Is it true: the energy is not preserved in wave equation with transverse force The wave equation with transverse force is given as following:

$u_{tt}-c^2u_{xx}+ku=0$ $(k>0)$

Define the total energy $$E(t)=\frac{1}{2} \int_{-\infty}^{\infty}((u_t)^2+c^2(u_x)^2)dx$$
There are three assumptions:


*

*$u(x,0)=0$ for all $-\infty < x < \infty$ 

*$\lim\limits_{\vert x \vert \rightarrow \infty}u_t(x,0)=0$ 

*$\lim\limits_{\vert x \vert \rightarrow \infty}u(x,t)=0$ for all $t>0$

I want to determine the sign of $E(\tau)-E(0)$ for some $\tau>0$.

Observe that $$\frac{d}{dt}E(t)=\frac{d}{dt}\frac{1}{2} \int_{-\infty}^{\infty}((u_t)^2+c^2(u_x)^2)dx=\frac{1}{2} \int_{-\infty}^{\infty}\frac{\partial}{\partial t}((u_t)^2+c^2(u_x)^2)dx$$$$=\int_{-\infty}^{\infty}(u_tu_{tt}+c^2u_xu_{xt})dx$$
Then by substituting $u_{tt}=c^2u_{xx}-ku$, we have $$\frac{d}{dt}E(t)=\int_{-\infty}^{\infty}((c^2u_{xx}-ku)u_t+c^2u_xu_{xt})dx$$$$=c^2\int_{-\infty}^{\infty}(u_{xx}u_t+u_xu_{xt})dx-\int_{-\infty}^{\infty} kuu_t \ dx$$$$=c^2\int_{-\infty}^{\infty}\frac{\partial}{\partial x}(u_xu_t)dx-\int_{-\infty}^{\infty} kuu_t \ dx$$$$=c^2(u_xu_t)\Big\vert_{x=-\infty}^{x=\infty}-\int_{-\infty}^{\infty} kuu_t \ dx$$$$=-\int_{-\infty}^{\infty} kuu_t \ dx$$
Thus, $$E(\tau)-E(0)=-\int_0^\tau \int_{-\infty}^{\infty} kuu_t \ dx dt$$
By Fubini's theorem (but I'm not sure that $\int_{-\infty}^{\infty} kuu_t \ dx$ converges),
$$E(\tau)-E(0)=-k\int_{-\infty}^{\infty}\int_0^\tau uu_t \ dt dx$$
Here, by integration by parts,
$$\int_0^\tau uu_t \ dt=u^2\Big\vert_{t=0}^{t=\tau}-\int_0^\tau uu_t \ dt$$$$=u^2(x,\tau)-\int_0^\tau uu_t \ dt$$
Thus $$E(\tau)-E(0)=-k\int_{-\infty}^{\infty}\frac{1}{2}u^2(x,\tau) dx<0$$

I am curious about three parts:

*

*Does the third assumption imply $(u_xu_t)\Big\vert_{x=-\infty}^{x=\infty}=0$? 

*Can I justify that Fubini's theorem holds for $\int_0^\tau \int_{-\infty}^{\infty} kuu_t \ dx dt=\int_{-\infty}^{\infty}\int_0^\tau kuu_t \ dt dx$? 

*Is the second assumption necessary? (It is mentioned in my textbook, but I didn't use it in my proof.)

I would appreciate it if someone gives me any intuitive answer.
 A: *

*In general $\lim_{x \to \pm \infty} u (x,t) = 0$ does not imply that $\lim_{x \to \pm \infty} u_x (x,t) = 0$ holds true. This is nicely explained in this post. In particular, $u_x(x, t)$ might approach $\pm \infty$ (or oscillate between $\infty$ and $-\infty$) for $x \to \pm \infty$. A good example is $u(x, t) = \sin\big(x^3\big) / x $ with $u_x(t, t) = \cos\big(x^3\big) 3x - \sin(x^3)/x^2$.

*You already provided the link to Wikipedia, where it is stated that $$\int_0^\tau \int_{\mathbb{R}} \vert k u u_t \vert \mathrm d(x, t) \overset{!}{<} \infty$$ has to be fulfilled to allow the application of Fubini's theorem. You know that $\lim_{x \to \pm \infty} u(x, t) = 0$ holds for all $t$, so this is no problem. Intuitively, it is somewhat clear that $\lim_{x \to \pm \infty} u(x, t) = 0 \: \forall t >0 $ implies that also $\lim_{x \to \pm \infty} u_t(x, t) = 0 \: \forall t >0 $, since for $\vert x \vert$ large enough $u$ effectively no longer varies with $t$ anymore. A rigorous proof would require that you can interchange the limits
$$ \lim_{x \to \pm \infty} \lim_{h \to 0} \underbrace{\frac{u(x, t+h) - u(x, t)}{h}}_{=:g(x, h)},$$
which can be done if the conditions of the Moore-Osgood theorem are met. See one of these [1], [2] resources for a more in-depth treatment. So in our case, you need that

$\lim_{x \to \pm \infty} g(x, h) $ converges uniformly for $h \neq 0$ and $\lim_{h \to 0} g(x, h)$ converges point-wise for $x \neq \pm \infty$.
If this condition is met, you can interchange the limits and obtain that $u_t(x \to \pm \infty, t) = 0$. In this case, the integral is bounded and you are good with applying Fubini.


*I see also no immediate consequence because of the second statement. In contrast, I see the difficulties with some missing information on the derivatives $u_t, u_x$. Assuming e.g. compact support of $u(t, x)$ would solve these, for instance.

