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Could someone please help me with the following question? I got stuck somewhere.

Given a function $u(t,x)$ satisfying the relationship: $$ u_{tt} + au_t \ = \ c^2u_{xx} \qquad \text{ for some } a>0 $$ And the requirement that holds for sufficiently big $x$: $$ \exists\alpha>0, \ \exists C(x)>0, \ \max \{|u_t(t,x)|,|u_x(t,x)|\} \leq\frac{ C(t)}{|x|^\alpha} $$ Show that the energy function $E(T)$ as defined below is a nonincreasing solution. $$ E(t) \ = \ \frac12 \int_{-\infty}^\infty u_t^2 + c^2u_x^2 dx $$

**What I tried **

I just took its derivative:

$$ \frac{\partial}{\partial t}E(t) \ = \ \frac12 \int_{-\infty}^\infty \frac{\partial}{\partial t}\left[u_t^2 + c^2u_x^2\right] dx \ = \ \int_{-\infty}^\infty u_tu_{tt} + c^2u_xu_{tx}dx $$ And $u_{tt} \ = \ -au_t +c^2u_{xx}$, so $$ \frac{\partial}{\partial t}E(t) \ = \ \int_{-\infty}^\infty u_t(-au_t +c^2u_{xx}) + c^2u_xu_{tx}dx \ = \ \int_{-\infty}^\infty u_t(-au_t +c^2u_{xx}) + c^2u_xu_{tx}dx \ = \ c^2\int_{-\infty}^\infty u_xu_{xt}+u_tu_{xx}dx - a \int_{-\infty}^\infty u_t^2dx $$ How could I show that this is greater than zero? In tried to find primitives but I failed.

Koenraad van Duin
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    You made a typo in your second to last line, it should read..

    \begin{align} c^{2} \int_{\mathbb{R}} u_{t} u_{xx} + u_{x} u_{tx} dx - a \int_{\mathbb{R}} u_{t}^{2} dx = c^{2} \int_{\mathbb{R}} \partial_{x} (u_{t} u_{x}) dx - a \int_{\mathbb{R}} u_{t}^{2} dx \end{align}

    Then use the fact that

    $$\int_{\mathbb{R}} u_{t}^{2} dx$$

    is always positive.

     – Matthew Cassell Jun 07 '15 at 14:48

1 Answers1

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$$ \frac{d}{dt}E(t) \ = \ c^2\int_{-\infty}^\infty u_{x}u_{xt}+u_tu_{xx}dx - a \int_{-\infty}^\infty u_t^2dx $$

$$ \frac{d}{dt}E(t) \ = \ c^2\int_{-\infty}^\infty \frac{\partial (u_t u_x)}{\partial x}dx - a \int_{-\infty}^\infty u_t^2dx $$

$$ \frac{d}{dt}E(t) \ = \ c^2 \left( (u_t u_x)_{x=+\infty} - (u_t u_x)_{x=-\infty} \right) - a \int_{-\infty}^\infty u_t^2dx $$ Here, we see that possibly something is missing in the wording of the problem : It should be stated that $u(x,t)$ is a function which has FINITE limits at infinity , especially if this function is the model of a physical phenomena or process. If we suppose that this assumption is true, then the derivative at infinity is zero. So, $\left( (u_t u_x)_{x=+\infty} - (u_t u_x)_{x=-\infty} \right)=0$

$$ \frac{d}{dt}E(t) = - a \int_{-\infty}^\infty u_t^2dx $$ The integral is $\geq 0$ and $a>0$ . Hense $\frac{\partial}{\partial t}E(t)\leq 0$ and $E(t)$ is not increassing.

JJacquelin
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  • Yes, you where right. Let me add that requirement. – Koenraad van Duin Jun 08 '15 at 12:21
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    How does $\lim_{|x|\to\infty} u(x,t) = 0$ imply $\lim_{|x|}u_t(x,t) =0$? I know $\lim_{|x|\to\infty} u_x(x,t) = 0$. – one potato two potato Mar 26 '22 at 01:49
  • In general $\lim_{x \to \pm \infty} u (x,t) = C$ does not imply that $\lim_{x \to \pm \infty} u_x (x,t) = 0$ holds true. This is explained here. 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 $. So how do you know that $u_tu_x \Big \vert_{x \to - \infty}^{x \to + \infty} = 0$ holds without any additional control on $u_t, u_x$? – Dan Doe Mar 31 '22 at 09:14