Two PDEs f,gf,g are compatible if they have a common solution. We must show that,
[ f , g ] = ∂ ( f , g ) ∂ ( x , p ) + p ∂ ( f , g ) ∂ ( z , p ) + ∂ ( f , g ) ∂ ( y , q ) + q ∂ ( f , g ) ∂ ( z , q ) = 0 [f,g]=\frac{\partial (f,g)}{\partial (x,p)}+p\frac{\partial (f,g)}{\partial (z,p)}+\frac{\partial (f,g)}{\partial (y,q)}+q\frac{\partial (f,g)}{\partial (z,q)}=0 [ f , g ] = ∂ ( x , p ) ∂ ( f , g ) + p ∂ ( z , p ) ∂ ( f , g ) + ∂ ( y , q ) ∂ ( f , g ) + q ∂ ( z , q ) ∂ ( f , g ) = 0
z=px+qy
Let f=z−px−qy=0
2xy(p^2+q^2)=z(yp+xq)
Let g=2xy(p^2+q^2)-z(yp+xq)=0
∂ f ∂ x = − p , ∂ f ∂ y = − q , ∂ f ∂ z = 1 , ∂ f ∂ p = − x , ∂ f ∂ q = − y ∂ g ∂ x = 2 y ( p 2 + q 2 ) − z q , ∂ g ∂ y = 2 x ( p 2 + q 2 ) − z p , ∂ g ∂ z = − ( y p + x q ) , ∂ g ∂ p = 4 x y p − y z , ∂ g ∂ q = 4 x y q − x z . \frac{\partial f}{\partial x}=-p,\frac{\partial f}{\partial y}=-q,\frac{\partial f}{\partial z}=1,\frac{\partial f}{\partial p}=-x,\frac{\partial f}{\partial q}=-y\\ \frac{\partial g}{\partial x}=2y(p^2+q^2)-zq,\frac{\partial g}{\partial y}=2x(p^2+q^2)-zp,\frac{\partial g}{\partial z}=-(yp+xq),\frac{\partial g}{\partial p}=4xyp-yz,\frac{\partial g}{\partial q}=4xyq-xz. ∂ x ∂ f = − p , ∂ y ∂ f = − q , ∂ z ∂ f = 1 , ∂ p ∂ f = − x , ∂ q ∂ f = − y ∂ x ∂ g = 2 y ( p 2 + q 2 ) − z q , ∂ y ∂ g = 2 x ( p 2 + q 2 ) − z p , ∂ z ∂ g = − ( y p + x q ) , ∂ p ∂ g = 4 x y p − yz , ∂ q ∂ g = 4 x y q − x z .
∂ ( f , g ) ∂ ( x , p ) = ∣ ∂ f ∂ x ∂ g ∂ x ∂ f ∂ p ∂ g ∂ p ∣ = ∣ − p 2 y ( p 2 + q 2 ) − x 4 x y p − y z ∣ = 2 x y q 2 + y z p − 2 x y p 2 \frac{\partial (f,g)}{\partial (x,p)}=\begin{vmatrix} \frac{\partial f}{\partial x} & \frac{\partial g}{\partial x} \\ \frac{\partial f}{\partial p} & \frac{\partial g}{\partial p} \end{vmatrix}=\begin{vmatrix} -p & 2y(p^2+q^2) \\ -x & 4xyp-yz \end{vmatrix}=2xyq^2+yzp-2xyp^2 ∂ ( x , p ) ∂ ( f , g ) = ∣ ∣ ∂ x ∂ f ∂ p ∂ f ∂ x ∂ g ∂ p ∂ g ∣ ∣ = ∣ ∣ − p − x 2 y ( p 2 + q 2 ) 4 x y p − yz ∣ ∣ = 2 x y q 2 + yz p − 2 x y p 2
∂ ( f , g ) ∂ ( z , p ) = ∣ ∂ f ∂ z ∂ g ∂ z ∂ f ∂ p ∂ g ∂ p ∣ = ∣ 1 − ( y p + x q ) − x 4 x y p − y z ∣ = 3 x y p − y z − x 2 q \frac{\partial (f,g)}{\partial (z,p)}=\begin{vmatrix} \frac{\partial f}{\partial z} & \frac{\partial g}{\partial z} \\ \frac{\partial f}{\partial p} & \frac{\partial g}{\partial p} \end{vmatrix}=\begin{vmatrix} 1 & -(yp+xq) \\ -x & 4xyp-yz \end{vmatrix}=3xyp-yz-x^2q ∂ ( z , p ) ∂ ( f , g ) = ∣ ∣ ∂ z ∂ f ∂ p ∂ f ∂ z ∂ g ∂ p ∂ g ∣ ∣ = ∣ ∣ 1 − x − ( y p + x q ) 4 x y p − yz ∣ ∣ = 3 x y p − yz − x 2 q
∂ ( f , g ) ∂ ( y , q ) = ∣ ∂ f ∂ y ∂ g ∂ y ∂ f ∂ q ∂ g ∂ q ∣ = ∣ − q 2 x ( p 2 + q 2 ) − y 4 x y q − x z ∣ = 2 x y p 2 + x z q − 2 x y q 2 \frac{\partial (f,g)}{\partial (y,q)}=\begin{vmatrix} \frac{\partial f}{\partial y} & \frac{\partial g}{\partial y} \\ \frac{\partial f}{\partial q} & \frac{\partial g}{\partial q} \end{vmatrix}=\begin{vmatrix} -q & 2x(p^2+q^2) \\ -y & 4xyq-xz \end{vmatrix}=2xyp^2+xzq-2xyq^2 ∂ ( y , q ) ∂ ( f , g ) = ∣ ∣ ∂ y ∂ f ∂ q ∂ f ∂ y ∂ g ∂ q ∂ g ∣ ∣ = ∣ ∣ − q − y 2 x ( p 2 + q 2 ) 4 x y q − x z ∣ ∣ = 2 x y p 2 + x z q − 2 x y q 2
∂ ( f , g ) ∂ ( z , q ) = ∣ ∂ f ∂ z ∂ g ∂ z ∂ f ∂ q ∂ g ∂ q ∣ = ∣ 1 − ( y p + x q ) − y 4 x y q − x z ∣ = 3 x y q − x z − y 2 p \frac{\partial (f,g)}{\partial (z,q)}=\begin{vmatrix} \frac{\partial f}{\partial z} & \frac{\partial g}{\partial z} \\ \frac{\partial f}{\partial q} & \frac{\partial g}{\partial q} \end{vmatrix}=\begin{vmatrix} 1 & -(yp+xq) \\ -y & 4xyq-xz \end{vmatrix}=3xyq-xz-y^2p ∂ ( z , q ) ∂ ( f , g ) = ∣ ∣ ∂ z ∂ f ∂ q ∂ f ∂ z ∂ g ∂ q ∂ g ∣ ∣ = ∣ ∣ 1 − y − ( y p + x q ) 4 x y q − x z ∣ ∣ = 3 x y q − x z − y 2 p
f,g]=2xyq^2+yzp-2xyp^2+3xyp^2-yzp-x^2pq+2xyp^2+xzq-2xyq^2+3xyq^2-xzq-y^2pq\\ [f,g]=3xyp^2+3xyp^2-x^2pq-y^2pq\\ [f,g]=4xyp^2+4xyq^2-xyp^2-x^2pq-xyq^2-y^2pq\\ [f,g]=4(xy(p^2+q^2))-xp(yp+xq)-yq(xq+yp)\\ [f,g]=2z(xp+yq)-(xp+yq)(yp+xq) \text{ From (2)}\\ [f,g]=2z(xp+yq)-z(yp+xq) \text{ From (1)}\\ [f,g]=z(yp+xq)\neq0
#331389 on Nov 2022
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