Question 2.28: Determine a PDE whose complete solution is a family of spher......
Determine a PDE whose complete solution is a family of spheres with the same radius R and with centers in the xy-planc, i.e.,
(x-a)^2+(y-b)^2+z^2=R^2, (2.43)
where a and b are arbitrary constants. Find then the general and the singular solution of the obtained PDE.
Starting from the complete solution (2.43), we can reconstruct the PDE as follows. Differentiating equation (2.43) in x and y, respectively, we get
2(x-a)+2zp=0 and 2(y-b)+2zq=0,
which after the elimination of the constants a and b gives the PDE
z^2\left(p^2+q^2+1\right)=R^2 . (2.44 )
The general solution of (2.44) is given by the system of equations
(x-a)^2+(y-b(a))^2+z^2=R^2 \text { and }(x-a)+b^{\prime}(a)(y-b(a))=0 \text {, } (2.45)
where b = b(a) is an arbitrary function of class C^1 (R). (The last equation was obtained by putting b = b( a) in (2.43) and differentiation in a.) The geometric interpretation of (2.45) is that it represents an envelope of the one-parameter family of spheres given by (2.43) for b = b( a ).
The singular solution is given by the system
(x-a)^2+(y-b)^2+z^2=R^2, \quad 2(x-a)=0, \quad 2(y-b)=0
which, after the elimination of the constants a and b, gives simply z² = R². The geometric interpretation of the singular solution is that it represents a union of two parallel planes, which are the envelopes of a family of surfaces (spheres) that make the complete solution.