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

MA 108 - Ordinary Differential Equations

Santanu Dey

Department of Mathematics,Indian Institute of Technology Bombay,

Powai, Mumbai [email protected]

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

d

dy(f (y))

dy

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

where f is an unknown function of y .

Set v = f (y) .

Then,

dv

dx=

dv

dy· dydx

=d

dy(f (y))

dy

dx.

Hence the given equation is

dv

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

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



Consider

d

dy(f (y))

dy

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


Set v = f (y) .

Then,

dv

dx=

dv

dy· dydx

=d

dy(f (y))

dy

dx.


dv





Consider

d

dy(f (y))

dy

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


Set v = f (y) .

Then,

dv

dx=

dv

dy· dydx

=d

dy(f (y))

dy

dx.


dv





Consider

d

dy(f (y))

dy

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


Set v = f (y) .

Then,

dv

dx=

dv

dy· dydx

=

d

dy(f (y))

dy

dx.


dv





Consider

d

dy(f (y))

dy

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


Set v = f (y) .

Then,

dv

dx=

dv

dy· dydx

=d

dy(f (y))

dy

dx.


dv





Consider

d

dy(f (y))

dy

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


Set v = f (y) .

Then,

dv

dx=

dv

dy· dydx

=d

dy(f (y))

dy

dx.


dv




Example

Solve : cos ydy

dx+

1

xsin y = 1.

Set v = sin y .

Then,

dv

dx=

dv

dy· dydx

= cos ydy

dx.


dv

dx+

1

xv = 1, which is linear in v .

That is,

e∫

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

2+ C

sin y =x

2+

C

x.


Example

Solve : cos ydy

dx+

1

xsin y = 1.

Set v = sin y .Then,

dv

dx=

dv

dy· dydx

= cos ydy

dx.


dv

dx+

1


That is,

e∫

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

2+ C

sin y =x

2+

C

x.


Example

Solve : cos ydy

dx+

1

xsin y = 1.


dv

dx=

dv

dy· dydx

=

cos ydy

dx.


dv

dx+

1


That is,

e∫

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

2+ C

sin y =x

2+

C

x.


Example

Solve : cos ydy

dx+

1

xsin y = 1.


dv

dx=

dv

dy· dydx

= cos ydy

dx.


dv

dx+

1


That is,

e∫

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

2+ C

sin y =x

2+

C

x.


Example

Solve : cos ydy

dx+

1

xsin y = 1.


dv

dx=

dv

dy· dydx

= cos ydy

dx.


dv

dx+

1


That is,

e∫

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

2+ C

sin y =x

2+

C

x.


Example

Solve : cos ydy

dx+

1

xsin y = 1.


dv

dx=

dv

dy· dydx

= cos ydy

dx.


dv

dx+

1


That is,

e∫

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

2+ C

sin y =x

2+

C

x.


Example

Solve : cos ydy

dx+

1

xsin y = 1.


dv

dx=

dv

dy· dydx

= cos ydy

dx.


dv

dx+

1


That is,

e∫

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

2+ C

sin y =x

2+

C

x.


Example

Solve : cos ydy

dx+

1

xsin y = 1.


dv

dx=

dv

dy· dydx

= cos ydy

dx.


dv

dx+

1


That is,

e∫

1xdxv(x) =

∫e∫

1xdxdx + C

=⇒ x v(x) =x2

2+ C

sin y =x

2+

C

x.



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



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

dy

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).


dy

dx= − 1

f (x , y).




Working Rule


F (x , y , c) = 0.

Find the DEdy

dx= f (x , y).


dy

dx= − 1

f (x , y).




Working Rule


F (x , y , c) = 0.

Find the DEdy

dx= f (x , y).


dy

dx= − 1

f (x , y).




Working Rule


F (x , y , c) = 0.

Find the DEdy

dx= f (x , y).


dy

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

y

The slope of OT’s aredy

dx=

y

x=⇒ y = kx .

Hence the orthogonal trajectories are given by y = kx .


Example


x + ydy

dx= 0

=⇒ dy

dx= −x

y


dx=

y

x=⇒ y = kx .



Example


x + ydy

dx= 0 =⇒ dy

dx= −x

y


dx=

y

x=⇒ y = kx .



Example


x + ydy

dx= 0 =⇒ dy

dx= −x

y

The slope of OT’s are

dy

dx=

y

x=⇒ y = kx .



Example


x + ydy

dx= 0 =⇒ dy

dx= −x

y


dx=

y

x

=⇒ y = kx .



Example


x + ydy

dx= 0 =⇒ dy

dx= −x

y


dx=

y

x=⇒ y = kx .



Example


x + ydy

dx= 0 =⇒ dy

dx= −x

y


dx=

y

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



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.



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.



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.

∂f

∂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


∂f

∂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



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

∣∣= |x |

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




Consider


∂f



|f (x , y1)− f (x , y2)| =

∣∣x |y1| − x |y2|∣∣

= |x |∣∣|y1| − |y2|∣∣

≤ |x | |y1 − y2|≤ a|y1 − y2|




Consider


∂f



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

∣∣

= |x |∣∣|y1| − |y2|∣∣

≤ |x | |y1 − y2|≤ a|y1 − y2|




Consider


∂f



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

∣∣= |x |

∣∣|y1| − |y2|∣∣

≤ |x | |y1 − y2|≤ a|y1 − y2|




Consider


∂f



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

∣∣= |x |

∣∣|y1| − |y2|∣∣≤ |x | |y1 − y2|

≤ a|y1 − y2|




Consider


∂f



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

∣∣= |x |

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




Consider


∂f



|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

{a,

b

K

}.

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

{a,

b

K

}.











α = min

{a,

b

K

}.











α = min

{a,

b

K

}.











α = min

{a,

b

K

}.











α = min

{a,

b

K

}.











α = min

{a,

b

K

}.





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