x2Uxx-xyUxy+y2Uyy-xUx+yUy=8y/x (x≠0,y≠0) find the general solution.
"\\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})"
Comments
Dear Luna, thank you for correcting us.
The discriminant formula isn't (b^2)-4ac? Because if we did it like that the discriminant would be -3(x^2)(y^2). I don't get that
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