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Let $f= f(t,x) : \mathbb{R}_+ \times \mathbb{R}^d \to \mathbb{R}$ be a Lipschitz function such that $$ \partial_t f - |\nabla f|^2 = 0 \qquad \text{almost everywhere in } \mathbb{R}_+ \times \mathbb{R}^d. $$ As is well-known, this condition does not determine the function $f$ uniquely in terms of the initial condition $f(0,\cdot)$; but uniqueness is restored if we impose in addition that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is convex (or locally semiconvex). I want to assume instead that, for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave. Under this condition, is the function $f$ determined uniquely in terms of $f(0,\cdot)$? The answer is "no", because for instance (say in $d = 1$) the function $(t,x) \mapsto t-|x|$ satisfies the required properties, but one can check that this is not the solution given by the Hopf-Lax formula. However, I struggle to find a counter-example if there is no kink in the initial condition to start with. So here is my question.

Assume that $f$ is a Lipschitz function that satisfies the equation above almost everywhere; that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave; and that the mapping $x \mapsto f(0,x)$ is smooth. Does the initial condition $f(0,\cdot)$ determine the function $f$ uniquely?

Let $f= f(t,x) : \mathbb{R}_+ \times \mathbb{R}^d \to \mathbb{R}$ be a Lipschitz function such that $$ \partial_t f - |\nabla f|^2 = 0 \qquad \text{almost everywhere in } \mathbb{R}_+ \times \mathbb{R}^d. $$ As is well-known, this condition does not determine the function $f$ uniquely in terms of the initial condition $f(0,\cdot)$; but uniqueness is restored if we impose in addition that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is convex (or locally semiconvex). I want to assume instead that, for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave. Under this condition, is the function $f$ determined uniquely in terms of $f(0,\cdot)$? The answer is "no", because for instance (say in $d = 1$) the function $(t,x) \mapsto t-|x|$ satisfies the required properties, but one can check that this is not the solution given by the Hopf-Lax formula. However, I struggle to find a counter-example if there is no kink in the initial condition to start with. So here is my question.

Assume that $f$ is a Lipschitz function that satisfies the equation above at every point of differentiability of $f$; that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave; and that the mapping $x \mapsto f(0,x)$ is smooth. Does the initial condition $f(0,\cdot)$ determine the function $f$ uniquely?

Let $f= f(t,x) : \mathbb{R}_+ \times \mathbb{R}^d \to \mathbb{R}$ be a Lipschitz function such that $$ \partial_t f - |\nabla f|^2 = 0 \qquad \text{almost everywhere in } \mathbb{R}_+ \times \mathbb{R}^d. $$ As is well-known, this condition does not determine the function $f$ uniquely in terms of the initial condition $f(0,\cdot)$; but uniqueness is restored if we impose in addition that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is convex (or locally semiconvex). I want to assume instead that, for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave. Under this condition, is the function $f$ determined uniquely in terms of $f(0,\cdot)$? The answer is "no", because for instance the function $(t,x) \mapsto t-|x|$ satisfies the required properties, but one can check that this is not the solution given by the Hopf-Lax formula. However, I struggle to find a counter-example if there is no kink in the initial condition to start with. So here is my question.

Assume that $f$ is a Lipschitz function that satisfies the equation above almost everywhere; that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave; and that the mapping $x \mapsto f(0,x)$ is smooth. Does the initial condition $f(0,\cdot)$ determine the function $f$ uniquely?

Let $f= f(t,x) : \mathbb{R}_+ \times \mathbb{R}^d \to \mathbb{R}$ be a Lipschitz function such that $$ \partial_t f - |\nabla f|^2 = 0 \qquad \text{almost everywhere in } \mathbb{R}_+ \times \mathbb{R}^d. $$ As is well-known, this condition does not determine the function $f$ uniquely in terms of the initial condition $f(0,\cdot)$; but uniqueness is restored if we impose in addition that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is convex (or locally semiconvex). I want to assume instead that, for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave. Under this condition, is the function $f$ determined uniquely in terms of $f(0,\cdot)$? The answer is "no", because for instance (say in $d = 1$) the function $(t,x) \mapsto t-|x|$ satisfies the required properties, but one can check that this is not the solution given by the Hopf-Lax formula. However, I struggle to find a counter-example if there is no kink in the initial condition to start with. So here is my question.

Assume that $f$ is a Lipschitz function that satisfies the equation above almost everywhere; that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave; and that the mapping $x \mapsto f(0,x)$ is smooth. Does the initial condition $f(0,\cdot)$ determine the function $f$ uniquely?

Let $f= f(t,x) : \mathbb{R}_+ \times \mathbb{R} \to \mathbb{R}$ be a Lipschitz function such that $$ \partial_t f - (\partial_x f)^2 = 0 \qquad \text{almost everywhere in } \mathbb{R}_+ \times \mathbb{R}. $$ As is well-known, this condition does not determine the function $f$ uniquely in terms of the initial condition $f(0,\cdot)$; but uniqueness is restored if we impose in addition that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is convex (or locally semiconvex). I want to assume instead that, for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave. Under this condition, is the function $f$ determined uniquely in terms of $f(0,\cdot)$? The answer is "no", because for instance the function $(t,x) \mapsto t-|x|$ satisfies the required properties, but one can check that this is not the solution given by the Hopf-Lax formula. However, I struggle to find a counter-example if there is no kink in the initial condition to start with. So here is my question.

Assume that $f$ is a Lipschitz function that satisfies the equation above almost everywhere; that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave; and that the mapping $x \mapsto f(0,x)$ is smooth. Does the initial condition $f(0,\cdot)$ determine the function $f$ uniquely?

Let $f= f(t,x) : \mathbb{R}_+ \times \mathbb{R}^d \to \mathbb{R}$ be a Lipschitz function such that $$ \partial_t f - |\nabla f|^2 = 0 \qquad \text{almost everywhere in } \mathbb{R}_+ \times \mathbb{R}^d. $$ As is well-known, this condition does not determine the function $f$ uniquely in terms of the initial condition $f(0,\cdot)$; but uniqueness is restored if we impose in addition that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is convex (or locally semiconvex). I want to assume instead that, for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave. Under this condition, is the function $f$ determined uniquely in terms of $f(0,\cdot)$? The answer is "no", because for instance the function $(t,x) \mapsto t-|x|$ satisfies the required properties, but one can check that this is not the solution given by the Hopf-Lax formula. However, I struggle to find a counter-example if there is no kink in the initial condition to start with. So here is my question.

Assume that $f$ is a Lipschitz function that satisfies the equation above almost everywhere; that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave; and that the mapping $x \mapsto f(0,x)$ is smooth. Does the initial condition $f(0,\cdot)$ determine the function $f$ uniquely?