In page 35 of https://web.stanford.edu/class/math220b/handouts/heateqn.pdf, has the following step:
But I dont know why this is true.
My attemp:
I try to break it down with respect to $\partial_s$ and $\Delta_y$
For $\partial_s$:
$$-\int_{\epsilon}^{t}{\int_{\mathbb{R^n}}{\mathcal{H}(y,s)\partial_s f(x-y,t-s)dy}ds} $$
$$=-\int_{\mathbb{R^n}}\int_{\epsilon}^{t}{{\mathcal{H}(y,s)\partial_s f(x-y,t-s)ds}dy} $$
$$=\int_{\mathbb{R^n}}{-F(y,t,x)dy} $$
where $F(y,t,x)=\int_{\epsilon}^{t}{\mathcal{H}(y,s)\partial_s f(x-y,t-s)ds}$
Using intergation by part on varialbe $s$ of $$\int_{a}^b{g\cdot h^\prime dx}=g \cdot h\Big\rvert_{a}^{b} - \int_{a}^b{g^\prime\cdot h~dx}$$
we get $$-F(y,t,x)=-~~\mathcal{H}(y,s) f(x-y,t-s)\Big\rvert_{s=\epsilon}^{s=t} + \int_{\epsilon}^{t}{\partial_s \mathcal{H}(y,s) f(x-y,t-s)ds} $$
If we place back $\int_{\mathbb{R^n}}{ds}$, the first term match with the last term in photo, and $\partial_s$ term also confirm with that in photo.
Then, it means intergation by part of $\Delta_y$ part should give 0.
But I dont see if the following can give 0 additional term:
1.$\mathcal{H}\Delta f=-f\Delta \mathcal{H} + \Delta(\mathcal{H}f)-2\nabla \mathcal{H} \cdot \nabla f $
2.$\mathcal{H}\Delta f=\nabla \cdot (\mathcal{H} \nabla f) - \nabla \mathcal{H} \cdot \nabla f$
So my question is: how to prove the $\Delta_y$ part ?
With the comment from whpowell96, I am able to answer my own question.
$(1)-2\cdot(2)$ gives $\mathcal{H}\Delta f=f \Delta \mathcal{H} - \nabla \cdot \nabla (\mathcal{H} f) + 2\nabla \cdot (\mathcal{H} \nabla f)$
As $f$ has compact support, outside some sufficient large ball, both $f$ and $\nabla f$ become $0$. Then apply divergent thm. to 2nd and 3rd term (consider intergating over $\partial B_r$ as $r \to \infty$), intergal of both gives $0$. And yield $\mathcal{H}\Delta f=f \Delta \mathcal{H}$