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transcript
MA 108 - Ordinary Differential Equations
Santanu Dey
Department of Mathematics,Indian Institute of Technology Bombay,
Powai, Mumbai 76dey@math.iitb.ac.in
March 10, 2014
Santanu Dey Lecture 7
Outline of the lecture
Lipschitz continuity
Existence & uniqueness
Picard’s iteration
Second order linear equations
Santanu Dey Lecture 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 (1)
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, (1) yieldsU ′(x)−MU(x) ≤ 0.
Santanu Dey Lecture 7
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
(1)
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, (1) yieldsU ′(x)−MU(x) ≤ 0.
Santanu Dey Lecture 7
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 (1)
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, (1) yieldsU ′(x)−MU(x) ≤ 0.
Santanu Dey Lecture 7
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 (1)
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, (1) yieldsU ′(x)−MU(x) ≤ 0.
Santanu Dey Lecture 7
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 (1)
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, (1) yieldsU ′(x)−MU(x) ≤ 0.
Santanu Dey Lecture 7
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 (1)
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, (1) yieldsU ′(x)−MU(x) ≤ 0.
Santanu Dey Lecture 7
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 (1)
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, (1) yieldsU ′(x)−MU(x) ≤ 0.
Santanu Dey Lecture 7
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 (1)
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, (1) yieldsU ′(x)−MU(x) ≤ 0.
Santanu Dey Lecture 7
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 (1)
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, (1) yieldsU ′(x)−MU(x) ≤ 0.
Santanu Dey Lecture 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
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 7
Second order differential equations
Recall that a general second order linear ODE is of the form
a2(x)d2y
dx2+ a1(x)
dy
dx+ a0(x)y = g(x).
An ODE of the form
d2y
dx2+ p(x)
dy
dx+ q(x)y = r(x)
is called a second order linear ODE in standard form.Though there is no formula to find all the solutions of such anODE, we study the existence, uniqueness and number of solutionsof such ODE’s.
Santanu Dey Lecture 7
Second order differential equations
Recall that a general second order linear ODE is of the form
a2(x)d2y
dx2+ a1(x)
dy
dx+ a0(x)y = g(x).
An ODE of the form
d2y
dx2+ p(x)
dy
dx+ q(x)y = r(x)
is called a second order linear ODE in standard form.
Though there is no formula to find all the solutions of such anODE, we study the existence, uniqueness and number of solutionsof such ODE’s.
Santanu Dey Lecture 7
Second order differential equations
Recall that a general second order linear ODE is of the form
a2(x)d2y
dx2+ a1(x)
dy
dx+ a0(x)y = g(x).
An ODE of the form
d2y
dx2+ p(x)
dy
dx+ q(x)y = r(x)
is called a second order linear ODE in standard form.Though there is no formula to find all the solutions of such anODE, we study the existence, uniqueness and number of solutionsof such ODE’s.
Santanu Dey Lecture 7
Homogeneous Linear Second Order DE
If r(x) ≡ 0 in the equation
d2y
dx2+ p(x)
dy
dx+ q(x)y = r(x),
that is,d2y
dx2+ p(x)
dy
dx+ q(x)y = 0,
then the ODE is said to be homogeneous.Otherwise it is called non-homogeneous.
Santanu Dey Lecture 7
Homogeneous Linear Second Order DE
If r(x) ≡ 0 in the equation
d2y
dx2+ p(x)
dy
dx+ q(x)y = r(x),
that is,d2y
dx2+ p(x)
dy
dx+ q(x)y = 0,
then the ODE is said to be homogeneous.Otherwise it is called non-homogeneous.
Santanu Dey Lecture 7
Homogeneous Linear Second Order DE
If r(x) ≡ 0 in the equation
d2y
dx2+ p(x)
dy
dx+ q(x)y = r(x),
that is,d2y
dx2+ p(x)
dy
dx+ q(x)y = 0,
then the ODE is said to be homogeneous.
Otherwise it is called non-homogeneous.
Santanu Dey Lecture 7
Homogeneous Linear Second Order DE
If r(x) ≡ 0 in the equation
d2y
dx2+ p(x)
dy
dx+ q(x)y = r(x),
that is,d2y
dx2+ p(x)
dy
dx+ q(x)y = 0,
then the ODE is said to be homogeneous.Otherwise it is called non-homogeneous.
Santanu Dey Lecture 7
Homogeneous Linear Second Order DE
If r(x) ≡ 0 in the equation
d2y
dx2+ p(x)
dy
dx+ q(x)y = r(x),
that is,d2y
dx2+ p(x)
dy
dx+ q(x)y = 0,
then the ODE is said to be homogeneous.Otherwise it is called non-homogeneous.
Santanu Dey Lecture 7
Initial Value Problem- Existence/Uniqueness
An initial value problem of a second order homogeneous linearODE is of the form:
y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a, y ′(x0) = b,
where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .
Santanu Dey Lecture 7
Initial Value Problem- Existence/Uniqueness
An initial value problem of a second order homogeneous linearODE is of the form:
y ′′ + p(x)y ′ + q(x)y = 0,
y(x0) = a, y ′(x0) = b,
where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .
Santanu Dey Lecture 7
Initial Value Problem- Existence/Uniqueness
An initial value problem of a second order homogeneous linearODE is of the form:
y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a,
y ′(x0) = b,
where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .
Santanu Dey Lecture 7
Initial Value Problem- Existence/Uniqueness
An initial value problem of a second order homogeneous linearODE is of the form:
y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a, y ′(x0) = b,
where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .
Santanu Dey Lecture 7
Initial Value Problem- Existence/Uniqueness
An initial value problem of a second order homogeneous linearODE is of the form:
y ′′ + p(x)y ′ + q(x)y = 0, y(x0) = a, y ′(x0) = b,
where p(x) and q(x) are assumed to be continuous on an openinterval I with x0 ∈ I , has a unique solution y(x) in the interval I .
Santanu Dey Lecture 7