Math 677. Fall 2009.
Homework #3 Solutions.
Part I. Exercises are taken from ”Diffential Equations and Dynamical Sys-
tems” by Perko, 3rd edition.
Problem Set 1: # 4
IVP ẋ = x3 , x(0) = 2 has solution in the form x(t) = √ 2 , which exists for
1−8t
all t ∈ (−∞, 81 ) and lim− x(t) = ∞.
t→ 18
Problem Set 3: # 2(a)
IVP ẋ = f (x), x(0) = y with f (x) = (−x1 , −x2 + x21 , x3 + x21 )T has solution
in the form
1
u(t, y) = (y1 e−t , y2 e−t + y12 (e−t − e−2t ), y2 et + y12 (et − e−2t ))T .
3
It follows that
e−t 0 0
∂u
Φ(t, y) = (t, y) = 2y1 (e−t − e−2t ) e−t 0
∂y 2
y (et − e−2t ) 0 et
3 1
−1 0 0
Df (x) = 2y1 e−t −1 0
2y1 e−t 0 1
By direct calculation, A(t, y)Φ(t, y) = Φ̇(t, y) and Φ(0, y) = I.
Problem Set 4: # 2(c)
1
IVP ẋ1 = , ẋ2 = x1 , x1 (0) = 1, x2 (0) = 1 can be solved exactly to obtain
2x1
√ 1 2
x1 (t) = 1 + t, x2 (t) = + (1 + t)3/2 . Maximal interval of existence of the
3 3
solution is (−1, ∞) and ~x → (0, 31 ) as t → −1− .
Problem Set 4: # 5
To show: if x1 = limt→β − x(t) exists and x1 ∈ E, then β = ∞. In addition,
f (x1 ) = 0 and x(t) = x1 is a solution for IVP with x(0) = x1 .
1
Proof: Define
x(t), 0 ≤ t ≤ β
u(t) =
x1 , t=β
It is easy to show that u(t) is continuous on [0, β] (use ǫ−δ argument), which
is a compact set. Consider K to be the image of u(t) for t ∈ [0, β]. Clearly
K ⊂ E is compact as an image of compact set under continuous mapping,
so by Corollary 2 β = ∞.
By continuity of f , we have f (x1 ) = f ( lim− x(t)) = lim f (x(t)) =
t→β t→∞
dx(t) dx1
lim = = 0.
t→∞ dt dt
Part II. p
(a) The IVP ẏ = |y|, y(t0 ) = 0 has at least two solutions y1 (t) = 0 and
1
(t − t0 )2 , t ≥ t0
y2 (t) = 4
− 41 (t − t0 )2 , t ≤ t0
This does not contradict the Existence and Uniqueness theorem since f =
p
|y| is not Lipschitz. In the perturbations
p y2 p
(i) y ′ = |y| + ǫ, y(t0 ) = 0, (ii) y ′ = |y|, y(t0 ) = 0,
y 2 + ǫ2
though, both right-hand-side functions are Lipschitz. Th unique solution of
(i) is given by
1 √
(t − t0 + 2 √ǫ)2 − ǫ, t ≥ t0
y2 (t) = 4
− 41 (t − t0 − 2 ǫ)2 + ǫ, t ≤ t0
while (ii) has only trivial solution. This demonstrates that starting from
an ODE which does not satisfy any uniqueness condition we can get two
drastically different families of solutions by making two different smooth
perturbations. The ODE with no uniqueness can be considered a branching
point in the space of ODEs.
(b)R Picard map for the IVP ẋ = f (t, x), x(t0 ) = x0 is given by (Ax)(t) =
t
x0 + t0 f (τ, x(τ ))dτ and the corresponding sequence x, Ax, A2 x, . . . is the
sequence of Picard approximations. Use these approximations to solve the
2
IVP ẋ = x, x(0) = 1. In this example, it is easy to compute
x0 = 1;
Rt
x1 = Ax0 = 1 + 0 dτ = 1 + t;
Rt
x2 = 1 + 0 (1 + τ )dτ = 1 + t + t2 /2;
...
xn = A x0 = 1 + t + . . . + tn /n!
n
so that limn→∞ An x0 = et .
(c) Contraction is a Lipschitz map with constant L = 1. Suppose f is
Lipschitz with some constant L in general, then ∀ǫ > 0 take δ = ǫ/L. It
follows that ||y − x|| < δ yields ||f (y) − f (x)|| < L||y − x|| < ǫ, so we proved
continuity.
