2 x 2 ∂ 2 z ∂ x 2 − 5 x y ∂ 2 z ∂ x ∂ y + 2 y 2 ∂ 2 z ∂ y 2 + 2 [ x ∂ z ∂ x + y ∂ z ∂ y ] = 0 2x^2\dfrac{\partial^2 z}{\partial x^2}-5xy\dfrac{\partial^2 z}{\partial x \partial y} + 2 y^2\dfrac{\partial^2 z}{\partial y^2} + 2[ x \dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y}]=0\\ 2 x 2 ∂ x 2 ∂ 2 z − 5 x y ∂ x ∂ y ∂ 2 z + 2 y 2 ∂ y 2 ∂ 2 z + 2 [ x ∂ x ∂ z + y ∂ y ∂ z ] = 0
Reduce the equation to the canonical form:
A = 2 x 2 B = − 2.5 x y C = 2 y 2 Δ = B 2 − A C = 6.25 x 2 y 2 − 4 x 2 y 2 = 2.25 y 2 A=2x^2\\
B=-2.5xy\\
C=2y^2\\
\Delta= B^2-AC=6.25x^2y^2-4x^2y^2=2.25 y^2 A = 2 x 2 B = − 2.5 x y C = 2 y 2 Δ = B 2 − A C = 6.25 x 2 y 2 − 4 x 2 y 2 = 2.25 y 2
So, we have hyperbolic type of equation.
Now let introduce a replacement.
2 x 2 ( d y ) 2 + 5 x y ( d x d y ) + 2 y 2 ( d x ) 2 = 0 ∣ d i v i d e b y ( d x ) 2 2 x 2 ( d y d x ) 2 + 5 x y ( d y d x ) + 2 y 2 = 0 y 1 ′ = − 5 x y + D 4 x 2 = − y 2 x = > η = 2 x + y y 2 ′ = − 5 x y − D 4 x 2 = − 2 y x = > ψ = x + 2 y 2x^2(dy)^2+5xy(dxdy)+2y^2(dx)^2=0|divide \space by \space (dx)^2\\
2x^2(\dfrac{dy}{dx})^2 + 5xy(\dfrac{dy}{dx}) + 2y^2=0\\
y'_1=\dfrac{-5xy+\sqrt{D}}{4x^2}=\dfrac{-y}{2x}=> \eta=2x+y\\
y'_2=\dfrac{-5xy-\sqrt{D}}{4x^2}=\dfrac{-2y}{x}=>\psi=x+2y\\ 2 x 2 ( d y ) 2 + 5 x y ( d x d y ) + 2 y 2 ( d x ) 2 = 0∣ d i v i d e b y ( d x ) 2 2 x 2 ( d x d y ) 2 + 5 x y ( d x d y ) + 2 y 2 = 0 y 1 ′ = 4 x 2 − 5 x y + D = 2 x − y => η = 2 x + y y 2 ′ = 4 x 2 − 5 x y − D = x − 2 y => ψ = x + 2 y
Now rewrite equation in new parameters:
∂ z ∂ x = 2 ∂ z ∂ η + ∂ z ∂ ψ ∂ z ∂ y = ∂ z ∂ η + 2 ∂ z ∂ ψ ∂ 2 z ∂ x 2 = 4 ∂ 2 z ∂ η 2 + 4 ∂ 2 z ∂ ψ ∂ η + ∂ 2 z ∂ ψ 2 ∂ 2 z ∂ y 2 = ∂ 2 z ∂ η 2 + 4 ∂ 2 z ∂ ψ ∂ η + 4 ∂ 2 z ∂ ψ 2 ∂ 2 z ∂ x ∂ y = 2 ∂ 2 z ∂ η 2 + 4 ∂ 2 z ∂ ψ ∂ η + 2 ∂ 2 z ∂ ψ 2 O u r e q u a t i o n : 8 η 2 − 8 η ψ + 2 ψ 2 9 ( 4 z η η + 4 z η ψ + z ψ ψ ) − 5 5 ψ η − 2 ψ 2 − 2 η 2 9 ( 2 ∂ 2 z ∂ η 2 + 4 ∂ 2 z ∂ ψ ∂ η + 2 ∂ 2 z ∂ ψ 2 ) + 8 ψ 2 − 8 η ψ + 2 η 2 9 ( ∂ 2 z ∂ η 2 + 4 ∂ 2 z ∂ ψ ∂ η + 4 ∂ 2 z ∂ ψ 2 ) + 2 [ 2 η − ψ 3 ( 2 ∂ z ∂ η + ∂ z ∂ ψ ) + 2 ψ − η 3 ( ∂ z ∂ η + 2 ∂ z ∂ ψ ) = 0 \dfrac{\partial z}{\partial x}=2\dfrac{\partial z}{\partial \eta}+\dfrac{\partial z}{\partial \psi}\\
\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial \eta}+2\dfrac{\partial z}{\partial \psi}\\
\dfrac{\partial^2 z}{\partial x^2}=4\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi
\partial \eta}+\dfrac{\partial^2 z}{\partial \psi^2}\\
\dfrac{\partial^2 z}{\partial y^2}=\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi \partial \eta}+4\dfrac{\partial^2 z}{\partial \psi^2}\\
\dfrac{\partial^2 z}{\partial x \partial y}=2\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi \partial \eta}+2\dfrac{\partial^2 z}{\partial \psi^2}\\
Our \space equation:\\
\dfrac{8\eta^2 -8\eta\psi +2\psi^2}{9}(4z_{\eta\eta}+4z_{\eta \psi}+z_{\psi \psi})-5\dfrac{5\psi\eta-2\psi^2-2\eta^2}{9}(2\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi \partial \eta}+2\dfrac{\partial^2 z}{\partial \psi^2})+\dfrac{8\psi^2 -8\eta\psi +2\eta^2}{9}(\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi \partial \eta}+4\dfrac{\partial^2 z}{\partial \psi^2})+2[\dfrac{2\eta-\psi}{3}(2\dfrac{\partial z}{\partial \eta}+\dfrac{\partial z}{\partial \psi})+\dfrac{2\psi-\eta}{3}(\dfrac{\partial z}{\partial \eta}+2\dfrac{\partial z}{\partial \psi})=0 ∂ x ∂ z = 2 ∂ η ∂ z + ∂ ψ ∂ z ∂ y ∂ z = ∂ η ∂ z + 2 ∂ ψ ∂ z ∂ x 2 ∂ 2 z = 4 ∂ η 2 ∂ 2 z + 4 ∂ ψ ∂ η ∂ 2 z + ∂ ψ 2 ∂ 2 z ∂ y 2 ∂ 2 z = ∂ η 2 ∂ 2 z + 4 ∂ ψ ∂ η ∂ 2 z + 4 ∂ ψ 2 ∂ 2 z ∂ x ∂ y ∂ 2 z = 2 ∂ η 2 ∂ 2 z + 4 ∂ ψ ∂ η ∂ 2 z + 2 ∂ ψ 2 ∂ 2 z O u r e q u a t i o n : 9 8 η 2 − 8 η ψ + 2 ψ 2 ( 4 z ηη + 4 z η ψ + z ψψ ) − 5 9 5 ψ η − 2 ψ 2 − 2 η 2 ( 2 ∂ η 2 ∂ 2 z + 4 ∂ ψ ∂ η ∂ 2 z + 2 ∂ ψ 2 ∂ 2 z ) + 9 8 ψ 2 − 8 η ψ + 2 η 2 ( ∂ η 2 ∂ 2 z + 4 ∂ ψ ∂ η ∂ 2 z + 4 ∂ ψ 2 ∂ 2 z ) + 2 [ 3 2 η − ψ ( 2 ∂ η ∂ z + ∂ ψ ∂ z ) + 3 2 ψ − η ( ∂ η ∂ z + 2 ∂ ψ ∂ z ) = 0
#340153 on Dec 2023
Comments