CHAPTER 1 Examples of the General Existence Theorem and Gronwall’s Inequality

1.1. Higher order scalar equations.

Consider the n order scalar initial value problem

y^{(n)}(t) = f(t, y(t), y′(t), …, y^{(n−1)}(t)),  (1.1)
y(0) = y_{0}, y′(0) = y_{1}, …, y^{(n−1)}(0) = y_{n−1}.

We can transform this into a single first order vector equation by letting

Y (t) := [ y(t) ; y′(t) ; ⋮ ; y^{(n−1)}(t)]

and for a vertor X = [x_{0}, x_{1}, …, x_{n−1}]^{t} (there A^{t} is the transpose of A).

F (t, X) := [ x_{1} ; x_{2} ; ⋮ ; x_{n−1} ; f(t, x_{0}, …, x_{n−1})]

Set Y_{0} := [y_{0}, y_{1}, …, y_{n−1}]^{t}. Then consider the first order intial value problem for the vector Y (t).

Y′(t) = F (t, Y (t)), Y′(0) = Y_{0}.  (1.2)

Exercise 1.1.1. 1. Give a brief indication why the scalar initial value problem (1.1) and the vector initial value problem (1.2) are equivalent.

2. Use our general theorem for the exisntace and uniquness to the initial value problem for first order vector equations to give general existence and uniqueness theorem for the higher order equation (1.1). State the hypotheses carefully. (Note that this only involves a translating from the higher order case to the vector case; you should not have to go through the proof again.)

3. Take the second version of Gronwall’s inequality given in class and use it to give a version of Gronwall’s inequality for the higher order scalar equation (1.1).

1.2. Linear Systems

In this section let X be a Banach space (which you can assume to be R^{n} if you wish). Recall that a linear map A : X → X is bounded iff there is a constant C so that ∥Ax∥ ≤ C∥x∥ for all x ∈ X. (It is not too hard to check that A is bounded iff it is continous.) If A is bounded then the operator norm ∥A∥_{op} of A is given by

∥A∥_{op} := sup_{∥x∥=1} ∥Ax∥ = sup_{x≠0} \frac{∥Ax∥}{∥x∥}.

Then

∥Ax∥ ≤ ∥A∥_{op}∥x∥

which is about the only property of the operator norm we will use. Note if we are assuming that X = R^{n} then a linear map A : R^{n} → R^{n} can be viewed as a matrix in the usual manner. We now consider the initial value problem

y′(t) = Ay(t), y(0) = y_{0}.  (1.3)

While we could reduce this to our general theory, it is much more fun to see how the proof works in this case. Let I be the identity map on X. We first note (1.3) is equivalent to the integral equation

y(t) = y_{0} + ∫_{0}^{t} Ay(s) ds = y_{0} + A ∫_{0}^{t} y(s) ds,

and then form the Picard iterates

y_{1}(t) = y_{0} + ∫_{0}^{t} Ay_{0} ds = y_{0} + tAY_{0} = (I + tA)y_{0},
y_{2}(t) = y_{0} + ∫_{0}^{t} Ay_{1}(s) ds = (I + tA + \frac{(tA)^{2}}{2} ) y_{0},
⋮ ⋮
y_{n+1}(t) = y_{0} + ∫_{0}^{t} Ay_{n}(s) ds.

Exercise 1.2.1. 1. Give an explicit formula for y_{n}(t).

2. Define e^{tA} by the power series

e^{tA} = ∑_{k=0}^{∞} \frac{(tA)^{k}}{k!} .

Show that this series converges for all t ∈ R and that

lim_{n→∞} y_{n}(t) = e^{tA}y_{0}.

This shows that

y(t) = e^{tA}y_{0}

is the unique solution to the initial value problem (1.3)

3. Show that e^{(s+t)A} = e^{sA}e^{tA}. (Hint: This can be done directly by using the series. But if, like me, you would like to avoid messy calculations fix any y_{0} ∈ X and s ∈ R and set y(t) = e^{tA}e^{sA}y_{0} and z(t) = e^{(s+t)A}y_{0}. Show that y(t) and z(t) satisfy the same initial value problem.)

Let B(X, X) be the set of all bounded linear maps A : X → X (when X = R^{n} this is just the set of all matrices). Then B(X, X) is a vector space and with the norm ∥A∥_{op} it is a Banach space. Let t ↦ A(t) be a continous map form R to B(X, X). Then we want to consider the linear equation

y′(t) = A(t)y(t) + g(t), y(0) = y_{0}.  (1.4)

where g : R → X is any continous map.

Exercise 1.2.2. Assume that there is a constant C so that ∥A(t)∥ ≤ C for all t.

1. Show that the initial value problem (1.4) has a unique solution that that it is defined for all t ∈ R. (Hint: Let f(t, x) := A(t)x+g(t) and show that ∥f(t, x_{2}) − f(t, x_{1})∥ ≤ C∥x_{2} − x_{1}∥. Therefore our general result applies.)

2. What does our first version of Gronwall’s inequality say in this setting?

Remark 1.2.3. For the existence and uniqueness in the last exercise the existence of the constant C so that ∥A(t)∥ ≤ C is not at all important, in only matters that t ↦ A(t) is continous. It is not hard to reduce the general case to the version of the exercise so we will not bother with it.