Suppose all second partial derivatives of [itex]F = F (x, y)[/itex] are continuous and [itex]F_{xx} + F_{yy} = 0[/itex] on an open rectangle [itex]R[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Show that [tex]F_ydx - F_xdy = 0[/tex] is exact on [itex]R[/itex], and therefore there’s a function [itex]G[/itex] such that

[tex]G_x = −F_y[/tex] and [tex]Gy = F_x[/tex] in [itex]R[/itex].

≈≈≈≈≈≈≈≈To prove that [itex]F_ydx + F_xdy = 0[/itex] is exact on [itex]R[/itex],Could you help me prove the existence of [itex]G[/itex] ? Thank you.....

I have [tex]F_{xx} + F_{yy} = 0[/tex]

which is [tex]F_{xx}=-F_{yy}[/tex]

Integrating both sides and cancel out the constants I obtain

[tex]F_x=-F_y[/tex]

This proves the function [itex]F_ydx - F_xdy = 0[/itex] is exact on [itex]R[/itex]

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# Proof of function existence

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