Let's see that $\frac{d}{dt}E[u](t)≤0$, with equality if and only if $\Delta u(x,t)=0.$
$$\frac{d}{dt}E[u](t)=\frac{1}{2}\frac{d}{dt}\int_{U}\sum_{i=1}^{n}(\partial_iu(x,t))^2dx=
\int_{U}\sum_{i=1}^{n}\partial_iu(x,t)\partial_t\partial_iu(x,t)dx=\\
\sum_{i=1}^{n}\int_{U}\partial_iu(x,t)\partial_i\partial_tu(x,t)dx\stackrel{(1)}{=}\\
-\sum_{i=1}^{n}\int_{U}\partial_i\partial_iu(x,t)\partial_tu(x,t)dx+
\sum_{i=1}^{n}\int_{\partial U}\partial_iu(x,t)\partial_tu(x,t)\nu_i(x) dS(x)\stackrel{(2)}{=}\\
-\int_{U}(\Delta u(x,t))^2dx+\sum_{i=1}^{n}\int_{\partial U}\partial_iu(x,t)\partial_tg(x)\nu_i(x) dS(x)=-\int_{U}(\Delta u(x,t))^2dx\leq0.$$
Where the equality $(1)$ is just integration by parts and in $(2)$ we are using the fact that $u$ satisfies the heat equation and the boundary condition.
For the second part let $u$, $v$ be solutions of the initial boundary value problem.
Using the first part we know that the energy is nonincreasing. If we show that
$E[u-v](0)=0$ then $E[u-v](t)=0$ for every $t$, being $E$ nonnegative.
$$E[u-v](0)=\frac{1}{2}\int_{U}\vert Du(x,0)-Dv(x,0)\vert^2 dx= \frac{1}{2}\int_{U}\vert Dh(x)-Dh(x)\vert^2 dx=0.$$
This means that $Du(x,t)=Dv(x,t)$ for every $x$ and $t$ and also $\partial_t u(x,t)=\partial_t v(x,t)$, thanks to the first part. Thus $u(x,t)=v(x,t)+c$ for some constant $c$. Using the initial value you get $c=0$
Let $u$ and $v$ be two solutions with the stated properties. Let $\eta$ be a standard mollifier on $\mathbb{R}^d$ and let $u_\varepsilon :=u \ast \eta_\epsilon$ and $v_\varepsilon := v \ast \eta_\epsilon$, where we only convolve with respect to the $x$-variable.
Claim: We have $u_\varepsilon, v_\varepsilon \in C ^\infty((0, \infty) \times \mathbb{R} ^d) \cap C ( [0, \infty) \times \mathbb{R} ^d)$, $\partial_t u_\varepsilon - \Delta u_\varepsilon = \partial_t v_\varepsilon - \Delta v_\varepsilon = 0$ and $u_\varepsilon(0) = v_\varepsilon(0) = u_0 \ast \eta_\epsilon \in C_b ( \mathbb{R} ^d)$.
Proof of the Claim: As $u$ and $v$ are bounded, we can pass the differentiation into the integrals and get
\begin{equation*}
\partial_t u_\varepsilon - \Delta u_\varepsilon = (u_t - \Delta u) \ast \eta_\epsilon = 0
\end{equation*}
and similarly $\partial_t v_\varepsilon - \Delta v_\varepsilon = 0$. Furthermore, $u_\varepsilon, v_\varepsilon \in C ^\infty((0, \infty) \times \mathbb{R} ^d)$ and $u_0 \ast \eta_\epsilon \in C_b ( \mathbb{R} ^d)$. Thus, we are left to show continuity in $t = 0$. This follows by the weak$^*$ convergence: Let $x \in \mathbb{R} ^d$. Then
\begin{align*}
u_\varepsilon (t,x) - u_\varepsilon (0,x)
& = \int_{ \mathbb{R} ^d }\! ( u(t,y) - u_0 (y) ) \eta_\epsilon (x - y) \, dy
\\
& \to 0
\end{align*}
as $\eta_\epsilon (x - \cdot) \in L^1(\mathbb{R} ^d)$ and $u(t,\cdot) \to u_0$ weak$^*$ in $L^\infty ( \mathbb{R}^d )$ as $t \to 0$. In the same fashion, we see that $v_\varepsilon$ is continuous in $t = 0$, which proofs the claim.
The heat equation for regular initial data in $C_b(\mathbb{R}^d)$ has a unique bounded classical solution, see Thm. 6 in Evans, Chapter 2.3. Hence, $u_\varepsilon = v_\varepsilon$ for all $\varepsilon > 0$. As $u_\varepsilon \to u$ and $v_\varepsilon \to v$ locally uniformly in $(0,\infty) \times \mathbb{R} ^d$ as $\varepsilon \downarrow 0$, we get $u = v$ in $(0,\infty) \times \mathbb{R} ^d$.
Best Answer
The solution $u(t,x)$ of the heat equation is analytic in $x$ as a function of $x$ for any $t>0$, so it cannot have compact support. This is independent of the smoothies or size of the initial data.
For the Schrödinger equation, the same is true if the initial data decays fast enough at $\infty$, in particular,if it has compact support.