Let $u:\mathbb{R}^n\times(0,+\infty)\to\mathbb{R}$ solves the following heat equation:
$$u_t(x,t)-\triangle u=0,\quad (x,t)\in\mathbb{R}\times(0,+\infty)$$
(a) Show that for each $\lambda\in\mathbb{R}$, $u_\lambda(x,t)=u(\lambda x,\lambda^2 t)$ also solves the heat equation.
(b) Use (a) to show that
$$v(x,t)=x\cdot Du(x,t)+2t u_t(x,t)$$
solves the heat equation.
Here $D$ is the differential
$$Du(x,t)=(u_{x_1}(x,t),u_{x_2}(x,t),\dots,u_{x_n}(x,t))$$
So
$$x\cdot Du(x,t)=\sum_{k=1}^{n}x_ku_{x_k}(x,t)$$
Part (a) is quite straight forward. My question is on part (b), I didn't see where and how to apply part (a). I think I only need a hint, then I can figure this out. Thanks.
Consider $v_{\lambda}(x,t):= \frac{d}{d\lambda} u_{\lambda}(x,t)$. What can you say about $v_1$?