0
$\begingroup$

How does one evaluate the following integral? $$\frac{1}{2}\frac{d}{dt}\int\limits_{-\infty}^\infty \left[ (v_t)^2 + c^2(v_x)^2 \right]dx - \frac{d}{dx}\int\limits_{-\infty}^\infty c^2 v_tv_xdx=-\nu\int\limits_{-\infty}^\infty(v_t)^2dx$$ where $c$ and $\nu$ are constants, and where any derivatives of $v$ vanish whenever $\left| x \right|\to \infty$.

I have no clue how to integrate a partial derivative with respect to one variable over another variable. Some hints would be appreciated.

$\endgroup$
  • $\begingroup$ I suspect you have made a mistake, since the second integral on the left is constant in $x$ so its $d/dx$ is 0 $\endgroup$ – Calvin Khor Jun 2 '16 at 21:30
2
$\begingroup$

$$\frac{1}{2}\frac{d}{dt}\left[v_t^2+c^2 v_x^2\right]-\frac{d}{dx}c^2 v_t v_x = v_t v_{tt}+ c^2 v_x v_{xt}-c^2 v_x v_{tx}-c^2 v_t v_{xx}$$ but assuming that Schwarz' condition $v_{xt}=v_{tx}$ holds, the RHS equals: $$ v_t v_{tt}-c^2v_t v_{xx} = \frac{1}{2}\frac{d}{dt} v_t^2-c^2 v_t v_{xx}=v_t\color{red}{(v_{tt}-c^2 v_{xx})}. $$ The red term has something to do with the Minkowski metric.

$\endgroup$
  • $\begingroup$ Unfortunately, I don't know what Minkowski metric is. $\endgroup$ – sequence Jun 2 '16 at 21:43
  • $\begingroup$ @sequence: strange, it looked like a relativistic preservation of energy. $\endgroup$ – Jack D'Aurizio Jun 2 '16 at 21:45
  • $\begingroup$ This is a first course in PDEs. @Jack D'Aurizio $\endgroup$ – sequence Jun 2 '16 at 21:49
  • $\begingroup$ I think this most commonly comes up as the energy integral of a dissipative wave equation $\endgroup$ – Triatticus Jun 2 '16 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.