MA108-Lecture 5-D3

MA 108-ODE- D3
Lecture 5
Debanjana Mitra
Department of Mathematics
Indian Institute of Technology Bombay
Powai, Mumbai - 76
May 08, 2023

Existence - Uniqueness Theorem
Recall: Existence - Uniqueness Theorem
Let R be a rectangle containing (x0 , y0 ): R : |x − x0 | < a, |y − y0 | < b.

I f (x, y ) be continuous at all points (x, y ) ∈ R in and
I bounded in R, that is, |f (x, y )| ≤ K , ∀(x, y ) ∈ R.
Then, the IVP y 0 = f (x, y ), y (x0 ) = y0 has at least one solution y (x) defined
for all x in the interval |x − x0 | < α, where

b
α = min a, .
K
In addition to the above conditions, if f satisfies the Lipschitz condition with

respect to y in R, that is,
|f (x, y1 ) − f (x, y2 )| ≤ M|y1 − y2 | ∀(x, y1 ), (x, y2 ) in R,
then,the IVP admits a unique solution on the interval (x0 − α, x0 + α). 1 .
1 Existence - Peano, Existence & uniqueness -Picard

Warm up!
Example. Consider y 0 = y 2 , y (1) = −1. Find α in the existence &

uniqueness theorem.
(1 − a, −1 + b) (1 + a, −1 + b)
(1, −1)
(1 − a, −1 − b) (1 + a, −1 − b)
f (x, y ) = y 2 , fy = 2y are continuous in the closed rectangle

R : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x, y )| = |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2 (1)

n o
b
Now, α = min a, (b+1) 2 .
Example (contd..)
Consider
b
F (b) = .
(b + 1)2
F 0 (b) = (b+1)
1−b
3 =⇒ the maximum value of F (b) for b > 0 occurs at
b = 1 (Why?); and we find F (1) = 14 and F (b) ≤ 1/4 for any b > 0.
Hence, for any given a > 0 and b > 0, α = min{a, F (b)} ≤ 41 .
1
In particular, for any a ≥ 1/4 and any b > 0, the best possible α = 4 and
the theorem gives that the IVP has a unique solution in
|x − 1| < 1/4 =⇒ 3/4 < x < 5/4 .
Example - Remarks
1. The theorem guarantees existence and uniqueness only in a very

small interval!
2. The theorem DOES NOT give the largest interval where the solution
exits.
3. What is the solution in this case by separation of variables and

where is it valid? Can you think of extending the solution to a larger
interval?
Ans. y (x) = −1x , solution defined on (0, ∞).
Example. Consider the IVP: y 0 = f (x, y ), y (0) = 0, where
f (x, y ) = y 2 + cos(x 2 ), ∀ (x, y ) with |x| < 1, |y | < 1. Does the IVP
has a solution on the interval (− 12 , 21 )? If yes, is the solution unique on
the interval?
Ans. Yes. For the rectangle R := {(x, y ) ∈ R2 | |x| < 1, |y | < 1}, all
the hypothesis of existence and uniqueness theorem holds. i.e., f is
continuous and bounded on R, with |f (x, y )| ≤ K for all (x, y ) ∈ R,
where K = 2. The function f satisfies Lipschitz condition with respect to
y on R:
|f (x, y1 )−f (x, y2 )| = |y12 −y22 | ≤ M|y1 −y2 |, ∀ (x, y1 ) ∈ R, (x, y2 ) ∈ R,
for some positive condition M, independent of x, y .

So, the theorem is applicable and the IVP admits a unique solution on
(−α, α), where recall α = min{a, b/k}, and here a = 1, b = 1, K = 2.
Hence, IVP has a solution on (− 12 , 12 ) and the solution is unique on the
interval.
Existence and Uniqueness
Example: Consider the IVP
10 2/5
y0 = xy , y (x0 ) = y0 . (2)
3
(i) For what points (x0 , y0 ), does the Theorem imply that (2) has a
solution?
(ii) For what points (x0 , y0 ), does the Theorem imply that (2) has a
unique solution on some open interval that contains x0 ?
Ans.(i) Since f (x, y ) = 10
3 xy
2/5
is continuous for all (x, y ), it follows that
the above IVP has a solution for every (x0 , y0 ).
(ii) f is not Lipschitz with respect to y on any rectangle R containing the
points (x, 0) for any x ∈ R2 . Otherwise f satisfies the Lipschitz condition
on any rectangle R not containing (x, 0).
Therefore, if y0 6= 0, there is an open rectangle on which f satisfies the
Lipschitz condition with respect to y , and hence and hence the above
IVP has a unique solution on some interval that contains x0 .
If y0 = 0, then on any rectangle containing (x0 , 0), f does not satisfy the
Lipschitz condition with respect to y , and thus Theorem for the
uniqueness is not applicable to this IVP if y0 = 0.
Linear first order ODEs
Consider the linear equation
y 0 + p(t)y = q(t), (3)
where p(·) and q(·) are continuous functions defined on an interval I .

(i) There is a general solution, containing an arbitrary constant, that
includes all solutions of the differential equation, i.e.,
R
Z R
− p(t)dt p(t)dt
y (t) = e e · q(t)dt + c .
A particular solution that satisfies a given initial condition can be

picked out by choosing the proper value for the arbitrary constant.
(ii) Let t0 ∈ I . The solution of the linear equation (3) subject to the
initial condition y (t0 ) = y0 exists for all t ∈ I and the solution can
be obtained from the above expression determining c by y (t0 ) = y0 .
The solution is unique. (Check! why?)
Linear ODE contd...
Let φ and ψ be two solutions of the above IVP. Set
w (t) = φ(t) − ψ(t), ∀t ∈ I.
Then w (·) satisfies
w 0 (t) + p(t)w (t) = 0, ∀t ∈ I, w (t0 ) = 0.

R
p(t) dt
Using Integrating factor denoting µ(t) = e , we deduce
d
µ(t)w (t) = 0, ∀t ∈ I, w (t0 ) = 0,
dt
and hence µ(t)w (t) = µ(t0 )w (t0 ) = 0 for all t ∈ I .
Thus, from above it follows w (t) = 0 for all t ∈ I , as µ(t) 6= 0 for all t
and
φ(t) = ψ(t), ∀ t ∈ I .
Picard’s iteration method
2
AIM : To solve
y 0 = f (x, y ), y (x0 ) = y0 (4)
METHOD
1. Integrate both sides of (4) to obtain
Z x
y (x) − y (x0 ) = f (t, y (t)) dt
x0
Z x
y (x) = y0 + f (t, y (t)) dt (5)
x0
Note that any solution of (4) is a solution of (5) and vice-versa.
2 Picard used this in his existence-uniqueness proof

Picard’s method
2. Solve (5) by iteration:

Z x
φ1 (x) = y0 + f (t, y0 ) dt
x
Z 0x
φ2 (x) = y0 + f (t, φ1 (t)) dt
x0
..
. Z x
φn (x) = y0 + f (t, φn−1 (t)) dt
x0
3. Under the assumptions of existence-uniqueness theorem, the

sequence of approximations converges to the solution φ(x) of (4).
That is,
φ(x) = lim φn (x).
n→∞

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MA 108-ODE- D3

May 08, 2023

Let R be a rectangle containing (x0 , y0 ): R : |x − x0 | < a, |y − y0 | < b.

In addition to the above conditions, if f satisfies the Lipschitz condition with

|f (x, y1 ) − f (x, y2 )| ≤ M|y1 − y2 | ∀(x, y1 ), (x, y2 ) in R,

then,the IVP admits a unique solution on the interval (x0 − α, x0 + α). 1 .

1 Existence - Peano, Existence & uniqueness -Picard

Example. Consider y 0 = y 2 , y (1) = −1. Find α in the existence &

f (x, y ) = y 2 , fy = 2y are continuous in the closed rectangle

|f (x, y )| = |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2 (1)

1. The theorem guarantees existence and uniqueness only in a very

3. What is the solution in this case by separation of variables and

|f (x, y1 )−f (x, y2 )| = |y12 −y22 | ≤ M|y1 −y2 |, ∀ (x, y1 ) ∈ R, (x, y2 ) ∈ R,

for some positive condition M, independent of x, y .

y 0 + p(t)y = q(t), (3)

where p(·) and q(·) are continuous functions defined on an interval I .

A particular solution that satisfies a given initial condition can be

Let φ and ψ be two solutions of the above IVP. Set

w (t) = φ(t) − ψ(t), ∀t ∈ I.

Then w (·) satisfies

w 0 (t) + p(t)w (t) = 0, ∀t ∈ I, w (t0 ) = 0.

Note that any solution of (4) is a solution of (5) and vice-versa.

2 Picard used this in his existence-uniqueness proof

2. Solve (5) by iteration:

3. Under the assumptions of existence-uniqueness theorem, the

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