[proofplan] We encode the variational expression as a smooth function $\Phi(t,x,y)$ and use the first-order optimality condition at the unique minimizer $Y(t,x)$. Since $\nabla_y \Phi(t,x,Y(t,x))=0$, every chain-rule contribution coming from differentiating the minimizer map $Y$ vanishes. The remaining explicit derivatives give $\nabla_x u(t,x) = (x-Y(t,x))/t$ and $\partial_t u(t,x) = -|x-Y(t,x)|^2/(2t^2)$, and substitution proves the Hamilton-Jacobi equation at the specified point. [/proofplan]

[step:Introduce the objective function and record the first-order optimality condition] Define the function \begin{align*} \Phi: V \times \mathbb{R}^n \to \mathbb{R} \end{align*} by \begin{align*} \Phi(t,x,y) = g(y) + \frac{|x-y|^2}{2t}. \end{align*} Since $g \in C^1(\mathbb{R}^n)$ and $t>0$ on $V$, the map $\Phi$ is $C^1$ in all variables. For each $(t,x) \in V$, the point $Y(t,x)$ is a minimizer of the differentiable function $y \mapsto \Phi(t,x,y)$ on the [open set](/page/Open%20Set) $\mathbb{R}^n$. Therefore its gradient with respect to $y$ vanishes: \begin{align*} \nabla_y \Phi(t,x,Y(t,x)) = 0. \end{align*} Computing the $y$-gradient gives \begin{align*} \nabla g(Y(t,x)) - \frac{x-Y(t,x)}{t} = 0. \end{align*} Thus, for every $(t,x) \in V$, \begin{align*} \nabla g(Y(t,x)) = \frac{x-Y(t,x)}{t}. \end{align*} [/step]

[step:Differentiate the value function and cancel the minimizer derivatives]Let $Y_i:V \to \mathbb{R}$ denote the $i$-th component of $Y$, and write \begin{align*} Y(t,x) = (Y_1(t,x),\dots,Y_n(t,x)). \end{align*} Since $Y \in C^1(V;\mathbb{R}^n)$ and $\Phi \in C^1(V \times \mathbb{R}^n)$, the composite function \begin{align*} u:V \to \mathbb{R} \end{align*} given by \begin{align*} u(t,x) = \Phi(t,x,Y(t,x)) \end{align*} is $C^1$ on $V$. Fix $(t,x) \in V$. For each spatial index $j \in \{1,\dots,n\}$, the chain rule gives \begin{align*} \partial_{x_j}u(t,x) = \partial_{x_j}\Phi(t,x,Y(t,x)) + \sum_{i=1}^n \partial_{y_i}\Phi(t,x,Y(t,x))\,\partial_{x_j}Y_i(t,x). \end{align*} The first-order optimality condition gives $\partial_{y_i}\Phi(t,x,Y(t,x))=0$ for every $i$, so the sum vanishes. Hence \begin{align*} \partial_{x_j}u(t,x) = \partial_{x_j}\Phi(t,x,Y(t,x)). \end{align*} Differentiating the explicit quadratic term with respect to $x_j$ gives \begin{align*} \partial_{x_j}u(t,x) = \frac{x_j-Y_j(t,x)}{t}. \end{align*} Therefore \begin{align*} \nabla_x u(t,x) = \frac{x-Y(t,x)}{t}. \end{align*} The same chain-rule cancellation applies to the time derivative: \begin{align*} \partial_t u(t,x) = \partial_t\Phi(t,x,Y(t,x)) + \sum_{i=1}^n \partial_{y_i}\Phi(t,x,Y(t,x))\,\partial_tY_i(t,x). \end{align*} Again the sum is zero, and differentiating the explicit factor $1/(2t)$ gives \begin{align*} \partial_t u(t,x) = -\frac{|x-Y(t,x)|^2}{2t^2}. \end{align*}[/step]

[guided]The important point is that $u$ depends on $(t,x)$ in two ways: directly through the variables in $\Phi(t,x,y)$, and indirectly through the minimizer $y=Y(t,x)$. We therefore must account for both dependences and then use the minimizing condition to remove the indirect terms. Let $Y_i:V \to \mathbb{R}$ be the $i$-th component of the $C^1$ map $Y:V \to \mathbb{R}^n$, so that \begin{align*} Y(t,x) = (Y_1(t,x),\dots,Y_n(t,x)). \end{align*} Since $\Phi$ is $C^1$ and $Y$ is $C^1$, the composite map \begin{align*} u:V \to \mathbb{R} \end{align*} defined by \begin{align*} u(t,x) = \Phi(t,x,Y(t,x)) \end{align*} is $C^1$. Fix $(t,x) \in V$ and a spatial index $j \in \{1,\dots,n\}$. The chain rule separates the direct $x_j$-dependence of $\Phi$ from the dependence through the minimizer: \begin{align*} \partial_{x_j}u(t,x) = \partial_{x_j}\Phi(t,x,Y(t,x)) + \sum_{i=1}^n \partial_{y_i}\Phi(t,x,Y(t,x))\,\partial_{x_j}Y_i(t,x). \end{align*} The reason the Hopf-Lax formula becomes simple is that the second term is multiplied by the $y$-gradient of $\Phi$ at the minimizing point. Because $Y(t,x)$ minimizes the differentiable function $y \mapsto \Phi(t,x,y)$ on all of $\mathbb{R}^n$, the first-order optimality condition gives \begin{align*} \nabla_y\Phi(t,x,Y(t,x)) = 0. \end{align*} Equivalently, $\partial_{y_i}\Phi(t,x,Y(t,x))=0$ for every $i \in \{1,\dots,n\}$. Thus the whole sum involving $\partial_{x_j}Y_i(t,x)$ vanishes, and only the explicit derivative remains: \begin{align*} \partial_{x_j}u(t,x) = \partial_{x_j}\Phi(t,x,Y(t,x)). \end{align*} Now $\Phi(t,x,y)=g(y)+|x-y|^2/(2t)$, and $g(y)$ has no direct dependence on $x_j$. Differentiating the quadratic term with respect to $x_j$ gives \begin{align*} \partial_{x_j}u(t,x) = \frac{x_j-Y_j(t,x)}{t}. \end{align*} Since this holds for every spatial component $j$, we obtain the vector identity \begin{align*} \nabla_x u(t,x) = \frac{x-Y(t,x)}{t}. \end{align*} The time derivative is handled in exactly the same chain-rule framework. Applying the chain rule in the $t$ variable gives \begin{align*} \partial_t u(t,x) = \partial_t\Phi(t,x,Y(t,x)) + \sum_{i=1}^n \partial_{y_i}\Phi(t,x,Y(t,x))\,\partial_tY_i(t,x). \end{align*} The first-order optimality condition again makes every coefficient $\partial_{y_i}\Phi(t,x,Y(t,x))$ equal to zero, so the derivative of the minimizer map $Y$ does not contribute. Therefore \begin{align*} \partial_t u(t,x) = \partial_t\Phi(t,x,Y(t,x)). \end{align*} When differentiating $\Phi(t,x,y)$ with respect to $t$, the variables $x$ and $y$ are held fixed. Since \begin{align*} \frac{\partial}{\partial t}\left(\frac{|x-y|^2}{2t}\right) = -\frac{|x-y|^2}{2t^2}, \end{align*} we get \begin{align*} \partial_t u(t,x) = -\frac{|x-Y(t,x)|^2}{2t^2}. \end{align*}[/guided]

[step:Substitute the derivative formulas into the Hamilton-Jacobi expression] Evaluating the two derivative identities at $(t_0,x_0)$ gives \begin{align*} \nabla_x u(t_0,x_0) = \frac{x_0-Y(t_0,x_0)}{t_0}. \end{align*} and \begin{align*} \partial_t u(t_0,x_0) = -\frac{|x_0-Y(t_0,x_0)|^2}{2t_0^2}. \end{align*} Hence \begin{align*} \partial_t u(t_0,x_0) + \frac{1}{2}|\nabla_x u(t_0,x_0)|^2 = -\frac{|x_0-Y(t_0,x_0)|^2}{2t_0^2} + \frac{1}{2}\left|\frac{x_0-Y(t_0,x_0)}{t_0}\right|^2. \end{align*} Using homogeneity of the Euclidean norm, \begin{align*} \left|\frac{x_0-Y(t_0,x_0)}{t_0}\right|^2 = \frac{|x_0-Y(t_0,x_0)|^2}{t_0^2}. \end{align*} Therefore \begin{align*} \partial_t u(t_0,x_0) + \frac{1}{2}|\nabla_x u(t_0,x_0)|^2 = 0. \end{align*} This is precisely the Hamilton-Jacobi equation for the quadratic Hamiltonian at $(t_0,x_0)$. [/step]