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### Transcript of Ordinary Differential Equations - Stanford University Ordinary Diï¬€erential Equations Arvind...

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Ordinary Differential Equations

Arvind Saibaba arvindks@stanford.edu

Institute for Computational and Mathematical Engineering Stanford University

September 16, 2010

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Lorenz Attractor

dx

dt = σ(y − x)

dy

dt = x(ρ− z)− y

dz

dt = xy − βz

σ is Prandtl number and ρ is Rayleigh number.

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Quantum mechanics

Time dependent Schrödinger’s equation

i~ ∂

∂t Ψ = ĤΨ

=

(

− ~ 2

2m ∇2 + V (r)

)

Ψ

Time independent Schrödinger’s equation

ĤΨ = ÊΨ

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Vampire Population

dv

dt = −av + dvh

dh

dt = nv − dvh

a is the death rate of vampires due to contact with sunlight, crucifixes, garlic and vampire hunters.

a

aJ. Optimization Theory and Applications: Vol. 75, No.3, 1992

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

What is an ODE?

In general, an n−th order ODE can be written as

F (x , y , y ′, y ′′, . . . , y (n)) = 0

We shall assume that the differential equations can be solved explicitly for y (n) in terms of the remaining qunatities

y (n) = f (x , y , y ′, . . . , y (n−1))

A differential equation is said to be linear if it is linear in y and all its derivatives. Thus, an n−th order ODE can be written as

Pn[y ] = p0(x)y (n) + · · ·+ pn(x)y = r(x)

If r(x) = 0, it is called homogenous.

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

An Existence Theorem

Theorem Let y0 ∈ B, an open subset of R

n, I ⊂ R an interval containing t0. Suppose F is continuous on I × B and satisfies the following Lipschitz estimate in y:

||F (t, y1)− F (t, y2)|| ≤ L||y1 − y2||

for t ∈ I , yj ∈ B. Then the equation

dy

dt = F (t, y) y(t0) = y0

has a unique solution on some t interval containing t0.

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Separation of Variables

If, the equation

dy

dx = f (x , y) can be written as

dy

dx = g(x)h(y)

then, the ODE is separable. If h(y) 6= 0, then

∫ 1

h(y) dy =

g(x)dx

is the desired solution, y is either an implicit or explicit function of x , upto an unknown constant of integration.

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Example

Solve dy

dx = (1 + y2)ex

Solution:

1

1 + y2 dy = exdx (1)

∫ 1

1 + y2 dy =

exdx

tan−1(y) = ex + C

As such, this is an implicit solution. Taking tan on both sides

y = tan(ex + C )

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Exact Equations

Theorem If the functions M(x , y) and N(x , y) along with their partial derivatives My (x , y) and Nx(x , y) be continuous in the rectangle S : |x − x0| < a, |Y − y0| < b (0 < a, b < ∞). Then the ODE

M(x , y)dx + N(x , y)dy = 0 or M(x , y) + N(x , y)y ′ = 0

is exact iff Mx = Ny

If the ODE is exact the implicit solution is

M + Ny ′ = ux + uyy ′ = 0 ⇒ u(x , y) = c

then we must have uxy = My and uyx = Nx . Since My and Nx are continuous, we must have that uxy = uyx

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

u(x , y) =

∫ x

x0

M(s, y)ds + g(y)

We shall obtain g(y) from the other equation uy = N. We have

∂y

∫ x

x0

M(s, y)ds + g ′(y) = N(x , y)

Thus,

g(y) =

∫ y

y0

N(x , t)−

∫ x

x0

M(s, y)ds +

∫ x

x0

M(s, y0)ds + g(y0)

So that the solution is given by

∫ y

y0

N(x , t) +

∫ x

x0

M(s, y0)ds = c

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Example

Solve the ODE

(y + 2xey ) + x(1 + xey )y ′ = 0

Solution Here, M = y + 2xey and N = x(1 + xey ). We have My = Nx = 1+ 2xe

y , ∀(x , y) ∈ S = R2. Thus, the given ODE is exact in R

2. Taking (x0, y0) = (0, 0) we have

∫ y

0

(x + x2et)dt +

∫ x

0

2sds = xy + x2ey = c

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Integrating Factors

Consider the following ODE

dy

dx + p(x)y = r(x)

Set r(x) = 0, to obtain the homogenous solution yh(x)

dyh dx

+ p(x)yh = 0 (2) ∫

dyh yh

= −

p(x)dx

⇒ yh = c0exp

(

p(x)dx

)

for a well defined constant c0

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Integrating Factors contd. For r(x) 6= 0, try solution of form

yp = u(x)exp

(

p(x)dx

)

By chain rule,

du

dx =

{ dyp dx

+ p(x)yp

}

exp

(∫

p(x)dx

)

(3)

= r(x)exp

(∫

p(x)dx

)

u(x) =

r(x)exp

(∫

p(x)dx

)

dx + C

So that,

yp(x) = exp

(

p(x)dx

){∫

r(x)exp

(∫

p(x)dx

)

dx + C

}

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Example

dy

dx −

4

x y = x5ex

I .F . ≡ exp

(

p(x)dx

)

(4)

= e4 log(x) = x4

Now try yp = u(x)x −4

du

dx = xex (5)

u = xex − ex

To complete the solution

y = x4 [(x − 1)ex + C ]

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Duhamel’s Principle

Consider the first order linear ODE

dy

dt = a(t)y + b(t) y(0) = y0

where, a(t) and b(t) are continuous real valued functions. Define

A(t) =

∫ t

0

a(s)ds

The above ODE can be written as

eA(t) d

dt

(

e−A(t)y )

= b(t)

which yields

y(t) = eA(t)y0 + e A(t)

∫ t

0

e−A(s)b(s)ds

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Linearity of Solutions

Consider the homogenous second-order ODE

p0(x)y ′′ + p1(x)y

′ + p2(x)y = 0 (6)

where p0(x)(> 0), p1(x) and p2(x) are continuous in [a, b]. We have

Theorem There exist two solutions y1(x) and y2(x) of eqn. 6 which are linearly independent in [a, b]. Because of linearity, for arbitrary constants c1 and c2,

y(x) = c1y1(x) + c2y2(x)

is also a solution to eqn. 6.

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Linear Independence

Define the Wronskian W (x) as

W (x) =

∣ ∣ ∣ ∣

y1(x) y2(x) y ′1(x) y

2(x)

∣ ∣ ∣ ∣ = y1(x)y

2(x)− y2(x)y ′

1(x)

Theorem Two solutions y1(x) and y2(x) of eqn. 6 are linearly independent if the Wronskian, as defined above, is non-zero for some x0 ∈ [a, b].

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

Variation of Parameters Suppose, one solution y1(x) of eqn. 6 is known. Substitute y(x) = u(x)y1(x),

p0(uy1) ′′ + p1(uy1)

′ + p2(uy1) = 0 (7)

p0y1u ′′ + (2p0y

1 + p1y1)u ′ + (p0y

′′

1 + p1y ′

1 + p2y1) ︸ ︷︷ ︸

=0. Why?

= 0

Let v = u′ and let y1 6= 0

p0y1v ′ + (2p0y

1 + p1y1)v = 0

It can be shown that

v(x) = 1

y21 (x) exp

(

∫ x p1(t)

p0(t) dt

)

So that,

y2(x) = y1(x)

∫ x 1

y21 (t) exp

(

∫ t p1(s)

p0(s) ds

)

dt

• Motivation Introduction First-Order ODE’s Second Order ODE’s Miscellaneous

VOP Example

It is easy to verify that y1(x) = x 2 is a solution of the differential equation

x2y ′′ − 2xy ′ + 2y = 0 x 6= 0

For the second solution,

y2(x) = y1(x)

∫ x 1

y21 (t) exp

(

∫ t p1(s)

p0(s) ds

)

dt (8)

= x2 ∫ x 1

t4 exp

(

∫ t (

− 2s

s2

)

ds