# PDE, wave equation and energy function.

We are given that

$$u_{tt} - \Delta u = 0, \tag 1$$

or

$$u_{tt} = \Delta u = \nabla^2 u = \nabla \cdot \nabla u; \tag 2$$

we define

$$E(t) = \displaystyle \int_{\Bbb R^2} (u_t^2 + \nabla u \cdot \nabla u) \; dx dy; \tag 3$$

(note that $$u_t^2 = \vert u_t \vert^2$$).

Then

$$E_t(t) = \dfrac{dE(t)}{dt} = \displaystyle \int_{\Bbb R^2} (2u_t u_{tt} + \nabla u_t \cdot \nabla u + \nabla u \cdot \nabla u_t) \; dx dy$$ $$= 2 \displaystyle \int_{\Bbb R^2} (u_t u_{tt} + \nabla u \cdot \nabla u_t) \; dx dy; \tag 4$$

via (2) this becomes

$$E_t(t) = 2 \displaystyle \int_{\Bbb R^2} (u_t \nabla \cdot \nabla u + \nabla u \cdot \nabla u_t) \; dx dy; \tag 5$$

we note that

$$\nabla \cdot (u_t \nabla u) = \nabla u_t \cdot \nabla u + u_t \nabla \cdot \nabla u, \tag 6$$

so that

$$\nabla u_t \cdot \nabla u = \nabla \cdot (u_t \nabla u) - u_t \nabla \cdot \nabla u; \tag 7$$

subtstituting (7) into (5) we find

$$E_t(t)$$ $$= 2 \displaystyle \int_{\Bbb R^2} (u_t \nabla \cdot \nabla u + \nabla \cdot (u_t \nabla u) - u_t \nabla \cdot \nabla u) \; dx dy$$ $$= 2 \displaystyle \int_{\Bbb R^2} \nabla \cdot (u_t \nabla u) \; dx dy; \tag 8$$

we compute the last integral in this equation via the divergence theorem applied to the circle $$S_R$$ centered at the origin which bounds the ball $$B_R$$ of radius $$R$$, also centered at $$(0, 0)$$:

$$2 \displaystyle \int_{\Bbb R^2} \nabla \cdot (u_t \nabla u) \; dx dy = 2\lim_{R \to \infty} \int_{B_R} \nabla \cdot (u_t \nabla u) \; dx dy$$ $$= 2\displaystyle \lim_{R \to \infty} \int_{S_R} u_t \nabla u \cdot \vec n_R \; ds, \tag 9$$

where $$\vec n_R$$ is the outward pointing unit normal on $$S_R$$ and $$ds$$ is the linear measure along $$S_R$$.

It now appears that, to complete the calculation of (9), a stronger assumption than the given

$$\nabla u \to 0 \; \text{as} \; R \to \infty \tag{10}$$

is required, for we need to ensure that

$$\displaystyle \lim_{R \to \infty} \int_{S_R} u_t \nabla u \cdot \vec n_R \; ds = 0, \tag{11}$$

which is not necessarily guaranteed by (10); for example, if $$\vert \nabla u \vert$$ falls off as, say, $$R^{-1/2}$$ as $$R \to \infty$$, and $$u_t$$ approaches a non-zero constant,

$$\displaystyle \lim_{R \to \infty} u_t \to c \ne 0, \tag{12}$$

then the integral occurring in (11) may in fact become unbounded as $$R \to \infty$$. A more stringent set of assumptions is therefore needed. For the present purposes we hypothesize that $$\nabla u$$ falls off as $$1/R^{1 + \epsilon}$$, $$\epsilon > 0$$ and $$u_t$$ is bounded for large $$R$$; then it is easy to see that (11) holds and thus that

$$\displaystyle \int_{\Bbb R^2} \nabla \cdot (u_t \nabla u) \; dx dy = 0; \tag{13}$$

from this in accord with (8)

$$E_t(t) = 0, \tag{14}$$

and hence that $$E(t)$$ is a constant for all $$t$$.

By directly differentiating the energy weobtain $$\dfrac{d}{dt}E(t)=\int u_tu_{tt}+\int \nabla u\cdot\nabla u_t$$ Then, by using the equation on the first integral and the Divergence's Theorem on the second one (and the fact that $$\nabla u$$ decay sufficiently fast at $$+\infty$$) we obtain $$\dfrac{d}{dt}E(t)=\int u_t\Delta u-\int u_t\Delta u=0.$$