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
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
Santanu Dey Lecture 4
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.
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