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

Ordinary Differential Equations

Arvind Saibabaarvindks@stanford.edu

Institute for Computational and Mathematical EngineeringStanford 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 Schrodinger’s equation

i~∂

∂tΨ = HΨ

=

(

−~2

2m∇2 + V (r)

)

Ψ

Time independent Schrodinger’sequation

HΨ = EΨ

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 withsunlight, 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 explicitlyfor 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 itsderivatives. 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

TheoremLet y0 ∈ B, an open subset of Rn, I ⊂ R an interval containing t0.Suppose F is continuous on I × B and satisfies the following Lipschitzestimate 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

Solvedy

dx= (1 + y2)ex

Solution:

1

1 + y2dy = exdx (1)

∫1

1 + y2dy =

∫

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

TheoremIf the functions M(x , y) and N(x , y) along with their partial derivativesMy (x , y) and Nx(x , y) be continuous in the rectangleS : |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 arecontinuous, we must have that uxy = uyx

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

ProofStart with the equation uX = M. Integrating both sides

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

SolutionHere, M = y + 2xey and N = x(1 + xey ). We haveMy = Nx = 1+ 2xey , ∀(x , y) ∈ S = R

2. Thus, the given ODE is exact inR

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)

dyhdx

+ p(x)yh = 0 (2)∫

dyhyh

= −

∫

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=

{dypdx

+ 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

xy = 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 + eA(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

TheoremThere exist two solutions y1(x) and y2(x) of eqn. 6 which are linearlyindependent in [a, b]. Because of linearity, for arbitrary constants c1 andc2,

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)

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

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

Variation of ParametersSuppose, one solution y1(x) of eqn. 6 is known. Substitutey(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) = x2 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

t4exp

(

−

∫ t (

−2s

s2

)

ds

)

dt

= x2∫ x 1

t4t2dt

= −x

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

Inhomogenous equations

TheoremLet y1(x) and y2(x) be two linearly independent solutions to thehomogenous equation 6. Let yp be a particular solution to the ODE.Then, the general solution to the ODE is given by

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

Further, yp can be computed as

yp(x) = −y1(x)

∫ x r(t)y2(t)

W (t)dt + y2(x)

∫ x r(t)y1(t)

W (t)dt

where, W (x) is the Wronskian defined before.

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

Constant coefficient case

Consider,ay ′′ + by ′ + cy = 0 (9)

We try solutions of the form y = erx . This results in the algebraicquadratic equation known as the characteristic equation.

ar2 + br + c = 0

• Real, distinct roots r1, r2.er1x and er2x are two solutions of eqn. 9 and the general solution is

y(x) = c1er1x + c2e

r2x

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

Constant coefficient case contd.

• Repeated, real roots r1 = r2 = r .One solution is, of course, erx . To find the other solution, we usethe method of Variation of Parameters. It is easy to show thaty2 = xerx , so that

y(x) = erx(c1 + c2x)

• Complex, conjugate roots r1 = µ+ iν, r2 = µ− iν Using Euler’sformula e iθ = cos θ + sin θ, we can write the general solution as

y(x) = c1eµx cos νx + c2e

µx sin νx

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

Euler-Cauchy equations

x2y ′′ + axy ′ + by = 0 x > 0

Also, known as equidimensional ODE’s. Plugging y = xm, we have

x2m(m − 1)xm−2 + axmxm−1 + bxm = 0

or,m(m − 1) + am + b = 0

Depending on the roots, we have three cases

1. Real, distinct roots, m1,m2 : y(x) = c1xm1 + c2x

m2

2. Real, repeated roots, m1 = m2 = m: y(x) = xm1(c1 + c2 log x)

3. Complex conjugate roots, m1 = µ+ iν,m2 = µ− iν:

y(x) = xµ(c1 cos(ν log x) + x2 sin(ν log x))

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

Higher Order Homogenous Constant Coefficient ODE’s

Consider the ODE

a0dny

dtn+ a1

dn−1y

dtn−1+ · · ·+ any = 0

Define,

z1 = y (10)

z2 = z ′1 = y ′

......

zn = z ′n−1 = y (n−1)

z ′n = −a1a0

zn −a2a0

zn−1 · · · −ana0

z1

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

System of ODEs

z1z2...

zn−1

zn

′

=

0 10 1

. . .. . .

0 1− an

a0− an−1

a0. . . − a2

a0− a1

a0

︸ ︷︷ ︸

A

z1z2...

zn−1

zn

︸ ︷︷ ︸

z

Can be written asdz

dt= Az z(t = 0) = z0

which has the solution z(t) = eAtz0

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

Exponential of a Matrix

eAt = I+ At +(At)2

2!+

(At)3

3!+ . . .

This series always converges and

(eAs

) (eAt

)= eA(s+t)

(eAt

) (e−At

)= I

If the matrix A is diagonalizable, i.e. A = SΛS−1

eAt = I + SΛS−1t + SΛ2S−1 t2

2!+ SΛ3S−1 t

3

3!+ . . . (11)

= S

(

I + Λt +(Λt)2

2!+

(Λt)3

3!+ . . .

)

S−1

= SeΛtS−1

Otherwise, appeal to Jordan Canonical form.

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

Euler’s method

We want to approximate the solution of the equation

y ′(t) = f (t, y(t)) y(t0) = y0

Define the points

t0 = a, ti = a+ ih, i = 1, 2, . . . ,N h =b − a

N

Starting with the differential equation, we replace the derivative with afinite difference approximation

y ′(t) =y(ti+1)− y(ti )

h+O(h)

So, using this approximation we have the forward Euler method whichdefines the approximate solution {yi}

Ni=0

yi+1 = yi + hf (ti , yi )

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

Properties

Define the error, ei = y(ti )− yi , i = 1, . . . , n. For the forward Eulermethod we expect

ei = O(h)

and assuming the function is Lipschitz continuous, one can also derivethe following error bound

|ei+1| ≤Mh

2L[eL(ti+1−a) − 1]

where,M = max a ≤ t ≤ b|y ′′(t)|

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

Boundary Value problems

Consider the Poisson’s equation

−u′′(x) = f (x) x ∈ [0, 1] u(0) = u(1) = 0

We start with a central difference approximation to the second derivative.Discretizing the domain into equal intervals with N + 1 points,xi = ih, h = 1/N, i = 0, . . . ,N

u′′(xi ) = −f (xi ) =ui+1 − 2ui + ui−1

h2+O(h2)

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

This results in a tridiagonal system of equations

1

h2

2 −1−1 2 −1

. . .. . .

. . .

−1 2

u1u2. . .un

=

f (x1)f (x2). . .

f (xN)

The solution to the problem using Finite Differences

−κu′′ + βu′ = f