MA 108 - Ordinary Differential Equationsdey/diffeqn_autumn13/lecture4.pdf · Hence the orthogonal trajectories are given by y = kx . Santanu Dey Lecture 4. De nitions 1 Let f be a
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MA 108 - Ordinary Differential Equations
Santanu Dey
Department of Mathematics,Indian Institute of Technology Bombay,
Powai, Mumbai 76dey@math.iitb.ac.in
September 27, 2013
Santanu Dey Lecture 4
Outline of the lecture
Orthogonal Trajectories
Lipschitz continuity
Existence & uniqueness
Picard’s iteration
Santanu Dey Lecture 4
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .
Then,
dv
dx=
dv
dy· dydx
= cos ydy
dx.
Hence the given equation is
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.
Santanu Dey Lecture 4
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dydx
= cos ydy
dx.
Hence the given equation is
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.
Santanu Dey Lecture 4
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dydx
=
cos ydy
dx.
Hence the given equation is
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.
Santanu Dey Lecture 4
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dydx
= cos ydy
dx.
Hence the given equation is
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.
Santanu Dey Lecture 4
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dydx
= cos ydy
dx.
Hence the given equation is
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.
Santanu Dey Lecture 4
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dydx
= cos ydy
dx.
Hence the given equation is
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.
Santanu Dey Lecture 4
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dydx
= cos ydy
dx.
Hence the given equation is
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.
Santanu Dey Lecture 4
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dydx
= cos ydy
dx.
Hence the given equation is
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.
Santanu Dey Lecture 4
Orthogonal Trajectories
If two families of curves always intersect each other at right angles,then they are said to be orthogonal trajectories of each other.
Santanu Dey Lecture 4
Orthogonal Trajectories
If two families of curves always intersect each other at right angles,then they are said to be orthogonal trajectories of each other.
Santanu Dey Lecture 4
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!)
Santanu Dey Lecture 4
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!)
Santanu Dey Lecture 4
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!)
Santanu Dey Lecture 4
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!)
Santanu Dey Lecture 4
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!)
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 are
dy
dx=
y
x=⇒ y = kx .
Hence the orthogonal trajectories are given by y = kx .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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 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|
for every (x , y1), (x , y2) which belong to D. The constant Mis called the Lipschitz constant.
Santanu Dey Lecture 4
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 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 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.
Santanu Dey Lecture 4
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 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|
for every (x , y1), (x , y2) which belong to D. The constant Mis called the Lipschitz constant.
Santanu Dey Lecture 4
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 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|
for every (x , y1), (x , y2) which belong to D. The constant Mis called the Lipschitz constant.
Santanu Dey Lecture 4
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 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|
for every (x , y1), (x , y2) which belong to D. The constant Mis called the Lipschitz constant.
Santanu Dey Lecture 4
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 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|
for every (x , y1), (x , y2) which belong to D. The constant Mis called the Lipschitz constant.
Santanu Dey Lecture 4
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].
Santanu Dey Lecture 4
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].
Santanu Dey Lecture 4
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].
Santanu Dey Lecture 4
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].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.
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.
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.
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.
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.
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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 .
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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! )
Santanu Dey Lecture 4
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! )
Santanu Dey Lecture 4
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! )
Santanu Dey Lecture 4
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! )
Santanu Dey Lecture 4
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! )
Santanu Dey Lecture 4
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! )
Santanu Dey Lecture 4
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! )
Santanu Dey Lecture 4
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).
Santanu Dey Lecture 4
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).
Santanu Dey Lecture 4
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).
Santanu Dey Lecture 4
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).
Santanu Dey Lecture 4
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).
Santanu Dey Lecture 4
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).
Santanu Dey Lecture 4
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).
Santanu Dey Lecture 4
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).
Santanu Dey Lecture 4
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).
Santanu Dey Lecture 4
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
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
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
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
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
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
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
A quick check!
1 Is f (x) = sin x Lipschitz continuous over R?
Yes.
2 Is f (x) = x2 globally Lipschitz continuous over R ? No.However, it is Lipschitz continuous over any closed interval ofR. We say that it is locally Lipschitz continuous over R.
3 Is f (x) =1
x2globally Lipschitz continuous on [α,∞) for any
α > 0? Yes.
Santanu Dey Lecture 4
A quick check!
1 Is f (x) = sin x Lipschitz continuous over R? Yes.
2 Is f (x) = x2 globally Lipschitz continuous over R ?
No.However, it is Lipschitz continuous over any closed interval ofR. We say that it is locally Lipschitz continuous over R.
3 Is f (x) =1
x2globally Lipschitz continuous on [α,∞) for any
α > 0? Yes.
Santanu Dey Lecture 4
A quick check!
1 Is f (x) = sin x Lipschitz continuous over R? Yes.
2 Is f (x) = x2 globally Lipschitz continuous over R ? No.
However, it is Lipschitz continuous over any closed interval ofR. We say that it is locally Lipschitz continuous over R.
3 Is f (x) =1
x2globally Lipschitz continuous on [α,∞) for any
α > 0? Yes.
Santanu Dey Lecture 4
A quick check!
1 Is f (x) = sin x Lipschitz continuous over R? Yes.
2 Is f (x) = x2 globally Lipschitz continuous over R ? No.However, it is Lipschitz continuous over any closed interval ofR.
We say that it is locally Lipschitz continuous over R.
3 Is f (x) =1
x2globally Lipschitz continuous on [α,∞) for any
α > 0? Yes.
Santanu Dey Lecture 4
A quick check!
1 Is f (x) = sin x Lipschitz continuous over R? Yes.
2 Is f (x) = x2 globally Lipschitz continuous over R ? No.However, it is Lipschitz continuous over any closed interval ofR. We say that it is locally Lipschitz continuous over R.
3 Is f (x) =1
x2globally Lipschitz continuous on [α,∞) for any
α > 0? Yes.
Santanu Dey Lecture 4
A quick check!
1 Is f (x) = sin x Lipschitz continuous over R? Yes.
2 Is f (x) = x2 globally Lipschitz continuous over R ? No.However, it is Lipschitz continuous over any closed interval ofR. We say that it is locally Lipschitz continuous over R.
3 Is f (x) =1
x2globally Lipschitz continuous on [α,∞) for any
α > 0?
Yes.
Santanu Dey Lecture 4
A quick check!
1 Is f (x) = sin x Lipschitz continuous over R? Yes.
2 Is f (x) = x2 globally Lipschitz continuous over R ? No.However, it is Lipschitz continuous over any closed interval ofR. We say that it is locally Lipschitz continuous over R.
3 Is f (x) =1
x2globally Lipschitz continuous on [α,∞) for any
α > 0? Yes.
Santanu Dey Lecture 4
A quick check!
1 Is f (x) = sin x Lipschitz continuous over R? Yes.
2 Is f (x) = x2 globally Lipschitz continuous over R ? No.However, it is Lipschitz continuous over any closed interval ofR. We say that it is locally Lipschitz continuous over R.
3 Is f (x) =1
x2globally Lipschitz continuous on [α,∞) for any
α > 0? Yes.
Santanu Dey Lecture 4
Example 1
Consider
y ′ = y1/3 y(0) = 0 in R : |x | ≤ a, |y | ≤ b.
f (x , y) is continuous in R and hence existence is guaranteed.
But φ1(x) = 0 and φ2(x) =
(2
3x)3/2 if x ≥ 0
0 if x ≤ 0are solutions in
−∞ < x <∞.Does this imply Lipschitz condition won’t be satisfied?
|f (x , y1)− f (x , y2)||y1 − y2|
=|y1/31 − y
1/32 |
|y1 − y2|.
Choosing y1 = δ, y2 = −δ, we see that the quotient is unboundedfor small values of δ and hence Lipschitz condition is not satisfied.Solution exists, but not unique.
Santanu Dey Lecture 4
Example 1
Consider
y ′ = y1/3 y(0) = 0 in R : |x | ≤ a, |y | ≤ b.
f (x , y) is continuous in R and hence existence is guaranteed.
But φ1(x) = 0 and φ2(x) =
(2
3x)3/2 if x ≥ 0
0 if x ≤ 0are solutions in
−∞ < x <∞.Does this imply Lipschitz condition won’t be satisfied?
|f (x , y1)− f (x , y2)||y1 − y2|
=|y1/31 − y
1/32 |
|y1 − y2|.
Choosing y1 = δ, y2 = −δ, we see that the quotient is unboundedfor small values of δ and hence Lipschitz condition is not satisfied.Solution exists, but not unique.
Santanu Dey Lecture 4
Example 1
Consider
y ′ = y1/3 y(0) = 0 in R : |x | ≤ a, |y | ≤ b.
f (x , y) is continuous in R and hence existence is guaranteed.
But φ1(x) = 0
and φ2(x) =
(2
3x)3/2 if x ≥ 0
0 if x ≤ 0are solutions in
−∞ < x <∞.Does this imply Lipschitz condition won’t be satisfied?
|f (x , y1)− f (x , y2)||y1 − y2|
=|y1/31 − y
1/32 |
|y1 − y2|.
Choosing y1 = δ, y2 = −δ, we see that the quotient is unboundedfor small values of δ and hence Lipschitz condition is not satisfied.Solution exists, but not unique.
Santanu Dey Lecture 4
Example 1
Consider
y ′ = y1/3 y(0) = 0 in R : |x | ≤ a, |y | ≤ b.
f (x , y) is continuous in R and hence existence is guaranteed.
But φ1(x) = 0 and φ2(x) =
(2
3x)3/2 if x ≥ 0
0 if x ≤ 0are solutions in
−∞ < x <∞.
Does this imply Lipschitz condition won’t be satisfied?
|f (x , y1)− f (x , y2)||y1 − y2|
=|y1/31 − y
1/32 |
|y1 − y2|.
Choosing y1 = δ, y2 = −δ, we see that the quotient is unboundedfor small values of δ and hence Lipschitz condition is not satisfied.Solution exists, but not unique.
Santanu Dey Lecture 4
Example 1
Consider
y ′ = y1/3 y(0) = 0 in R : |x | ≤ a, |y | ≤ b.
f (x , y) is continuous in R and hence existence is guaranteed.
But φ1(x) = 0 and φ2(x) =
(2
3x)3/2 if x ≥ 0
0 if x ≤ 0are solutions in
−∞ < x <∞.Does this imply Lipschitz condition won’t be satisfied?
|f (x , y1)− f (x , y2)||y1 − y2|
=|y1/31 − y
1/32 |
|y1 − y2|.
Choosing y1 = δ, y2 = −δ, we see that the quotient is unboundedfor small values of δ and hence Lipschitz condition is not satisfied.Solution exists, but not unique.
Santanu Dey Lecture 4
Example 1
Consider
y ′ = y1/3 y(0) = 0 in R : |x | ≤ a, |y | ≤ b.
f (x , y) is continuous in R and hence existence is guaranteed.
But φ1(x) = 0 and φ2(x) =
(2
3x)3/2 if x ≥ 0
0 if x ≤ 0are solutions in
−∞ < x <∞.Does this imply Lipschitz condition won’t be satisfied?
|f (x , y1)− f (x , y2)||y1 − y2|
=
|y1/31 − y1/32 |
|y1 − y2|.
Choosing y1 = δ, y2 = −δ, we see that the quotient is unboundedfor small values of δ and hence Lipschitz condition is not satisfied.Solution exists, but not unique.
Santanu Dey Lecture 4
Example 1
Consider
y ′ = y1/3 y(0) = 0 in R : |x | ≤ a, |y | ≤ b.
f (x , y) is continuous in R and hence existence is guaranteed.
But φ1(x) = 0 and φ2(x) =
(2
3x)3/2 if x ≥ 0
0 if x ≤ 0are solutions in
−∞ < x <∞.Does this imply Lipschitz condition won’t be satisfied?
|f (x , y1)− f (x , y2)||y1 − y2|
=|y1/31 − y
1/32 |
|y1 − y2|.
Choosing y1 = δ, y2 = −δ, we see that the quotient is unboundedfor small values of δ and hence Lipschitz condition is not satisfied.Solution exists, but not unique.
Santanu Dey Lecture 4
Example 1
Consider
y ′ = y1/3 y(0) = 0 in R : |x | ≤ a, |y | ≤ b.
f (x , y) is continuous in R and hence existence is guaranteed.
But φ1(x) = 0 and φ2(x) =
(2
3x)3/2 if x ≥ 0
0 if x ≤ 0are solutions in
−∞ < x <∞.Does this imply Lipschitz condition won’t be satisfied?
|f (x , y1)− f (x , y2)||y1 − y2|
=|y1/31 − y
1/32 |
|y1 − y2|.
Choosing y1 = δ, y2 = −δ, we see that the quotient is unboundedfor small values of δ and hence Lipschitz condition is not satisfied.
Solution exists, but not unique.
Santanu Dey Lecture 4
Example 1
Consider
y ′ = y1/3 y(0) = 0 in R : |x | ≤ a, |y | ≤ b.
f (x , y) is continuous in R and hence existence is guaranteed.
But φ1(x) = 0 and φ2(x) =
(2
3x)3/2 if x ≥ 0
0 if x ≤ 0are solutions in
−∞ < x <∞.Does this imply Lipschitz condition won’t be satisfied?
|f (x , y1)− f (x , y2)||y1 − y2|
=|y1/31 − y
1/32 |
|y1 − y2|.
Choosing y1 = δ, y2 = −δ, we see that the quotient is unboundedfor small values of δ and hence Lipschitz condition is not satisfied.Solution exists, but not unique.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)| = |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2 (1)
Now, α = min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)| = |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2 (1)
Now, α = min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)| = |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2 (1)
Now, α = min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)|
= |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2 (1)
Now, α = min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)| = |y |2
≤ |(−b − 1)|2 ≤ (b + 1)2 (1)
Now, α = min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)| = |y |2 ≤ |(−b − 1)|2
≤ (b + 1)2 (1)
Now, α = min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)| = |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2
(1)
Now, α = min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)| = |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2 (1)
Now, α =
min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)| = |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2 (1)
Now, α = min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2
Consider y ′ = y2, y(1) = −1. Find α in the existence &uniqueness theorem.
(1,−1)
(1 − a,−1 − b) (1 + a,−1 − b)
(1 − a,−1 + b) (1 + a,−1 + b)
f (x , y) = y2, fy = 2y are continuous in the closed rectangleR : |x − 1| ≤ a, |y + 1| ≤ b.
|f (x , y)| = |y |2 ≤ |(−b − 1)|2 ≤ (b + 1)2 (1)
Now, α = min
{a,
b
(b + 1)2
}.
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?);
and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 ,
F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)}
= F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2
≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4.
Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4,
α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.
Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒
3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 (contd..)
Consider
F (b) =b
(b + 1)2.
F ′(b) =1− b
(b + 1)3=⇒ the maximum value of F (b) for b > 0
occurs at b = 1 (Why?); and we find F (1) =1
4.
Hence, if a ≥ 1/4 , F (b) =b
(b + 1)2≤ a for all b > 0 and
α = min{a,F (b)} = F (b) =b
(b + 1)2≤ 1/4, whatever be a.
If a < 1/4 , then certainly α < 1/4. Thus in any case, α ≤ 1/4 .
For b = 1, a ≥ 1/4, α = min{a, 1/4} = 1/4.Thus the best possible α from the theorem gives that the IVP has
a unique solution in |x − 1| ≤ 1/4 =⇒ 3/4 ≤ x ≤ 5/4 .
Santanu Dey Lecture 4
Example 2 - Remarks
1 The theorem guarantees existence and uniqueness only in avery small interval!
2 The theorem DOES NOT give the largest interval where thesolution is unique.
3 What is the solution in this case by separation of variables andwhere is it valid? Can you think of extending the solution to alarger interval?
Santanu Dey Lecture 4
Example 2 - Remarks
1 The theorem guarantees existence and uniqueness only in avery small interval!
2 The theorem DOES NOT give the largest interval where thesolution is unique.
3 What is the solution in this case by separation of variables andwhere is it valid? Can you think of extending the solution to alarger interval?
Santanu Dey Lecture 4
Example 2 - Remarks
1 The theorem guarantees existence and uniqueness only in avery small interval!
2 The theorem DOES NOT give the largest interval where thesolution is unique.
3 What is the solution in this case by separation of variables andwhere is it valid? Can you think of extending the solution to alarger interval?
Santanu Dey Lecture 4
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