Answer to Question #158607 in Differential Equations for Cypress

Question

x²U_xx-xyU_xy+y²U_yy-xU_x+yU_y=8y/x (x≠0,y≠0) find the general solution.

Answer 1

$\xi=\xi(x,y),\eta=\eta(x,y)$

$w(\xi,\eta)=u(x(\xi,\eta),y(\xi,\eta))\implies u(x,y)=w(\xi(x,y),\eta(x,y))$

$aw_{\xi\xi}+bw_{\xi\eta}+cw_{\eta\eta}=\varphi(\xi,\eta,w,w_{\xi},w_{\eta})$

$A=x^2,B=-xy,C=y^2$

$B^2-4AC=x^2y^2-4x^2y^2=-3x^2y^2<0$

This is an elliptic equation.

The roots of the characteristic polynomial:

$\lambda_1=\frac{B-i\sqrt{4AC-B^2}}{2A}=\frac{-xy-i\sqrt{4x^2y^2-x^2y^2}}{2x^2}=-\frac{y(1+i\sqrt{3})}{2x}$

$\lambda_2=\frac{B+i\sqrt{4AC-B^2}}{2A}=-\frac{y(1-i\sqrt{3})}{2x}$

$\frac{dy}{dx}=-\frac{y(1+i\sqrt{3})}{2x}\implies lny=-\frac{1+i\sqrt{3}}{2}lnx+lnc_1$

$c_1=yx^{(1+i\sqrt{3})/2}=\alpha$

$\frac{dy}{dx}=-\frac{y(1-i\sqrt{3})}{2x}$

$c_2=yx^{(1-i\sqrt{3})/2}=\beta$

$\xi=\frac{\alpha+\beta}{2}=y\sqrt{x}cos(\sqrt{3}lnx/2)$

$\eta=\frac{\alpha-\beta}{2i}=y\sqrt{x}sin(\sqrt{3}lnx/2)$

$\xi_x=\frac{y}{2\sqrt{x}}(cos(\sqrt{3}lnx/2)-\sqrt{3}sin(\sqrt{3}lnx/2))$

$\eta_x=\frac{y}{2\sqrt{x}}(sin(\sqrt{3}lnx/2)+\sqrt{3}cos(\sqrt{3}lnx/2))$

$\xi_y=\sqrt{x}cos(\sqrt{3}lnx/2)$

$\eta_y=\sqrt{x}sin(\sqrt{3}lnx/2)$

$\xi_{xx}=-\frac{y}{4x^{3/2}}(cos(\sqrt{3}lnx/2)-\sqrt{3}sin(\sqrt{3}lnx/2))+$

$+\frac{y\sqrt{3}}{4x^{3/2}}(-sin(\sqrt{3}lnx/2)-\sqrt{3}cos(\sqrt{3}lnx/2))=$

$=\frac{y}{2x^{3/2}}(cos(\sqrt{3}lnx/2)+\sqrt{3}sin(\sqrt{3}lnx/2))$

$\eta_{xx}=\frac{y}{2x^{3/2}}(\sqrt{3}cos(\sqrt{3}lnx/2)-sin(\sqrt{3}lnx/2))$

$\xi_{yy}=\eta_{yy}=0$

$\xi_{xy}=\frac{1}{2\sqrt{x}}(cos(\sqrt{3}lnx/2)-\sqrt{3}sin(\sqrt{3}lnx/2))$

$\eta_{xy}=\frac{1}{2\sqrt{x}}(sin(\sqrt{3}lnx/2)+\sqrt{3}cos(\sqrt{3}lnx/2))$

Let: $k=cos(\sqrt{3}lnx/2), m=sin(\sqrt{3}lnx/2)$

$u_x=w_{\xi}\xi_x+w_{\eta}\eta_x$

$u_x=\frac{y}{2\sqrt{x}}(k-m\sqrt{3})w_{\xi}+\frac{y}{2\sqrt{x}}(m+k\sqrt{3})w_{\eta}$

$u_y=w_{\xi}\xi_y+w_{\eta}\eta_y$

$u_y=k\sqrt{x}w_{\xi}+m\sqrt{x}w_{\eta}$

$u_{xx}=w_{\xi\xi}\xi^2_x+2w_{\xi\eta}\xi_x\eta_x+w_{\eta\eta}\eta^2_x+w_{\xi}\xi_{xx}+w_{\eta}\eta_{xx}$

$u_{xx}=\frac{y^2}{4x}(k-m\sqrt{3})^2w_{\xi\xi}+\frac{y^2}{2x}(k-m\sqrt{3})(m+k\sqrt{3})w_{\xi\eta}+$

$+\frac{y^2}{4x}(m+k\sqrt{3})^2w_{\eta\eta}+\frac{y}{2x^{3/2}}(k+m\sqrt{3})w_{\xi}+\frac{y}{2x^{3/2}}(k\sqrt{3}-m)w_{\eta}$

$u_{yy}=w_{\xi\xi}\xi^2_y+2w_{\xi\eta}\xi_y\eta_y+w_{\eta\eta}\eta^2_y+w_{\xi}\xi_{yy}+w_{\eta}\eta_{yy}$

$u_{yy}=xk^2w_{\xi\xi}+2xkmw_{\xi\eta}+xm^2w_{\eta\eta}$

$u_{xy}=w_{\xi\xi}\xi_x\xi_y+w_{\xi\eta}(\xi_x\eta_y+\xi_y\eta_x)+w_{\eta\eta}\eta_x\eta_y+w_{\xi}\xi_{xy}+w_{\eta}\eta_{xy}$

$u_{xy}=\frac{y}{2}k(k-m\sqrt{3})w_{\xi\xi}+\frac{y}{2}(m(k-m\sqrt{3})+k(m+k\sqrt{3}))w_{\xi\eta}+$

$+\frac{y}{2}m(m+k\sqrt{3})w_{\eta\eta}+\frac{1}{2\sqrt{x}}(k-m\sqrt{3})w_{\xi}+\frac{1}{2\sqrt{x}}(m+k\sqrt{3})w_{\eta}$

Substituting the give values into the initial equation:

$\frac{3}{4}xy^2(w_{\xi\xi}+w_{\eta\eta})=\frac{8y}{x}-ym\sqrt{x}(w_{\xi}\sqrt{3}-w_{\eta})$

We also have:

$\xi^2+\eta^2=xy^2$

$x=e^{2tan^{-1}(\eta/\xi)/\sqrt{3}}$

$\frac{y}{x}=\frac{\sqrt{\xi^2+\eta^2}}{e^{tan^{-1}(\eta/\xi)/\sqrt{3}}}$

So, canonical form of equation:

$w_{\xi\xi}+w_{\eta\eta}=\frac{32}{3e^{tan^{-1}(\eta/\xi)/\sqrt{3}}\sqrt{\xi^2+\eta^2}}-\frac{4\eta}{3(\xi^2+\eta^2)}(w_{\xi}\sqrt{3}-w_{\eta})$