I have this problem:
Let $K(y, t) = \frac{-1}{(4 \pi)^{\frac{n}{2}}} \cdot t^{\frac{-n}{2}} \cdot e^{\frac{- \mid y \mid^2}{4t}}$. If $v \in C^2 (\mathbb{R}^n \times (0, \infty))$, prove that for $t > 0$ the following is valid:
$v(x, t) = \displaystyle\int_{0}^{t} \displaystyle\int_{}^{} K(x-y, t-s) \cdot (\triangle v(y, s) – v_s (y, s)) \, dy \, ds$ and $v(x, 0) = 0$.
I know that $-K(y, t)$ is the fundamental solution of the heat equation. Then:
$\displaystyle\int_{0}^{t} \displaystyle\int_{}^{} K(x-y, t-s) \cdot (\triangle v(y, s) – v_s (y, s)) \, dy \, ds =$
$\displaystyle\int_{0}^{t} \displaystyle\int_{}^{} \frac{-1}{(4 \pi)^{\frac{n}{2}}} \cdot t^{\frac{-n}{2}} \cdot e^{\frac{- \mid y \mid^2}{4t}} \cdot (\triangle v(y, s) – v_s (y, s)) \, dy \, ds =$
$\displaystyle\int_{0}^{t} \displaystyle\int_{}^{} \frac{1}{(4 \pi)^{\frac{n}{2}}} \cdot t^{\frac{-n}{2}} \cdot e^{\frac{- \mid y \mid^2}{4t}} \cdot (- \triangle v(y, s) + v_s (y, s)) \, dy \, ds =$
$\displaystyle\int_{0}^{t} \displaystyle\int_{}^{} \frac{1}{(4 \pi)^{\frac{n}{2}}} \cdot t^{\frac{-n}{2}} \cdot e^{\frac{- \mid y \mid^2}{4t}} \cdot (v_s (y, s) - \triangle v(y, s)) \, dy \, ds$.
How can I follow this demo?or does this not work? I'm lost :(