xuxx+2(xy)1/2uxy+yuyy−ux=0...(1)(1)a=x,b=(xy)1/2,c=y⇒b2−ac=0⇒(1)isparabolic.
Characteristic polynomial is given by,
dxdy=ab=(y/x)1/2⇒32y3/2−32x3/2=c[cisaconstant]
Define,
η(x,y)=32y3/2−32x3/2ηx=−xηy=y
Choose,
ζ(x,y)=−x⇒J=ζxηx−ζyηy=x>0
w(ζ,η)=u(x,y)∴ux=wζζy+wηηx=−wζ−wηxuy=yuxx=wζζ+x(x+1)wζη+wηη−2x1wηuxy=−ywζη−xywηηuyy=ywηη+2y1wη
Now from (1), the required canonical form is,
xwζζ+(xx+x2x−2yx)wζη+(x−y)2wηη+2y+xwη+wζ=0xwζζ+(xx+x2x−2yx)wζη+(x−y)2wηη+2y+xwη+wζ=0
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