# 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:

1. $$u(x,0)=0$$ for all $$-\infty < x < \infty$$
2. $$\lim\limits_{\vert x \vert \rightarrow \infty}u_t(x,0)=0$$
3. $$\lim\limits_{\vert x \vert \rightarrow \infty}u(x,t)=0$$ 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:

1. Does the third assumption imply $$(u_xu_t)\Big\vert_{x=-\infty}^{x=\infty}=0$$?
2. Can I justify that Fubini's theorem holds for $$\int_0^\tau \int_{-\infty}^{\infty} kuu_t \ dx dt=\int_{-\infty}^{\infty}\int_0^\tau kuu_t \ dt dx$$?
3. 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.

• Intuitively from a physics perspective : If your system has the extra transverse force term $ku$ there will also be potential Energy coming from that. It can't be true that the total conserved energy is the same as if $k=0$. Mar 27 at 7:37
• Related question Mar 29 at 17:06

1. 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 $$u_x(t, t) = \cos\big(x^3\big) 3x - \sin(x^3)/x^2$$.
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.
1. 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.