MA 108 - Ordinary Differential Equations

Santanu Dey

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai 76dey@math.iitb.ac.in

October 4, 2013

Santanu Dey Lecture 5

Outline of the lecture

Picard’s iteration

Second order linear equations

Santanu Dey Lecture 5

Picard’s iteration method

1 AIM : To solve

y ′ = f (x , y), y(x0) = y0 (1)

METHOD

1. Integrate both sides of (1) to obtain

y(x)− y(x0) =

∫ x

x0

f (t, y(t)) dt

y(x) = y0 +

∫ x

x0

f (t, y(t)) dt (2)

Note that any solution of (1) is a solution of (2) andvice-versa.

1Picard used this in his existence-uniqueness proofSantanu Dey Lecture 5

Picard’s iteration method

1 AIM : To solve

y ′ = f (x , y), y(x0) = y0 (1)

METHOD

1. Integrate both sides of (1) to obtain

y(x)− y(x0) =

∫ x

x0

f (t, y(t)) dt

y(x) = y0 +

∫ x

x0

f (t, y(t)) dt (2)

Note that any solution of (1) is a solution of (2) andvice-versa.

1Picard used this in his existence-uniqueness proofSantanu Dey Lecture 5

Picard’s iteration method

1 AIM : To solve

y ′ = f (x , y), y(x0) = y0 (1)

METHOD

1. Integrate both sides of (1) to obtain

y(x)− y(x0) =

∫ x

x0

f (t, y(t)) dt

y(x) = y0 +

∫ x

x0

f (t, y(t)) dt

(2)

Note that any solution of (1) is a solution of (2) andvice-versa.

1Picard used this in his existence-uniqueness proofSantanu Dey Lecture 5

Picard’s iteration method

1 AIM : To solve

y ′ = f (x , y), y(x0) = y0 (1)

METHOD

1. Integrate both sides of (1) to obtain

y(x)− y(x0) =

∫ x

x0

f (t, y(t)) dt

y(x) = y0 +

∫ x

x0

f (t, y(t)) dt (2)

Note that any solution of (1) is a solution of (2) andvice-versa.

1Picard used this in his existence-uniqueness proofSantanu Dey Lecture 5

Picard’s method

2. Solve (2) by iteration:

y1(x) = y0 +

∫ x

x0

f (t, y0) dt

y2(x) = y0 +

∫ x

x0

f (t, y1(t)) dt

...

yn(x) = y0 +

∫ x

x0

f (t, yn−1(t)) dt

3. Under the assumptions of existence-uniqueness theorem, thesequence of approximations converges to the solution y(x) of(1). That is,

y(x) = limn→∞

yn(x).

Santanu Dey Lecture 5

Picard’s method

2. Solve (2) by iteration:

y1(x) = y0 +

∫ x

x0

f (t, y0) dt

y2(x) = y0 +

∫ x

x0

f (t, y1(t)) dt

...

yn(x) = y0 +

∫ x

x0

f (t, yn−1(t)) dt

3. Under the assumptions of existence-uniqueness theorem, thesequence of approximations converges to the solution y(x) of(1). That is,

y(x) = limn→∞

yn(x).

Santanu Dey Lecture 5

Picard’s method

2. Solve (2) by iteration:

y1(x) = y0 +

∫ x

x0

f (t, y0) dt

y2(x) = y0 +

∫ x

x0

f (t, y1(t)) dt

...

yn(x) = y0 +

∫ x

x0

f (t, yn−1(t)) dt

3. Under the assumptions of existence-uniqueness theorem, thesequence of approximations converges to the solution y(x) of(1). That is,

y(x) = limn→∞

yn(x).

Santanu Dey Lecture 5

Picard’s method

2. Solve (2) by iteration:

y1(x) = y0 +

∫ x

x0

f (t, y0) dt

y2(x) = y0 +

∫ x

x0

f (t, y1(t)) dt

...

yn(x) = y0 +

∫ x

x0

f (t, yn−1(t)) dt

3. Under the assumptions of existence-uniqueness theorem, thesequence of approximations converges to the solution y(x) of(1). That is,

y(x) = limn→∞

yn(x).

Santanu Dey Lecture 5

Picard’s method

2. Solve (2) by iteration:

y1(x) = y0 +

∫ x

x0

f (t, y0) dt

y2(x) = y0 +

∫ x

x0

f (t, y1(t)) dt

...

yn(x) = y0 +

∫ x

x0

f (t, yn−1(t)) dt

3. Under the assumptions of existence-uniqueness theorem, thesequence of approximations converges to the solution y(x) of(1).

That is,y(x) = lim

n→∞yn(x).

Santanu Dey Lecture 5

Picard’s method

2. Solve (2) by iteration:

y1(x) = y0 +

∫ x

x0

f (t, y0) dt

y2(x) = y0 +

∫ x

x0

f (t, y1(t)) dt

...

yn(x) = y0 +

∫ x

x0

f (t, yn−1(t)) dt

3. Under the assumptions of existence-uniqueness theorem, thesequence of approximations converges to the solution y(x) of(1). That is,

y(x) = limn→∞

yn(x).

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.

1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) =

1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt =

1 +x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) =

1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt

= 1 +x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) =

1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n.

(By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) =

limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x)

= ex2/2.

Santanu Dey Lecture 5

Example : Picard’s

Solve : y ′ = xy , y(0) = 1 using Picard’s iteration method.1 The integral equation is

y(x) = 1 +

∫ x

x0

ty dt.

2 The successive approximations are :

y1(x) = 1 +

∫ x

0t · 1 dt = 1 +

x2

2.

y2(x) = 1 +

∫ x

0t(1 +

t2

2) dt = 1 +

x2

2+

x4

2 · 4.

...

yn(x) = 1 + (x2

2) +

1

2!(x2

2)2 + · · ·+ 1

n!(x2

2)n. (By induction )

3 y(x) = limn→∞

yn(x) = ex2/2.

Santanu Dey Lecture 5

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?

(No, consider f (x) =√

x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .? (n=3).

Santanu Dey Lecture 5

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?(No, consider f (x) =

√x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .? (n=3).

Santanu Dey Lecture 5

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?(No, consider f (x) =

√x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .?

(n=3).

Santanu Dey Lecture 5

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?(No, consider f (x) =

√x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .? (n=3).

Santanu Dey Lecture 5

Exercises

1 Does uniform continuity =⇒ Lipschitz continuity ?(No, consider f (x) =

√x x ∈ [0, 2].)

2 The value of n such that the curves xn + yn = C are theorthogonal trajectories of the family

y =x

1− Kx

is . . . . . . . . .? (n=3).

Santanu Dey Lecture 5

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 5

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 5

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 5

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 5

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 5

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 5

Uniqueness of solution

Suppose φ1 and φ2 are both solutions of

y ′ = f (x , y), y(x0) = y0 .

Thus, both these satisfy the integral equation

φi (x) = y0 +

∫ x

x0

f (t, φi (t)) dt i = 1, 2.

For x ≥ x0,

φ1(x)− φ2(x) =

∫ x

x0

(f (t, φ1(t))− f (t, φ2(t))) dt.

Thus,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t))− f (t, φ2(t))| dt.

Since f satisfies Lipschitz condition w.r.t. the second variable, wehave

|f (t, φ1(t))− f (t, φ2(t))| ≤ M|φ1(t)− φ2(t)|.

Santanu Dey Lecture 5

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (3)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (3) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 5

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt

(3)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (3) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 5

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (3)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt.

Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (3) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 5

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (3)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0,

U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (3) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 5

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (3)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (3) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 5

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (3)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (3) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 5

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (3)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (3) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 5

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (3)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (3) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 5

Uniqueness - proof contd..

That is,

|φ1(x)− φ2(x)| ≤∫ x

x0

|f (t, φ1(t)− f (t, φ2(t)| dt

≤∫ x

x0

M|φ1(t)− φ2(t)| dt (3)

Set U(x) =

∫ x

x0

|φ1(t)− φ2(t)| dt. Then,

U(x0) = 0, U(x) ≥ 0, ∀x ≥ x0.

Further, U(x) is differentiable and

U ′(x) = |φ1(x)− φ2(x)|.

Hence, (3) yieldsU ′(x)−MU(x) ≤ 0.

Santanu Dey Lecture 5

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0 =⇒ U ′(x) ≡ 0 =⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 5

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0

=⇒ U ′(x) ≡ 0 =⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 5

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0 =⇒ U ′(x) ≡ 0

=⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 5

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0 =⇒ U ′(x) ≡ 0 =⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 5

Uniqueness Proof contd...

Multiplying the above equation by e−Mx gives

(e−MxU(x))′ ≤ 0 for x ≥ x0.

Integrating this from x0 to x we get U(x) ≤ 0 for x ≥ x0. So

U(x) = 0 ∀x ≥ x0 =⇒ U ′(x) ≡ 0 =⇒ φ1(x) ≡ φ2(x)

which contradicts the initial hypothesis.Use a similar argument to show for x ≤ x0.

Thus, φ1(x) ≡ φ2(x).

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations

- Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separable

Reducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separable

Exact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factors

Reducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theorem

Picard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5

Summary - First Order Equations

Linear Equations - Solution

Reducible to linear - Bernoulli

Non-linear equations

Variable separableReducible to variable separableExact equations - Integrating factorsReducible to Exact

Existence & Uniqueness results for IVP :

y ′ = f (x , y), y(x0) = y0

Peano’s existence theoremPicard’s existence-uniqueness theorem

Picard’s iteration method

Santanu Dey Lecture 5