Math 6341

Problems a) and b) illustrate advantages of the different pictures of the solutions (characteristics as functions of s vs. constants of the motion):

1. Solve
   1. Solving on characteristics. The characteristic equations are:
      
      the solutions of which are:
      
      Observe that the solution will blow up when s=1/x₀ or s=1/ysub>0, but that is not a true singularity, because for those values of s, x or y, respectively, goes to infinity.
   2. Solving by constants of the motion. By looking at the solutions from part (i), we see that one constant of the motion is
      
      Another one we can come up with from the solutions (or directly from the characteristic equations) is:
      
      Of course, the second constant does not involve u, so it is not a solution by itself; in order to obtain one, we could combine it with ₁, for instance to get
      
      Therefore a general implicit solution will be of the form:
      
      Here F can be any function with nonzero partial derivatives.
   3. Solving an initial-value problem, for instance of the form
      u(t,t) = g(t):
      The initial values on the characteristic curves are then x₀ =t, y₀ =-t, u₀ =g(t), so from the solution in (i), s = -2(x+y)/xy and t = 2xy/(y-x). Therefore, by substituting:
      
2. Solve x (y-u) ux + y(u-x)uy = (x-y) u.
   The characteristic equations are
   
   These are more difficult to solve than for the previous problem, because this equation is nonlinear. Still, since there is a symmetry in the characteristic equations when we permute the variables x,y,u, it is not too difficult to find combinations of the variables which are easier to deal with. Specifically:
   
   which implies that one constant of the motion is
   
   Similarly,
   
   The integral of the left side of this is ln(xyu), which will be a constant of the motion, but a simpler constant of the motion will be:
   
   The general solution will thus be of the form F(x+y+u, xyu) = 0.

   Having this, it is not necessary to solve the characteristic equations themselves, but in case you want to, you can substitute into them from these cosntants of the motion. For instance, (x-y) u can be rewritten entirely in terms of u and the constants by using:

   
   where
   
3. Let's solve this equation with initial conditions, such as
   
   or, equivalently, x(t,s=0) = t, y(t,s=0)= 2t/(t2-1). We see that
   
   and
   
   Hence for s=0,
   
   This constant is therefore the one defining the solution away from the initial data as well. Hence
   x+y+u = xyu.
   This can be solved for u explicitly:
   
   Observe that this solution blows up on the hyperbola xy=1.