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If the solution to

$u_t=u_{xx}+u_{yy}=0 u_t=u_{xx}+u_{yy}$

reaches an equilibrium solution, then $u_{t}=0$ at that equilibrium, so $u_{xx}+u_{yy}=0$. The author has shown that a necessary condition for $u$ to be minimizer of $E(u)$ is $u_{xx}+u_{yy}=0$.

This isn't steepest descent in the way that it is normally presented as an optimization algorithm for minimizing a function $f(x)$, but it is conceptually the same.

Of course you don't want to simply minimize $E(u)$ without respecting the original image- you want to somehow balance the minimization of $E(u)$ with keeping the original image. By starting the time dependent PDE with the original image and then stopping after a finite time, you can achieve this balance.

1

If the solution to

$u_t=u_{xx}+u_{yy}=0$

reaches an equilibrium solution, then $u_{t}=0$ at that equilibrium, so $u_{xx}+u_{yy}=0$. The author has shown that a necessary condition for $u$ to be minimizer of $E(u)$ is $u_{xx}+u_{yy}=0$.

This isn't steepest descent in the way that it is normally presented as an optimization algorithm for minimizing a function $f(x)$, but it is conceptually the same.

Of course you don't want to simply minimize $E(u)$ without respecting the original image- you want to somehow balance the minimization of $E(u)$ with keeping the original image. By starting the time dependent PDE with the original image and then stopping after a finite time, you can achieve this balance.