I wonder if this can be solved using a Feynman-Kac type argument. Let $X_{t,s}(a)$ be a path labeled at the time $t$ (i.e. $X_{t,t}(a)=a$) going back to $s$. Let $X_{t,s}(a)$ satisfy the following stochastic equation: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(x,s))}\ \hat{{\rm{d}}} W_s $$ where $W_s$ is a 1-dimensional Wiener process and $\hat{{\rm{d}}}$ indicates that this is a backwards Ito differential $\sigma$ is some arbitrary smooth function of $u$. Then, the following is true:

$\textbf{Claim:}$ A function $u(x,t)$ is a solution to the equation $$ \partial_t u = \sigma(u) \triangle u $$ if and only if the pair $(u,X)$ satisfies the following stochastic system: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(x,s))}\ \hat{{\rm{d}}} W_s $$ $$ u(x,t)= \mathbb{E}\left[u_0(X_{t,0}(a))\right] $$ where the expectation is taken over Brownian motions.

Here I am assuming we are solving the heat equation on the real line with no boundary conditions but it is straight-forward to incorporate boundaries. Of course, you can choose $\sigma$ to be the function you desire (for example the one in your question). I hope this helps.

I wonder if this can be solved using a Feynman-Kac type argument. Let $X_{t,s}(a)$ be a path labeled at the time $t$ (i.e. $X_{t,t}(a)=a$) going back to $s$. Let $X_{t,s}(a)$ satisfy the following stochastic equation: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(X_{t,s}(a),s))}\ \hat{{\rm{d}}} W_s $$ where $W_s$ is a 1-dimensional Wiener process and $\hat{{\rm{d}}}$ indicates that this is a backwards Ito differential $\sigma$ is some arbitrary smooth function of $u$. Then, the following is true:

$\textbf{Claim:}$ A function $u(x,t)$ is a solution to the equation $$ \partial_t u = \sigma(u) \triangle u $$ if and only if the pair $(u,X)$ satisfies the following stochastic system: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(X_{t,s}(a),s))}\ \hat{{\rm{d}}} W_s $$ $$ u(x,t)= \mathbb{E}\left[u_0(X_{t,0}(a))\right] $$ where the expectation is taken over Brownian motions.

Here I am assuming we are solving the heat equation on the real line with no boundary conditions but it is straight-forward to incorporate boundaries. Of course, you can choose $\sigma$ (with some technical conditions). I hope this helps.

I wonder if this can be solved using a Feynman-Kac type argument. Let $X_{t,s}(a)$ be a path labeled at the time $t$ (i.e. $X_{t,t}(a)=a$) going back to $s$. Let $X_{t,s}(a)$ satisfy the following stochastic equation: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(x,s))}\ \hat{{\rm{d}}} W_s $$ where $W_s$ is a 1-dimensional Wiener process and $\hat{{\rm{d}}}$ indicates that this is a backwards Ito differential $\sigma$ is some arbitrary smooth function of $u$. Then, the following is true:

$\textbf{Claim:}$ A function $u(x,t)$ is a solution to the equation $$ \partial_t u = \sigma(u) \triangle u $$ if and only if the pair $(u,X)$ satisfies the following stochastic system: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(x,s))}\ \hat{{\rm{d}}} W_s $$ $$ u(x,t)= \mathbb{E}\left[u_0(X_{t,0}(a))\right] $$ where the expectation is taken over Brownian motions.

I wonder if this can be solved using a Feynman-Kac type argument. Let $X_{t,s}(a)$ be a path labeled at the time $t$ (i.e. $X_{t,t}(a)=a$) going back to $s$. Let $X_{t,s}(a)$ satisfy the following stochastic equation: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(x,s))}\ \hat{{\rm{d}}} W_s $$ where $W_s$ is a 1-dimensional Wiener process and $\hat{{\rm{d}}}$ indicates that this is a backwards Ito differential $\sigma$ is some arbitrary smooth function of $u$. Then, the following is true:

$\textbf{Claim:}$ A function $u(x,t)$ is a solution to the equation $$ \partial_t u = \sigma(u) \triangle u $$ if and only if the pair $(u,X)$ satisfies the following stochastic system: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(x,s))}\ \hat{{\rm{d}}} W_s $$ $$ u(x,t)= \mathbb{E}\left[u_0(X_{t,0}(a))\right] $$ where the expectation is taken over Brownian motions.

Here I am assuming we are solving the heat equation on the real line with no boundary conditions but it is straight-forward to incorporate boundaries. Of course, you can choose $\sigma$ to be the function you desire (for example the one in your question). I hope this helps.