MA 108 - Ordinary Differential Equations
Santanu Dey
Department of Mathematics,Indian Institute of Technology Bombay,
Powai, Mumbai [email protected]
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