Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 6 pps

Since the function is even, there are no sine terms in the Fourier series... We expand the solution in a series of the eigenfunctions.. Thus the Fourier sine series isIn Figure 28.10 the

We use the definition of the Fourier coefficients to evaluate the integrals in the last sum.

Z π

−π

x sin(nx) dx

= 1π

3 Consider f (x) = x2 Since the function is even, there are no sine terms in the Fourier series The coefficients inthe cosine series are

a0 = 2π

Z π 0

Z π 0

x2cos(nx) dx

= 4(−1)

n

n2 Thus the Fourier series is

Now we integrate the series for f (x) = x2.

Z x 0

2 We integrate SN0

SN(x) − SN(0) = x −

Z x 0

(−1)Ncos N +12
ξ

cos 2ξ
dξ

x − SN =

Z x 0

(x − π) = −nπ

sin N +12
(ξ − π)
sin ξ−π2
dξ

We shift the limits of integration

E0 =

Z π π/(N +1/2)

sin N +12
ξ

sin ξ2
dξ

We add and subtract an integral over [0 π/(N + 1/2)].

E0 =

Z π 0

sin N +12
ξ

sin 2ξ
dξ −

Z π/(N +1/2) 0

sin ((2N + 1) ξ)sin (ξ) dξ

= 12

Z π

−π

sin ((2N + 1) ξ)sin (ξ) dξ

= 1

2 −Z

C

= z2N +1
(z − 1/z)/(ı2)

dzız

= =



−Z

We approximate the second integral.

Z π/(N +1/2) 0

sin N +12
ξ

sin ξ2
dξ =

22N + 1

Z π 0

sin(x)sin 2N +1x
dx

≈ 2

Z π 0

sin(x)

x dx

= 2

Z π 0

1x

≈ 3.70387

In the limit as N → ∞, the overshoot is

|π − 3.70387| ≈ 0.56

Solution 28.3

1 The eigenfunctions of the self-adjoint problem

−y00 = λy, y(0) = y(1) = 0,are

φn= sin(nπx), n ∈ Z+

We find the series expansion of the inhomogeneity f (x) = 1.

sin(nπx) dx

fn = 2



−cos(nπx)nπ

1 0

4nπ(2 − π2n2)sin(nπx)

2 Now we solve the boundary value problem directly.

y00+ 2y = 1 y(0) = y(1) = 0The general solution of the differential equation is

y = c1cos√

2x

+ c2sin√

2x

+1

y(x) sin(nπx) dx

an=

Z 1 0

1 − cos√

2x+ cos

√2
− 1sin √

2
sin

√

2x

!sin(nπx) dx

We obtain the same series as in the first part.

Solution 28.4

1 The eigenfunctions of the self-adjoint problem

−y00 = λy, y0(0) = y0(π) = 0,are

Z π 0

sin(x) dx

f0 = 4π

fn= 2π

Z π 0

sin(x) cos(nx) dx

fn= 2(1 + (−1)

n)π(1 − n2)

We substitute the series into the differential equation.

4π(1 − n2)cos(nx)

4π(1 − n2)(2 − n2)cos(nx)

2 We expand the solution in a series of the eigenfunctions

4π(1 − n2)cos(nx)

It is not possible to solve for the a2 coefficient That equation is

(0)a2 = − 4

3π.This problem is to be expected, as this boundary value problem does not have a solution The solution of thedifferential equation is

y = c1cos(2x) + c2sin(2x) +1

3sin(x)

The boundary conditions give us an inconsistent set of constraints.

y0(0) = 0, y0(π) = 0

c2+1

3 = 0, c2− 1

3 = 0Thus the problem has no solution

Solution 28.5

Cosine Series The coefficients in the cosine series are

a0 = 2π

Z π 0

Z π 0

x2cos(nx) dx

= 4(−1)

n

n2 Thus the Fourier cosine series is

-3 -2 -1 1 2 3

2 4 6 8

-10 -5

5 10

Figure 28.10: The Fourier Cosine and Sine Series of f (x) = x2

Sine Series The coefficients in the sine series are

bn= 2π

Z π 0

Thus the Fourier sine series is

In Figure 28.10 the odd periodic extension of f (x) and the sum of the first five terms in the sine series are plotted.Since the odd periodic extension of f (x) is not continuous, the series is not differentiable

(eınx+ e−ınx) +n

1

(eı(n−2)x+ e−ı(n−2)x) + · · ·

+

n(n − 1)/2

(eıx+ e−ıx)

#

= 1

2n

n0

cos((n − 2m)x)

n(n − k)/2

cos(kx)

(eınx+ e−ınx) +n

1

(eı(n−2)x+ e−ı(n−2)x) + · · ·

+

nn/2 − 1

(eı2x+ e−i2x) +

nn/2

#

= 1

2n

n0



2 cos(2x) +

nn/2



= 1

2n

nn/2

cos((n − 2m)x)

= 1

2n

nn/2

n(n − k)/2

cos(kx)



Solution 28.7

We expand f (x) in a cosine series The coefficients in the cosine series are

a0 = 2π

Z π 0

Z π 0

x2cos(nx) dx

= 4(−1)

n

n2 Thus the Fourier cosine series is

We substitute x = π into the Fourier series.

Z π 0

f (x) sin(nx) dx

= 2π

Z π 0

cos x − 1 +2x

π

sin(nx) dx

= 2(1 + (−1)

n)π(n3− n)

an=

π(n 3 −n) for even n

2 From our work in the previous part, we see that the Fourier coefficients decay as 1/n3 The Fourier sine seriesconverges to the odd periodic extension of the function, ˆf (x) We can determine the rate of decay of the Fouriercoefficients from the smoothness of ˆf (x) For −π < x < π, the odd periodic extension of f (x) is defined

f (x) is continuously differentiable, C1 However, since

ˆ00

(0+) = −1, and ˆ00(0−) = 1ˆ

f (x) is not C2 Since ˆf (x) is C1 we know that the Fourier coefficients decay as 1/n3

Solution 28.9

Cosine Series The even periodic extension of f (x) is a C0, continuous, function (See Figure 28.11 Thus thecoefficients in the cosine series will decay as 1/n2 The Fourier cosine coefficients are

a0 = 2π

Z π 0

x sin x dx

= 2

a1 = 2π

Z π 0

x sin x cos x dx

= −12

an= 2π

Z π 0

x sin x cos(nx) dx

= 2(−1)

n+1

n2− 1 , for n ≥ 2The Fourier cosine series is

Figure 28.11: The even periodic extension of x sin x

Sine Series The odd periodic extension of f (x) is a C1, continuously differentiable, function (See Figure 28.12.Thus the coefficients in the cosine series will decay as 1/n3 The Fourier sine coefficients are

a1 = 1π

Z π 0

x sin x sin x dx

= π2

an= 2π

Z π 0

x sin x sin(nx) dx

= −4(1 + (−1)

n)nπ(n2− 1)2 , for n ≥ 2The Fourier sine series is

ˆ

f (x) = π

2sin x −

4π

∞

X

n=2

(1 + (−1)n)n(n2− 1)2 cos(nx)

1

Figure 28.12: The odd periodic extension of x sin x

Solution 28.10

If ν = n is an integer, then the Fourier cosine series is the single term cos(|n|x) We assume that ν 6= n

We note that the even periodic extension of cos(νx) is C0 so that the series converges to cos(νx) for −π ≤ x ≤ π

and the coefficients decay as 1/n2 We compute the Fourier cosine coefficients.

a0 = 2π

Z π 0

cos(νx) dx

= 2 sin(πν)πν

an= 2π

Z π 0

cos(νx) cos(nx) dx

= (−1)n

1

ν − n +

1

ν + n

sin(πν)The Fourier cosine series is

ν − n +

1

ν + n

sin(πν) cos(nx)

We substitute x = 0 into the Fourier cosine series

ν − n+

1

ν + n

sin(πν)

πsin πν =

ν − n +

1

ν + n

sin(πν)(−1)n

ν − n +

1

ν + n



Note that neither cot(πν) nor 1/ν is integrable at ν = 0 We write the last formula so each side is integrable.

+ ln n + θ

n



ln sin(πθ)πθ

ln

cosx2

Since | eıx| ≤ 1 and eıx 6= −1 for =(x) ≥ 0, x 6= (2k + 1)π, we can expand the last term in a Taylor series inthat domain.

= − ln 2 −

ln

cosx2



dx =

Z π 0

π 0

Z π 0

sin((x + ξ)/2)sin((x − ξ)/2)



sin((x + ξ)/2)sin((x − ξ)/2)

which has the solutions,

We expand the solution and the inhomogeneity in the eigenfunctions

b − a

Z b a

f (x) sin nπ(x − a)

b − a

dx

Since the solution y(x) satisfies the same homogeneous boundary conditions as the eigenfunctions, we can differentiatethe series We substitute the series expansions into the differential equation

Thus the solution of the problem has the series representation,

Solution 28.13

The eigenfunction problem associated with this problem is

φ00+ λ2φ = 0, φ(a) = φ(b) = 0,

which has the solutions,

We expand the solution and the inhomogeneity in the eigenfunctions

b − a

Z b a

f (x) sin nπ(x − a)

b − a

dx

Since the solution y(x) does not satisfy the same homogeneous boundary conditions as the eigenfunctions, we candifferentiate the series We multiply the differential equation by an eigenfunction and integrate from a to b We useintegration by parts to move derivatives from y to the eigenfunction

y00+ αy = f (x)

Z b a

y00(x) sin(λmx) dx + α

Z b a

y(x) sin(λmx) dx =

Z b a

f (x) sin(λmx) dx[y0sin(λmx)]ba−

Z b a

Thus the solution of the problem has the series representation,

sin nπ(x − a)

b − a



2 We consider the principal branch of the logarithm.

= log (1 + r cos x + ır sin x)

= log |1 + r cos x + ır sin x| + ı arg (1 + r cos x + ır sin x)

= logp(1 + r cos x)2+ (r sin x)2+ ı arctan (1 + r cos x, r sin x)

A = 1

2log 1 + 2r cos x + r

2
, B = arctan (1 + r cos x, r sin x)

1 · sin x dx = [− cos x]π0 = 2

Thus the system is not orthogonal on the interval [0, π] Consider the interval [a, a + π].

Z a+π a

1 · sin x dx = [− cos x]a+πa = 2 cos a

Z a+π a

1 · cos x dx = [sin x]a+πa = −2 sin a

Since there is no value of a for which both cos a and sin a vanish, the system is not orthogonal for any interval

of length π

2 First note that

Z π 0

cos nx dx = 0 for n ∈ N

If n 6= m, n ≥ 1 and m ≥ 0 then

Z π 0

cos nx cos mx dx = 1

2

Z π 0

cos((n − m)x) + cos((n + m)x)
dx = 0

Thus the set {1, cos x, cos 2x, } is orthogonal on [0, π] Since

Z π 0

dx = π

Z π 0

cos2(nx) dx = π

2,the set

(r1

π,

r2

πcos x,

r2

πcos 2x,

)

is orthonormal on [0, π]

If n 6= m, n ≥ 1 and m ≥ 1 then

Z π 0

sin nx sin mx dx = 1

2

Z π 0

cos((n − m)x) − cos((n + m)x)
dx = 0

Thus the set {sin x, sin 2x, } is orthogonal on [0, π] Since

Z π 0

sin2(nx) dx = π

2,the set

(r2

π sin x,

r2

Z π 0

= π2

an = 2π

Z π 0

x cos(nx) dx

= 2π



xsin(nx)n

π 0

− 2π

Z π 0

sin(nx)

= −2π

 cos(nx)

n2

π 0

∞

X

n=1 odd n

1

n2 cos(nx) for x ∈ [−π, π]Define RN(x) = f (x) − SN(x) We seek an upper bound on |RN(x)|

|RN(x)| =

4π

∞

X

n=N +1 odd n

1

n2 cos(nx)

...