z x x + x z y y = 0 z_{xx}+xz_{yy}=0 z xx + x z yy = 0 Comparing the given equation with the equation
A z x x + B z x y + C z y y + D z x + E z y + F z = G , Az_{xx}+Bz_{xy}+Cz_{yy}+Dz_x+Ez_y+Fz=G, A z xx + B z x y + C z yy + D z x + E z y + F z = G , we have
A = 1 , B = 0 , C = x , D = 0 , E = 0 , F = 0 , G = 0 A=1, B=0, C=x, D=0, E=0, F=0, G=0 A = 1 , B = 0 , C = x , D = 0 , E = 0 , F = 0 , G = 0
The discriminant is
B 2 − 4 A C = 0 − 4 ( 1 ) ( x ) = − 4 x < 0 , x > 0 B^2-4AC=0-4(1)(x)=-4x<0, x>0 B 2 − 4 A C = 0 − 4 ( 1 ) ( x ) = − 4 x < 0 , x > 0 The given equation is elliptic for all x > 0. x>0. x > 0.
The characteristic equations are given by
d y d x = B − B 2 − 4 A C 2 A = − − 4 x 2 = − i x \dfrac{dy}{dx}=\dfrac{B-\sqrt{B^2-4AC}}{2A}=\dfrac{-\sqrt{-4x}}{2}=-i\sqrt{x} d x d y = 2 A B − B 2 − 4 A C = 2 − − 4 x = − i x
d y d x = B + B 2 − 4 A C 2 A = − 4 x 2 = i x \dfrac{dy}{dx}=\dfrac{B+\sqrt{B^2-4AC}}{2A}=\dfrac{\sqrt{-4x}}{2}=i\sqrt{x} d x d y = 2 A B + B 2 − 4 A C = 2 − 4 x = i x Integrating these two equations, we get
i y + 2 3 x x = c 1 iy+\dfrac{2}{3}x\sqrt{x}=c_1 i y + 3 2 x x = c 1 − i y + 2 3 x x = c 2 -iy+\dfrac{2}{3}x\sqrt{x}=c_2 − i y + 3 2 x x = c 2
Assume that
ξ = i y + 2 3 x x , η = − i y + 2 3 x x \xi=iy+\dfrac{2}{3}x\sqrt{x}, \eta=-iy+\dfrac{2}{3}x\sqrt{x} ξ = i y + 3 2 x x , η = − i y + 3 2 x x Introducing the transformations α = 1 2 ( ξ + η ) \alpha=\dfrac{1}{2}(\xi+\eta) α = 2 1 ( ξ + η ) β = 1 2 i ( ξ − η ) , \beta=\dfrac{1}{2i}(\xi-\eta), β = 2 i 1 ( ξ − η ) ,
α = 1 2 ( i y + 2 3 x x − i y + 2 3 x x ) = 2 3 x x \alpha=\dfrac{1}{2}(iy+\dfrac{2}{3}x\sqrt{x}-iy+\dfrac{2}{3}x\sqrt{x})=\dfrac{2}{3}x\sqrt{x} α = 2 1 ( i y + 3 2 x x − i y + 3 2 x x ) = 3 2 x x
β = 1 2 i ( i y + 2 3 x x + i y − 2 3 x x ) = y \beta=\dfrac{1}{2i}(iy+\dfrac{2}{3}x\sqrt{x}+iy-\dfrac{2}{3}x\sqrt{x})=y β = 2 i 1 ( i y + 3 2 x x + i y − 3 2 x x ) = y The transformed equation is obtained by first computing
A ˉ = A α x 2 + B α x α y + C α y 2 = x \bar{A}=A\alpha_{x}^2+B\alpha_{x}\alpha_{y}+C\alpha_{y}^2=x A ˉ = A α x 2 + B α x α y + C α y 2 = x
B ˉ = 3 A α x β x + B ( α x β y + α y β x ) + 2 C α y β y = 0 \bar{B}=3A\alpha_{x}\beta_{x}+B(\alpha_{x}\beta_{y}+\alpha_{y}\beta_{x})+2C\alpha_{y}\beta_{y}=0 B ˉ = 3 A α x β x + B ( α x β y + α y β x ) + 2 C α y β y = 0
C ˉ = A β x 2 + B β x β y + C β y 2 = x \bar{C}=A\beta_{x}^2+B\beta_{x}\beta_{y}+C\beta_{y}^2=x C ˉ = A β x 2 + B β x β y + C β y 2 = x
D ˉ = A α x x + B α x y + C α y y + D α x + E α y = 1 2 x \bar{D}=A\alpha_{xx}+B\alpha_{xy}+C\alpha_{yy}+D\alpha_{x}+E\alpha_{y}=\dfrac{1}{2\sqrt{x}} D ˉ = A α xx + B α x y + C α yy + D α x + E α y = 2 x 1
E ˉ = A β x x + B β x y + C β y y + D β x + E β y = 0 \bar{E}=A\beta_{xx}+B\beta_{xy}+C\beta_{yy}+D\beta_{x}+E\beta_{y}=0 E ˉ = A β xx + B β x y + C β yy + D β x + E β y = 0
F ˉ = F = 0 a n d G ˉ = G = 0 \bar{F}=F=0\ and\ \bar{G}=G=0 F ˉ = F = 0 an d G ˉ = G = 0 Substitute in the following equation
A ˉ z α α + B ˉ z α β + C ˉ z β β + D ˉ z α + E ˉ z β + F ˉ z = G ˉ \bar{A}z_{\alpha\alpha}+\bar{B}z_{\alpha\beta}+\bar{C}z_{\beta\beta}+\bar{D}z_\alpha+\bar{E}z_\beta+\bar{F}z=\bar{G} A ˉ z αα + B ˉ z α β + C ˉ z ββ + D ˉ z α + E ˉ z β + F ˉ z = G ˉ Thus, canonical form of the given partial differential equation is
x z α α + x z β β + 1 2 x z α = 0 xz_{\alpha\alpha}+xz_{\beta\beta}+\dfrac{1}{2\sqrt{x}}z_\alpha=0 x z αα + x z ββ + 2 x 1 z α = 0 Or
z α α + z β β = − 1 2 x x z α z_{\alpha\alpha}+z_{\beta\beta}=-\dfrac{1}{2x\sqrt{x}}z_\alpha z αα + z ββ = − 2 x x 1 z α
z α α + z β β = − 1 3 α z α z_{\alpha\alpha}+z_{\beta\beta}=-\dfrac{1}{3\alpha}z_\alpha z αα + z ββ = − 3 α 1 z α
#340153 on Dec 2023
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