
#235
Oct1211, 01:38 AM

P: 621

An Analytical Endeavor:
Equation: [tex]{Cot}{(}{\theta}{)}{=}{[}\frac{{\partial}^{2}{\theta}}{{\partial x}^{2}}{+}\frac{{\partial}^{2}{\theta}}{{\partial y}^{2}}{]}[/tex]  (1) We look for a solution of the form: [tex]\theta{=}{A}{(}{}{ln}{Sin}{x}{}\frac{1}{2}{x}^{2}{)}{+}{B}{(}{}{ln}{Sin}{y}{}\frac{1}{2}{y}^{2}{)}{+}{\lambda}{x}^{2}[/tex] (2) A and B are constants Lambda is a parameter independent of x and y. Using the above trial in relation (1) we have [tex]{A}{Cot}^{2}{x}{+}{B}{cot}^{2}{y}{+}{2}{\lambda} {=}{Cot}\theta[/tex] (3) We may eliminate lambda between (2) and (3) to find Particular Integrals[surfaces theta=f(x,y)] [Even if one has to use numerical methods, the equation will be an ordinary one instead of a“differential Equation”] Instead of (2) one may use a trial of the form below, to obtain a greater variety of particular integrals: [tex]\theta{=}{A}{f1}{(}{x}{)}{+}{B}{f2}{(}{y}{)}{+}{\lambda}{x}^{2}[/tex] (4) f1 and f1 are known ,arbitrary functionswell behaved ones of course,in terms of continuity ,differentiability etc. The homogeneous part of equation (1) is simply Laplace’s equation in two dimensions. We have the familiar solutions. [This posting will be revised] 



#236
Oct1211, 07:38 PM

Mentor
P: 16,477

So plug your transformation equation back into the metric and calculate the curvature.



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