Let $u,v\in{C}^2(\mathbb{R}^n\times\mathbb{R})$ be two solutions to the heat equation such that $u(x,t)=v(x,t)$ for all $(x,t)$ in $\mathbb{R}^n\times\mathbb{R}$ with $|x|>1$ or $t<0$. Show that $u=v$ for all $(x,t)$.
So I have to prove that $u=v$ on $|x|\leq1$ and $t\geq0$. The only approaches I can think of would be applying the comparison principle, but for that I would need information for $u$ and $v$ on $|x|=1$ and $t=0$. Or I could define $w=u-v$ and try to prove that $w=0$ on $|x|\leq1$ and $t\geq0$, but I wouldn't know how to continue from here.
I would be very grateful on any ideas on how to solve this problem.
If $u,v\in C^2(\mathbb{R}^n\times \mathbb{R})$, then $w = u-v\in C^2(\mathbb{R}^n\times \mathbb{R})$ as well and $w(x,t)=0$ for $t<0$ or $|x|>1$. By the continuity of $u$ and $v$, and hence $w$, this also implies that $w(x,0) = 0 \ \forall x$ and that $w(x,t)|_{|x|=1} = 0$ $\forall t$. We then have uniqueness by the energy identity:
$$ \begin{aligned} w_t - \Delta w &= 0, \ x\in\mathbb{R}^n, \ t>0 \\ w(x,0) &= 0 \\ \implies \int_{\mathbb{R}^n} ww_t -w\Delta w dx &= \int_{|x|<1} ww_t -w\Delta w dx = \int_{|x|<1}ww_t +|\nabla w|^2 dx = 0 \\ \implies \int_{|x|<1} \frac{1}{2}\frac{\partial}{\partial t} (w^2)dx &\leq 0 \end{aligned} $$ This implies that $\int_{|x|<1} \frac{1}{2}w^2dx$ is nonincreasing in $t$, so we have $$ \int_{|x|<1} \frac{1}{2}w^2dx \leq \int_{|x|<1} \frac{1}{2}w(x,0)^2dx = 0, $$ so $w(x,t) = 0$ for $|x|\leq1$ and $t>0$, and hence, for all $x$ and $t$, as desired.