Answer to Question #153501 in Differential Equations for Ashweta Padhan

Question

Reduce the equation into canonical form and find the solution

xu_xx +2(xy)^1/2 u_xy +yu_yy -u_x =0

Answer 1

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

Characteristic polynomial is given by,

$\frac{dy}{dx}=\frac{b}{a}=(y/x)^{1/2}\\\;\\ \Rightarrow \frac{2}{3}y^{3/2}-\frac{2}{3}x^{3/2}=c \;\;[c\;is \; a \;constant]\\$

Define,

$\eta (x,y)=\frac{2}{3}y^{3/2}-\frac{2}{3}x^{3/2}\\ \eta_x=-\sqrt x\\ \eta_y=\sqrt y$

Choose,

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

$w(\zeta,\eta)=u(x,y)\\ \therefore u_x=w_\zeta \zeta_y+w_\eta \eta_x=-w_\zeta-w_\eta\sqrt x\\ u_y=\sqrt y\\ u_{xx}=w_{\zeta \zeta}+\sqrt x(x+1) w_{\zeta\eta}+w_{\eta\eta}-\frac{1}{2\sqrt x}w_\eta\\ u_{xy}=-\sqrt yw_{\zeta\eta}-\sqrt{xy}w_{\eta\eta}\\ u_{yy}=yw_{\eta\eta}+\frac{1}{2\sqrt y}w_\eta$

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

$\pmb{xw_{\zeta\zeta}+(x\sqrt x +x^2\sqrt x -2y\sqrt x )w_{\zeta\eta}+(x-y)^2w_{\eta\eta}+\frac{\sqrt y +\sqrt x}{2}w_\eta+w_\zeta=0 }$