^1$, then $\nabla u = \nabla v$ and $u_t = v_t$ there, so $v_t + H(\nabla v, x_0) = u_t + H(\nabla u, x_0) = 0$.

(b) We need a $C^1$ function $v$ with $u - v$ having a local maximum at $(x_0, t_0)$ and the same derivatives as $u$ there; then the subsolution test gives $\leq 0$, and the analogous construction from below gives $\geq 0$. Write $z = (x, t)$ and subtract an affine function, so that $z_0 = 0$, $u(0) = 0$ and $u$ has derivative $0$ at $0$. Then $u(z) = |z|\rho_1(z)$ with $\rho_1$ continuous, $\rho_1(0) = 0$. Let $\rho_2(r) = \max_{|z|\leq r}|\rho_1(z)|$, continuous, non-decreasing, $\rho_2(0) = 0$, and

Then $|v(z)| \leq |z|\rho_2(2|z|) + |z|^2 = o(|z|)$, so $v(0) = 0$ and $v$ is differentiable at $0$ with derivative $0$; for $z \neq 0$, $\nabla v(z) = \big(2\rho_2(2|z|) - \rho_2(|z|)\big)\frac{z}{|z|} + 2z$, which is continuous and tends to $0$, so $v \in C^1$. Finally, since $\rho_2$ is non-decreasing, $\int_{|z|}^{2|z|}\rho_2 \geq |z|\rho_2(|z|)$, so for $z \neq 0$

$u - v$ has a strict maximum at $0$.

(c) Let $v \in C^\infty$ with $u - v$ having a local maximum at $(x_0, t_0)$; replacing $v$ by $v + |x - x_0|^4 + (t - t_0)^4$, the maximum is strict. Then $u^\varepsilon - v$ has a maximum over a small closed ball around $(x_0, t_0)$ at points $(x_\varepsilon, t_\varepsilon)$ which, by uniform convergence and strictness, tend to $(x_0, t_0)$ as $\varepsilon \to 0$ (along a sequence). At an interior maximum of the smooth function $u^\varepsilon - v$: $\nabla u^\varepsilon = \nabla v$, $u^\varepsilon_t = v_t$, and $\Delta u^\varepsilon \leq \Delta v$. So

Let $\varepsilon \to 0$: $v_t + H(\nabla v, x_0) \leq 0$. The supersolution test is the same with minima. (Test functions in $C^\infty$ suffice, by the remark after the definition.)

Proof. Subsolution. Let $u - v$ have a local maximum at $(x_0, t_0)$. For $q \in \mathbb{R}^n$ and small $h > 0$, Lemma 9.2 (with the competitor $y = x_0 - hq$ at time $t_0 - h$) gives $u(x_0, t_0) \leq hL(q) + u(x_0 - hq, t_0 - h)$. Since $u - v$ is maximal at $(x_0, t_0)$, $u(x_0 - hq, t_0 - h) - u(x_0, t_0) \leq v(x_0 - hq, t_0 - h) - v(x_0, t_0)$. Combining, $v(x_0, t_0) - v(x_0 - hq, t_0 - h) \leq hL(q)$. Divide by $h$ and let $h \to 0$: $v_t + q\cdot\nabla v \leq L(q)$ for every $q$, so $v_t + H(\nabla v) \leq 0$.

Supersolution. Let $u - v$ have a local minimum at $(x_0, t_0)$. With $z$, $q = \frac{x_0 - z}{t_0}$ as in the proof of Theorem 9.4, $u(x_0, t_0) - u(x_0 - hq, t_0 - h) \geq hL(q)$. By minimality, $v(x_0, t_0) - v(x_0 - hq, t_0 - h) \geq u(x_0, t_0) - u(x_0 - hq, t_0 - h) \geq hL(q)$. So $v_t + q\cdot\nabla v \geq L(q)$ and $v_t + H(\nabla v) \geq v_t + q\cdot\nabla v - L(q) \geq 0$.

Proof. Suppose $\sigma = \sup(u - v) > 0$. The difficulty is that $u$ and $v$ are not differentiable, so we cannot compare their equations at a maximum point of $u - v$. The trick, due to Kruzhkov in a related setting and to Crandall and Lions here, is to double the variables: maximise a function of two independent points, penalising their distance, so that $u$ and $v$ are each touched by a smooth function at its own point.

The auxiliary function. For small $\varepsilon > 0$ and $\lambda = \frac{\sigma}{8T}$, let

on $(\mathbb{R}^n\times[0, T])^2$. It is continuous and tends to $-\infty$ at infinity (as $u$ and $v$ are bounded), so it has a maximum at some $(x_0, y_0, t_0, s_0)$. Choose $(\bar x, \bar t)$ with $u(\bar x, \bar t) - v(\bar x, \bar t) \geq \frac{3\sigma}{4}$. Then for $\varepsilon$ small,

The two points are close. From $\Phi(x_0, y_0, t_0, s_0) \geq \Phi(0, 0, 0, 0)$ and the boundedness of $u$, $v$: $\varepsilon(|x_0|^2 + |y_0|^2) \leq C$, so $\varepsilon|x_0|, \varepsilon|y_0| \leq C\sqrt\varepsilon$; and $|x_0 - y_0|^2 + (t_0 - s_0)^2 \leq C\varepsilon^2$. Better, comparing with $\Phi(x_0, x_0, t_0, t_0)$:

and the right side tends to $0$ as $\varepsilon \to 0$, because $v$ is uniformly continuous, $|x_0 - y_0|, |t_0 - s_0| \leq C\varepsilon$, and $\varepsilon\big||x_0|^2 - |y_0|^2\big| \leq \varepsilon|x_0 - y_0|(|x_0| + |y_0|) \leq C\sqrt\varepsilon$.

Neither point is at time zero. If $t_0 = 0$, then, writing $\omega$ for a modulus of continuity of $u$ and $v$,

since $s_0 = |t_0 - s_0| \leq C\varepsilon$. This is impossible for small $\varepsilon$. The same holds if $s_0 = 0$. So $t_0, s_0 > 0$ for small $\varepsilon$.

Testing the subsolution. The function $(x, t) \mapsto \Phi(x, y_0, t, s_0)$ has a maximum at $(x_0, t_0)$; that is, $u - \phi$ has a maximum there, with

a smooth function. The subsolution test gives

Testing the supersolution. The function $(y, s) \mapsto -\Phi(x_0, y, t_0, s)$ has a minimum at $(y_0, s_0)$; that is, $v - \psi$ has a minimum there, with $\psi(y, s) = u(x_0, t_0) - \lambda(t_0 + s) - \frac{|x_0 - y|^2 + (t_0 - s)^2}{\varepsilon^2} - \varepsilon(|x_0|^2 + |y|^2)$. Since $\psi_s = -\lambda + \frac{2(t_0 - s_0)}{\varepsilon^2}$ and $\nabla\psi = \frac{2(x_0 - y_0)}{\varepsilon^2} - 2\varepsilon y_0$ there,

The contradiction. Subtract (3) from (2): with $p = \frac{2(x_0 - y_0)}{\varepsilon^2}$,

Now $u - \phi$ has a maximum at $(x_0, t_0)$ and $u$ is Lipschitz in $x$ with constant $\Lambda$, so $|\nabla_x\phi(x_0, t_0)| = |p + 2\varepsilon x_0| \leq \Lambda$ (a $C^1$ function touching a $\Lambda$-Lipschitz function from above has gradient at most $\Lambda$ at the contact point). So $|p| \leq \Lambda + C\sqrt\varepsilon$, and both arguments of $H$ lie in a fixed ball, on which $H$ is uniformly continuous; and they differ by $2\varepsilon(|x_0| + |y_0|) \leq C\sqrt\varepsilon$. So the right side tends to $0$ as $\varepsilon \to 0$, while $2\lambda = \frac{\sigma}{4T} > 0$. Contradiction.

Uniqueness. Two bounded Lipschitz viscosity solutions with the same initial data are each a subsolution and a supersolution, so comparison in both directions gives equality. The Hopf–Lax function is Lipschitz and a viscosity solution (Proposition 9.9); it is bounded when $g$ is bounded (and the boundedness assumption can be removed with more care).

Proof. ($\leq$) Fix $\alpha$ on $[t, t + h]$, reaching $x_1 = x(t + h)$. For $\delta > 0$ choose a control $\beta$ on $[t + h, T]$ with $C_{x_1,t+h}[\beta] \leq u(x_1, t + h) + \delta$. The concatenation of $\alpha$ and $\beta$ is a control from $(x, t)$, so $u(x, t) \leq \int_t^{t+h}r\,ds + u(x_1, t + h) + \delta$. ($\geq$) For any control $\alpha$ on $[t, T]$, its restriction to $[t + h, T]$ is a control from $(x(t + h), t + h)$, so $C_{x,t}[\alpha] \geq \int_t^{t+h}r\,ds + u(x(t + h), t + h)$. Take the infimum over $\alpha$.