Despite the seeming generality of the uniqueness theorem, there are initial value problems...

Despite the seeming generality of the uniqueness theorem, there are initial value problems which have more than one solution. Consider the differential equation y′ =  Notice that y(t) = 0 is a solution with the initial condition y(0) = 0. (Of course by  we mean the nonnegative square root.)

a) This equation is separable. Use this to find a solution to the equation with the initial value y(t0) = 0 assuming that y ≥ 0. You should get the answer y(t) = (t − t0)2/4. Notice, however, that this is a solution only for t ≥ t0. Why?

b) Show that the function

is continuous, has a continuous first derivative, and satisfies the differential equation y′ = .

c) For any t0 ≥ 0 the function defined in part b) satisfies the initial condition y(0) = 0. Why doesn’t this violate the uniqueness part of the theorem?

d) Find another solution to the initial value problem in a) by assuming that y ≤ 0.

e) You might be curious (as were the authors) about what dfield6 will do with this equation. Find out. Use the rectangle defined by −1 ≤ t ≤ 1 and −1 ≤ y ≤ 1 and plot the solution of y′ =  with initial value y(0) = 0. Also, plot the solution for y(0) = 10−50 (the MATLAB notation for 10−50 is le−50). Plot a few other solutions as well. Do you see evidence of the non-uniqueness observed in part c)?