MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture4_D4.pdf · Orthogonal...

transcript

MA 108 - Ordinary Differential Equations

Santanu Dey

Department of Mathematics,Indian Institute of Technology Bombay,

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

March 6, 2014

Santanu Dey Lecture 4

Outline of the lecture

Bernoulli equation

Orthogonal Trajectories

Lipschitz continuity

Existence & uniqueness

Equations reducible to linear equations

Consider

dy(f (y))

dx+ P(x)f (y) = Q(x),

where f is an unknown function of y .

Set v = f (y) .

dy· dydx

dy(f (y))

Hence the given equation is

dx+ P(x)v = Q(x), which is linear in v .

Remark : Bernoulli DE is a special case when f (y) = y1−n.

Consider

dy(f (y))

dx+ P(x)f (y) = Q(x),

Set v = f (y) .

dy· dydx

dy(f (y))

Consider

dy(f (y))

dx+ P(x)f (y) = Q(x),

Set v = f (y) .

dy· dydx

dy(f (y))

Consider

dy(f (y))

dx+ P(x)f (y) = Q(x),

Set v = f (y) .

dy· dydx

dy(f (y))

Consider

dy(f (y))

dx+ P(x)f (y) = Q(x),

Set v = f (y) .

dy· dydx

dy(f (y))

Consider

dy(f (y))

dx+ P(x)f (y) = Q(x),

Set v = f (y) .

dy· dydx

dy(f (y))

Example

Solve : cos ydy

xsin y = 1.

Set v = sin y .

dy· dydx

= cos ydy

xv = 1, which is linear in v .

That is,

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

sin y =x

Example

Solve : cos ydy

xsin y = 1.

Set v = sin y .Then,

dy· dydx

= cos ydy

That is,

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

sin y =x

Example

Solve : cos ydy

xsin y = 1.

dy· dydx

cos ydy

That is,

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

sin y =x

Example

Solve : cos ydy

xsin y = 1.

dy· dydx

= cos ydy

That is,

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

sin y =x

Example

Solve : cos ydy

xsin y = 1.

dy· dydx

= cos ydy

That is,

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

sin y =x

Example

Solve : cos ydy

xsin y = 1.

dy· dydx

= cos ydy

That is,

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

sin y =x

Example

Solve : cos ydy

xsin y = 1.

dy· dydx

= cos ydy

That is,

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

sin y =x

Example

Solve : cos ydy

xsin y = 1.

dy· dydx

= cos ydy

That is,

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

sin y =x

If two families of curves always intersect each other at right angles,then they are said to be orthogonal trajectories of each other.

Working Rule

To find the OT of a family of curves

F (x , y , c) = 0.

Find the DEdy

dx= f (x , y).

Slopes of the OT’s are given by

dx= − 1

f (x , y).

Obtain a one parameter family of curves G (x , y , c) = 0 assolutions of the above DE.

( Leaving a part certain trajectories that are vertical lines!)

Working Rule

F (x , y , c) = 0.

Find the DEdy

dx= f (x , y).

dx= − 1

f (x , y).

Working Rule

F (x , y , c) = 0.

Find the DEdy

dx= f (x , y).

dx= − 1

f (x , y).

Working Rule

F (x , y , c) = 0.

Find the DEdy

dx= f (x , y).

dx= − 1

f (x , y).

Working Rule

F (x , y , c) = 0.

Find the DEdy

dx= f (x , y).

dx= − 1

f (x , y).

Example

Find the set of OT’s of the family of circles x2 + y2 = c2.

x + ydy

dx= 0 =⇒ dy

dx= −x

The slope of OT’s aredy

x=⇒ y = kx .

Hence the orthogonal trajectories are given by y = kx .

Example

x + ydy

=⇒ dy

dx= −x

x=⇒ y = kx .

Example

x + ydy

dx= 0 =⇒ dy

dx= −x

x=⇒ y = kx .

Example

x + ydy

dx= 0 =⇒ dy

dx= −x

The slope of OT’s are

x=⇒ y = kx .

Example

x + ydy

dx= 0 =⇒ dy

dx= −x

=⇒ y = kx .

Example

x + ydy

dx= 0 =⇒ dy

dx= −x

x=⇒ y = kx .

Example

x + ydy

dx= 0 =⇒ dy

dx= −x

x=⇒ y = kx .

Definitions

1 Let f be a real function defined on D, where D is either adomain or a closed domain of the xy plane. The function f issaid to be bounded in D if there exists a positive number Msuch that

|f (x , y)| ≤ M

for all (x , y) in D.

2 Let f be defined and continuous on a closed rectangleR : a ≤ x ≤ b, c ≤ y ≤ d . Then, f is bounded in R.

3 Let f be defined on D, where D is either a domain or a closeddomain of the xy - plane. The function f is said to satisfyLipschitz condition (with respect to y) in D if ∃ a constantM > 0 such that

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

for every (x , y1), (x , y2) which belong to D. The constant Mis called the Lipschitz constant.

Definitions

|f (x , y)| ≤ M

for all (x , y) in D.2 Let f be defined and continuous on a closed rectangle

R : a ≤ x ≤ b, c ≤ y ≤ d .

Then, f is bounded in R.3 Let f be defined on D, where D is either a domain or a closed

domain of the xy - plane. The function f is said to satisfyLipschitz condition (with respect to y) in D if ∃ a constantM > 0 such that

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

Definitions

|f (x , y)| ≤ M

R : a ≤ x ≤ b, c ≤ y ≤ d . Then, f is bounded in R.

3 Let f be defined on D, where D is either a domain or a closeddomain of the xy - plane. The function f is said to satisfyLipschitz condition (with respect to y) in D if ∃ a constantM > 0 such that

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

Definitions

|f (x , y)| ≤ M

R : a ≤ x ≤ b, c ≤ y ≤ d . Then, f is bounded in R.3 Let f be defined on D, where D is either a domain or a closed

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

Definitions

|f (x , y)| ≤ M

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

Definitions

|f (x , y)| ≤ M

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

Definitions

|f (x , y)| ≤ M

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

Understanding the Lipschitz condition - y = g(x)

Consider

|g(x2)− g(x1)| ≤ M|x2 − x1| ∀x1, x2 in the domain of g .

This condition in the form|g(x2)− g(x1)||x2 − x1|

≤ M can be interpreted

as follows :At each point (a, g(a)), the entire graph of g lies between the lines

y = g(a)−M(x − a) &y = g(a) + M(x − a).

Example : x2 is Lipschitz in [1, 2].

Consider

as follows :

At each point (a, g(a)), the entire graph of g lies between the lines

y = g(a)−M(x − a) &y = g(a) + M(x − a).

Consider

y = g(a)−M(x − a) &y = g(a) + M(x − a).

Consider

y = g(a)−M(x − a) &y = g(a) + M(x − a).

Example : x2 is Lipschitz in [1, 2].Santanu Dey Lecture 4

Understanding Lipschitz condition - z = f (x , y)

Let (x , y1) and (x , y2) be any two points in D having thesame abscissa x .

Consider the corresponding points

P1(x , y1, f (x , y1)) & P2(x , y2, f (x , y2))

on the surface z = f (x , y), and let α (0 ≤ α ≤ π/2) denotethe angle that the chord joining P1 and P2 makes with thexy - plane.

Then if the condition

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

holds in D, then tanα is bounded in absolute value.

That is, the chord joining P1 and P2 is bounded away frombeing perpendicular to the xy - plane.

Further, this bound is independent of the points (x , y1) and(x , y2) belonging to D.

P1(x , y1, f (x , y1)) & P2(x , y2, f (x , y2))

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

P1(x , y1, f (x , y1)) & P2(x , y2, f (x , y2))

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

P1(x , y1, f (x , y1)) & P2(x , y2, f (x , y2))

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

P1(x , y1, f (x , y1)) & P2(x , y2, f (x , y2))

|f (x , y1)− f (x , y2)| ≤ M|y1 − y2|

Lipschitz condition =⇒ Continuity ?

If f satisfies Lipschitz condition with respect to y in D, then foreach fixed x , the resulting function of y is a continuous function ofy , for all (x , y) in D.

Example : Let f (x , y) = y + [x ].

For fixed x ,

f (x , y1)− f (x , y2) = y1 + [x ]− y2 − [x ]

= y1 − y2

That is, |f (x , y1)− f (x , y2)| = |y1 − y2| ≤ 1 · |y1 − y2|But we know that f is discontinuous w.r.t. x for every integralvalue of x .

Note that the condition of Lipschitz continuity implies nothing con-cerning the continuity of f with respect to x .

For fixed x ,

f (x , y1)− f (x , y2) = y1 + [x ]− y2 − [x ]

= y1 − y2

For fixed x ,

f (x , y1)− f (x , y2) = y1 + [x ]− y2 − [x ]

= y1 − y2

For fixed x ,

f (x , y1)− f (x , y2)

= y1 + [x ]− y2 − [x ]

= y1 − y2

For fixed x ,

f (x , y1)− f (x , y2) = y1 + [x ]− y2 − [x ]

= y1 − y2

For fixed x ,

f (x , y1)− f (x , y2) = y1 + [x ]− y2 − [x ]

= y1 − y2

For fixed x ,

f (x , y1)− f (x , y2) = y1 + [x ]− y2 − [x ]

= y1 − y2

That is, |f (x , y1)− f (x , y2)| = |y1 − y2|

≤ 1 · |y1 − y2|But we know that f is discontinuous w.r.t. x for every integralvalue of x .

For fixed x ,

f (x , y1)− f (x , y2) = y1 + [x ]− y2 − [x ]

= y1 − y2

That is, |f (x , y1)− f (x , y2)| = |y1 − y2| ≤ 1 · |y1 − y2|

But we know that f is discontinuous w.r.t. x for every integralvalue of x .

For fixed x ,

f (x , y1)− f (x , y2) = y1 + [x ]− y2 − [x ]

= y1 − y2

For fixed x ,

f (x , y1)− f (x , y2) = y1 + [x ]− y2 − [x ]

= y1 − y2

Does Continuity w.r.t. second variable =⇒ Lipschitzcondtn. w.r.t. second variable?

Continuity w.r.t. second variable DOES NOT imply Lipschitzcondtn. w.r.t. second variable.

Example : Consider f (x , y) =√|y |.

f is continuous for all y .Note that f doesn’t satisfy Lipschitz condition in any region thatincludes y = 0 as for y1 = 0, y2 > 0, we have

|f (x , y1)− f (x , y2)||y1 − y2|

√y2|y2|

=1√y2

which can be made as large as we want by making y2 smaller.The Lipschitz condition requires that the quotient should bebounded by a fixed constant K .

|f (x , y1)− f (x , y2)||y1 − y2|

√y2|y2|

=1√y2

f is continuous for all y .

Note that f doesn’t satisfy Lipschitz condition in any region thatincludes y = 0 as for y1 = 0, y2 > 0, we have

|f (x , y1)− f (x , y2)||y1 − y2|

√y2|y2|

=1√y2

|f (x , y1)− f (x , y2)||y1 − y2|

√y2|y2|

=1√y2

|f (x , y1)− f (x , y2)||y1 − y2|

√y2|y2|

=1√y2

|f (x , y1)− f (x , y2)||y1 − y2|

√y2|y2|

=1√y2

|f (x , y1)− f (x , y2)||y1 − y2|

√y2|y2|

=1√y2

|f (x , y1)− f (x , y2)||y1 − y2|

√y2|y2|

=1√y2

which can be made as large as we want by making y2 smaller.

The Lipschitz condition requires that the quotient should bebounded by a fixed constant K .

|f (x , y1)− f (x , y2)||y1 − y2|

√y2|y2|

=1√y2

Sufficiency

Result : If f is such that∂f

∂yexists and is bounded for all

(x , y) ∈ D, then f satisfies Lipschitz condition w.r.t. y in D, wherethe Lipschitz constant

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

Proof : Mean value theorem

=⇒ f (x , y1)− f (x , y2) = (y1 − y2)∂f

∂y(x , ξ), ξ ∈ (y1, y2).

|f (x , y1)− f (x , y2)| = |y1 − y2||∂f

∂y(x , ξ)|

≤ |y1 − y2| l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

That is, f satisfies Lipschitz condition.

Sufficiency

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

f (x , y1)− f (x , y2) = (y1 − y2)∂f

∂y(x , ξ), ξ ∈ (y1, y2).

|f (x , y1)− f (x , y2)| = |y1 − y2||∂f

∂y(x , ξ)|

≤ |y1 − y2| l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

Sufficiency

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

=⇒ f (x , y1)− f (x , y2) = (y1 − y2)∂f

∂y(x , ξ),

ξ ∈ (y1, y2).

|f (x , y1)− f (x , y2)| = |y1 − y2||∂f

∂y(x , ξ)|

≤ |y1 − y2| l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

Sufficiency

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

=⇒ f (x , y1)− f (x , y2) = (y1 − y2)∂f

∂y(x , ξ), ξ ∈ (y1, y2).

|f (x , y1)− f (x , y2)| = |y1 − y2||∂f

∂y(x , ξ)|

≤ |y1 − y2| l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

Sufficiency

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

=⇒ f (x , y1)− f (x , y2) = (y1 − y2)∂f

∂y(x , ξ), ξ ∈ (y1, y2).

|f (x , y1)− f (x , y2)|

= |y1 − y2||∂f

∂y(x , ξ)|

≤ |y1 − y2| l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

Sufficiency

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

=⇒ f (x , y1)− f (x , y2) = (y1 − y2)∂f

∂y(x , ξ), ξ ∈ (y1, y2).

|f (x , y1)− f (x , y2)| = |y1 − y2||∂f

∂y(x , ξ)|

≤ |y1 − y2| l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

Sufficiency

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

=⇒ f (x , y1)− f (x , y2) = (y1 − y2)∂f

∂y(x , ξ), ξ ∈ (y1, y2).

|f (x , y1)− f (x , y2)| = |y1 − y2||∂f

∂y(x , ξ)|

≤ |y1 − y2| l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

Sufficiency

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

=⇒ f (x , y1)− f (x , y2) = (y1 − y2)∂f

∂y(x , ξ), ξ ∈ (y1, y2).

|f (x , y1)− f (x , y2)| = |y1 − y2||∂f

∂y(x , ξ)|

≤ |y1 − y2| l .u.b.(x ,y)∈D |∂f

∂y(x , y)|.

Example

Consider

f (x , y) = y2 defined in D : |x | ≤ a, |y | ≤ b.

fy = 2y is bounded in D. The Lipschitz contant is

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)| = l .u.b.(x ,y)∈D |2y | = 2b.

(Verify Lipschitz condition directly! )

Example

Consider

fy = 2y is bounded in D.

The Lipschitz contant is

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)| = l .u.b.(x ,y)∈D |2y | = 2b.

Example

Consider

l .u.b.(x ,y)∈D |∂f

∂y(x , y)| = l .u.b.(x ,y)∈D |2y | = 2b.

Example

Consider

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)| =

l .u.b.(x ,y)∈D |2y | = 2b.

Example

Consider

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)| = l .u.b.(x ,y)∈D |2y |

Example

Consider

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)| = l .u.b.(x ,y)∈D |2y | = 2b.

Example

Consider

M = l .u.b.(x ,y)∈D |∂f

∂y(x , y)| = l .u.b.(x ,y)∈D |2y | = 2b.

Bounded derivative - sufficient condition

Consider

f (x , y) = x |y | defined in D : |x | ≤ a, |y | ≤ b.

∂ydoesn’t exist for any point (x , 0) ∈ D. (Why?)

Now f satisfies Lipschitz condition :

|f (x , y1)− f (x , y2)| =∣∣x |y1| − x |y2|

∣∣= |x |

∣∣|y1| − |y2|∣∣≤ |x | |y1 − y2|≤ a|y1 − y2|

Existence of bounded derivative fy is a sufficient condition for Lips-chitz condition to hold true (not necessary).

Consider

∂ydoesn’t exist for any point (x , 0) ∈ D.

(Why?)

|f (x , y1)− f (x , y2)| =∣∣x |y1| − x |y2|

∣∣= |x |

∣∣|y1| − |y2|∣∣≤ |x | |y1 − y2|≤ a|y1 − y2|

Consider

|f (x , y1)− f (x , y2)| =∣∣x |y1| − x |y2|

∣∣= |x |

∣∣|y1| − |y2|∣∣≤ |x | |y1 − y2|≤ a|y1 − y2|

Consider

|f (x , y1)− f (x , y2)| =

∣∣x |y1| − x |y2|∣∣

= |x |∣∣|y1| − |y2|∣∣

≤ |x | |y1 − y2|≤ a|y1 − y2|

Consider

|f (x , y1)− f (x , y2)| =∣∣x |y1| − x |y2|

∣∣

= |x |∣∣|y1| − |y2|∣∣

≤ |x | |y1 − y2|≤ a|y1 − y2|

Consider

|f (x , y1)− f (x , y2)| =∣∣x |y1| − x |y2|

∣∣= |x |

∣∣|y1| − |y2|∣∣

≤ |x | |y1 − y2|≤ a|y1 − y2|

Consider

|f (x , y1)− f (x , y2)| =∣∣x |y1| − x |y2|

∣∣= |x |

∣∣|y1| − |y2|∣∣≤ |x | |y1 − y2|

≤ a|y1 − y2|

Consider

|f (x , y1)− f (x , y2)| =∣∣x |y1| − x |y2|

∣∣= |x |

∣∣|y1| − |y2|∣∣≤ |x | |y1 − y2|≤ a|y1 − y2|

Consider

|f (x , y1)− f (x , y2)| =∣∣x |y1| − x |y2|

∣∣= |x |

∣∣|y1| − |y2|∣∣≤ |x | |y1 − y2|≤ a|y1 − y2|

Existence - Uniqueness Theorem

Let R be a rectangle containing (x0, y0) in the domain D,

f (x , y) be continuous at all points (x , y) inR : |x − x0| < a, |y − y0| < b and

bounded in R, that is, |f (x , y)| ≤ K ∀(x , y) ∈ R.

Then, the IVP y ′ = f (x , y), y(x0) = y0 has at least one solution y(x)

defined for all x in the interval |x − x0| < α, where

α = min

In addition to the above conditions, if f satisfies the Lipschitz conditionwith respect to y in R, that is,

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

then, the solution y(x) defined at least for all x in the interval |x−x0| < α,with α defined above is unique 1.

1Existence - Peano, Existence & uniqueness -PicardSantanu Dey Lecture 4

α = min

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