642:515 Spring 00, Problem set 3

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  The first two problems give a bit more practice with Lipschitz functions,
  the third is just a "puzzle" type of problem, and the fourth one shows that,
  as far as the solution sets are concerned (but not time-dependence), one
  can always assume that solutions are defined globally in time.
  Problem 5 generalizes the Lipschitz condition, and 6 does a simple example
  of ODE's with discontinuous right-hand side.
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1. Suppose that g is a vector field, locally Lipschitz with respect to x on a
   set D, and f is a locally Lipschitz scalar function with respect to x on D.
   Prove that the product
      f(t,x) g(t,x)
   is also locally Lipschitz with respect to x on D.

2. Suppose that f is defined for all (t,x) in R x Rn  and it is locally
   Lipschitz with respect to x.  Assume, moreover, that f has compact support.
   Show that, then, f is (globally) Lipschitz with respect to x.

3. Consider the scalar ODE  x' = f(x) = x sin(1/x) [understanding f(0)=0].
   Show that f is not locally Lipschitz but, nonetheless, there is uniqueness
   of solutions.

4. Consider a system of ODE's  x' = f(x) , where f is locally Lipschitz and
   defined for all x in Rn.  Show that there exists a locally Lipschitz vector
   field g, also defined for all x in Rn, so that the following properties
   hold:

   a. for each z, the solution of x'=g(x), x(0)=z, exists for all t>0 ;

   b. the solution paths for x'=g(x) are the same as for the original system,
      that is: for each z, consider the solution x(t) of x'=f(x), x(0)=z
      and the solution y(t) of y'=g(y), y(0)=z; then the set of points
      {x(t), 0<t<T} equals the set {y(t), t>0}, where "T" is the sup of the
      set of t's such that x(t) is defined.

   [Hint: obtain g(x) from f(x) by multiplication by a suitable scalar factor
   which makes g bounded; show then that solutions correspond to each other by
   reparameterization.]

5. Consider the following property for a continuous function f defined on an
   open set D: for each compact subset K of D, there is a constant c so that:

                                           2
   < x - y , f(t,x) - f(t,y) >   <  c |x-y|    for all (t,x) and (t,y) in D
                                 -

   a. Show that any locally Lipschitz f satisfies this property.

   b. Provide an example of an f that satisfies this property but is not
      locally Lipschitz.

   c. Show that uniqueness of solutions of x'=f(t,x) holds if f satisfies
      this property.

6. When f is not continuous on x, solutions may or may not exist.  To
   illustrate, we take the following two equations on R:

            / -1 if x > 0                  / -1 if x < 0
   a. x' = |                      b. x' = |
            \ 1  otherwise                 \ 1  otherwise

   Show that, for one of these examples, solutions exist for all initial
   conditions and all times t>0, but for the other example there is some
   initial condition for which no solution exists on any interval [0,epsilon].
   (By a "solution" of x'=f(x), when f may not be continuous, let us mean a
   piecewise differentiable function x such that x'(t)=f(x(t)) at all points
   of differentiability.  A better definition is "an absolutely continuous
   function such that x'=f(x) a.e.", but in any case, for this particular
   problem it makes no difference what definition we use.)