A = 1 , B = − 1 , C = 1 A=1, B=-1, C=1 A = 1 , B = − 1 , C = 1
B 2 − 4 A C = 1 − 4 = − 3 < 0 B^2-4AC=1-4=-3<0 B 2 − 4 A C = 1 − 4 = − 3 < 0 The given partial differential equation is elliptic.
The characteristic equations are given by
d y d x = B − B 2 − 4 A C 2 A = − 1 − − 3 2 ( 1 ) \dfrac{dy}{dx}=\dfrac{B-\sqrt{B^2-4AC}}{2A}=\dfrac{-1-\sqrt{-3}}{2(1)} d x d y = 2 A B − B 2 − 4 A C = 2 ( 1 ) − 1 − − 3
= − 1 − i 3 2 =\dfrac{-1-i\sqrt{3}}{2} = 2 − 1 − i 3 and
d y d x = B + B 2 − 4 A C 2 A = − 1 + − 3 2 ( 1 ) \dfrac{dy}{dx}=\dfrac{B+\sqrt{B^2-4AC}}{2A}=\dfrac{-1+\sqrt{-3}}{2(1)} d x d y = 2 A B + B 2 − 4 A C = 2 ( 1 ) − 1 + − 3
= − 1 + i 3 2 =\dfrac{-1+i\sqrt{3}}{2} = 2 − 1 + i 3
Integrating these two equations, we get
y = x ( − 1 − i 3 2 ) + c 1 y=x(\dfrac{-1-i\sqrt{3}}{2})+c_1 y = x ( 2 − 1 − i 3 ) + c 1 and
y = x ( − 1 + i 3 2 ) + c 2 y=x(\dfrac{-1+i\sqrt{3}}{2})+c_2 y = x ( 2 − 1 + i 3 ) + c 2
Assume that
ξ = y + x ( 1 + i 3 2 ) \xi=y+x(\dfrac{1+i\sqrt{3}}{2}) ξ = y + x ( 2 1 + i 3 )
η = y + x ( 1 − i 3 2 ) \eta=y+x(\dfrac{1-i\sqrt{3}}{2}) η = y + x ( 2 1 − i 3 ) The transformation
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 = y + x \alpha=y+x α = y + x
β = 3 2 x \beta=\dfrac{\sqrt{3}}{2}x β = 2 3 x
α x = 1 , α y = 1 \alpha_x=1, \alpha_y=1 α x = 1 , α y = 1
α x x = 0 , α x y = 0 , α y y = 0 \alpha_{xx}=0, \alpha_{xy}=0, \alpha_{yy}=0 α xx = 0 , α x y = 0 , α yy = 0
β x = 3 2 , β y = 0 \beta_x=\dfrac{\sqrt{3}}{2}, \beta_y=0 β x = 2 3 , β y = 0
β x x = 0 , β x y = 0 , β y y = 0 \beta_{xx}=0, \beta_{xy}=0, \beta_{yy}=0 β xx = 0 , β x y = 0 , β yy = 0
A = 1 , B = − 1 , C = 1 , D = 1 , E = 0 , F = 0 , G = 0 A=1, B=-1, C=1,D=1, E=0, F=0, G=0 A = 1 , B = − 1 , C = 1 , D = 1 , E = 0 , F = 0 , G = 0
A ˉ = A α x 2 + B α x α y + C α y 2 = 1 − 1 + 1 = 1 \bar{A}=A\alpha_x^2+B\alpha_x\alpha_y+C\alpha_y^2=1-1+1=1 A ˉ = A α x 2 + B α x α y + C α y 2 = 1 − 1 + 1 = 1
B ˉ = 2 A α x β x + B ( α x β y + α y β x ) + 2 C α y β y \bar{B}=2A\alpha_x\beta_x+B(\alpha_x\beta_y+\alpha_y\beta_x)+2C\alpha_y\beta_y B ˉ = 2 A α x β x + B ( α x β y + α y β x ) + 2 C α y β y
= 2 ( 1 ) ( 3 2 ) − 1 ( 0 + ( 1 ) 3 2 ) + 2 ( 1 ) ( 0 ) = 3 2 =2(1)(\dfrac{\sqrt{3}}{2})-1(0+(1)\dfrac{\sqrt{3}}{2})+2(1)(0)=\dfrac{\sqrt{3}}{2} = 2 ( 1 ) ( 2 3 ) − 1 ( 0 + ( 1 ) 2 3 ) + 2 ( 1 ) ( 0 ) = 2 3
C ˉ = A β x 2 + B β x β y + 2 C α y β y = \bar{C}=A\beta_x^2+B\beta_x\beta_y+2C\alpha_y\beta_y= C ˉ = A β x 2 + B β x β y + 2 C α y β y =
= 3 4 + 0 + 0 = 3 4 =\dfrac{3}{4}+0+0=\dfrac{3}{4} = 4 3 + 0 + 0 = 4 3
D ˉ = A α x x + B α x y + C α y y + D α x + E α y = 1 \bar{D}=A\alpha_{xx}+B\alpha_{xy}+C\alpha_{yy}+D\alpha_x+E\alpha_y=1 D ˉ = A α xx + B α x y + C α yy + D α x + E α y = 1
E ˉ = A β x x + B β x y + C β y y + D β x + E β y = 3 2 \bar{E}=A\beta_{xx}+B\beta_{xy}+C\beta_{yy}+D\beta_x+E\beta_y=\dfrac{\sqrt{3}}{2} E ˉ = A β xx + B β x y + C β yy + D β x + E β y = 2 3
F ˉ = F = 0 , G ˉ = G = 0 \bar{F}=F=0, \bar{G}=G=0 F ˉ = F = 0 , G ˉ = G = 0
u α α + 3 2 u α β + 3 2 u β β + u α + 3 2 u β = 0 u_{\alpha\alpha}+\dfrac{\sqrt{3}}{2}u_{\alpha\beta}+\dfrac{\sqrt{3}}{2}u_{\beta\beta}+u_{\alpha}+\dfrac{\sqrt{3}}{2}u_\beta=0 u αα + 2 3 u α β + 2 3 u ββ + u α + 2 3 u β = 0
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