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.

  • $\begingroup$ 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

$$\frac{1}{2}\frac{d}{dt}\left[v_t^2+c^2 v_x^2\right]-\frac{d}{dx}c^2 v_t v_x = v_t v_{tt}+ c^2 v_x v_{xt}-c^2 v_x v_{tx}-c^2 v_t v_{xx}$$ 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.

  • $\begingroup$ 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

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