If $e(t_0) = \int_{U} g(x)^2\,d x $ for some $t_0>0$, what can you conclude about $u(x,t)$?

Question

Suppose that $U\subset \mathbb{R}^n$ is open and bounded with $C^1$ boundary, and $u=u(x,t)$ is a smooth solution of the initial-boundary value problem $$ \left\{ \begin{aligned} u_t(x,t) -\Delta u(x,t) &= 0 \quad\text{in} ~{U}\times (0,\infty)\\ \frac{\partial u}{\partial \nu} (x,t)&=0\quad \text{on}~ \partial U \times (0,\infty)\\ u(x,0) &= g(x) \end{aligned} \right. $$ where $\nu$ denotes the unit outward normal to $\partial U$. Define $$ I(t):=\int\limits_{U} u(x,t)d x\, \quad\text{and}\quad e(t):=\int\limits_{U} u(x,t)^2\,d x. $$

1. Show that $\displaystyle I(t) = \int\limits_{U} g(x)\,d x $ for all $t\geq 0$.
2. Show that $\displaystyle e(t) \leq \int\limits_{U} g(x)^2\,d x $ for all $t\geq 0$.
3. If $\displaystyle e(t_0) = \int\limits_{U} g(x)^2\,d x $ for some $t_0>0$, what can you conclude about $u(x,t)$?

My Solution
I manage to do the part one and two by using Green's identities. But what about the last question.
Here when proving part $2$, we get the result that $\frac{d}{dt}e(t)\leq 0$. That is it is a decreasing function. Hence for the given condition in part $3$, we can of course say that $\forall t\in[0,t_0]$,$\displaystyle e(t) = \int\limits_{U} g(x)^2\,d x $.

So is it same as saying no energy dissipation of $u(x,t)$ until time $t_0$ and then it decreases. Or is there any better conclusion that we can get from this result?
Thank you

Answer 1

With

$I = \displaystyle \int_U u \; dV, \tag 1$

we have

$I_t = \displaystyle \int_U u_t \; dV, \tag 2$

or, in light of

$u_t = \nabla^2 u = \nabla \cdot \nabla u, \tag{2.5}$

$I_t = \displaystyle \int_U \nabla \cdot \nabla u \; dV; \tag 3$

we invoke the divergence theorem, where $dS$ is the volume element on $\partial U$:

$I_t = \displaystyle \int_{\partial U} \nabla u \cdot \nu \; dS = \displaystyle \int_{\partial U} \dfrac{\partial u}{\partial \nu} \; dS = 0; \tag 4$

so $I$ is constant, whence

$I(t) = I(0) = \displaystyle \int_U u(x, 0) \; dV = \int_U g(x) \; dV. \tag 5$

Next,

$e(t) = \displaystyle \int_U u^2(x, t) \; dV, \tag 6$

$e_t = 2 \displaystyle \int_U uu_t \; dV, \tag 7$

$e_t = 2 \displaystyle \int_U u \nabla^2 u \; dV; \tag 8$

we now invoke the well-known identity

$\nabla \cdot (u \nabla u) = \nabla u \cdot \nabla u + u \nabla^2 u \tag 9$

yielding

$e_t = 2 \displaystyle \int_U u \nabla^2 u \; dV = 2\int_U ( \nabla \cdot (u \nabla u) - \nabla u \cdot \nabla u) \; dV$ $= 2\displaystyle \int_U \nabla \cdot (u \nabla u) \; dV - 2 \int_U \nabla u \cdot \nabla u \; dV; \tag{10}$

we evaluate the first integral on the right:

$\displaystyle \int_U \nabla \cdot (u \nabla u) \; dV = \int_{\partial U} u\nabla u \cdot \nu \; dS = \displaystyle \int_{\partial U} u \dfrac{\partial u}{\partial \nu} \; dS = 0; \tag{11}$

thus (10) becomes

$e_t = 2 \displaystyle \int_U u \nabla^2 u \; dV = -2\displaystyle \int_U \nabla u \cdot \nabla u \; dV \le 0; \tag{12}$

thus

$e(t) = \displaystyle \int_U u^2(x, t) \; dV \le \int_U u^2(x, 0) \; dV = \int_U g^2(x) \; dV. \tag{13}$

Next observe that since $u$ is smooth, (12) implies that

$e_t = 0 \Longleftrightarrow \nabla u = 0; \tag{14}$

now if for some $t_0 > 0$

$e(t_0) = \displaystyle \int_U g^2(x) \; dV = e(0), \tag{15}$

it follows that

$e_t(t) = 0, \; 0 \le t \le t_0, \tag{16}$

and then in accord with (14),

$\nabla u = 0, \; 0 \le t \le t_0; \tag{17}$

since

$u_t = \nabla^2 u = \nabla \cdot \nabla u, \tag{18}$

we infer that

$u_t = 0, \; 0 \le t \le t_0; \tag{19}$

by (17) and (19), $u$ is constant on $U \times [0, t_0]$. Since

$u(x, 0) = g(x), \tag{20}$

$g(x)$ is constant on $U$ as well. We see then that assumption (3) in the question implies that $u(x, t)$ is constant for $t \in [0, t_0]$, and since $u(x, t_0)$ is constant in $x$ we may, taking it as an initial condition at $t = t_0$, likely infer that $u(x, t_0)$ is constant for all $t \ge t_0$. Of course, certainty of such an observation requires perhaps more detailed information concerning the existence and uniqueness of solutions to (18), which discussion I will leave to another time and place.