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Show 28 viz: solutions exist, dv du We derive can dv for a more instead of (2Fuv) and dv + =0 (Fvv) formulation symmetric du 0 = noting that the by solving resulting (*) line two equations du from differ (**) (FUV±UV2_FUUFVV) the only by + dv Fvv multiplicative = 0 constants Fuu = Fvv Taking of one each of the solutions we obtain two line Fuv2-FuuFvv=0 (the the equations (FUV+/FUv2-FUUFVV) (FUV+/FUv2-FUUFVV) du (Fvv) (Fuu) du + + which of case the contain degeneracies when even parabolic cylender). In the dv = latter 0 have case we zero but lines which are At two the no 0 dv the the normals defined as same saddle to du (Fuu) + dv (Fuv) = 0 du (Fuv) + dv (Fvv) = 0 since line point the Fuv2=FuuFvv. itself the intersecting gradient level is curves are the still |
