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I'm running a potential flow solver, and I have values of the velocity potential $\phi$ on the boundaries. I'd like to compute the spatial evolution kinetic energy in the system. In particular, I'd like to find

$$T(x) = \int \int_{-\infty}^{\eta(x,t)} \frac{1}{2}(\phi_x^2+\phi_z^2) \ dz \ dt.$$

Can I calculate this integral using just boundary values of $\phi$ and its derivatives?

Note, the more commonly used measure is the temporal evolution of the kinetic energy,

$$Ke(t) = \int \int_{-\infty}^{\eta(x,t)}\frac{1}{2}(\phi_x^2+\phi_z^2) \ dz \ dx = \int_{\partial A}\phi(\phi_z \ dx - \phi_x \ dz), $$ with $\partial A$ the boundary of the domain, and we have employed a Green's identity to for $Ke(t)$ based just on values on the boundary.

A solution is to solve for $\phi$ in the interior and then simply compute the 2d integral for $T$, but the boundary is evolving in space and time, so that I'd prefer not to solve Laplace's equation at each time step.

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  • $\begingroup$ I'm assuming that on the right-hand-most side of your kinetic equation, when you write $y$, you mean $z$? $\endgroup$ – Geoff Oxberry Aug 18 '15 at 6:57
  • $\begingroup$ @GeoffOxberry Yes, thanks for pointing this out! $\endgroup$ – Nick P Aug 18 '15 at 7:02
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Can I calculate this integral using just boundary values of $\phi$ and its derivatives?

I hate to be the bearer of bad news, but probably not, no. I detail reasons why below:


It looks like you are invoking the Divergence Theorem when you convert the integral for kinetic energy from a volume integral to a surface integral.

You can't use similar logic here to convert the "volume" integral for the spatial evolution kinetic energy to a surface integral, because the form of your integral in two-dimensions doesn't satisfy the hypotheses of the Divergence Theorem. Also, attempting to apply the Divergence Theorem in a single-dimension (that is, the 2-D version only along the $z$ direction) doesn't really achieve what you want because it reduces to the Fundamental Theorem of Calculus, which won't be helpful. (If you already had the antiderivative of your argument, you wouldn't be asking this question.)

We can also appeal to basic intuition here: the Divergence Theorem is saying that the total change of a quantity in a volume is equal to changes induced by fluxes through the boundary surface. Here, you're not integrating over the whole surface, so some flux information is going to be missing.

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  • $\begingroup$ A coherent explanation of what I (sadly) suspected! I was hoping I was unaware of some manipulation. Alas, I'll take the long way. Thanks for the help. $\endgroup$ – Nick P Aug 18 '15 at 7:50

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