Question #140577
The differential equation
(1-x^2)(∂^2z/∂x^2 -2xy[∂^2z/(∂x.∂y)] + (1-y^2)(∂^2z/∂y^2) + 2(∂z/∂x) +3y(∂z/∂y)=0, is elliptic for (x,y) outside the circle x^2 + y^2=1
True or false with correct explanation
1
Expert's answer
2020-10-27T18:04:04-0400

Given differential equation is

(1x2)(2z/x2)2xy[2z/(x.y)]+(1y2)(2z/y2)+2(z/x)+3y(z/y)=0(1-x^2)(∂^2z/∂x^2) -2xy[∂^2z/(∂x.∂y)] + (1-y^2)(∂^2z/∂y^2) + 2(∂z/∂x) +3y(∂z/∂y)=0 .

Compare it with

A(2z/x2)+B[2z/(x.y)]+C(2z/y2)+D(z/x)+E(z/y)=0A(∂^2z/∂x^2)+B[∂^2z/(∂x.∂y)] + C(∂^2z/∂y^2) + D(∂z/∂x) +E(∂z/∂y)=0 , we have

A=1x2,B=2xy,C=1y2A= 1-x^2, B = -2xy, C =1-y^2 .

Now, B24AC=4x2y24(1x2)(1y2)=4(x2+y21)B^2-4AC = 4x^2 y^2 - 4(1-x^2)(1-y^2) = 4(x^2+y^2-1)

For (x,y) outside the circle x2+y2=1x^2 + y^2=1 , we have x2+y2>1x^2 + y^2 > 1 .

    B24AC>0\implies B^2-4AC > 0 .

Hence, the given differential equation is Hyperbolic equation.

Thus, given statement is false.


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