 Reverse Loewner flow

ID: reverse-loewner-flow

For a continuous driver

U

and terminal time

T

, the reverse Loewner flow solves

\partial_{s} h_{s} (z) = - \frac{2}{h _{s} ( z ) - ( U _{T - s} - U _{T} )}, h_{0} (z) = z .

It satisfies

h_{T} (z) = g_{T}^{- 1} (U_{T} + z) - U_{T}

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