Suppose we have a closed (compact without boundry) Riemannian manifold $(M,g)$, do we have uniqueness (assuming solutions exist on $[0,T]$ ) for the backward heat equation


$$\frac{\partial f }{\partial t}+\Delta\_gf=0$$
$$f(x,0)=f\_0(x),$$
where $f\_0$ is some Borel function on $M$.



Uniqueness of backward heat equation on closed manifold with given initial data
I'm assuming you are supposing that $f\in C^\infty(M)$ for each $t \in [0,T]$, or something along this line. I'm not sure what happens if you allow $f$ to have some singularities. Then, picking any $0 < T\_1 < T$, we can consider $f$ as solving a forwards parabolic PDE from $T\_1$ to $0$. This puts parabolic regularity at our disposal and will allow you to make the following argument rigorous.


Write $f(t) = \sum\_{j=1}^\infty c\_j(t) \varphi\_j$ where $\Delta\_g \varphi\_j + \lambda\_j\varphi\_j=0$ are the eigenfunctions for the Laplacian on $(M,g)$. Then, the backwards heat equation becomes, at the level of the constants $c\_j(t)$, 
$$
c\_j'(t) = \lambda\_j c\_j(t),
$$
so $c\_j(t) = c\_j(0) e^{\lambda\_j t}$. Thus, we see that
$$
f = \sum\_{j=1}^\infty c\_j(0) e^{\lambda\_j t} \varphi\_j.
$$
The parabolic regularity argument above shows that for $t \in [0,T)$, this series converges in $L^2(M)$ (along with all derivatives, but this doesn't matter). Thus, we see that $f$ is unique given $f(0)$ (which determines $c\_j(0)$.




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This method is probably the easiest way to prove what you ask, but certainly is not the most robust. One can prove much more with techniques like Carleman estimates: see, e.g. <http://arxiv.org/pdf/0704.1349v2.pdf>.


