Math 201 Assignment - novoseltsev/2011Winter201ES1/HW11.pdfآ  Math 201 Assignment #11 Problem 1...

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Transcript of Math 201 Assignment - novoseltsev/2011Winter201ES1/HW11.pdfآ  Math 201 Assignment #11 Problem 1...

  • Math 201 Assignment #11

    Problem 1 (10.5 2) Find a formal solution to the given initial-boundary value problem.

    ∂u

    ∂t =

    ∂2u

    ∂x2 , 0 < x < π, t > 0

    u(0, t) = u(π, t) = 0, t > 0

    u(x, 0) = x2, 0 < x < π

    Problem 2 (10.5 5) Find a formal solution to the given initial-boundary value problem.

    ∂u

    ∂t =

    ∂2u

    ∂x2 , 0 < x < π, t > 0

    ∂u

    ∂x (0, t) =

    ∂u

    ∂x (π, t) = 0, t > 0

    u(x, 0) = ex, 0 < x < 1

    Problem 3 (10.5 8) Find a formal solution to the given initial-boundary value problem.

    ∂u

    ∂t =

    ∂2u

    ∂x2 , 0 < x < π, t > 0

    u(0, t) = 0, u(π, t) = 3π, t > 0

    u(x, 0) = 0, 0 < x < π

    Problem 4 (10.5 10) Find a formal solution to the given initial-boundary value problem.

    ∂u

    ∂t = 3

    ∂2u

    ∂x2 + x, 0 < x < π, t > 0

    u(0, t) = u(π, t) = 0, t > 0

    u(x, 0) = sin(x), 0 < x < π

    Problem 5 (10.5 14) Find a formal solution to the given initial-boundary value problem.

    ∂u

    ∂t = 3

    ∂2u

    ∂x2 + 5, 0 < x < π, t > 0

    u(0, t) = u(π, t) = 1, t > 0

    u(x, 0) = 1, 0 < x < π

    1

  • Problem 6 (10.6 2) Find a formal solution.

    ∂2u

    ∂t2 = 16

    ∂2u

    ∂x2 , 0 < x < π, t > 0

    u (0, t) = u (π, t) = 0, t > 0

    u (x, 0) = sin2 x, 0 < x < 1

    ∂u

    ∂t (x, 0) = 1 − cos x, 0 < x < π.

    (Ans. P

    n=1 [an cos (4nt) + bn sin (4nt)] sin (nx),

    an =

    (

    − 8 π

    1

    n(n2−4) n odd

    0 n even , bn =

    (

    1 πn2

    n odd −

    1

    π(n2−1) n even .

    Problem 7 (10.6 8) Find a formal solution.

    ∂2u

    ∂t2 =

    ∂2u

    ∂x2 + x sin t, 0 < x < π, t > 0.

    u (0, t) = u (π, t) = 0, t > 0

    u (x, 0) = 0, 0 < x < π,

    ∂u

    ∂t (x, 0) = 0, 0 < x < π.

    (Ans. [sin t − t cos t] sin x + P

    n=2 2(−1)n

    n2(n2−1) [sin (nt) − n sin t] sin (nx) .)

    Problem 8 (10.6 10) Derive a formal formula for the solution.

    ∂2u

    ∂t2 = α2

    ∂2u

    ∂x2 , 0 < x < L, t > 0

    u (0, t) = U1 u (L, t) = U2, t > 0

    u (x, 0) = f (x) , 0 < x < L,

    ∂u

    ∂t (x, 0) = g (x) , 0 < x < L,

    where U1, U2 are constants. (Ans.

    u (x, t) = U1 + U2 − U1

    L x +

    ∞ X

    n=1

    »

    an cos

    nπαt

    L

    «

    + bn sin

    nπαt

    L

    «–

    sin “

    nπx

    L

    an = 2

    L

    Z

    L

    0

    f (x) −

    »

    U1 + U2 − U1

    L x

    –ff

    sin “

    nπx

    L

    dx,

    bn = 2

    nπα

    Z

    L

    0

    g (x) sin “

    nπx

    L

    dx.

    2

  • Problem 9 (10.6 14) Consider

    ∂2u

    ∂t2 = α2

    ∂2u

    ∂x2 , −∞ < x < ∞, t > 0

    u (x, 0) = f (x) , −∞ < x < ∞

    ∂u

    ∂t (x, 0) = g (x) , −∞ < x < ∞

    with f (x) = x2, g (x) = 0. (Ans. x2 + α2t2)

    Problem 10 (10.6 16) Same as above, but with f (x) = sin 3x, g (x) = 1. (Ans. sin (3x) cos (3αt) + t.)

    3

  • Solution 1. (10.5 2) Using separation of variables, assume

    u(x, t) = X(x)T (t)

    then

    ∂u

    ∂t = X(x)T ′(t)

    ∂2u

    ∂x2 = X ′′(x)T (t)

    Substituting back into the PDE we have

    X(x)T ′(t) = X ′′(x)T (t)

    ⇒ T ′(t)

    T (t) =

    X ′′(x)

    X(x) = K, K constant

    ⇒ {

    X ′′(x) − KX(x) = 0 T ′(t) − KT (t) = 0

    Consider the boundary conditions

    u(0, t) = u(π, t) = 0

    then X(0)T (t) = X(π)T (t) = 0

    so either T (t) = 0, the trivial solution, or

    X(0) = X(π) = 0

    Solving the boundary value problem

    X ′′(x) − KX(x) = 0, X(0) = X(π) = 0

    this is second order linear homogeneous equation with auxiliary equation r2−K = 0 We investigate three possible cases for K.

    Case 1: K > 0 ⇒ r = ± √

    K, real and distinct roots. This has the general solution

    X(x) = c1e √

    Kx + c2e − √

    Kx

    Using the boundary conditions:

    X(0) = c1 + c2 = 0 ⇒ c2 = −c1

    X(π) = c1e √

    Kπ − c1e− √

    Kπ = c1e − √

    Kπ (

    e2 √

    Kπ − 1 )

    = 0

    Since K > 0, we must have c1 = 0, which is the trivial solution.

    4

  • Case 2: K = 0 ⇒ r = 0, double root. This has the general solution

    X(x) = c1 + c2x

    Using the boundary conditions: X(0) = c1 = 0

    X(π) = c2π = 0 ⇒ c2 = 0 This case also gives us the trivial solution.

    Case 3: K < 0 ⇒ r = ± √ −Ki, complex roots.

    This has the general solution

    X(x) = c1 cos( √ −Kx) + c2 sin(

    √ −Kx)

    Using the boundary conditions: X(0) = c1 = 0

    X(π) = c2 sin( √ −Kπ) = 0

    Either c2 = 0 (the trivial solution) or

    sin( √ −Kπ) = 0

    ⇒ √ −Kπ = nπ ⇒ K = −n2, n = 1, 2, . . .

    These eigenvalues have the corresponding eigenfunctions

    Xn(x) = cn sin(nx)

    Using the eigenvalues in the equation for T (t):

    T ′(t) − KT (t) = 0 ⇒ T ′(t) + n2T (t) = 0

    ⇒ Tn(t) = ane−n 2t

    Putting the equations for X(x) and T (t) together and combining the constants, we have

    un(x, t) = Xn(x)Tn(t) = bn sin(nx)e −n2t

    Since un(x, t) is a solution to the PDE for any value of n, we take the infinite series

    u(x, t) =

    ∞ ∑

    n=1

    bn sin(nx)e −n2t

    From the general solution and initial condition, we have

    u(x, 0) = ∞ ∑

    n=1

    bn sin(nx)

    = x2

    5

  • which means we can choose the bn’s as the coefficients in the Fourier sine series for f(x) = x 2:

    bn = 2

    π

    ∫ π

    0

    x2 sin(nx)dx

    = 2

    π

    [−n2x2 cos(nx) + 2 cos(nx) + 2nx sin(nx) n3

    0

    = 2

    πn3 (

    (2 − n2π2)(−1)n − 2 )

    Thus the formal solution to the given initial-boundary value problem is

    u(x, t) = 2

    π

    ∞ ∑

    n=1

    1

    n3 (

    (2 − n2π2)(−1)n − 2 )

    sin(nx)e−n 2t

    6

  • Solution 2. (10.5 5) Following a similar argument as for 10.5 2, we use separation of variables and the boundary conditions to arrive at the system of ODEs:

    {

    X ′′(x) − KX(x) = 0, X ′(0) = X ′(π) = 0 T ′(t) − KT (t) = 0

    Consider the boundary value problem:

    X ′′(x) − KX(x) = 0, X ′(0) = X ′(π) = 0

    This has the auxiliary equation r2 − K = 0

    Again we consider three cases as above.

    Case 1: K > 0 Using a similar argument as in 10.5 2, the only solution in this case is the trivial solution.

    Case 2: K = 0 Here

    X(x) = c0 + c1x ⇒ X ′(x) = c1 Using the boundary conditions:

    X ′(0) = c1 = 0

    and c0 is arbitrary. So here we have the nontrivial solution

    X(x) = c0

    Case 3: K < 0 In this case the general solution is

    X(x) = c1 cos( √ −Kx) + c2 sin(

    √ −Kx)

    ⇒ X ′(x) = −c1 √ −K sin(

    √ −Kx) + c2

    √ −K cos(

    √ −Kx)

    Using the boundary conditions: X ′(0) = c2 = 0

    X ′(π) = −c1 √ −K sin(

    √ −Kπ) = 0

    Either c1 = 0 (trivial solution) or

    sin( √ −Kπ) = 0 ⇒ K = −n2

    These eigenvalues have the corresponding eigenfunctions

    Xn(x) = cn cos(nx)

    Combining cases 2 and 3, we have the eigenvalues and eigenfunctions

    K = −n2

    Xn(x) = cn cos(nx)

    7

  • for n = 0, 1, 2, . . .

    Returning to the equation for T (t):

    T ′(t) − KT (t) = 0 ⇒ T ′(t) + n2T (t) = 0

    ⇒ Tn(t) = ane−n 2t

    Putting the equations for X(x) and T (t) together and merging the constants, we have

    un(x, t) = an cos(nx)e −n2t, n = 0, 1, 2, . . .

    Taking the infinite series of these solutions, we have

    u(x, t) = a0 2

    +

    ∞ ∑

    n=1

    an cos(nx)e −n2t

    From the general solution and the initial condition, we have

    u(x, 0) = a0 2

    + ∞ ∑

    n=1

    an cos(nx)

    = ex

    which means we can choose the an’s as the coefficients in the Fourier cosine series for f(x) = e x:

    a0 = 2

    π

    ∫ π

    0

    exdx = 2

    π (eπ − 1)

    an = 2

    π

    ∫ π

    0

    ex cos(nx)dx

    = 2

    π

    (

    eπ(−1)n − 1 n2 + 1

    )

    Thus the formal solution to the given initial-boundary value problem is

    u(x, t) = 1

    π (eπ − 1) + 2

    π

    ∞ ∑

    n=1

    eπ(−1)n − 1 n2 + 1

    cos(nx)e−n 2t

    8

  • Solution 3. (10.5 8) We assume the solution consists of a steady-state solution v(x) and a transient solution w(x, t) so that

    u(x, t) = v(x) + w(x, t)

    where w(x, t) and