hw4, math 322, fall 2016
TRANSCRIPT
HW4, Math 322, Fall 2016
Nasser M. Abbasi
December 30, 2019
Contents
1 HW 4 21.1 Problem 2.5.24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Problem 3.2.2 (b,d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.1 Part b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Part d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.3 Problem 3.2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4 Problem 3.3.2 (d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5 Problem 3.3.3 (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.6 Problem 3.3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.6.1 Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.6.2 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.6.3 Part (c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.7 Problem 3.4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.7.1 Part (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.7.2 Part (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.8 Problem 3.4.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.9 Problem 3.4.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
1
2
1 HW 4
1.1 Problem 2.5.24
88
(a) 8' (x, 0) = 0,
(b) u(x,0) = 0,
(c) u(x,0) = 0,
(d) (x, 0) = 0,
Chapter 2. Method of Separation of Variables
"' (x, H) = 0, u(0, y) = f (y)
u(x, H) = 0, u(0,y) = f(y)
u(x, H) = 0, (0,y) = f(y)
Ou (x, H) = 0, a: (0, y) = f (y)
Show that the solution [part (d)] exists only if fH f (y) dy = 0.
2.5.16. Consider Laplace's equation inside a rectangle 0 < x < L, 0 < y < H, withthe boundary conditions
8uau
8u&"
8x(0,y) = 0, 8x(L, y) = g(y),
8y(x, 0)= 0, 8y (x, H) = f (x)
(a) What is the solvability condition and its physical interpretation?(b) Show that u(x, y) = A(x2 - y2) is a solution if f (x) and g(y) are
constants [under the conditions of part (a)].(c) Under the conditions of part (a), solve the general case [nonconstant
f (x) and g(y)]. [Hints: Use part (b) and the fact that f (x) = f +[f (x) - f.,.], where f.,. = L fL f (x) dx.]
2.5.17. Show that the mass density p(x, t) satisfies k + V (pu) = 0 due to con-servation of mass.
2.5.18. If the mass density is constant, using the result of Exercise 2.5.17, showthat
2.5.19. Show that the streamlines are parallel to the fluid velocity.
2.5.20. Show that anytime there is a stream function, V x u = 0.
2.5.21. From u and v=- ,derive u,-=rue=-2.5.22. Show the drag force is zero for a uniform flow past a cylinder including
circulation.
2.5.23. Consider the velocity ug at the cylinder. Where do the maximum andminimum occur?
2.5.24. Consider .the velocity ue at the cylinder. If the circulation is negative, showthat the velocity will be larger above the cylinder than below.
2.5.25. A stagnation point is a place where u = 0. For what values of the circulationdoes a stagnation point exist on the cylinder?
2.5.26. For what values of 0 will u,. = 0 off the cylinder? For these 6, where (forwhat values of r) will ue = 0 also?
2.5.27. Show that r/ = a 81T B satisfies Laplace's equation. Show that the streamlinesare circles. Graph the streamlines.
Introduction. The stream velocity οΏ½οΏ½ in Cartesian coordinates is
οΏ½οΏ½ = π’ π€ + π£ π₯
=πΞ¨ππ¦
π€ βπΞ¨ππ₯
π₯ (1)
Where Ξ¨ is the stream function which satisfies Laplace PDE in 2D β 2Ξ¨ = 0. In Polar coordinatesthe above becomes
οΏ½οΏ½ = π’ποΏ½οΏ½ + π’ποΏ½οΏ½
=1ππΞ¨ππ
οΏ½οΏ½ βπΞ¨ππ
οΏ½οΏ½ (2)
The solution to β 2Ξ¨ = 0 was found under the following conditions
1. When π very large, or in other words, when too far away from the cylinder or the wing, theflow lines are horizontal only. This means at π = β the π¦ component of οΏ½οΏ½ in (1) is zero. This
meansπΞ¨οΏ½π₯,π¦οΏ½
ππ₯ = 0. Therefore Ξ¨οΏ½π₯, π¦οΏ½ = π’0π¦ where π’0 is some constant. In polar coordinates thisimplies Ξ¨(π, π) = π’0π sinπ, since π¦ = π sinπ.
2. The second condition is that radial component of οΏ½οΏ½ is zero. In other words, 1ππΞ¨ππ = 0 when
π = π, where π is the radius of the cylinder.
3. In addition to the above two main condition, there is a condition that Ξ¨ = 0 at π = 0
Using the above three conditions, the solution to β 2Ξ¨ = 0 was derived in lecture Sept. 30, 2016, tobe
Ξ¨(π, π) = π1 ln οΏ½ πποΏ½ + π’0 οΏ½π β
π2
π οΏ½sinπ
Using the above solution, the velocity οΏ½οΏ½ can now be found using the definition in (2) as follows
1ππΞ¨ππ
=1ππ’0 οΏ½π β
π2
π οΏ½cosπ
πΞ¨ππ
=π1π+ π’0 οΏ½1 +
π2
π2 οΏ½sinπ
Hence, in polar coordinates
οΏ½οΏ½ = οΏ½ 1ππ’0 οΏ½π βπ2
ποΏ½ cosποΏ½ οΏ½οΏ½ β οΏ½ π1π + π’0 οΏ½1 +
π2
π2οΏ½ sinποΏ½ οΏ½οΏ½ (3)
3
Now the question posed can be answered. The circulation is given by
Ξ = οΏ½2π
0π’ππππ
But from (3) π’π = β οΏ½π1π + π’0 οΏ½1 +
π2
π2οΏ½ sinποΏ½, therefore the above becomes
Ξ = οΏ½2π
0β οΏ½π1π+ π’0 οΏ½1 +
π2
π2 οΏ½sinποΏ½ πππ
At π = π the above simplifies to
Ξ = οΏ½2π
0β οΏ½π1π+ 2π’0 sinποΏ½ πππ
= οΏ½2π
0βπ1 β 2ππ’0 sinπππ
= βοΏ½2π
0π1ππ β 2ππ’0οΏ½
2π
0sinπππ
But β«2π
0sinπππ = 0, hence
Ξ = βπ1οΏ½2π
0ππ
= β2π1π
Since Ξ < 0, then π1 > 0. Now that π1 is known to be positive, then the velocity is calculated at π = βπ2
and then at π = +π2 to see which is larger. Since this is calculated at π = π, then the radial velocity is
zero and only π’π needs to be evaluated in (3).
At π = βπ2
π’οΏ½ βπ2 οΏ½ = β οΏ½π1π+ π’0 οΏ½1 +
π2
π2 οΏ½sin οΏ½βπ
2οΏ½οΏ½
= β οΏ½π1πβ π’0 οΏ½1 +
π2
π2 οΏ½sin οΏ½π
2οΏ½οΏ½
= β οΏ½π1πβ π’0 οΏ½1 +
π2
π2 οΏ½οΏ½
At π = π
π’οΏ½ βπ2 οΏ½ = β οΏ½π1πβ 2π’0οΏ½
= βπ1π+ 2π’0 (4)
At π = +π2
π’οΏ½+π2 οΏ½ = β οΏ½π1π+ π’0 οΏ½1 +
π2
π2 οΏ½sin οΏ½π
2οΏ½οΏ½
= β οΏ½π1π+ π’0 οΏ½1 +
π2
π2 οΏ½οΏ½
4
At π = π
π’οΏ½ βπ2 οΏ½ = β οΏ½π1π+ 2π’0οΏ½
= βπ1πβ 2π’0 (5)
Comparing (4),(5), and since π1 > 0, then the magnitude of π’π at π2 is larger than the magnitude of
π’π at βπ2 . Which implies the stream flows faster above the cylinder than below it.
1.2 Problem 3.2.2 (b,d)
3.2. Convergence Theorem
L n7rx L n7rx n7rxan
Lf (x) cos L dx = L J cos L dx nir sin L
L L/2
= 1sinn7r - sin
nirnir 2
bn = 1 L f (x) sinnirx
L
1Ldx = sin n
xL nir LLJ
/2
nnCos - - cos n7r
n7r 2
95
(3.2.7)
(3.2.8)
We omit simplifications that arise by noting that sinn7r = 0, cosnir = (-1)", andso on.
EXERCISES 3.2
3.2.1. For the following functions, sketch the Fourier series off (x) (on the interval-L < x < L). Compare f (x) to its Fourier series:
(a) f(x) = 1 *(b) f(x) = x2
(c) f(x)=1+x *(d) f(x) = ex
(e) f (x) = { 2x x > 0 * (f) f (x) 1+x(g) f(x) x
0x < L/2x > L/2
x > O
3.2.2. For the following functions, sketch the Fourier series of f (x) (on the interval-L < x < L) and determine the Fourier coefficients:
*(a) f(x)=x (b) f(x) = e-x
*(c) f(x) = sin
L(d) f(x)
0 x < 0
l x x>0
(e) f(x) I jxj < L/20 jxI > L/2
(g) f(x) = I 1 x < 0
l 2 x>0
-1 nirxdx = - cos
* (f) f (x) = l 0
L
IL/2
L
L/2
x<0x>0
1.2.1 Part b
The following is sketch of periodic extension of πβπ₯ from π₯ = βπΏβ―πΏ (for πΏ = 1) for illustration. Thefunction will converge to πβπ₯ between π₯ = βπΏβ―πΏ and between π₯ = β3πΏβ―β πΏ and between π₯ = πΏβ―3πΏand so on. But at the jump discontinuities which occurs at π₯ = β― ,β3πΏ, βπΏ, πΏ, 3πΏ,β― it will converge tothe average shown as small circles in the sketch.
oo o o
-3 -2 -1 1 2 3
0.5
1.0
1.5
2.0
2.5
3.0
Periodic extension of exp(-x) from -1...1
5
By definitions,
π0 =1π οΏ½
π/2
βπ/2π (π₯) ππ₯
ππ =1π/2 οΏ½
π/2
βπ/2π (π₯) cos οΏ½π οΏ½
2ππ οΏ½
π₯οΏ½ ππ₯
ππ =1π/2 οΏ½
π/2
βπ/2π (π₯) sin οΏ½π οΏ½
2ππ οΏ½
π₯οΏ½ ππ₯
The period here is π = 2πΏ, therefore the above becomes
π0 =12πΏ οΏ½
πΏ
βπΏπ (π₯) ππ₯
ππ =1πΏ οΏ½
πΏ
βπΏπ (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
ππ =1πΏ οΏ½
πΏ
βπΏπ (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
These are now evaluated for π (π₯) = πβπ₯
π0 =12πΏ οΏ½
πΏ
βπΏπβπ₯ππ₯ =
12πΏ οΏ½
πβπ₯
β1 οΏ½πΏ
βπΏ=β12πΏ
(πβπ₯)πΏβπΏ =β12πΏ
οΏ½πβπΏ β ππΏοΏ½ =ππΏ β πβπΏ
2πΏ
Now ππ is found
ππ =1πΏ οΏ½
πΏ
βπΏπβπ₯ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
This can be done using integration by parts. β«π’ππ£ = π’π£ β β«π£ππ’. Let
πΌ = οΏ½πΏ
βπΏπβπ₯ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
and π’ = cos οΏ½πππΏ π₯οΏ½ , ππ£ = π
βπ₯,β ππ’ = βπππΏ sin οΏ½ππ
πΏ π₯οΏ½ , π£ = βπβπ₯, therefore
πΌ = [π’π£]πΏβπΏ βοΏ½πΏ
βπΏπ£ππ’
= οΏ½βπβπ₯ cos οΏ½πππΏπ₯οΏ½οΏ½
πΏ
βπΏβπππΏ οΏ½
πΏ
βπΏπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
= οΏ½βπβπΏ cos οΏ½πππΏπΏοΏ½ + ππΏ cos οΏ½ππ
πΏ(βπΏ)οΏ½οΏ½ β
πππΏ οΏ½
πΏ
βπΏπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
= οΏ½βπβπΏ cos (ππ) + ππΏ cos (ππ)οΏ½ β πππΏ οΏ½
πΏ
βπΏπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
Applying integration by parts again to β« πβπ₯ sin οΏ½πππΏ π₯οΏ½ ππ₯ where now π’ = sin οΏ½ππ
πΏ π₯οΏ½ , ππ£ = πβπ₯ β ππ’ =
6
πππΏ cos οΏ½ππ
πΏ π₯οΏ½ , π£ = βπβπ₯, hence the above becomes
πΌ = οΏ½βπβπΏ cos (ππ) + ππΏ cos (ππ)οΏ½ β πππΏοΏ½π’π£ βοΏ½π£ππ’οΏ½
= οΏ½βπβπΏ cos (ππ) + ππΏ cos (ππ)οΏ½ β πππΏ
ββββββββββββ
0
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½βπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½οΏ½
πΏ
βπΏ+πππΏ οΏ½
πΏ
βπΏπβπ₯ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
ββββββββββββ
= οΏ½βπβπΏ cos (ππ) + ππΏ cos (ππ)οΏ½ β πππΏ οΏ½
πππΏ οΏ½
πΏ
βπΏπβπ₯ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
= οΏ½βπβπΏ cos (ππ) + ππΏ cos (ππ)οΏ½ β οΏ½πππΏοΏ½2οΏ½
πΏ
βπΏπβπ₯ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
But β«πΏ
βπΏπβπ₯ cos οΏ½ππ
πΏ π₯οΏ½ ππ₯ = πΌ and the above becomes
πΌ = βπβπΏ cos (ππ) + ππΏ cos (ππ) β οΏ½πππΏοΏ½2πΌ
Simplifying and solving for πΌ
πΌ + οΏ½πππΏοΏ½2πΌ = cos (ππ) οΏ½ππΏ β πβπΏοΏ½
πΌ οΏ½1 + οΏ½πππΏοΏ½2οΏ½ = cos (ππ) οΏ½ππΏ β πβπΏοΏ½
πΌ οΏ½πΏ2 + π2π2
πΏ2 οΏ½ = cos (ππ) οΏ½ππΏ β πβπΏοΏ½
πΌ = οΏ½πΏ2
πΏ2 + π2π2 οΏ½ cos (ππ) οΏ½ππΏ β πβπΏοΏ½
Hence ππ becomes
ππ =1πΏ οΏ½
πΏ2
πΏ2 + π2π2 οΏ½ cos (ππ) οΏ½ππΏ β πβπΏοΏ½
But cos (ππ) = β1π hence
ππ = (β1)π οΏ½
πΏπ2π2 + πΏ2 οΏ½
οΏ½ππΏ β πβπΏοΏ½
Similarly for ππ
ππ =1πΏ οΏ½
πΏ
βπΏπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
This can be done using integration by parts. β«π’ππ£ = π’π£ β β«π£ππ’. Let
πΌ = οΏ½πΏ
βπΏπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
7
and π’ = sin οΏ½πππΏ π₯οΏ½ , ππ£ = π
βπ₯,β ππ’ = πππΏ cos οΏ½ππ
πΏ π₯οΏ½ , π£ = βπβπ₯, therefore
πΌ = [π’π£]πΏβπΏ βοΏ½πΏ
βπΏπ£ππ’
=
0
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½βπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½οΏ½
πΏ
βπΏ+πππΏ οΏ½
πΏ
βπΏπβπ₯ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
=πππΏ οΏ½
πΏ
βπΏπβπ₯ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
Applying integration by parts again to β« πβπ₯ cos οΏ½πππΏ π₯οΏ½ ππ₯ where now π’ = cos οΏ½ππ
πΏ π₯οΏ½ , ππ£ = πβπ₯ β ππ’ =
βπππΏ sin οΏ½ππ
πΏ π₯οΏ½ , π£ = βπβπ₯, hence the above becomes
πΌ =πππΏοΏ½π’π£ βοΏ½π£ππ’οΏ½
=πππΏ οΏ½οΏ½βπβπ₯ cos οΏ½ππ
πΏπ₯οΏ½οΏ½
πΏ
βπΏβπππΏ οΏ½
πΏ
βπΏπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
=πππΏ οΏ½βπβπΏ cos οΏ½ππ
πΏπΏοΏ½ + ππΏ cos οΏ½ππ
πΏπΏοΏ½ β
πππΏ οΏ½
πΏ
βπΏπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
=πππΏ οΏ½cos (ππ) οΏ½ππΏ β πβπΏοΏ½ β ππ
πΏ οΏ½πΏ
βπΏπβπ₯ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
But β«πΏ
βπΏπβπ₯ cos οΏ½ππ
πΏ π₯οΏ½ ππ₯ = πΌ and the above becomes
πΌ =πππΏοΏ½cos (ππ) οΏ½ππΏ β πβπΏοΏ½ β ππ
πΏπΌοΏ½
Simplifying and solving for πΌ
πΌ =πππΏ
cos (ππ) οΏ½ππΏ β πβπΏοΏ½ β οΏ½πππΏοΏ½2πΌ
πΌ + οΏ½πππΏοΏ½2πΌ =
πππΏ
cos (ππ) οΏ½ππΏ β πβπΏοΏ½
πΌ οΏ½1 + οΏ½πππΏοΏ½2οΏ½ =
πππΏ
cos (ππ) οΏ½ππΏ β πβπΏοΏ½
πΌ οΏ½πΏ2 + π2π2
πΏ2 οΏ½ =πππΏ
cos (ππ) οΏ½ππΏ β πβπΏοΏ½
πΌ = οΏ½πΏ2
πΏ2 + π2π2 οΏ½πππΏ
cos (ππ) οΏ½ππΏ β πβπΏοΏ½
Hence ππ becomes
ππ =1πΏ οΏ½
πΏ2
πΏ2 + π2π2 οΏ½πππΏ
cos (ππ) οΏ½ππΏ β πβπΏοΏ½
= οΏ½ππ
πΏ2 + π2π2 οΏ½ cos (ππ) οΏ½ππΏ β πβπΏοΏ½
But cos (ππ) = β1π hence
ππ = (β1)π οΏ½
πππΏ2 + π2π2 οΏ½ οΏ½π
πΏ β πβπΏοΏ½
8
Summary
π0 =ππΏ β πβπΏ
2πΏ
ππ = (β1)π οΏ½
πΏπ2π2 + πΏ2 οΏ½
οΏ½ππΏ β πβπΏοΏ½
ππ = (β1)π οΏ½
πππΏ2 + π2π2 οΏ½ οΏ½π
πΏ β πβπΏοΏ½
π (π₯) β π0 +βοΏ½π=1
ππ cos οΏ½π οΏ½2ππ οΏ½
π₯οΏ½ + ππ sin οΏ½π οΏ½2ππ οΏ½
π₯οΏ½
β π0 +βοΏ½π=1
ππ cos οΏ½πππΏπ₯οΏ½ + ππ sin οΏ½ππ
πΏπ₯οΏ½
The following shows the approximation π (π₯) for increasing number of terms. Notice the Gibbsphenomena at the jump discontinuity.
-3 -2 -1 0 1 2 3
0.0
0.5
1.0
1.5
2.0
2.5
x
f(x)
Fourier series approximation, number of terms 3Showing 3 periods extenstion of -L..L, with L=1
-3 -2 -1 0 1 2 3
0.0
0.5
1.0
1.5
2.0
2.5
x
f(x)
Fourier series approximation, number of terms 10Showing 3 periods extenstion of -L..L, with L=1
9
-3 -2 -1 0 1 2 3
0.0
0.5
1.0
1.5
2.0
2.5
3.0
x
f(x)
Fourier series approximation, number of terms 50Showing 3 periods extenstion of -L..L, with L=1
1.2.2 Part d
The following is sketch of periodic extension of π (π₯) from π₯ = βπΏβ―πΏ (for πΏ = 1) for illustration. Thefunction will converge to π (π₯) between π₯ = βπΏβ―πΏ and between π₯ = β3πΏβ―βπΏ and between π₯ = πΏβ―3πΏand so on. But at the jump discontinuities which occurs at π₯ = β― ,β3πΏ, βπΏ, πΏ, 3πΏ,β― it will converge tothe average 1
2 shown as small circles in the sketch.
oo o o
-3 -2 -1 1 2 3
0.20.40.60.81.0
Showing 3 periods extenstion of f(x) between -L..L, with L=1
By definitions,
π0 =1π οΏ½
π/2
βπ/2π (π₯) ππ₯
ππ =1π/2 οΏ½
π/2
βπ/2π (π₯) cos οΏ½π οΏ½
2ππ οΏ½
π₯οΏ½ ππ₯
ππ =1π/2 οΏ½
π/2
βπ/2π (π₯) sin οΏ½π οΏ½
2ππ οΏ½
π₯οΏ½ ππ₯
The period here is π = 2πΏ, therefore the above becomes
π0 =12πΏ οΏ½
πΏ
βπΏπ (π₯) ππ₯
ππ =1πΏ οΏ½
πΏ
βπΏπ (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
ππ =1πΏ οΏ½
πΏ
βπΏπ (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
10
These are now evaluated for given π (π₯)
π0 =12πΏ οΏ½
πΏ
βπΏπ (π₯) ππ₯
=12πΏ οΏ½οΏ½
0
βπΏπ (π₯) ππ₯ +οΏ½
πΏ
0π (π₯) ππ₯οΏ½
=12πΏ οΏ½
0 +οΏ½πΏ
0π₯ππ₯οΏ½
=12πΏ οΏ½
π₯2
2 οΏ½πΏ
0
=πΏ4
Now ππ is found
ππ =1πΏ οΏ½
πΏ
βπΏπ (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
=1πΏ οΏ½οΏ½
0
βπΏπ (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯ +οΏ½
πΏ
0π (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
=1πΏ οΏ½
πΏ
0π₯ cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
Integration by parts. Let π’ = π₯, ππ’ = 1, ππ£ = cos οΏ½πππΏ π₯οΏ½ , π£ =
sinοΏ½πππΏ π₯οΏ½
πππΏ
, hence the above becomes
ππ =1πΏ
ββββββββββββ
0
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½πππΏπ₯ sin οΏ½ππ
πΏπ₯οΏ½οΏ½
πΏ
0βοΏ½
πΏ
0
sin οΏ½πππΏ π₯οΏ½
πππΏ
ππ₯
ββββββββββββ
=1πΏ οΏ½βπΏππ οΏ½
πΏ
0sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
=1πΏ
βββββββββ
πΏππ
ββββββββ cos οΏ½ππ
πΏ π₯οΏ½
πππΏ
βββββββ
πΏ
0
ββββββββ
=1πΏ
ββββββοΏ½πΏπποΏ½
2
cos οΏ½πππΏπ₯οΏ½
πΏ
0
ββββββ
=πΏ
π2π2 cos οΏ½πππΏπ₯οΏ½
πΏ
0
=πΏ
π2π2 οΏ½cos οΏ½πππΏπΏοΏ½ β 1οΏ½
=πΏ
π2π2 [β1π β 1]
11
Now ππ is found
ππ =1πΏ οΏ½
πΏ
βπΏπ (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
=1πΏ οΏ½οΏ½
0
βπΏπ (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯ +οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
=1πΏ οΏ½
πΏ
0π₯ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
Integration by parts. Let π’ = π₯, ππ’ = 1, ππ£ = sin οΏ½πππΏ π₯οΏ½ , π£ =
β cosοΏ½πππΏ π₯οΏ½
πππΏ
, hence the above becomes
ππ =1πΏ
βββββββοΏ½β
πΏπππ₯ cos οΏ½ππ
πΏπ₯οΏ½οΏ½
πΏ
0+οΏ½
πΏ
0
cos οΏ½πππΏ π₯οΏ½
πππΏ
ππ₯
βββββββ
=1πΏ οΏ½βπΏππ
οΏ½πΏ cos οΏ½πππΏπΏοΏ½ β 0οΏ½ +
πΏππ οΏ½
πΏ
0cos οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
=1πΏ
ββββββββββββββββ
βπΏ2
ππ(β1)π +
πΏππ
0
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β‘β’β’β’β’β’β£sin οΏ½ππ
πΏ π₯οΏ½
πππΏ
β€β₯β₯β₯β₯β₯β¦
πΏ
0
ββββββββββββββββ
=πΏππ
οΏ½β (β1)ποΏ½
= (β1)π+1πΏππ
Summary
π0 =πΏ4
ππ =πΏ
π2π2 [β1π β 1]
ππ = (β1)π+1 πΏ
ππ
π (π₯) β π0 +βοΏ½π=1
ππ cos οΏ½π οΏ½2ππ οΏ½
π₯οΏ½ + ππ sin οΏ½π οΏ½2ππ οΏ½
π₯οΏ½
β π0 +βοΏ½π=1
ππ cos οΏ½πππΏπ₯οΏ½ + ππ sin οΏ½ππ
πΏπ₯οΏ½
The following shows the approximation π (π₯) for increasing number of terms. Notice the Gibbsphenomena at the jump discontinuity.
12
-3 -2 -1 0 1 2 3
0.0
0.2
0.4
0.6
0.8
x
f(x)
Fourier series approximation, number of terms 3Showing 3 periods extenstion of -L..L, with L=1
-3 -2 -1 0 1 2 3
0.0
0.2
0.4
0.6
0.8
1.0
x
f(x)
Fourier series approximation, number of terms 10Showing 3 periods extenstion of -L..L, with L=1
-3 -2 -1 0 1 2 3-0.2
0.0
0.2
0.4
0.6
0.8
1.0
x
f(x)
Fourier series approximation, number of terms 50Showing 3 periods extenstion of -L..L, with L=1
13
1.3 Problem 3.2.4
96 Chapter 3. Fourier Series
3.2.3. Show that the Fourier series operation is linear: that is, show that theFourier series of c1 f (x) + c2g(x) is the sum of cl times the Fourier series off (x) and c2 times the Fourier series of g(x).
3.2.4. Suppose that f (x) is piecewise smooth. What value does the Fourier seriesof f (x) converge to at the endpoint x = -L? at x = L?
3.3 Fourier Cosine and Sine SeriesIn this section we show that the series of sines only (and the series of cosines only)are special cases of a Fourier series.
3.3.1 Fourier Sine SeriesOdd functions. An odd function is a function with the property f (-x)- f (x). The sketch of an odd function for x < 0 will be minus the mirror image off (x) for x > 0, as illustrated in Fig. 3.3.1. Examples of odd functions are f (x) = x3(in fact, any odd power) and f (x) = sin 4x. The integral of an odd function overa symmetric interval is zero (any contribution from x > 0 will be canceled by acontribution from x < 0).
Figure 3.3.1 An odd function.
Fourier series of odd functions. Let us calculate the Fourier coeffi-cients of an odd function:
ao1 fL
1 J_Lf (x) dx = 0
L
an = L ff(x)cosdx=0.L
Both are zero because the integrand, f (x) cos nirx/L, is odd (being the product ofan even function cos n7rx/L and an odd function f (x)). Since an = 0, all the cosinefunctions (which are even) will not appear in the Fourier series of an odd function.The Fourier series of an odd function is an infinite series of odd functions (sines):
00
f (x) - bn sinn1x,
(3.3.1)n=1
It will converge to the average value of the function at the end points after making periodic extensionsof the function. Specifically, at π₯ = βπΏ the Fourier series will converge to
12οΏ½π (βπΏ) + π (πΏ)οΏ½
And at π₯ = πΏ it will converge to
12οΏ½π (πΏ) + π (βπΏ)οΏ½
Notice that if π (πΏ) has same value as π (βπΏ), then there will not be a jump discontinuity when periodicextension are made, and the above formula simply gives the value of the function at either end, sinceit is the same value.
1.4 Problem 3.3.2 (d)
114 Chapter 3. Fourier Series
(a)
(c)
f(x) = 1
f(x) _ {1 + x x > 0
(e) f(x) e-x x > 0
(b) f(x)=1+x
*(d) f(x) = ex
3.3.2. For the following functions, sketch the Fourier sine series of f (x) and deter-mine its Fourier coefficients.
(a) [Verify formula (3.3.13).]
(c) f(x)0
xx < L/2
x > L/2
1 x < L/6
(b) f (x) = 3 L/6 < x < L/20 x > L/2
* (d) f (x) 1 x < L/20 x > L/2
3.3.3. For the following functions, sketch the Fourier sine series of f (x). Also,roughly sketch the sum of a finite number of nonzero terms (at least thefirst two) of the Fourier sine series:
(a) f (x) = cos irx/L [Use formula (3.3.13).]
(b) f(x) _ { 1 x < L/20 x > L/2
(c) f (x) = x [Use formula (3.3.12).]
3.3.4. Sketch the Fourier cosine series of f (x) = sin irx/L. Briefly discuss.
3.3.5. For the following functions, sketch the Fourier cosine series of f (x) anddetermine its Fourier coefficients:
1 x < L/6(a) f (x) = x2 (b) f (x) = 3 L/6 < x < L/2 (c) f (x) =
0 x > L/2 fx x > L/2
3.3.6. For the following functions, sketch the Fourier cosine series of f (x). Also,roughly sketch the sum of a finite number of nonzero terms (at least thefirst two) of the Fourier cosine series:
(a) f (x) = x [Use formulas (3.3.22) and (3.3.23).]
(b) f (x) = 0 x<L/21 x > L12 [Use carefully formulas (3.2.6) and (3.2.7).]
_ 0 x < L/2(c) f (x)1 x > L/2 [Hint: Add the functions in parts (b) and (c).]
3.3.7. Show that ex is the sum of an even and an odd function.
π (π₯) =
β§βͺβͺβ¨βͺβͺβ©1 π₯ < πΏ
20 π₯ > πΏ
2
The first step is to sketch π (π₯) over 0β―πΏ. This is the result for πΏ = 1 as an example.
14
-3 -2 -1 1 2 3
0.2
0.4
0.6
0.8
1.0
Original f(x) function defined for 0..L
The second step is to make an odd extension of π (π₯) over βπΏβ―πΏ. This is the result.
-3 -2 -1 1 2 3
-1.0
-0.5
0.5
1.0
odd extension of f(x) defined for -L..L
The third step is to extend the above as periodic function with period 2πΏ (as normally would bedone) and mark the average value at the jump discontinuities. This is the result
o
o o
o
o o
o
o
o
-3 -2 -1 1 2 3
-1.0
-0.5
0.5
1.0
Showing 3 periods extenstion of f(x) between -L..L, with L=1
Now the Fourier sin series is found for the above function. Since the function π (π₯) is odd, then onlyππ will exist
π (π₯) ββοΏ½π=1
ππ sin οΏ½π οΏ½2π2πΏ οΏ½
π₯οΏ½
ββοΏ½π=1
ππ sin οΏ½πππΏπ₯οΏ½
Where
ππ =1πΏ οΏ½
πΏ
βπΏπ (π₯) sin οΏ½π οΏ½
2π2πΏ οΏ½
π₯οΏ½ ππ₯ =1πΏ οΏ½
πΏ
βπΏπ (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
15
Since π (π₯) sin οΏ½πππΏ π₯οΏ½ is even, then the above becomes
ππ =2πΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
=2πΏ οΏ½οΏ½
πΏ/2
01 Γ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯ +οΏ½
πΏ/2
00 Γ sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½
=2πΏ οΏ½
πΏ/2
0sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
=2πΏ
β‘β’β’β’β’β’β£β
cos οΏ½πππΏ π₯οΏ½
πππΏ
β€β₯β₯β₯β₯β₯β¦
πΏ/2
0
=β2ππ οΏ½
cos οΏ½πππΏπ₯οΏ½οΏ½
πΏ/2
0
=β2ππ οΏ½
cos οΏ½πππΏπΏ2οΏ½β 1οΏ½
=β2ππ οΏ½
cos οΏ½ππ2οΏ½ β 1οΏ½
=2ππ οΏ½
1 β cos οΏ½ππ2οΏ½οΏ½
Therefore
π (π₯) ββοΏ½π=1
2ππ
οΏ½1 β cos οΏ½ππ2οΏ½οΏ½ sin οΏ½ππ
πΏπ₯οΏ½
The following shows the approximation π (π₯) for increasing number of terms. Notice the Gibbsphenomena at the jump discontinuity.
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 3Showing 3 periods extenstion of -L..L, with L=1
16
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 10Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 50Showing 3 periods extenstion of -L..L, with L=1
1.5 Problem 3.3.3 (b)
114 Chapter 3. Fourier Series
(a)
(c)
f(x) = 1
f(x) _ {1 + x x > 0
(e) f(x) e-x x > 0
(b) f(x)=1+x
*(d) f(x) = ex
3.3.2. For the following functions, sketch the Fourier sine series of f (x) and deter-mine its Fourier coefficients.
(a) [Verify formula (3.3.13).]
(c) f(x)0
xx < L/2
x > L/2
1 x < L/6
(b) f (x) = 3 L/6 < x < L/20 x > L/2
* (d) f (x) 1 x < L/20 x > L/2
3.3.3. For the following functions, sketch the Fourier sine series of f (x). Also,roughly sketch the sum of a finite number of nonzero terms (at least thefirst two) of the Fourier sine series:
(a) f (x) = cos irx/L [Use formula (3.3.13).]
(b) f(x) _ { 1 x < L/20 x > L/2
(c) f (x) = x [Use formula (3.3.12).]
3.3.4. Sketch the Fourier cosine series of f (x) = sin irx/L. Briefly discuss.
3.3.5. For the following functions, sketch the Fourier cosine series of f (x) anddetermine its Fourier coefficients:
1 x < L/6(a) f (x) = x2 (b) f (x) = 3 L/6 < x < L/2 (c) f (x) =
0 x > L/2 fx x > L/2
3.3.6. For the following functions, sketch the Fourier cosine series of f (x). Also,roughly sketch the sum of a finite number of nonzero terms (at least thefirst two) of the Fourier cosine series:
(a) f (x) = x [Use formulas (3.3.22) and (3.3.23).]
(b) f (x) = 0 x<L/21 x > L12 [Use carefully formulas (3.2.6) and (3.2.7).]
_ 0 x < L/2(c) f (x)1 x > L/2 [Hint: Add the functions in parts (b) and (c).]
3.3.7. Show that ex is the sum of an even and an odd function.
This is the same problem as 3.3.2 part (d). But it asks to plot for π = 1 and π = 2 in the sum. Thesketch of the Fourier sin series was done above in solving 3.3.2 part(d) and will not be repeated
17
again. From above, it was found that
π (π₯) ββοΏ½π=1
π΅π sin οΏ½πππΏπ₯οΏ½
Where π΅π =2πποΏ½1 β cos οΏ½ππ2 οΏ½οΏ½. The following is the plot for π = 1β―10.
18
-1.0 -0.5 0.0 0.5 1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 1Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 2Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 3Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 4Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 5Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 6Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 7Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 8Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 9Showing 3 periods extenstion of -L..L, with L=1
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
f(x)
Fourier series approximation, number of terms 10Showing 3 periods extenstion of -L..L, with L=1
19
1.6 Problem 3.3.8
3.3. Cosine and Sine Series 115
3.3.8. (a) Determine formulas for the even extension of any f (x). Compare tothe formula for the even part of f (x).
(b) Do the same for the odd extension of f (x) and the odd part of f (x).
(c) Calculate and sketch the four functions of parts (a) and (b) if
= J x x>0f(x x2 x <0.
Graphically add the even and odd parts of f (x). What occurs? Simi-larly, add the even and odd extensions. What occurs then?
3.3.9. What is the sum of the Fourier sine series of f (x) and the Fourier cosineseries of f (x)? [What is the sum of the even and odd extensions of f (x)?]
2
*3.3.10. If f (x) = e_z x > 0 , what are the even and odd parts of f (x)?
3.3.11. Given a sketch of f(x), describe a procedure to sketch the even and odd
parts of f (x).
3.3.12. (a) Graphically show that the even terms (n even) of the Fourier sine seriesof any function on 0 < x < L are odd .(antisymmetric) around x = L/2.
(b) Consider a function f (x) that is odd around x = L/2. Show that theodd coefficients (n odd) of the Fourier sine series of f (x) on 0 < x < Lare zero.
*3.3.13. Consider a function f (x) that is even around x = L/2. Show that the evencoefficients (n even) of the Fourier sine series of f (x) on 0 < x < L are zero.
3.3.14. (a) Consider a function f (x) that is even around x = L/2. Show thatthe odd coefficients (n odd) of the Fourier cosine series of f (x) on0 < x < L are zero.
(b) Explain the result of part (a) by considering a Fourier cosine series off (x) on the interval 0 < x < L/2.
3.3.15. Consider a function f (x) that is odd around x = L/2. Show that the evencoefficients (n even) of the Fourier cosine series of f (x) on 0 < x < L arezero.
3.3.16. Fourier series can be defined on other intervals besides -L < x < L. Sup-pose that g(y) is defined for a < y < b. Represent g(y) using periodictrigonometric functions with period b - a. Determine formulas for the coef-ficients. [Hint: Use the linear transformation
a+b b-a2 + 2L
1.6.1 Part (a)
The even extension of π (π₯) isβ§βͺβͺβ¨βͺβͺβ©
π (π₯) π₯ > 0π (βπ₯) π₯ < 0
But the even part of π (π₯) is12οΏ½π (π₯) + π (βπ₯)οΏ½
1.6.2 Part (b)
The odd extension of π (π₯) isβ§βͺβͺβ¨βͺβͺβ©
π (π₯) π₯ > 0βπ (βπ₯) π₯ < 0
While the odd part of π (π₯) is12οΏ½π (π₯) β π (βπ₯)οΏ½
1.6.3 Part (c)
First a plot of π (π₯) is given
π (π₯) =
β§βͺβͺβ¨βͺβͺβ©π₯ π₯ > 0π₯2 π₯ < 0
20
-1.0 -0.5 0.0 0.5 1.00.0
0.2
0.4
0.6
0.8
1.0
x
f(x)
Plot of f(x)
A plot of even extension and the even part for π (π₯) Is given below
-1.0 -0.5 0.0 0.5 1.0
0.0
0.2
0.4
0.6
0.8
1.0
x
f(x)
Plot of even extension
-1.0 -0.5 0.0 0.5 1.0
0.0
0.2
0.4
0.6
0.8
1.0
x
f(x)
Plot of even part
A plot of odd extension and the odd part is given below
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
x
function
Plot of odd extension
-1.0 -0.5 0.0 0.5 1.0
-0.10
-0.05
0.00
0.05
0.10
x
function
Plot of odd part of f(x)
Adding the even part and the odd part gives back the original function
21
-1.0 -0.5 0.0 0.5 1.0
0.0
0.2
0.4
0.6
0.8
1.0
x
function
Plot of (odd +even parts of f(x))
Plot of adding the even extension and the odd extension is below
-1.0 -0.5 0.0 0.5 1.0
0.0
0.5
1.0
1.5
2.0
x
function
Plot of (even extension+odd extension of f(x))
1.7 Problem 3.4.3
3.4. Term-by-Term Differentiation 125
3.4.3. Suppose that f (x) is continuous [except for a jump discontinuity at x = xΒ°if (xa) = a and f (xo) = 01 and df /dx is piecewise smooth.
*(a) Determine the Fourier sine series of df /dx in terms of the Fourier cosineseries coefficients of f (x).
(b) Determine the Fourier cosine series of df /dx in terms of the Fouriersine series coefficients of f(x).
3.4.4. Suppose that f (x) and df /dx are piecewise smooth.
(a) Prove that the Fourier sine series of a continuous function f (x) canonly be differentiated term by term if f (0) = 0 and f (L) = 0.
(b) Prove that the Fourier cosine series of a continuous function f (x) canbe differentiated term by term.
3.4.5. Using (3.3.13) determine the Fourier cosine series of sin 7rx/L.
3.4.6. There are some things wrong in the following demonstration. Find themistakes and correct them.
In this exercise we attempt to obtain the Fourier cosine coefficients of ex:
00 nirxex=AΒ°+E A.cos r .
n=1
Differentiating yields
00 nir nirxe2=-L ,
n=1
the Fourier sine series of ex. Differentiating again yields
(3.4.22)
Β°O 2n7rex - ( L) An cos nLx (3.4.23)
n=1
Since equations (3.4.22) and (3.4.23) give the Fourier cosine series of ex,
they must be identical. Thus,
`40 0 (obviously wrong!).An =0
By correcting the mistakes, you should be able to obtain AΒ° and An withoutusing the typical technique, that is, An = 2/L f L ex cos nirx/L dx.
3.4.7. Prove that the Fourier series of a continuous function u(x, t) can be differ-entiated term by term with respect to the parameter t if 8u/8t is piecewisesmooth.
1.7.1 Part (a)
Fourier sin series of πβ² (π₯) is given by, assuming period is βπΏβ―πΏ
πβ² (π₯) βΌβοΏ½π=1
ππ sin οΏ½πππΏπ₯οΏ½
22
Where
ππ =2πΏ οΏ½
πΏ
0πβ² (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
Applying integration by parts. Let πβ² (π₯) = ππ£, π’ = sin οΏ½πππΏ π₯οΏ½, then π£ = π (π₯) , ππ’ = ππ
πΏ cos οΏ½πππΏ π₯οΏ½. Sinceπ£ = π (π₯) has has jump discontinuity at π₯0 as described, and assuming π₯0 > 0, then, and usingsin οΏ½ππ
πΏ π₯οΏ½ = 0 at π₯ = πΏ
ππ =2πΏ οΏ½
πΏ
0π’ππ£
=2πΏ οΏ½οΏ½[π’π£]π₯
β00 + [π’π£]πΏπ₯+0 οΏ½ βοΏ½
πΏ
0π£ππ’οΏ½
=2πΏ
ββββββοΏ½sin οΏ½π
ππΏπ₯οΏ½ π (π₯)οΏ½
π₯β0
0+ οΏ½sin οΏ½π
ππΏπ₯οΏ½ π (π₯)οΏ½
πΏ
π₯+0βπππΏ οΏ½
πΏ
0π (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
ββββββ
=2πΏ οΏ½
sin οΏ½πππΏπ₯β0οΏ½ π οΏ½π₯β0οΏ½ β sin οΏ½ππ
πΏπ₯+0π οΏ½π₯+0 οΏ½οΏ½ β
πππΏ οΏ½
πΏ
0π (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½ (1)
In the above, sin οΏ½πππΏ π₯οΏ½ = 0 and at π₯ = πΏ was used. But
π οΏ½π₯β0οΏ½ = πΌ
π οΏ½π₯+0 οΏ½ = π½
And since sin is continuous, then sin οΏ½πππΏ π₯
β0οΏ½ = sin οΏ½ππ
πΏ π₯+0 οΏ½ = sin οΏ½ππ
πΏ π₯0οΏ½. Equation (1) simplifies to
ππ =2πΏ οΏ½οΏ½πΌ β π½οΏ½ sin οΏ½ππ
πΏπ₯0οΏ½ β
πππΏ οΏ½
πΏ
0π (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½ (2)
On the other hand, the Fourier cosine series for π (π₯) is given by
π (π₯) βΌ π0 +βοΏ½π=1
ππ cos οΏ½πππΏπ₯οΏ½
Where
π0 =1πΏ οΏ½
πΏ
0π (π₯) ππ₯
ππ =2πΏ οΏ½
πΏ
0π (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
Therefore β«πΏ
0π (π₯) cos οΏ½ππ
πΏ π₯οΏ½ ππ₯ =πΏ2ππ. Substituting this into (2) gives
ππ =2πΏ οΏ½οΏ½πΌ β π½οΏ½ sin οΏ½ππ
πΏπ₯0οΏ½ β
πππΏ οΏ½
πΏ2πποΏ½οΏ½
=2πΏοΏ½πΌ β π½οΏ½ sin οΏ½ππ
πΏπ₯0οΏ½ β
2πΏπππΏ οΏ½
πΏ2πποΏ½
Hence
ππ =2πΏ sin οΏ½ππ
πΏ π₯0οΏ½ οΏ½πΌ β π½οΏ½ βπππΏ ππ (3)
Summary the Fourier sin series of πβ² (π₯) is
πβ² (π₯) βΌβοΏ½π=1
ππ sin οΏ½πππΏπ₯οΏ½
23
With ππ given by (3). The above is in terms of ππ, which is the Fourier cosine series of π (π₯), which iswhat required to show. In addition, the cos series of π (π₯) can also be written in terms of sin series ofπβ² (π₯). From (3), solving for ππ
ππ =πΏππππ β
2ππ
sin οΏ½πππΏπ₯0οΏ½ οΏ½πΌ β π½οΏ½
π (π₯) βΌ π0 +βοΏ½π=1
1π οΏ½
πΏπππ β
2π
sin οΏ½πππΏπ₯0οΏ½ οΏ½πΌ β π½οΏ½οΏ½ cos οΏ½ππ
πΏπ₯οΏ½
This shows more clearly that the Fourier series of π (π₯) has order of convergence in ππ as1π as expected.
1.7.2 Part (b)
Fourier cos series of πβ² (π₯) is given by, assuming period is βπΏβ―πΏ
πβ² (π₯) βΌβοΏ½π=0
ππ cos οΏ½πππΏπ₯οΏ½
Where
π0 =1πΏ οΏ½
πΏ
0πβ² (π₯) ππ₯
=1πΏ
βββββοΏ½
π₯β0
0πβ² (π₯) ππ₯ +οΏ½
πΏ
π₯+0πβ² (π₯) ππ₯
βββββ
=1πΏ οΏ½οΏ½π (π₯)οΏ½
π₯β00+ οΏ½π (π₯)οΏ½
πΏ
π₯+0οΏ½
=1πΏοΏ½οΏ½πΌ β π (0)οΏ½ + οΏ½π (πΏ) β π½οΏ½οΏ½
=οΏ½πΌ β π½οΏ½πΏ
+π (0) + π (πΏ)
πΏAnd for π > 0
ππ =2πΏ οΏ½
πΏ
0πβ² (π₯) cos οΏ½ππ
πΏπ₯οΏ½ ππ₯
Applying integration by parts. Let πβ² (π₯) = ππ£, π’ = cos οΏ½πππΏ π₯οΏ½, then π£ = π (π₯) , ππ’ =
βπππΏ sin οΏ½πππΏ π₯οΏ½. Since
π£ = π (π₯) has has jump discontinuity at π₯0 as described, then
ππ =2πΏ οΏ½
πΏ
0π’ππ£
=2πΏ οΏ½οΏ½[π’π£]π₯
β00 + [π’π£]πΏπ₯+0 οΏ½ βοΏ½
πΏ
0π£ππ’οΏ½
=2πΏ
ββββββοΏ½cos οΏ½ππ
πΏπ₯οΏ½ π (π₯)οΏ½
π₯β0
0+ οΏ½cos οΏ½ππ
πΏπ₯οΏ½ π (π₯)οΏ½
πΏ
π₯+0+πππΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
ββββββ
=2πΏ οΏ½
cos οΏ½πππΏπ₯β0οΏ½ π οΏ½π₯β0οΏ½ β π (0) + cos (ππ) π (πΏ) β cos οΏ½ππ
πΏπ₯+0 οΏ½ π οΏ½π₯+0 οΏ½ +
πππΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½ (1)
But
π οΏ½π₯β0οΏ½ = πΌ
π οΏ½π₯+0 οΏ½ = π½
24
And since cos is continuous, then cos οΏ½πππΏ π₯
β0οΏ½ = cos οΏ½ππ
πΏ π₯+0 οΏ½ = cos οΏ½ππ
πΏ π₯0οΏ½, therefore (1) becomes
ππ =2πΏ οΏ½
cos (ππ) π (πΏ) β π (0) + cos οΏ½πππΏπ₯0οΏ½ οΏ½πΌ β π½οΏ½ +
πππΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯οΏ½ (2)
On the other hand, the Fourier π ππ series for π (π₯) is given by
π (π₯) βΌβοΏ½π=0
ππ sin οΏ½πππΏπ₯οΏ½
Where
ππ =2πΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
Therefore β«πΏ
0π (π₯) sin οΏ½ππ
πΏ π₯οΏ½ ππ₯ =πΏ2ππ. Substituting this into (2) gives
ππ =2πΏ οΏ½
cos (ππ) π (πΏ) β π (0) + cos οΏ½πππΏπ₯0οΏ½ οΏ½πΌ β π½οΏ½ +
πππΏπΏ2πποΏ½
=2πΏ
cos (ππ) π (πΏ) β 2πΏπ (0) +
2πΏ
cos οΏ½πππΏπ₯0οΏ½ οΏ½πΌ β π½οΏ½ +
2πΏππ2ππ
=2πΏοΏ½(β1π) π (πΏ) β π (0)οΏ½ +
2πΏ
cos οΏ½πππΏπ₯0οΏ½ οΏ½πΌ β π½οΏ½ +
πππΏππ
Hence
ππ =2πΏοΏ½(β1π) π (πΏ) β π (0)οΏ½ + 2
πΏ cos οΏ½πππΏ π₯0οΏ½ οΏ½πΌ β π½οΏ½ +
πππΏ ππ (3)
Summary the Fourier cos series of πβ² (π₯) is
πβ² (π₯) βΌβοΏ½π=0
ππ cos οΏ½πππΏπ₯οΏ½
π0 =οΏ½πΌ β π½οΏ½πΏ
+π (0) + π (πΏ)
πΏ
ππ =2πΏοΏ½(β1π) π (πΏ) β π (0)οΏ½ +
2πΏ
cos οΏ½πππΏπ₯0οΏ½ οΏ½πΌ β π½οΏ½ +
πππΏππ
The above is in terms of ππ, which is the Fourier π ππ series of π (π₯), which is what required to show.
25
1.8 Problem 3.4.9
126 Chapter 3. Fourier Series
3.4.8. Considerau _ a2uat kaxe
subject to
au/ax(0,t) = 0, au/ax(L,t) = 0, and u(x,0) = f(x).
Solve in the following way. Look for the solution as a Fourier cosine se-ries. Assume that u and au/ax are continuous and a2u/axe and au/at arepiecewise smooth. Justify all differentiations of infinite series.
*3.4.9 Consider the heat equation with a known source q(x, t):
2
= k jx2 + q(x, t) with u(0, t) = 0 and u(L, t) = 0.
Assume that q(x, t) (for each t > 0) is a piecewise smooth function of x.Also assume that u and au/ax are continuous functions of x (for t > 0) anda2u/axe and au/at are piecewise smooth. Thus,
u(x, t) _ N,(t) sin .Lx.n=1
What ordinary differential equation does satisfy? Do not solve thisdifferential equation.
3.4.10. Modify Exercise 3.4.9 if instead au/ax(0, t) = 0 and au/ax(L, t) = 0.
3.4.11. Consider the nonhomogeneous heat equation (with a steady heat source):
2
at kax2 +g(x).
Solve this equation with the initial condition
u(x,0) = f(x)
and the boundary conditions
u(O,t) = 0 and u(L, t) = 0.
Assume that a continuous solution exists (with continuous derivatives).[Hints: Expand the solution as a Fourier sine series (i.e., use the methodof eigenfunction expansion). Expand g(x) as a Fourier sine series. Solvefor the Fourier sine series of the solution. Justify all differentiations withrespect to x.]
*3.4.12. Solve the following nonhomogeneous problem:
2
= k5-2 + e-t + e- 2tcos 3Lx [assume that 2 # k(37r/L)2]
The PDE isππ’ππ‘
= ππ2π’ππ₯2
+ π (π₯, π‘) (1)
Since the boundary conditions are homogenous Dirichlet conditions, then the solution can be writtendown as
π’ (π₯, π‘) =βοΏ½π=1
ππ (π‘) sin οΏ½πππΏπ₯οΏ½
Since the solution is assumed to be continuous with continuous derivative, then term by termdiοΏ½erentiation is allowed w.r.t. π₯
ππ’ππ₯
=βοΏ½π=1
πππΏππ (π‘) cos οΏ½ππ
πΏπ₯οΏ½
π2π’ππ₯2
= ββοΏ½π=1
οΏ½πππΏοΏ½2ππ (π‘) sin οΏ½π
ππΏπ₯οΏ½ (2)
Also using assumption that ππ’ππ‘ is smooth, then
ππ’ππ‘
=βοΏ½π=1
πππ (π‘)ππ‘
sin οΏ½πππΏπ₯οΏ½ (3)
Substituting (2,3) into (1) givesβοΏ½π=1
πππ (π‘)ππ‘
sin οΏ½πππΏπ₯οΏ½ = βπ
βοΏ½π=1
οΏ½πππΏοΏ½2ππ (π‘) sin οΏ½π
ππΏπ₯οΏ½ + π (π₯, π‘) (4)
Expanding π (π₯, π‘) as Fourier sin series in π₯. Hence
π (π₯, π‘) =βοΏ½π=1
ππ (π‘) sin οΏ½πππΏπ₯οΏ½
Where now ππ (π‘) are time dependent given by (by orthogonality)
ππ (π‘) =2πΏ οΏ½
πΏ
0π (π₯, π‘) sin οΏ½ππ
πΏπ₯οΏ½
26
Hence (4) becomesβοΏ½π=1
πππ (π‘)ππ‘
sin οΏ½πππΏπ₯οΏ½ = β
βοΏ½π=1
π οΏ½πππΏοΏ½2ππ (π‘) sin οΏ½π
ππΏπ₯οΏ½ +
βοΏ½π=1
π (π‘)π sin οΏ½πππΏπ₯οΏ½
Applying orthogonality the above reduces to one term only
πππ (π‘)ππ‘
sin οΏ½πππΏπ₯οΏ½ = βπ οΏ½
πππΏοΏ½2ππ (π‘) sin οΏ½π
ππΏπ₯οΏ½ + π (π‘)π sin οΏ½ππ
πΏπ₯οΏ½
Dividing by sin οΏ½πππΏ π₯οΏ½ β 0
πππ (π‘)ππ‘
= βπ οΏ½πππΏοΏ½2ππ (π‘) + ππ (π‘)
πππ (π‘)ππ‘
+ π οΏ½πππΏοΏ½2π΅π (π‘) = ππ (π‘) (5)
The above is the ODE that needs to be solved for ππ (π‘). It is first order inhomogeneous ODE. Thequestion asks to stop here.
1.9 Problem 3.4.11
126 Chapter 3. Fourier Series
3.4.8. Considerau _ a2uat kaxe
subject to
au/ax(0,t) = 0, au/ax(L,t) = 0, and u(x,0) = f(x).
Solve in the following way. Look for the solution as a Fourier cosine se-ries. Assume that u and au/ax are continuous and a2u/axe and au/at arepiecewise smooth. Justify all differentiations of infinite series.
*3.4.9 Consider the heat equation with a known source q(x, t):
2
= k jx2 + q(x, t) with u(0, t) = 0 and u(L, t) = 0.
Assume that q(x, t) (for each t > 0) is a piecewise smooth function of x.Also assume that u and au/ax are continuous functions of x (for t > 0) anda2u/axe and au/at are piecewise smooth. Thus,
u(x, t) _ N,(t) sin .Lx.n=1
What ordinary differential equation does satisfy? Do not solve thisdifferential equation.
3.4.10. Modify Exercise 3.4.9 if instead au/ax(0, t) = 0 and au/ax(L, t) = 0.
3.4.11. Consider the nonhomogeneous heat equation (with a steady heat source):
2
at kax2 +g(x).
Solve this equation with the initial condition
u(x,0) = f(x)
and the boundary conditions
u(O,t) = 0 and u(L, t) = 0.
Assume that a continuous solution exists (with continuous derivatives).[Hints: Expand the solution as a Fourier sine series (i.e., use the methodof eigenfunction expansion). Expand g(x) as a Fourier sine series. Solvefor the Fourier sine series of the solution. Justify all differentiations withrespect to x.]
*3.4.12. Solve the following nonhomogeneous problem:
2
= k5-2 + e-t + e- 2tcos 3Lx [assume that 2 # k(37r/L)2]The PDE is
ππ’ππ‘
= ππ2π’ππ₯2
+ π (π₯) (1)
Since the boundary conditions are homogenous Dirichlet conditions, then the solution can be writtendown as
π’ (π₯, π‘) =βοΏ½π=1
ππ (π‘) sin οΏ½πππΏπ₯οΏ½
Since the solution is assumed to be continuous with continuous derivative, then term by term
27
diοΏ½erentiation is allowed w.r.t. π₯ππ’ππ₯
=βοΏ½π=1
πππΏππ (π‘) cos οΏ½ππ
πΏπ₯οΏ½
π2π’ππ₯2
= ββοΏ½π=1
οΏ½πππΏοΏ½2ππ (π‘) sin οΏ½π
ππΏπ₯οΏ½ (2)
Also using assumption that ππ’ππ‘ is smooth, then
ππ’ππ‘
=βοΏ½π=1
πππ (π‘)ππ‘
sin οΏ½πππΏπ₯οΏ½ (3)
Substituting (2,3) into (1) givesβοΏ½π=1
πππ (π‘)ππ‘
sin οΏ½πππΏπ₯οΏ½ = βπ
βοΏ½π=1
οΏ½πππΏοΏ½2ππ (π‘) sin οΏ½π
ππΏπ₯οΏ½ + π (π₯) (4)
Using hint given in the problem, which is to expand π (π₯) as Fourier sin series. Hence
π (π₯) =βοΏ½π=1
ππ sin οΏ½πππΏπ₯οΏ½
Where
ππ =2πΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½
Hence (4) becomesβοΏ½π=1
πππ (π‘)ππ‘
sin οΏ½πππΏπ₯οΏ½ = β
βοΏ½π=1
π οΏ½πππΏοΏ½2ππ (π‘) sin οΏ½π
ππΏπ₯οΏ½ +
βοΏ½π=1
ππ sin οΏ½πππΏπ₯οΏ½
Applying orthogonality the above reduces to one term only
πππ (π‘)ππ‘
sin οΏ½πππΏπ₯οΏ½ = βπ οΏ½
πππΏοΏ½2ππ (π‘) sin οΏ½π
ππΏπ₯οΏ½ + ππ sin οΏ½ππ
πΏπ₯οΏ½
Dividing by sin οΏ½πππΏ π₯οΏ½ β 0
πππ (π‘)ππ‘
= βπ οΏ½πππΏοΏ½2ππ (π‘) + ππ
πππ (π‘)ππ‘
+ π οΏ½πππΏοΏ½2ππ (π‘) = ππ (5)
This is of the form π¦β² + ππ¦ = ππ, where π = π οΏ½πππΏοΏ½2. This is solved using an integration factor π = πππ‘,
where πππ‘οΏ½πππ‘π¦οΏ½ = πππ‘ππ, giving the solution
π¦ (π‘) =1π οΏ½
πππππ‘ +ππ
Hence the solution to (5) is
ππ (π‘) πποΏ½ πππΏ οΏ½
2π‘ = οΏ½πποΏ½
πππΏ οΏ½
2π‘ππππ‘ + π
ππ (π‘) πποΏ½ πππΏ οΏ½
2π‘ =
πΏ2πποΏ½πππΏ οΏ½
2π‘
ππ2π2 ππ + π
ππ (π‘) =πΏ2
ππ2π2 ππ + ππβποΏ½ πππΏ οΏ½
2π‘
28
Where π above is constant of integration. Hence the solution becomes
π’ (π₯, π‘) =βοΏ½π=1
ππ (π‘) sin οΏ½πππΏπ₯οΏ½
=βοΏ½π=1
οΏ½πΏ2
ππ2π2 ππ + ππβποΏ½ πππΏ οΏ½
2π‘οΏ½ sin οΏ½ππ
πΏπ₯οΏ½
At π‘ = 0, π’ (π₯, 0) = π (π₯), therefore
π (π₯) =βοΏ½π=1
οΏ½πΏ2
ππ2π2 ππ + ποΏ½ sin οΏ½πππΏπ₯οΏ½
ThereforeπΏ2
ππ2π2 ππ + π =2πΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯
Solving for π gives
π =2πΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯ β
πΏ2
ππ2π2 ππ
This completes the solution. Everything is now known. Summary
π’ (π₯, π‘) =βοΏ½π=1
ππ (π‘) sin οΏ½πππΏπ₯οΏ½
ππ (π‘) = οΏ½πΏ2
ππ2π2 ππ + ππβποΏ½ πππΏ οΏ½
2π‘οΏ½
ππ =2πΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½
π =2πΏ οΏ½
πΏ
0π (π₯) sin οΏ½ππ
πΏπ₯οΏ½ ππ₯ β
πΏ2
ππ2π2 ππ