Ordinary Differential Equations - Stanford University Ordinary Differential Equations Arvind...

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  • 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

    Proof Start 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

    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