Question #153501

Reduce the equation into canonical form and find the solution

xuxx +2(xy)1/2 uxy +yuyy -ux =0


1
Expert's answer
2021-01-04T18:53:50-0500

xuxx+2(xy)1/2uxy+yuyyux=0    ...(1)a=x,b=(xy)1/2,c=yb2ac=0(1)    is  parabolic.xu_{xx}+2(xy)^{1/2}u_{xy}+yu_{yy}-u_x=0 \;\;...\pmb{(1)}\\ a=x, b=(xy)^{1/2},c=y\\ \Rightarrow b^2-ac=0\\ \Rightarrow (1) \;\; is \; parabolic.\\

Characteristic polynomial is given by,

dydx=ba=(y/x)1/2  23y3/223x3/2=c    [c  is  a  constant]\frac{dy}{dx}=\frac{b}{a}=(y/x)^{1/2}\\\;\\ \Rightarrow \frac{2}{3}y^{3/2}-\frac{2}{3}x^{3/2}=c \;\;[c\;is \; a \;constant]\\


Define,

η(x,y)=23y3/223x3/2ηx=xηy=y\eta (x,y)=\frac{2}{3}y^{3/2}-\frac{2}{3}x^{3/2}\\ \eta_x=-\sqrt x\\ \eta_y=\sqrt y

Choose,

ζ(x,y)=xJ=ζxηxζyηy=x>0\zeta (x,y)=-x\\ \Rightarrow J=\zeta_x\eta_x-\zeta_y\eta_y=\sqrt x >0


w(ζ,η)=u(x,y)ux=wζζy+wηηx=wζwηxuy=yuxx=wζζ+x(x+1)wζη+wηη12xwηuxy=ywζηxywηηuyy=ywηη+12ywηw(\zeta,\eta)=u(x,y)\\ \therefore u_x=w_\zeta \zeta_y+w_\eta \eta_x=-w_\zeta-w_\eta\sqrt x\\ u_y=\sqrt y\\ u_{xx}=w_{\zeta \zeta}+\sqrt x(x+1) w_{\zeta\eta}+w_{\eta\eta}-\frac{1}{2\sqrt x}w_\eta\\ u_{xy}=-\sqrt yw_{\zeta\eta}-\sqrt{xy}w_{\eta\eta}\\ u_{yy}=yw_{\eta\eta}+\frac{1}{2\sqrt y}w_\eta


Now from (1), the required canonical form is,

xwζζ+(xx+x2x2yx)wζη+(xy)2wηη+y+x2wη+wζ=0\pmb{xw_{\zeta\zeta}+(x\sqrt x +x^2\sqrt x -2y\sqrt x )w_{\zeta\eta}+(x-y)^2w_{\eta\eta}+\frac{\sqrt y +\sqrt x}{2}w_\eta+w_\zeta=0 }



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