![Page 1: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/1.jpg)
MA 108 - Ordinary Differential Equations
Santanu Dey
Department of Mathematics,Indian Institute of Technology Bombay,
Powai, Mumbai [email protected]
March 4, 2014
Santanu Dey Lecture 5
![Page 2: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/2.jpg)
Outline of the lecture
Bernoulli equation
Orthogonal Trajectories
Lipschitz continuity
Existence & uniqueness
Santanu Dey Lecture 5
![Page 3: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/3.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x (linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 4: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/4.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x (linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 5: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/5.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx
=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x (linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 6: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/6.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x (linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 7: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/7.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x
(linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 8: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/8.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x (linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 9: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/9.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x (linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 10: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/10.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x (linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 11: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/11.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x (linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 12: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/12.jpg)
Example - Bernoulli
Solve :dy
dx+ y = xy3.
Let v = y−2 .
dv
dx= −2y−3 dy
dx=⇒ −1
2
dv
dx+ v = x
That is,dv
dx− 2v = −2x (linear equation in v)
Integrating factor is e−2x .
ve−2x = −∫
2xe−2x dx + C
=2xe−2x
−2−
∫�2e−2x
�2+ C
= xe−2x +e−2x
2+ C
=⇒ 1
y2= x +
1
2+ Ce2x .
Santanu Dey Lecture 5
![Page 13: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/13.jpg)
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· dy
dx=
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 5
![Page 14: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/14.jpg)
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· dy
dx=
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 5
![Page 15: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/15.jpg)
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· dy
dx=
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 5
![Page 16: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/16.jpg)
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· dy
dx=
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 5
![Page 17: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/17.jpg)
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· dy
dx=
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 5
![Page 18: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/18.jpg)
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· dy
dx=
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 5
![Page 19: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/19.jpg)
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .
Then,
dv
dx=
dv
dy· dy
dx= cos y
dy
dx.
Hence the given equation is
dv
dx+
1
xv = 1, which is linear in v .
That is,
eR
1x
dxv(x) =
∫e
R1x
dxdx + C
=⇒ x v(x) =x2
2+ C
sin y =x
2+
C
x.
Santanu Dey Lecture 5
![Page 20: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/20.jpg)
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dy
dx= cos y
dy
dx.
Hence the given equation is
dv
dx+
1
xv = 1, which is linear in v .
That is,
eR
1x
dxv(x) =
∫e
R1x
dxdx + C
=⇒ x v(x) =x2
2+ C
sin y =x
2+
C
x.
Santanu Dey Lecture 5
![Page 21: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/21.jpg)
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dy
dx=
cos ydy
dx.
Hence the given equation is
dv
dx+
1
xv = 1, which is linear in v .
That is,
eR
1x
dxv(x) =
∫e
R1x
dxdx + C
=⇒ x v(x) =x2
2+ C
sin y =x
2+
C
x.
Santanu Dey Lecture 5
![Page 22: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/22.jpg)
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dy
dx= cos y
dy
dx.
Hence the given equation is
dv
dx+
1
xv = 1, which is linear in v .
That is,
eR
1x
dxv(x) =
∫e
R1x
dxdx + C
=⇒ x v(x) =x2
2+ C
sin y =x
2+
C
x.
Santanu Dey Lecture 5
![Page 23: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/23.jpg)
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dy
dx= cos y
dy
dx.
Hence the given equation is
dv
dx+
1
xv = 1, which is linear in v .
That is,
eR
1x
dxv(x) =
∫e
R1x
dxdx + C
=⇒ x v(x) =x2
2+ C
sin y =x
2+
C
x.
Santanu Dey Lecture 5
![Page 24: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/24.jpg)
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dy
dx= cos y
dy
dx.
Hence the given equation is
dv
dx+
1
xv = 1, which is linear in v .
That is,
eR
1x
dxv(x) =
∫e
R1x
dxdx + C
=⇒ x v(x) =x2
2+ C
sin y =x
2+
C
x.
Santanu Dey Lecture 5
![Page 25: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/25.jpg)
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dy
dx= cos y
dy
dx.
Hence the given equation is
dv
dx+
1
xv = 1, which is linear in v .
That is,
eR
1x
dxv(x) =
∫e
R1x
dxdx + C
=⇒ x v(x) =x2
2+ C
sin y =x
2+
C
x.
Santanu Dey Lecture 5
![Page 26: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/26.jpg)
Example
Solve : cos ydy
dx+
1
xsin y = 1.
Set v = sin y .Then,
dv
dx=
dv
dy· dy
dx= cos y
dy
dx.
Hence the given equation is
dv
dx+
1
xv = 1, which is linear in v .
That is,
eR
1x
dxv(x) =
∫e
R1x
dxdx + C
=⇒ x v(x) =x2
2+ C
sin y =x
2+
C
x.
Santanu Dey Lecture 5
![Page 27: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/27.jpg)
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 5
![Page 28: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/28.jpg)
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 5
![Page 29: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/29.jpg)
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 5
![Page 30: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/30.jpg)
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 5
![Page 31: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/31.jpg)
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 5
![Page 32: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/32.jpg)
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 5
![Page 33: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/33.jpg)
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 5
![Page 34: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/34.jpg)
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 5
![Page 35: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/35.jpg)
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 5
![Page 36: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/36.jpg)
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 5
![Page 37: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/37.jpg)
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 5
![Page 38: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/38.jpg)
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 5
![Page 39: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/39.jpg)
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 5
![Page 40: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/40.jpg)
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 5
![Page 41: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/41.jpg)
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 5
![Page 42: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/42.jpg)
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 5
![Page 43: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/43.jpg)
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 5
![Page 44: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/44.jpg)
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 5
![Page 45: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/45.jpg)
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 5
![Page 46: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/46.jpg)
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 5
![Page 47: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/47.jpg)
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 5
![Page 48: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/48.jpg)
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 5
![Page 49: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/49.jpg)
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 5
![Page 50: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/50.jpg)
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 5
![Page 51: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/51.jpg)
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 5
![Page 52: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/52.jpg)
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 5
![Page 53: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/53.jpg)
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 5
![Page 54: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/54.jpg)
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 5
![Page 55: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/55.jpg)
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 5
![Page 56: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/56.jpg)
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 5
![Page 57: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/57.jpg)
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 5
![Page 58: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/58.jpg)
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 5
![Page 59: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/59.jpg)
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 5
![Page 60: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/60.jpg)
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 5
![Page 61: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/61.jpg)
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 5
![Page 62: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/62.jpg)
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 5
![Page 63: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/63.jpg)
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 5
![Page 64: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/64.jpg)
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 5
![Page 65: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/65.jpg)
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 5
![Page 66: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/66.jpg)
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 5
![Page 67: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/67.jpg)
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 5
![Page 68: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/68.jpg)
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 5
![Page 69: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/69.jpg)
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 5
![Page 70: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/70.jpg)
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 5
![Page 71: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/71.jpg)
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 5
![Page 72: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/72.jpg)
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 5
![Page 73: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/73.jpg)
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 5
![Page 74: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/74.jpg)
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 5
![Page 75: MA 108 - Ordinary Differential Equationsdey/diffeqn_spring14/lecture5_D2.pdf · Santanu Dey Lecture 5. Outline of the lecture Bernoulli equation Orthogonal Trajectories Lipschitz](https://reader034.vdocument.in/reader034/viewer/2022052106/604141ea2a0056514b31c712/html5/thumbnails/75.jpg)
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 5