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Consider $(M,g_{t})$ equipped with a $1$-parameter family of warped metrics for real parameter $t>0$

$$g_{t} = \frac{1}{\phi_t(u)^{2}}\ du^{2} + \phi_t(u)\ dv^{2}$$

and suppose that the warping function obeys the linear equation

$$ t \frac{\partial^2}{\partial t^2}\phi_t(u)=-u \frac{\partial}{\partial u}\phi_t(u) $$

for warping function $$ \phi_t(u)=\exp\big(t/\log u \big) $$

Does this imply that $g_t$ satisfies the same linear PDE as well?

I suspect $g_t$ obeys the same PDE but am not sure what to do with the $v$ variable in the metric and where the $v$ variable comes into the PDE. I believe $t$ should act as a time parameter and $u,v$ acting as space variables.

J. Zimmerman
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  • Might be missing something silly here, but are the variables $u,v$ time-dependent? If not, you could just try to derive and see if stuff work out as expected. If they are, I'm not sure we can answer that question without knowing what the variable's evolution equations are. – NG_ May 30 '24 at 11:21

1 Answers1

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No this cannot be true due to the $1/\phi_t(u)$ factor. The reason boils down to the fact that the reciprocal satisfies a drastically different pde which is well-posed however the factor $\phi_t(u)$ resembles a kind of backwards type heat equation (albeit with space and time swapped), which is ill-posed and of course not physically practical (if the order of the time variable is greater than the space variable, then physically you can violate causality for ex.) If there was more homogeneity in $g_t$ i.e.

$$ g_t= \phi_t(u)du^2+\phi_t(u)dv^2 $$

then it would be possible as both warping functions satisfy the same pde, although again, the pde is not especially well-posed so it working with it might be awkward.

Aside: The pde's are both linear but one is ill posed and non-parabolic (due to the negative sign) and the other is linear and parabolic.

J. Zimmerman
  • 842