Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 5 ppt

This process will yield a first order, ordinary differential equation for each of the an’s.. For the second method: Make the change of variables vx, t = ux, t − µx, where µx is the equil

Because of the inhomogeneous boundary conditions, the convergence of the series will not be uniform You candifferentiate the series with respect to t, but not with respect to x Multiply the partial differential equation by theeigenfunction sin(nπx/L) and integrate from x = 0 to x = L Use integration by parts to move derivatives in x from

u to the eigenfunctions This process will yield a first order, ordinary differential equation for each of the an’s

For the second method: Make the change of variables v(x, t) = u(x, t) − µ(x), where µ(x) is the equilibriumtemperature distribution to obtain a problem with homogeneous boundary conditions

Hint 37.18

Use separation of variables to find eigen-solutions of the partial differential equation that satisfy the homogeneousboundary conditions There will be two eigen-solutions for each eigenvalue Expand u(x, t) in a series of the eigen-solutions Use the two initial conditions to determine the constants

Hint 37.19

Expand the solution in a series of eigenfunctions in x Determine these eigenfunctions by using separation of variables

on the homogeneous partial differential equation You will find that the answer has the form,

Substitute this series into the partial differential equation to determine ordinary differential equations for each of the

un’s The boundary conditions on u(x, y) will give you boundary conditions for the un’s Solve these ordinary differentialequations with Green functions

Hint 37.20

Solve this problem by expanding the solution in a series of eigen-solutions that satisfy the partial differential equationand the homogeneous boundary conditions Use the initial conditions to determine the coefficients in the expansion.Hint 37.21

Use separation of variables to find eigen-solutions that satisfy the partial differential equation and the homogeneousboundary conditions The solution is a linear combination of the eigen-solutions The whole solution will be exponentiallydecaying if each of the eigen-solutions is exponentially decaying

nπyb



qmn(t) = 4

ab

Z a 0

Z b 0

q(x, y, t) sinmπx

a

sinnπyb

sin

nπyb



fmn = 4

ab

Z a 0

Z b 0

f (x, y) sinmπx

a

sinnπyb



dy dx

We substitute the expansion of the solution into the diffusion equation and the initial condition to determine initialvalue problems for the coefficients in the expansion.

2

+nπb

2

umn(t)

sinmπxa

sinnπyb



u0mn(t) + κ



mπa

2

+nπb

2

umn(t) = qmn(t)u(x, y, 0) = f (x, y)

2

+nπb

2(t − τ )

2

+nπb

2t

First we solve the problem for Θ to determine the eigenvalues and eigenfunctions The Rayleigh quotient is

Θ = c1cos√

λθ+ c2sin√

λθThe solution that satisfies the left boundary condition is



= 0

λn = (2n − 1)2, Θn= cos ((2n − 1)θ) , n ∈ Z+Now we solve the differential equation for R Since this is an Euler equation, we make the substitition R = rα

r2R00n+ rR0n− (2n − 1)2Rn = 0α(α − 1) + α − (2n − 1)2 = 0

α = ±(2n − 1)

Rn= c1r2n−1+ c2r1−2nThe solution which is bounded in 0 ≤ r ≤ 1 is

Rn = r2n−1

The solution of Laplace’s equation is a linear combination of the eigensolutions.

Z π/2 0

Now we solve the differential equation for R Since this is an Euler equation, we make the substitition R = rα.

r2R00n+ rR0n− (2n)2Rn= 0α(α − 1) + α − (2n)2 = 0

α = ±2n

R0 = c1+ c2ln(r), Rn= c1r2n+ c2r−2n, n ∈ Z+The solutions which are bounded in 0 ≤ r ≤ 1 are

Rn= r2n.The solution of Laplace’s equation is a linear combination of the eigensolutions

g(θ) dθ = 0

This is the condition for the existence of a solution of the problem If this is satisfied, we can solve for thecoefficients in the expansion u0 is arbitrary

un= 4π

Z π/2 0

g(θ) cos (2nθ) dθ, n ∈ Z+

We use the initial condition to determine the coefficients.

Z 1 0

X00+ λX = 0, X0(0) = X0(π) = 0

The eigenfunctions form the familiar cosine series

λn= n2, n ∈ Z0+, X0 = 1

2, Xn= cos(nx)Next we solve the differential equation for T (t)

We use the initial condition to determine the coefficients in the series.

Z π 0

x dx = π

un = 2π

Z π 0

a0(r) = 1, an(r) = rn, bn(r) = rn.Thus φ(r, θ) has the form

We apply the boundary condition at r = R.

c0 = 1

π

Z 2π 0

φ(R, α) dα, cn = 1

πRn

Z 2π 0

φ(R, α) cos(nα) dα, dn = 1

πRn

Z 2π 0

nZ 2π 0

φ(R, α) cos(n(θ − α)) dα

φ(r, θ) = 1

2π

Z 2π 0

φ(R, α) dα + 1

π

Z 2π 0

n

eın(θ−α)

!dα

φ(r, θ) = 1

2π

Z 2π 0

φ(R, α) dα + 1

π

Z 2π 0

φ(r, θ) = 1

2π

Z 2π 0

φ(R, α) dα + 1

π

Z 2π 0

φ(r, θ) = 1

2π

Z 2π 0

φ(R, α) dα + 1

π

Z 2π 0

2− r2

R2+ r2 − 2Rr cos(θ − α)dαSolution 37.7

In order that the solution is continuously differentiable, (which it must be in order to satisfy the differential equation),

we impose the boundary conditions

u(0, t) = u(2π, t), uθ(0, t) = uθ(2π, t)

We apply the separation of variables u(θ, t) = Θ(θ)T (t).

e−ınθf (θ) dθ

2 sin(nπx) sinh(nπ(1 − y))

We use the inhomogeneous boundary condition to determine coefficients

sin(nπξ)f (ξ) dξ

Θ00+ λΘ = 0, Θ(0) = Θ(2π), Θ0(0) = Θ0(2π)

r2R00+ rR0 − λR = 0, R is boundedThe eigensolutions for Θ form the familiar Fourier series

R = rα

r2R00n+ rR0n− λnRn = 0α(α − 1) + α − λn= 0

α = ±pλn

First we consider the case λ0 = 0 The solution is

Rn= rn.The solution for u is a linear combination of the eigensolutions

Z 2π 0

f (θ) cos(nθ) dθ, bn= 1

π

Z 2π 0

f (θ) sin(nθ) dθ

Solution 37.10

A normal mode of frequency ω is periodic in time

v(r, θ, t) = u(r, θ) eıωt

We substitute this form into the wave equation to obtain a Helmholtz equation, (also called a reduced wave equation).

1r

Rn(r) = c1Jn(kr) + c2Yn(kr)

Since the Bessel function of the second kind, Yn(kr), is unbounded at r = 0, the solution has the form

Rn(r) = cJn(kr)

Applying the second boundary condition gives us the admissable frequencies

Jn(k) = 0

knm= jnm, Rnm = Jn(jnmr), n ∈ Z0+, m ∈ Z+Here jnm is the mth positive root of Jn We combining the above results to obtain the normal modes of oscillation

v0m = 1

2J0(j0mr) e

vnm = cos(nθ + α)Jnm(jnmr) eıcjnm t, n, m ∈ Z+Some normal modes are plotted in Figure 37.2 Note that cos(nθ + α) represents a linear combination of cos(nθ) andsin(nθ) This form is preferrable as it illustrates the circular symmetry of the problem

Note that in order to satisfy φ(0, t) = φ(l, t) = 0, the Xn must satisfy the same homogeneous boundary conditions,

Xn(0) = Xn(l) = 0 This gives us a Sturm-Liouville problem for X(x)

Thus we seek a solution of the form

2

Tn(t) sin

nπxl



Tn0 = −

anπl

2

Tn

Now we substitute Equation 37.10 into the initial condition for φ to determine initial conditions for the Tn.

sinnπxl

φ(x, 0) dx

Tn(0) = 2

l

Z l/2 0

sinnπxl



x dx + 2

l

Z l/2 0

sinnπxl

(l − x) dx

φ = 4l

2

sin (2n − 1)πx

l



That the coefficients decay as 1/n2 corroborates that φ(x, 0) is C0.

The derivatives of φ with respect to x are

l



For any fixed t > 0, the coefficients in the series for ∂∂xnφ decay exponentially These series are uniformly convergent in

x Thus for any fixed t > 0, φ is C∞ in x

Solution 37.12

ut= κuxx, 0 < x < L, t > 0u(0, t) = T0, u(L, t) = T1, u(x, 0) = f (x),Method 1 We solve this problem with an eigenfunction expansion in x To find an appropriate set of eigenfunctions,

we apply the separation of variables, u(x, t) = X(x)T (t) to the partial differential equation with the homogeneousboundary conditions, u(0, t) = u(L, t) = 0

λn = nπx

L , Xn= sin

nπxL

, n ∈ N

We expand the solution of the partial differential equation in terms of these eigenfunctions.



dx − κhuxsinmπx

L

iL 0

+ κmπL

Z L 0

uxcosmπx

L



dx = 0L

+ κmπL

2Z L 0

u sinmπx

L



dx = 0L

2Z L 0



!sin

mπxL



dx = 0L

Now we have a first order differential equation for each of the an’s We obtain initial conditions for each of the an’sfrom the initial condition for u(x, t).



Note that the series does not converge uniformly due to the 1/n term

Method 2 For our second method we transform the problem to one with homogeneous boundary conditions sothat we can use the partial differential equation to determine the time dependence of the eigen-solutions We make thechange of variables v(x, t) = u(x, t) − µ(x) where µ(x) is some function that satisfies the inhomogeneous boundaryconditions If possible, we want µ(x) to satisfy the partial differential equation as well For this problem we can chooseµ(x) to be the equilibrium solution which satisfies

µ00(x) = 0, µ(0)T0, µ(L) = T1.This has the solution

we obtain the problem

vt= κvxx, 0 < x < L, t > 0v(0, t) = 0, v(L, t) = 0, v(x, 0) = f (x) −



T0 +T1− T0

.Now we substitute the separation of variables v(x, t) = X(x)T (t) into the partial differential equation

, n ∈ N

The ordinary differential equation for T becomes,

Tn0 = −κnπ

L

2

Tn,which, (up to a multiplicative constant), has the solution,



e−κ(nπ/L)2t, n ∈ N

Let v(x, t) have the series expansion,

The coefficients in the expansion are the Fourier sine coefficients of f (x) − T0+T1 −T 0

L x

an = 2L

Z L 0

dx

an = fn− 2(T0− (−1)

nT1)nπ

With the coefficients defined above, the solution for u(x, t) is

sinnπxL



e−κ(nπ/L)2t

Since the coefficients in the sum decay exponentially for t > 0, we see that the series is uniformly convergent for positive

t It is clear that the two solutions we have obtained are equivalent

l sin

 (2n − 1)πx2l

, n ∈ Z+

We expand the solution and inhomogeneity in Equation 37.5 in a series of the eigenvalues.

!

Solution 37.14

Separation of variables leads to the eigenvalue problem

β00+ λβ = 0, β(0) = 0, β(l) + cβ0(l) = 0

First we consider the case λ = 0 A set of solutions of the differential equation is {1, x} The solution that satisfiesthe left boundary condition is β(x) = x The right boundary condition imposes the constraint l + c = 0 Since c ispositive, this has no solutions λ = 0 is not an eigenvalue.

Now we consider λ 6= 0 A set of solutions of the differential equation is {cos(√

λl= −

√λcFor large λ, the we can determine approximate solutions

βn(x) = sin

√

λnx
q

βn(x)w(x, t) dx

Since φ satisfies the same homgeneous boundary conditions as β, we substitute the series into Equation 37.5 todetermine differential equations for the Tn(t).

at = −c(al + 1)t

a = − c

1 + clu(x, t) =



1 − cx

1 + cl

t

Now we define ψ to be the difference of φ and u.

wnexp −a2λn(t − τ )
dτ,

wn(t) =

Z l 0

βn(x) cx

1 + cldx.

This determines the solution for φ

ψt = a2ψxx,ψ(0, t) = ψ(l, t) + ψx(l, t) = 0, ψ(x, 0) = x

βn(x) x

l + 1dxThis expansion is useful for large t because the coefficients decay exponentially with increasing t

Now we solve this problem with the Laplace transform.

φt= a2φxx, φ(0, t) = 0, φ(l, t) + φx(l, t) = 1, φ(x, 0) = 0

s ˆφ = a2φˆxx, φ(0, s) = 0,ˆ φ(l, s) + ˆˆ φx(l, s) = 1

sˆ

We apply the right boundary condition to determine the constant

ˆ

φ =

sinh

√ sx a



ssinh

√ sl a

+

√ s

a cosh

√ sl a



We expand this in a series of simpler functions of s.

ˆ

φ =

2 sinh

√ sx a



s

exp

√ sl a



− exp−

√ sl a

+

√ s a

exp

√ sl a

+ exp



−

√ sl a



s exp

√ sl a



1

1 +

√ s

a −1 −

√ s a

exp−2

√ sl a



ˆ

φ =exp

√ sx a



− exp−

√ sx a

exp

√ sl a

exp



−2

√ sl a



ˆ

φ =exp

√ s(x−l) a



− exp

√ s(−x−l) a



s1 +

√ s a

!

By expanding

(1 −√

s/a)n(1 +√



Taking the first term in each series yields

ˆ

φ ∼ a

s3/2

exp



−

√s(l − x)a



− erfc l + x

2a√t

First we consider the case of a double root when λ = 1/4 The solutions of Equation 37.11are {x−1/2, x−1/2ln x}.The solution that satisfies the left boundary condition is X = x−1/2ln x Since this does not satisfy the right boundarycondition, λ = 1/4 is not an eigenvalue.

Now we consider λ 6= 1/4 The solutions of Equation 37.11 are

1

p

λ − 1/4 ln 2 = nπ, n ∈ Z+.This gives us the eigenvalues and eigenfunctions

We normalize the eigenfunctions

Z 2 1

sin2(nπξ) dξ = ln 2

2

Xn(x) =

r2



We expand φ in a series of the eigensolutions.

Xn(x)f (x) dxSolution 37.18

utt = c2uxx, 0 < x < L, t > 0,u(0, t) = 0, ux(L, t) = 0,u(x, 0) = f (x), ut(x, 0) = 0,

We substitute the separation of variables u(x, t) = X(x)T (t) into the partial differential equation

The problem for X is a regular Sturm-Liouville eigenvalue problem From the Rayleigh quotient,



n − 12

, n ∈ N

The eigenvalues and eigenfunctions are

λn = (2n − 1)π

2L , Xn = sin

 (2n − 1)πx2L

, n ∈ N

The differential equation for T becomes

T00 = −c2 (2n − 1)π

2L

2

T,which has the two linearly independent solutions,

Tn(1) = cos (2n − 1)cπt

2L

, Tn(2) = sin (2n − 1)cπt

2L