uniqueness heat equation Consider the heat equation,
$(1)$  $u_t=u_{xx}+f(x,t)$, $0<x<1$, $t>0$ 
$(2)$  $u(x,0)=\phi(x)$
$(3)$   $u(0,t)=g(t)$, $u(1,t)=h(t)$
When one wants to Show the uniqueness of solution of problem $(1)-(3)$, s/he can use so-called energy method or use maximum principle. 
My Question: What is the difference between these method? Is the class of solutions change for each method?
Thanks in advance!...
 A: The methods are different, but they can lead to the same conclusion, the first thing to do is assume that $u_1,u_2$ satisfy the problem, and let $w=u_1-u_2$, then $w$ solves
$$w_t-w_{xx}=0$$
$$w(x,0)=0$$
$$w(0,t)=w(1,t)=0$$
To use the energy method we multiply the equation by $w$, integrate and use integration by parts (noting that $w$ is $0$ on the boundary) to obtain:
$$\int_0^1ww_t+|w_x|^2\,dx=0$$
and $ww_t=\frac{1}{2}\frac{\partial }{\partial t}(w^2)$, and since $|w_x|^2\ge 0$ we get
$$\int_0^1\frac{1}{2}\frac{\partial}{\partial t}(w^2)\,dx\le 0$$
$$\frac{d}{dt}(\frac{1}{2}\int_0^1w^2\,dx)\le 0$$
We let $E(t)=\frac{1}{2}\int_0^1w^2\,dx$, we this is the "energy", and we see that it is decreasing in time (since the derivative is negative) so $E(t)\le E(0)$, i.e.
$$\int_0^1w^2(x,t)\,dx\le\int_0^1w^2(x,0)\,dx=0$$
Thus $w=0$, i.e. $u_1=u_2$.
The maximum principle is a different approach, I assume you mean to use the strong maximum principle, where again we take the same $w$, and the strong max principle tells us that if there is a point $(x_0,t_0)\in(0,1)\times(0,T)=:Q(T)$ such that $w(x_0,t_0)=\sup_{\overline{Q}(T)}w$, then $w$ is constant.
Now for $w$ by the SMP we see that the maximums of $w$ occur at $x=0,x=1$ or $t=0$, in both cases $w$ is zero here, and we find that $\sup w=0$, thus $w\le0$. i.e. $u_1\le u_2$, now apply this again with $w=u_2-u_1$, and you obtain $u_2\le u_1$, combining these you get $u_1=u_2$.
