Gauss curvature of surface represented by real part of an analytic function-
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From: JEMebius on 8 Jun 2010 14:45 gudi wrote: > If u(x,y) ,v(x,y) are real/imaginary parts of an analytical function > f( x + i y), under what conditions does surface Z = u(x,y) ( in Monge > form ) have a positive Gauss curvature K? For most of the functions I > find K is negative. Is it always so? Please give examples when K >0. > TIA, > > Narasimham The real and imaginary parts u and v of an analytic function u + vi = f(x + yi) are two-dimensional harmonic functions u(x, y) and v(x, y) of x and y, i.e. they satisfy the 2D Laplace equation d^2u/dx^2 + d^2u/dy^2 = 0; analogously for v(x, y). Harmonic functions have the mean-value property: the mean value of U(x, y) taken over a circle with centre (X, Y) within the domain of U equals U(X, Y), the value at the centre. This implies that the 3D graph z = u(x, y) cannot have positive Gaussian curvature at any point. For instance: consider the neighbourhood of a local maximum or minimum. See http://en.wikipedia.org/wiki/Harmonic_function , in particular http://en.wikipedia.org/wiki/Harmonic_function#Mean_value_property Ciao: Johan E. Mebius
From: gudi on 10 Jun 2010 15:49 On Jun 8, 11:45 pm, JEMebius <jemeb...(a)xs4all.nl> wrote: > gudi wrote: > > If u(x,y) ,v(x,y) are real/imaginary parts of an analytical function > > f( x + i y), under what conditions does surface Z = u(x,y) ( in Monge > > form ) have a positive Gauss curvature K? For most of the functions I > > find K is negative. Is it always so? Please give examples when K >0. > > TIA, > > > Narasimham > > The real and imaginary parts u and v of an analytic function u + vi = f(x + yi) are > two-dimensional harmonic functions u(x, y) and v(x, y) of x and y, i.e. they satisfy the > 2D Laplace equation d^2u/dx^2 + d^2u/dy^2 = 0; analogously for v(x, y). > > Harmonic functions have the mean-value property: > the mean value of U(x, y) taken over a circle with centre (X, Y) within the domain of U > equals U(X, Y), the value at the centre. The circle is the curved loop of intersection between U(x,y) and a cylinder with axis parallel to z-axis? > This implies that the 3D graph z = u(x, y) cannot have positive Gaussian curvature at any > point. For instance: consider the neighbourhood of a local maximum or minimum. > > See http://en.wikipedia.org/wiki/Harmonic_function, in particular http://en.wikipedia.org/wiki/Harmonic_function#Mean_value_property > > Ciao: Johan E. Mebius Thanks, Narasimham
From: gudi on 11 Jun 2010 07:58
On Jun 8, 11:45 pm, JEMebius <jemeb...(a)xs4all.nl> wrote: > gudi wrote: > > If u(x,y) ,v(x,y) are real/imaginary parts of an analytical function > > f( x + i y), under what conditions does surface Z = u(x,y) ( in Monge > > form ) have a positive Gauss curvature K? For most of the functions I > > find K is negative. Is it always so? Please give examples when K >0. > > TIA, > > > Narasimham > > The real and imaginary parts u and v of an analytic function u + vi = f(x + yi) are > two-dimensional harmonic functions u(x, y) and v(x, y) of x and y, i.e. they satisfy the > 2D Laplace equation d^2u/dx^2 + d^2u/dy^2 = 0; analogously for v(x, y). > > Harmonic functions have the mean-value property: > the mean value of U(x, y) taken over a circle with centre (X, Y) within the domain of U > equals U(X, Y), the value at the centre. > > This implies that the 3D graph z = u(x, y) cannot have positive Gaussian curvature at any > point. For instance: consider the neighbourhood of a local maximum or minimum. > > Seehttp://en.wikipedia.org/wiki/Harmonic_function, in particularhttp://en..wikipedia.org/wiki/Harmonic_function#Mean_value_property > > Ciao: Johan E. Mebius Even after seeing the above references,I cannot find from them how the mean value property implies to a conclusion that every 2 dimensional harmonic function z = u(x, y) should have a negative Gaussian curvature. Can you please help explain? Thanking in advance, Best Regards, Narasimham |