Question

yZxx+(x+y)Zxy+xZyy

Answer 1

Solution

a=y

b=x+y

c=x

Hence;

$b^2-4ac=x^2-2xy+y^2=(x-y)^2>0$

So the PDE is hyperbolic and we solve as follows;

Take ;

$ar_x^2+br_xr_y+cr_y^2=0$

$yr_x^2+(x+y)r_xr_y+xr_y^2=0$

Factor;

$r_x(yr_x+xr_y)+r_y(yr_x+xr_y)$ =0

$(yr_x+xr_y)(r_x+r_y)=0$

Since s also satisfies the same equation ,we take;

$yr_x+xr_y=0$

Which gives ;

$r=x^2-y^2$

$s_x+s_y=0$

Which gives;

$s=x-y$

Calculate the first derivatives;

$Z_x=2xZ_r+Z_s$

$Z_y=-2yZ_r-Z_s$

Calculating the second derivatives;

$Z_{xx}=4x^2Z_{rr}+4xZ_{rs}+Z_{ss}+2Z_r$

$Z_{xy}=-4xyZ_{rr}-2xZ_{rs}-2yZ_{rs}-Z_{ss}$

$Z_{yy}=4y^2Z_{rr}+4yZ_{rs}+Z_{ss}-2Z_r$

Substitute to the given questions to obtain;

$4x^2yZ_{rr}+4xyZ_{rs}+yZ_{ss}+2yZ_r-4x^2yZ_{rr}-2x^2Z_{rs}-2yxZ_{rs}-xZ_{ss}-4xy^2Z_{rr}-2xyZ_{rs}-2y^2Z_{rs}-yZ_{ss}+4y^2xZ_{rr}+4yxZ_{rs}+xZ_{ss}-2xZ_r$

Simplify to obtain;

$2xZ_r-2yZ_r+2x^2Z_{rs}+2y^2Z_{rs}-4xyZ_{rs}$

Divide by 2 and factorise to obtaining;

$(x-y)^2Z_{rs}+(x-y)Z_r=0$

But x-y=s

By substitution;

$s^2Z_{rs}+sZ_r=0$

Rewrite as;

$Z_{rs}+\frac1sZ_r=0$