Question 2.36: a) Find and solve the PDE of the form z = F(x,y,p,q), satisf......
a) Find and solve the PDE of the form z = F(x,y,p,q), satisfied by the conic surfaces with a peak in a fixed point V(a, b, c).
b) Assume now a = b = c = 0. Then find a function \Phi=\Phi(x, y, p, q), such that it is in involution with F, see (2.39).
[F, \Phi]=\left|\begin{array}{ll}\frac{\partial F}{\partial p} & \frac{\partial F}{\partial x}+p \frac{\partial F}{\partial z} \\\frac{\partial \Phi}{\partial p} & \frac{\partial \Phi}{\partial x}+p \frac{\partial \Phi}{\partial z}\end{array}\right|+\left|\begin{array}{ll}\frac{\partial F}{\partial q} & \frac{\partial F}{\partial y}+q \frac{\partial F}{\partial z} \\\frac{\partial \Phi}{\partial q} & \frac{\partial \Phi}{\partial y}+q \frac{\partial \Phi}{\partial z}\end{array}\right| (2.39)
The given condition can be written as a scalar product
(x – a,y – b,z – c)· (p,q, -1) = 0,
or
p( x – a) + q(y – b) = z – c. (2.49)
The solution of this linear first order PDE is
G\left(\frac{z-c}{x-a}, \frac{y-b}{x-a}\right)=0
where G is an arbitrary function of the class C^1\left( \mathbf{R} ^2\right).
b) For a = b = c = 0 we get from (2.49)
F(x, y,p, q) = px + qy.
We are looking for a function Φ which is in involution with F, i.e., the Poisson bracket (F, Φ, see (2.40), is equal to zero:
(F, \Phi)=\left|\begin{array}{ll}\frac{\partial F}{\partial p} & \frac{\partial F}{\partial x} \\\frac{\partial \Phi}{\partial p} & \frac{\partial \Phi}{\partial x}\end{array}\right|+\left|\begin{array}{ll}\frac{\partial F}{\partial q} & \frac{\partial F}{\partial y} \\\frac{\partial \Phi}{\partial q} & \frac{\partial \Phi}{\partial y}\end{array}\right| (2.40)
\begin{aligned}(F, \Phi) & =\left|\begin{array}{cc}x & p \\\frac{\partial \Phi}{\partial p} & \frac{\partial \Phi}{\partial x}\end{array}\right|+\left|\begin{array}{cc}y & q \\\frac{\partial \Phi}{\partial q} & \frac{\partial \Phi}{\partial y}\end{array}\right| \\& =x \frac{\partial \Phi}{\partial x}-p \frac{\partial \Phi}{\partial p}+y \frac{\partial \Phi}{\partial y}-q \frac{\partial \Phi}{\partial q} \\& =0 .\end{aligned}
Using this equality, we get a linear first order PDE in Φ, whose solution is easily found to be
\Phi(x, y, p, q)=\Psi(x p, y q, x / y)=0 \quad\left(\Psi \in C^1\left( \mathbf{R} ^3\right)\right)