Answer to Question #155302 in Differential Equations for Nirob islam

Question

Show that the partial differential equations

z = px + qy and 2xy(p² + q²) = z(yp + xq)are compatible

Answer 1

Two PDEs $f,g$ are compatible if they have a common solution. We must show that,

$[f,g]=\frac{\partial (f,g)}{\partial (x,p)}+p\frac{\partial (f,g)}{\partial (z,p)}+\frac{\partial (f,g)}{\partial (y,q)}+q\frac{\partial (f,g)}{\partial (z,q)}=0$

$z=px+qy..................................................(1)\\ \text{Let }f=z-px-qy=0\\ 2xy(p^2+q^2)=z(yp+xq)............................(2)\\ \text{Let }g=2xy(p^2+q^2)-z(yp+xq)=0$.

We shall compute the expression for $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z},\frac{\partial f}{\partial p},\frac{\partial f}{\partial q}$ . The same thing for $g$.

$\frac{\partial f}{\partial x}=-p,\frac{\partial f}{\partial y}=-q,\frac{\partial f}{\partial z}=1,\frac{\partial f}{\partial p}=-x,\frac{\partial f}{\partial q}=-y\\ \frac{\partial g}{\partial x}=2y(p^2+q^2)-zq,\frac{\partial g}{\partial y}=2x(p^2+q^2)-zp,\frac{\partial g}{\partial z}=-(yp+xq),\frac{\partial g}{\partial p}=4xyp-yz,\frac{\partial g}{\partial q}=4xyq-xz.$

$\frac{\partial (f,g)}{\partial (x,p)}=\begin{vmatrix} \frac{\partial f}{\partial x} & \frac{\partial g}{\partial x} \\ \frac{\partial f}{\partial p} & \frac{\partial g}{\partial p} \end{vmatrix}=\begin{vmatrix} -p & 2y(p^2+q^2) \\ -x & 4xyp-yz \end{vmatrix}=2xyq^2+yzp-2xyp^2$

$\frac{\partial (f,g)}{\partial (z,p)}=\begin{vmatrix} \frac{\partial f}{\partial z} & \frac{\partial g}{\partial z} \\ \frac{\partial f}{\partial p} & \frac{\partial g}{\partial p} \end{vmatrix}=\begin{vmatrix} 1 & -(yp+xq) \\ -x & 4xyp-yz \end{vmatrix}=3xyp-yz-x^2q$

$\frac{\partial (f,g)}{\partial (y,q)}=\begin{vmatrix} \frac{\partial f}{\partial y} & \frac{\partial g}{\partial y} \\ \frac{\partial f}{\partial q} & \frac{\partial g}{\partial q} \end{vmatrix}=\begin{vmatrix} -q & 2x(p^2+q^2) \\ -y & 4xyq-xz \end{vmatrix}=2xyp^2+xzq-2xyq^2$

$\frac{\partial (f,g)}{\partial (z,q)}=\begin{vmatrix} \frac{\partial f}{\partial z} & \frac{\partial g}{\partial z} \\ \frac{\partial f}{\partial q} & \frac{\partial g}{\partial q} \end{vmatrix}=\begin{vmatrix} 1 & -(yp+xq) \\ -y & 4xyq-xz \end{vmatrix}=3xyq-xz-y^2p$

$[f,g]=2xyq^2+yzp-2xyp^2+3xyp^2-yzp-x^2pq+2xyp^2+xzq-2xyq^2+3xyq^2-xzq-y^2pq\\ [f,g]=3xyp^2+3xyp^2-x^2pq-y^2pq\\ [f,g]=4xyp^2+4xyq^2-xyp^2-x^2pq-xyq^2-y^2pq\\ [f,g]=4(xy(p^2+q^2))-xp(yp+xq)-yq(xq+yp)\\ [f,g]=2z(xp+yq)-(xp+yq)(yp+xq) \text{ From (2)}\\ [f,g]=2z(xp+yq)-z(yp+xq) \text{ From (1)}\\ [f,g]=z(yp+xq)\neq0$

Hence, $f,g$ are not compatible.