Worksheet 1 - uni-regensburg.desem04141/hm16solJun16.pdf0 Z d3r(x2 + y2) = ˆ 0 Z R 0 dr Z ˇ 0 d Z 2ˇ 0 d˚J(r; ;˚) h (rsin cos˚)2 + (rsin sin˚)2 i = ˆ 0 Z R 0 dr Z ˇ 0 d Z

Worksheet 1

Problem 1

Symmetry implies that

4∑n=1

rn ≡

x1 + 2x2

03z1 + `

= 0,

resulting in z1 = −13` and x2 = −1

2x1,

r1 =

x1

0−1

3`

, r2,3 =

−12x1

±y2

−13`

.

To find x1 and y2, notice that

|rn| = ` (n = 1, 2, 3, 4).

Consequently,

x1 =

√`2 − 1

9`2 =

2√

2

3` = 0.9428 `, y2 =

√`2 − 1

4x2

1 −1

9`2 =

√2

3` = 0.8165 `.

For the side length, we get

k = |r4 − r1| =√x2

1 +(4

3`)2

=2√

6

3` = 1.633 `.

The bond angle θ is obtained by

cos θ =r4 · r1

|r4| |r1|= −1

3⇒ θ = 109.471.

Problem 2

a× b = (a1u1 + a2u2 + a3u3)× (b1u1 + b2u2 + b3u3)

= a1b1 (u1 × u1)︸ ︷︷ ︸0

+ a1b2 (u1 × u2)︸ ︷︷ ︸u3

+ a1b3 (u1 × u3)︸ ︷︷ ︸−u2

+

+ a2b1 (u2 × u1)︸ ︷︷ ︸−u3

+ a2b2 (u2 × u2)︸ ︷︷ ︸0

+ a2b3 (u2 × u3)︸ ︷︷ ︸u1

+

+ a3b1 (u3 × u1)︸ ︷︷ ︸u2

+ a3b2 (u3 × u2)︸ ︷︷ ︸−u1

+ a3b3 (u3 × u3)︸ ︷︷ ︸0

= (a2b3 − a3b2)u1 + (a3b1 − a1b3)u2 + (a1b2 − a2b1)u3.

1

Worksheet 2

Problem 1

(b)

TTTTT (t) =

− sin(12αt2)

+ cos(12αt2)

0

, NNNN (t) =

− cos(12αt2)

− sin(12αt2)

0

.

(c)

v(t) ≡ d

dtr(t) =

−Rαt sin(12αt2)

+Rαt cos(12αt2)

0

= Rαt TTTTT (t).

(d)

a(t) ≡ d

dtv(t) = Rα TTTTT (t) + R(αt)2NNNN (t)

= (Rα) TTTTT (t) +v(t)2

RNNNN (t).

This is the vector sum of linear and centripetal accelerations.

Problem 2

(a) Setting T (x, y) = T and solving for y = fT (x), we find for |x| < 1

fT (x) =1

A1−x2 + 1

, 0 ≤ A ≡ k(T0

T− 1)≤ ∞.

• T = T0 or A = 0: y = fT0(x) ≡ 1 (”upper” rim of the rectangle);alternatively, we have T = T0 also for |x| = 1 (”left” and ”right” rims).

• T = 0 or A =∞: y = f0(x) ≡ 0 (”lower” rim).• 0 < T < T0 or ∞ > A > 0: The graph of the symmetric function fT (x), with

f ′T (x) = − 2Ax

(A+ 1− x2)2,

has zeros for x→ ∓1 with slope ± 2A

and reaches its maximum 1A+1

at x = 0.

(b) The wanted partial derivatives are

Tx(x, y) ≡ ∂T (x, y)

∂x=

2kxy(1− y)

[ky + (1− x2)(1− y)]2,

Ty(x, y) ≡ ∂T (x, y)

∂y=

k(1− x2)

[ky + (1− x2)(1− y)]2.

For k = 2, the gradient GT (x, y) =(Tx(x,y)Ty(x,y)

)assumes the ”values”

GT (0.3, 0.5) =

(0.1420.860

), GT (0.5, 0.5) =

(0.2640.793

), GT (0.8, 0.5) =

(0.5750.517

).

2

Problem 5

(a) ∫ 4

0

dx

∫ 1+x2

1−x4

dy (ax+ by) =

∫ 4

0

dx

[axy +

b

2y2

]y=1+x2

y=1−x4

=

∫ 4

0

dx

[3a

4x2 +

b

2

(1 +

x

2

)2

−(

1− x

4

)2]

=

∫ 4

0

dx

[3a

4x2 +

b

2

(3

2x+

3

16x2)]

=

∫ 4

0

dx

[(3a

4+

3b

32

)x2 +

3b

4x

]= 16a+ 8b.

(b) Since the points P1, P2 and P3 are distributed within the triangle Ω rather uniformly,we have 1

3

∑3i=1 f(xi, yi) ≈ 〈f(r)〉r∈Ω , and thus∫

Ω

d2r f(x, y) ≡ AΩ ·⟨f(r)

⟩r∈Ω ≈ AΩ ·

1

3

3∑i=1

f(xi, yi)

= 6 · 1

3(7.5a+ 3.5b) = 15a+ 7b.

(c) Generally, since f is linear, we have exactly∫Ω

d2r f(x, y) = AΩ · f(xM , yM),

6 · f(83, 4

3) = 6 · (8

3a+ 4

3b) = 16a+ 8b.

(d) Part (a): ∫ 4

0

dx

∫ 1+x2

1−x4

dy (ax2 + bxy + cy2) =

∫ 4

0

dx

[ax2y +

b

2xy2 +

c

3y3

]y=1+x2

y=1−x4

=

∫ 4

0

dx

[3a

4x3 +

b

2x

(1 +

x

2

)2

−(

1− x

4

)2

+c

3

(1 +

x

2

)3

−(

1− x

4

)3]

.

Using the trinomic formula, we obtain

... =

∫ 4

0

dx

[3a

4x3 +

b

2x

(1 + x+

x2

4

)−(

1− x

2+x2

16

)+c

3

(1 +

3

2x+

3

4x2 +

x3

8

)−(

1− 3

4x+

3

16x2 − x3

64

)3]

=

∫ 4

0

dx

[(3a

4+

3b

32+

3c

64

)x3 +

(3b

4+

3c

16

)x2 +

3c

4x

]= (48a+ 6b+ 3c) + (16b+ 4c) + 6c

= 48a+ 22b+ 13c.

Part (b) and (c), respectively:

6 · 1

3

3∑i=1

f(xi, yi) = 44.5a+ 17.5b+ 10.5c,

6 · f(xM , yM) =128

3a+

64

3b+

32

3c.

3

Problem 6

(a) The area of Ω is

AΩ =

∫ 2

−2

dx (4− x2) = 16− 16

3=

32

3.

(b) ∫Ω

d2r f(r) =

∫ 2

−2

dx

∫ 4−x2

0

dy (x2 + y2) =

∫ 2

−2

dx[x2y +

y3

3

]y=4−x2

y=0

=

∫ 2

−2

dx[y(x2 +

y2

3

)]y=4−x2

y=0

=

∫ 2

−2

dx (4− x2)(x2 +

(4− x2)2

3

)=

1

3

∫ 2

−2

dx (4− x2)(x4 − 5x2 + 16)

=1

3

∫ 2

−2

dx (−x6 + 9x4 − 36x2 + 64) =1664

35= 47 +

19

35.

(c) The wanted average value is⟨f(r)

⟩r∈Ω

=1

AΩ

∫Ω

d2r f(r) =3

32· 1664

35=

156

35= 4 +

16

35.

Estimate:

1

12· 2[f(0.5|0.5) + f(0.5|1.5) + f(0.5|2.5) + f(0.5|3.5) + f(1.5|0.5) + f(1.5|1.5)

]=

1

6

[0.5 + 2.5 + 6.5 + 12.5 + 2.5 + 4.5

]=

29

6= 4 +

5

6.

(d) Bonus: Maximum: f(0, 4) = 16, Minimum: f(0, 0) = 0.

4

Problem 7

(a) ∫Ω

d3r xz =

∫ a

0

dx x

∫ √a2−x20

dy

∫ √a2−x2−y2

0

dz z

=

∫ a

0

dx x

∫ √a2−x20

dy[z2

2

]z=√a2−x2−y2

z=0

=1

2

∫ a

0

dx x

∫ √a2−x20

dy (a2 − x2 − y2)

=1

2

∫ a

0

dx x[(a2 − x2)y − y3

3

]y=√a2−x2

y=0

=1

2

∫ a

0

dx x[(a2 − x2)3/2 − (a2 − x2)3/2

3

]=

1

6

∫ a

0

dx 2x (a2 − x2)3/2 =1

6

[− 2

5(a2 − x2)5/2

]x=a

x=0=

a5

15.

(b) ∫Ω

d3r xyz =

∫ a

0

dx x

∫ √a2−x20

dy y

∫ √a2−x2−y2

0

dz z

=1

2

∫ a

0

dx x

∫ √a2−x20

dy y(a2 − x2 − y2) [as in part (a)]

=1

2

∫ a

0

dx x[(a2 − x2)

y2

2− y4

4

]y=√a2−x2

y=0

=1

2

∫ a

0

dx x[(a2 − x2)2

2− (a2 − x2)2

4

]=

1

16

∫ a

0

dx 2x (a2 − x2)2 =1

16

[− (a2 − x2)3

3

]x=a

x=0=

a6

48.

5

Worksheet 3

Problem 1

(a) (i) Since arcsin(1) = π2

and arcsin(−1) = −π2, we obtain∫ R

−Rdx

∫ √R2−x2

−√R2−x2

dy 1 ≡∫ R

−Rdx 2√R2 − x2 =

[R2 arcsin

x

R

]R−R

= πR2.

(ii) In planar polar coordinates, we have [notice the Jacobian J(r, φ) = r!]∫ R

0

dr

∫ 2π

0

dφ r · 1 =

∫ R

0

dr 2πr =[2π

r2

2

]R0

= πR2.

(b) Now, the r-integral has a φ-dependent upper limit r(φ) = 3 + cos 4φ > 0,∫ 2π

0

dφ

∫ r(φ)

0

dr r · 1 =

∫ 2π

0

dφr(φ)2

2=

1

2

∫ 2π

0

dφ (9 + 6 cos 4φ+ cos2 4φ).

Since cos2 4φ = 12

ddφ

(14

sin 4φ cos 4φ+ φ), this becomes

... =1

2

18π +

[6

sin 4φ

4

]2π

0+

1

2

[sin 4φ cos 4φ

4+ φ]2π

0

=

1

2

18π + 0 + π

=

19π

2.

Problem 4.1

(a) The distance of a point r = (x, y, z) from the z-axis is a(r) =√x2 + y2.

(b) For a sphere Ω of radius R, centered at the origin, we obtain

I =

∫Ω

d3r ρ(r) a(r)2

= ρ0

∫Ω

d3r (x2 + y2)

= ρ0

∫ R

0

dr

∫ π

0

dθ

∫ 2π

0

dφJ(r, θ, φ)[(r sin θ cosφ)2 + (r sin θ sinφ)2

]= ρ0

∫ R

0

dr

∫ π

0

dθ

∫ 2π

0

dφ (r2 sin θ)[r2 sin2 θ

(cos2 φ+ sin2 φ

)]= ρ0

∫ R

0

dr

∫ π

0

dθ

∫ 2π

0

dφ r4 sin3 θ

= ρ0

∫ R

0

dr

∫ π

0

dθ 2π r4 sin3 θ.

Writing sin3 θ = sin θ(1− cos2 θ) = sin θ − sin θ cos2 θ, we find

I = 2πρ0

∫ R

0

dr r4[− cos θ +

cos3 θ

3

]π0

= 2πρ0

∫ R

0

dr r4[2− 2

3

]= 2πρ0

R5

5

4

3=

2

5

(4π

3R3ρ0

)R2 =

2

5MR2.

6

(c) For a cylinder Ω of radius R, height H, and centered at the origin, the moment ofinertia for rotation about the x-axis is

I =

∫Ω

d3r ρ(r) a(r)2 =

∫Ω

d3rm

VΩ(y2 + z2).

In cylindrical coordinates (s, z, φ), with x = s cosφ, y = s sinφ, d3r = s ds dz dφ,this integral reads

I =m

VΩ

∫ H2

−H2

dz

∫ 2π

0

dφ

∫ R

0

ds s (s2 sin2 φ+ z2)

=m

VΩ

∫ H2

−H2

dz

∫ 2π

0

dφ(R4

4sin2 φ +

R2

2z2).

Using sin2 φ = 12

ddφ

(φ− sinφ cosφ), we obtain

I =m

VΩ

∫ H2

−H2

dz(R4

4π + 2π

R2

2z2)

=m

VΩ

[πR4

4z + πR2 z

3

3

]H2

−H2

=m

VΩπR2H︸ ︷︷ ︸m

(R2

4+H2

12

).

Problem 4.2

(a)

(b)

Ix = ρ

∫ +H/2

−H/2dz

∫ R

0

ds

∫ 2π

0

dφ s[(s sinφ)2 + z2

]= ρ

∫ +H/2

−H/2dz

∫ R

0

ds

∫ 2π

0

dφ[s3 sin2 φ + sz2

],

mit sin2 φ = 12(1− cos 2φ) also

Ix = ρ

∫ +H/2

−H/2dz

∫ R

0

ds[s3 1

2

(φ− 1

2sin 2φ

)+ sz2 φ

]φ=2π

φ=0

= ρ

∫ +H/2

−H/2dz

∫ R

0

ds[πs3 + 2πz2s

]= ρ

∫ +H/2

−H/2dz[πs4

4+ πz2s2

]s=Rs=0

= ρπR2

∫ +H/2

−H/2dz[R2

4+ z2

]= ρπR2

[R2

4z +

z3

3

]z=+H/2

z=−H/2= ρπR2H︸ ︷︷ ︸

M

[R2

4+H2

12

].

7

Problem 5

(a) ∫Ω

d3r xz =

∫ a

0

dr

∫ π/2

0

dθ

∫ π/2

0

dφJ(r, θ, φ)[(r sin θ cosφ)(r cos θ)

]=

∫ a

0

dr

∫ π/2

0

dθ

∫ π/2

0

dφ r4 sin2 θ cos θ cosφ

= ...

(b) ∫Ω

d3r xyz =

∫ a

0

dr

∫ π/2

0

dθ

∫ π/2

0

dφJ(r, θ, φ)[(r sin θ cosφ)(r sin θ sinφ)(r cos θ)

]=

∫ a

0

dr

∫ π/2

0

dθ

∫ π/2

0

dφ r5 sin3 θ cos θ cosφ sinφ

=

∫ a

0

dr

∫ π/2

0

dθ r5 sin3 θ cos θ[sin2 φ

2

]π/20

=1

2

∫ a

0

dr

∫ π/2

0

dθ r5 sin3 θ cos θ

=1

2

∫ a

0

dr r5[sin4 θ

4

]π/20

=1

8

∫ a

0

dr r5 =a6

48.

8

Worksheet 4

Problem 1

(a) I = a2

2ln(1 + b). In the case a = b = 1, we estimate

I =ln 2

2= 0.347 ≈

f(14, 1

4) + f(3

4, 1

4) + f(1

4, 3

4) + f(3

4, 3

4)

4

=1

4

( 14

+ 34

54

+14

+ 34

74

)=

1

4

(4

5+

4

7

)= 0.343.

(b) I = ab (area of Ω !),

(c) I = abc (volume of 3D plot !).

Problem 2

(a) We have v(r) = ∇f(r), where f(r) = xyz.

(b) We have p(r) = ∇g(r), where g(r) = xy + 12z2.

(c) In contrast, q(r) is no gradient, since for x 6= 0 we have

x =∂q1

∂y6= ∂q2

∂x= 0.

Problem 3

(a)

∇×B(r) =

∂y0− ∂zx∂z(−y)− ∂x0∂xx− ∂y(−y)

=

002

.

(b) Since ∂yB3 = ∂zB2 = ∂zB1 = ∂xB3 = 0, we now have

∇×B(r) =

00

∂xB2 − ∂yB1

.

Evaluating

∂xB2 =∂

∂x

x

x2 + y2=

(x2 + y2)− x · 2x(x2 + y2)2

=y2 − x2

(x2 + y2)2,

∂yB1 =∂

∂y

−yx2 + y2

=−(x2 + y2)− (−y) · 2y

(x2 + y2)2=

y2 − x2

(x2 + y2)2,

we obtain ∇×B(r) = 0 (at least for x2 + y2 > 0).

9

Problem 4

(a) The vectors are

F(1, 0) =

(01

), F(2, 0) =

(0

0.5

),

F(0, 1) =

(10

), F(1, 1) =

(0.50.5

), F(2, 1) =

(0.20.4

),

F(0, 2) =

(0.50

), F(1, 2) =

(0.40.2

), F(2, 2) =

(0.250.25

).

They are plotted as red arrows in the following diagram.

(b) Um die Kurve in das Diagramm einzuzeichnen, berechnet man den Ortsvektor r(φ)fur einige Werte des Kurvenparameters φ. Im vorliegenden Fall handelt es sich umeinen Viertelkreis um den Ursprung mit Radius R = 2. Er ist eine der drei blauenKurven im Diagramm.Da die Tangential-Projektion der roten Pfeile in jedem Kurvenpunkt in Richtungwachsenden Kurvenparameters φ zu zeigen scheinen (man mußte noch mehr solchePfeile einzeichnen), sollte

∫Γ1

dr · F(r) > 0 sein.

(c) ∫Γ1

dr · F(r) =

∫ π/4

0

dφ r(φ) · F(r(φ)

)=

∫ π/4

0

dφ

(−R sinφ+R cosφ

)·(R sinφ/R2

R cosφ/R2

)=

∫ π/4

0

dφ[− sin2 φ+ cos2 φ

]=

∫ π/4

0

dφ cos 2φ =[1

2sin 2φ

]π/40

=1

2.

(d) ∫Γ2

dr · F(r) =

∫ 2

1

du

(11

)·(

1/2u1/2u

)=

∫ 2

1

du1

u= ln 2.

10

(e) ∫Γ3

dr · F(r) =

∫ 2

1/2

du

(1

−1/u2

)·(

u/(1 + u4)u3/(1 + u4)

)=

∫ 2

1/2

du 0 = 0.

Problem 5

(a)

∇f(r) =

(∂x∂y

) (x3

3+ xy2

)=

(x2 + y2

2xy

).

(b)

(i)

∫ 1

0

du

(6u6u2

)·(

9u4 + 4u6

12u5

)=

∫ 1

0

du[54u4 + 24u7 + 72u7

]=

54

6+

24

8+

72

8= 21.

(ii)

∫ 1

0

du

(−18u2

5

)·(

36u6 + 25u2

−60u4

)=

∫ 1

0

du[− 648u8 − 450u4 − 300u4

]= −648

9− 450

5− 300

5= −222.

(iii)

∫ 1

0

du

(3

16u

)·(

9u2 + 64u4

48u3

)=

∫ 1

0

du[27u2 + 192u4 + 768u4

]=

27

3+

192

5+

768

5= 201.

(c) We have rA = (00), in all three cases. Consequently, f(rA) = 0.

(i): rB = (32) and thus f(rB) = 21.

(ii): rB = (−65 ) and thus f(rB) = −222.

(iii): rB = (38) and thus f(rB) = 201.

In other words: The value of the line integral is always the difference between thevalues of the scalar field f(r) at the end and starting points of the curve,∫

Γ

dr ·[∇f(r)

]= f(rB)− f(rA).

11

Worksheet 5

Problem 1

(a)

hs(α) ≡ ∂r(α)

∂s=

cosφsinφ

0

, hφ(α) =

−s sinφ+s cosφ

0

, hz(α) =

001

.

(b) Drawing: One eighth of a circle with radius 3.

(c) Due to the formula given in the lecture, we have (possibly apart from a sign)∫Σ

dA · F(r) =

∫ π/2

π/4

dφ

∫ a

−adz[hφ(α)× hz(α)

]· F(r(s0, φ, z)

)=

∫ π/2

π/4

dφ

∫ a

−adz

s0 cosφs0 sinφ

0

· 5s0 cosφ− 7z

3s0 cosφ+ 5s0 sinφ6s0 sinφ+ 2z

=

∫ π/2

π/4

dφ

∫ a

−adz(

5s20 cos2 φ− 7s0z cosφ+ 3s2

0 sinφ cosφ+

+5s20 sin2 φ+ 0 + 0

)=

∫ π/2

π/4

dφ

∫ a

−adz(

5s20 − 7s0z cosφ+ 3s2

0 sinφ cosφ)

=

∫ π/2

π/4

dφ[5s2

0z −7

2s0z

2 cosφ+ 3s20z sinφ cosφ

]z=az=−a

=

∫ π/2

π/4

dφ(

10s20a + 0 + 6s2

0a sinφ cosφ)

=[10s2

0aφ + 3s20a sin2 φ

]φ=π/2

φ=π/4=

5π + 3

2s2

0a.

Problem 2

(a) We have a · b = a · c = b · c = 0, and |a| =√

22 + 12 + (−2)2 = 3 = |b| = |c|.

(b) Writing

r(u, v, w) ≡ r(α) =

x(u, v, w)y(u, v, w)z(u, v, w)

=

2u+ v + 2wu+ 2v − 2w−2u+ 2v + w

,

we find

hu ≡∂r(α)

∂u=

21−2

= a, hv = b, hw = c.

12

(c) Σ is the square with the corners

r1 = 0, r2 = a, r3 = b, r4 = a + b.

(d) Due to the formula given in the lecture, we have (possibly apart from a sign)

I =

∫ 1

0

du

∫ 1

0

dv[hu × hv

]· F(r(u, v, 0)

)=

∫ 1

0

du

∫ 1

0

dv

21−2

×1

22

· 5(2u+ v)− 7(−2u+ 2v)

3(2u+ v) + 5(u+ 2v)6(u+ 2v) + 2(−2u+ 2v)

=

∫ 1

0

du

∫ 1

0

dv

6−63

· 24u− 9v

11u+ 13v2u+ 16v

=

∫ 1

0

du

∫ 1

0

dv(

84u− 84v)

=

∫ 1

0

du[84uv − 42v2

]v=1

v=0

=

∫ 1

0

du(

84u− 42)

= 42− 42 = 0.

Problem 3

(a)

∇ · J(r) =∂J1(x, y)

∂x+∂J2(x, y)

∂y= c

a2 − x2 + y2

(a2 + x2 + y2)2+ 0.

(b) Wertetabelle fur J1(r) = x1+x2+y2

:

y\x −3 −2 −1 0 1 2 3

2 −0.21 −0.22 −0.17 0 0.17 0.22 0.211 −0.27 −0.33 −0.33 0 0.33 0.33 0.270 −0.30 −0.40 −0.50 0 0.50 0.40 0.30

(c) Eine 2D Stromdichte hat strenggenommen die Dimension [J] = 1mgm s

;folglich haben wir: [a] = 1 m, [c] = 1mg

s.

Interpretiert man J als 3D Stromdichte, mit [J] = 1 mgm2 s

,so folgt dagegen: [a] = 1 m, [c] = 1mg

s m.

(d) Punkte der xy-Ebene mit ∇ · J(x, y) = 0 genugen der Gleichung

a2 − x2 + y2 = 0 ⇔ y = ±√x2 − a2.

Sie liegen also auf zwei getrennten Hyperbelasten mit x ≥ a bzw. x ≤ −a.Im Gebiet zwischen diesen, etwa auf der y-Achse, ist ∇ · J(x, y) > 0 (Quellen).In den beiden Gebieten außerhalb der Hyperbelaste gilt dagegen ∇ · J(x, y) < 0(Senken).

13

Problem 4

(a) Since the vector hθ(α)× hφ(α) is pointing out of the spherical surface, we have∫Σ

dA · F(r) =

∫ π

0

dθ

∫ φ0

0

dφ[hθ(α)× hφ(α)

]· F(r(α)

)=

∫ π

0

dθ

∫ φ0

0

dφR2

sin2 θ cosφsin2 θ sinφcos θ sin θ

· R sin θ(2 cosφ+ 3 sinφ)

R sin θ(5 cosφ− 4 sinφ)7R cos θ

= R3

∫ π

0

dθ

∫ φ0

0

dφ(

2 sin3 θ cos2 φ + (3 + 5) sin3 θ cosφ sinφ +

−4 sin3 θ sin2 φ + 7 cos2 θ sin θ)

= R3

∫ π

0

dθ

∫ φ0

0

dφ(

sin3 θ(2 + 8 cosφ sinφ − 6 sin2 φ

)+

+ 7 cos2 θ sin θ)

= R3

∫ π

0

dθ[

sin3 θ(

2φ + 4 sin2 φ − 6

2

(φ− 1

2sin 2φ

))+

+ 7φ cos2 θ sin θ]φ=φ0

φ=0

= R3

∫ π

0

dθ(

sin3 θ− φ0 + 4 sin2 φ0 +

3

2sin 2φ0

+ 7φ0 cos2 θ sin θ

)= R3

[ 1

4

(− 3 cos θ +

1

3cos 3θ

)...− 7

3φ0 cos3 θ

]θ=πθ=0

= R3[(3

2+

1

12· (−2)

)...

+14

3φ0

]= R3

[ 4

3

− φ0 + 4 sin2 φ0 +

3

2sin 2φ0

+

14

3φ0

]= R3

[10

3φ0 +

16

3sin2 φ0 + 2 sin 2φ0

].

In the particular case φ0 = 2π, this becomes∫Σ

dA · F(r) =20π

3R3.

(b) Since ∇ · F(r) = 2− 4 + 7 = 5 is a constant, we obtain∫Ω

d3r[∇ · F(r)

]≡ VΩ · 5 =

20π

3R3

(where VΩ = 4π3R3 is the volume of the sphere Ω), the same result!

Problem 5: Gauss’ theorem (II)

Given are the vector field F(r) and, in spherical coordinates (r, θ, φ), the volume regionΩ (”upper“ hemisphere with radius R),

F(r) =

5x− 3y2x+ 8y

4Lx2

, Ω =

r(r, θ, φ)∣∣∣ 0 ≤ r ≤ R, 0 ≤ θ ≤ π

2, 0 ≤ φ ≤ 2π

.

14

(a) The flux integral of F(r) out of the surface ∂Ω of Ω is∮∂Ω

dA · F(r) =

∫Σ1

dA · F(r) +

∫Σ2

dA · F(r).

• Σ1 is the upper half of the curved spherical surface,∫Σ1

dA · F(r) =

∫ π/2

0

dθ

∫ 2π

0

dφ[hθ × hφ

]· F(r(r, θ, φ)

)∣∣∣r=R

=

∫ π/2

0

dθ

∫ 2π

0

dφ

R cos θ cosφR cos θ sinφ−R sin θ

× −R sin θ sinφ

R sin θ cosφ0

· F=

∫ π/2

0

dθ

∫ 2π

0

dφ

R2 sin2 θ cosφR2 sin2 θ sinφR2 sin θ cos θ

· R sin θ(5 cosφ− 3 sinφ)

R sin θ(2 cosφ+ 8 sinφ)4LR2 sin2 θ cos2 φ

=

∫ π/2

0

dθ

∫ 2π

0

dφR3 sin3 θ(

5− sinφ cosφ+ 3 sin2 φ)

+

+ 4R

L

∫ π/2

0

dθ

∫ 2π

0

dφR3 sin3 θ cos θ cos2 φ

= R3(

10π − 0 + 3π)∫ π/2

0

dθ sin3 θ + 4R

LR3π

[sin4 θ

4

]π/20

.

Substituting u = cos θ, du = − sin θdθ, we find∫ π/2

0dθ sin3 θ =

∫ 1

0du(1− u2) = 2

3,∫

Σ1

dA · F(r) =26π

3R3 + π

R

LR3 ≡ π

(26

3+R

L

)R3.

• Σ2 is the planar disk with radius R that forms the ”floor” of the hemisphere Ω ,∫Σ2

dA · F(r) =

∫ R

0

dr

∫ 2π

0

dφ[hφ × hr

]· F(r(r, θ, φ)

)∣∣∣θ=π

2

=

∫ R

0

dr

∫ 2π

0

dφ

00−r

· r(5 cosφ− 3 sinφ)

r(2 cosφ+ 8 sinφ)4Lr2 cos2 φ

= − 4

L

∫ R

0

dr r3π = −π R4

L.

In summary, ∮∂Ω

dA · F(r) =26π

3R3.

(b) The volume integral of the divergence of the vector field is∫Ω

d3r[∇ · F(r)

]= VΩ

⟨∇ · F(r)

⟩r∈Ω

.

Since the integrand is a constant, ∇ · F(r) = 5 + 8 + 0 = 13, we simply have∫Ω

d3r[∇ · F(r)

]= VΩ · 13 = 13 · 2π

3R3,

where VΩ = 2π3R3 is the volume of the hemisphere.

Since the result is the same as in part (a), we have verified Gauss’ theorem.

15

Worksheet 6

Problem 1: Chain rule for partial derivatives

Problem 2: Estimating flux integrals

(a) In the case δ π2, the curvature of Σ becomes negligible. Consequently, the unit

normal vector nΣ (r) is approximately the same at different points r ∈ Σ ,

nΣ (r) ≡ nΣ (R, φ, z) ≈ nΣ (R,α, z) =

cosαsinα

0

. (1)

For a constant vector field, e.g. E(r) = E0, we may estimate∫Σ

dA · E(r) ≡ AΣ

⟨nΣ (r) · E(r)

⟩r∈Σ

≈ AΣ nΣ (R,α, z) · E0 =

AΣ︷ ︸︸ ︷(2δRH)

cosαsinα

0

· 3−84

= (2δRH)(3 cosα− 8 sinα). (2)

[For the vector field V(r), we have for r ∈ Σ

nΣ (r) ·V(r) =

cosφsinφ

0

· R cosφ f(z)

R sinφ f(z)g(z)

= Rf(z),

without using the estimate (1). Now, a careful analysis yields⟨nΣ (r) ·V(r)

⟩r∈Σ

= R⟨f(z)

⟩z∈[0,H]

= R1

H

∫ H

0

dz f(z), ... ].

(b) The rigorous evaluation of the integral (2) yields∫Σ

dA · F(r) =

∫ H

0

dz

∫ α+δ

α−δdφ[hφ(s, φ, z)× hz(s, φ, z)

]· F(r(s, φ, z)

)∣∣∣s=R

=

∫ H

0

dz

∫ α+δ

α−δdφ

R cosφR sinφ

0

· F1(R, φ, z)

F2(R, φ, z)F3(R, φ, z)

=

∫ H

0

dz

∫ α+δ

α−δdφR

[F1(R, φ, z) cosφ + F2(R, φ, z) sinφ

].

With F(r) = E0, we have F1(R, φ, z) = 3 and F2(R, φ, z) = −8,∫Σ

dA · F(r) = RH

∫ α+δ

α−δdφ (3 cosφ− 8 sinφ)

= RH[3 sinφ+ 8 cosφ

]φ=α+δ

φ=α−δ

= RH 2 sin δ(3 cosα− 8 sinα

),

where we have used the addition theorems for sine and cosine in the last step.For δ π

2, we have sin δ ≈ δ and recover the result of part (a).

16

Problem 3: Velocity field of a rotating solid

The circle ∂Σ , centerd at r0 and with radius R, is parametrized by

∂Σ =

r(φ)∣∣∣ 0 ≤ φ ≤ 2π

, r(φ) = r0 + R(a cosφ+ b sinφ)

≡ r0 + Ru(φ),

where a and b are unit vectors with a × b = nΣ , the unit normal vector of the disk Σ .The wanted circulation is the line integral∮∂Σ

dr · v(r) =

∫ 2π

0

dφ r(φ) · v(r(φ)

)=

∫ 2π

0

dφRu(φ) ·ωωωω ×

[r0 +Ru(φ)

]=

∫ 2π

0

dφR[b cosφ− a sinφ

]·ωωωω × r0 + Rωωωω × a cosφ + Rωωωω × b sinφ

.

Upon evaluation of the dot product, the integrand becomes a sum of six terms.Since

∫ 2π

0dφ sinφ =

∫ 2π

0dφ cosφ =

∫ 2π

0dφ sinφ cosφ = 0, only two of them survive,∮

∂Σ

dr · v(r) =

∫ 2π

0

dφR2b · (ωωωω × a) cos2 φ − R2a · (ωωωω × b) sin2 φ

= πR2

b · (ωωωω × a) − a · (ωωωω × b)

= πR2 2ωωωω · (a× b),

where we have used the vector identities A · (B×C) = B · (C×A) and B×A = −A×B.Since a× b = nΣ , our proof is complete.

Problem 4: Example for Stokes’ Theorem

(a) Σ is the paralellogram with the corners

rA = r0 − ap− bq, rC = r0 + ap + bq,

rB = r0 + ap− bq, rD = r0 − ap + bq.

Consequently, we have

AΣnΣ = (rB−rA)× (rC−rB) = 2ap× 2bq = 4abp× q = 4ab

0−11

,

and, since nΣ is a unit vector,

AΣ = 4√

2ab, nΣ =1√2

0−11

.

17

(b) ∂Σ consists of the four sides of Σ , given by

ΓAB =

rAB(ξ) = r0 + ξp− bq∣∣∣ − a ≤ ξ ≤ a

, rAB(ξ) =

2 + ξ − b5− b4− b

,

ΓBC =

rBC(ξ) = r0 + ap + ξq∣∣∣ − b ≤ ξ ≤ b

, rBC(ξ) =

2 + a+ ξ5 + ξ4 + ξ

,

ΓCD =

rCD(ξ) = r0 − ξp + bq∣∣∣ − a ≤ ξ ≤ a

, rCD(ξ) =

2− ξ + b5 + b4 + b

,

ΓDA =

rDA(ξ) = r0 − ap− ξq∣∣∣ − b ≤ ξ ≤ b

, rDA(ξ) =

2− a− ξ5− ξ4− ξ

.

Now we have∮∂Σ

dr · F(r) = (∫ΓAB

+∫ΓBC

+∫ΓCD

+∫ΓDA

)dr · F(r).We begin with ∫

ΓAB

dr · F(r) =

∫ a

−adξ rAB(ξ) · F

(rAB(ξ)

).

Since rAB(ξ) = p = (1, 0, 0), this is∫ΓAB

dr · F(r) =

∫ a

−adξ F1

(rAB(ξ)

)=

∫ a

−adξ[4(2 + ξ − b)− 8(5− b) + 7(4− b)

]=

∫ a

−adξ[− 4 + 4ξ − 3b

]= −2a(4 + 3b).

Since rBC(ξ) = q = (1, 1, 1), we similarly get∫ΓBC

dr · F(r) =

∫ b

−bdξ[F1

(rBC(ξ)

)+ F2

(rBC(ξ)

)+ F3

(rBC(ξ)

)]=

∫ b

−bdξ[15(2 + a+ ξ)− 5(5 + ξ) + 4(4 + ξ)

]=

∫ b

−bdξ[21 + 15a+ 14ξ

]= 2b(21 + 15a).

The two remaining integrals are treated in a very similar way.Since rCD(ξ) = −p = (−1, 0, 0), we have∫

ΓCD

dr · F(r) = −∫ a

−adξ F1

(rCD(ξ)

)= −

∫ a

−adξ[4(2− ξ + b)− 8(5 + b) + 7(4 + b)

]= −

∫ a

−adξ[− 4− 4ξ + 3b

]= 2a(4− 3b).

18

Since rDA(ξ) = −q = (−1,−1,−1), we similarly get∫ΓDA

dr · F(r) = −∫ b

−bdξ[F1

(rDA(ξ)

)+ F2

(rDA(ξ)

)+ F3

(rDA(ξ)

)]= −

∫ b

−bdξ[15(2− a− ξ)− 5(5− ξ) + 4(4− ξ)

]= −

∫ b

−bdξ[21− 15a− 14ξ

]= −2b(21− 15a).

Eventually, the sum of these four integrals is∮∂Σ

dr · F(r) = 48ab.

(c)

limAΣ→0

1

AΣ

∮∂Σ

dr · F(r) = lima,b→0

48ab

4√

2ab= 6√

2.

This is in fact equal to

nΣ · [∇× F(r)]|r=r0 =1√2

0−11

× 7−111

.

(d) Since ∇× F(r) = (7,−1, 11) is a constant vector, we have∫Σ

dA ·[∇× F(r)

]= AΣnΣ ·

[∇× F(r)

]= 4ab

0−11

· 7−111

= 48ab,

confirming Stokes’ theorem, since∮∂Σ

dr · F(r) = 48ab, due to part (b).

Problem 5

The circulation around ∂Σ ≡ r(r, θ, φ) | r = R, θ = π2, 0 ≤ φ ≤ 2π is∮

∂Σ

dr · F(r) =

∫ 2π

0

dφ

−R sinφ+R cosφ

0

· 4R cosφ− 8R sinφ+ 7 · 0

3R cosφ+ 5R sinφ− 9 · 0...

=

∫ 2π

0

dφ[− 4R2 sinφ cosφ+ 8R2 sin2 φ+ 3R2 cos2 φ+ 5R2 cosφ sinφ

]= 0 + 8R2π + 3R2π + 0 = 11R2π.

The flux of ∇× F(r) = (7,−1, 11) through Σ is∫Σ

dA ·[∇× F

]=

∫ π/2

0

dθ

∫ 2π

0

dφ

R cos θ cosφR cos θ sinφ−R sin θ

×−R sin θ sinφ

+R sin θ cosφ0

· 7−111

=

∫ π/2

0

dθ

∫ 2π

0

dφ

R2 sin2 θ cosφR2 sin2 θ sinφR2 cos θ sin θ

· 7−111

=

∫ π/2

0

dθ

∫ 2π

0

dφ[7R2 sin2 θ cosφ−R2 sin2 θ sinφ+ 11R2 cos θ sin θ

]= 0 + 0 + 2π · 11R2

[1

2sin2 θ

]π/20

= 11R2π.

19

Worksheet 7

Problem 3: Integral theorems

(a) All four statements are true.

(b) In cylindrical coordinates, we have

hφ × hz =

s cosφs sinφ

0

, hs × hφ =

00s

.

We first evaluate the flux of F(r) out of the side surface S,∫S

dA · F(r) =

∫ H

0

dz

∫ 2π

0

dφ

s cosφs sinφ

0

· s(A cosφ+B sinφ) + Cz

s(L cosφ+M sinφ) +Nz...

∣∣∣∣∣s=R

= R2

∫ H

0

dz

∫ 2π

0

dφA cos2 φ+ (B + L) cosφ sinφ+M sin2 φ

= R2H(A+M)π.

In a second step, the flux of F(r) out of the top/bottom surfacees T,B is∫T,B

dA · F(r) =

∫ R

0

ds

∫ 2π

0

dφ

00±s

· ...

...s(U cosφ+ V sinφ) +Wz

∣∣∣∣∣z=H,0

= ±∫ R

0

ds 2πWsz∣∣∣z=H,0

=

πWR2H (T ),

0 (B).

In summary,∮∂Ω

dA·F(r) = πR2H(A+M+W ) ≡ VΩ(A+M+W ), thus confirmingGauss’ law, since ∇ · F(r) = A+M +W is a constant,∫

Ω

d3r[∇ · F(r)

]≡ VΩ

⟨∇ · F(r)

⟩r∈Ω = VΩ (A+M +W ) =

∮∂Ω

dA · F(r).

(c) In terms of the parametrizations

∂T =

r(R, φ,H)∣∣∣ 0 ≤ φ ≤ 2π

, ∂B =

r(R, φ, 0)

∣∣∣ 0 ≤ φ ≤ 2π,

the circulations around T,B are∮∂T,∂B

dr · F(r) =

∫ 2π

0

dφ

−R sinφR cosφ

0

· R(A cosφ+B sinφ) + Cz

R(L cosφ+M sinφ) +Nz...

∣∣∣∣∣z=H,0

=

∫ 2π

0

dφ−R2B sin2 φ + R2L cos2 φ

∣∣∣z=H,0

= πR2(L−B).

The curl of F(r) is a constant vector,

∇× F(r) =

V −NC − UL−B

.

Consequently, we are confirming Stokes’ law∮∂T,∂B

dr ·F(r) =∫T,B

dA ·[∇×F(r)

],∫

T,B

dA ·[∇× F(r)

]≡ AT,B

⟨NNNN (r) ·

[∇× F(r)

]⟩r∈T,B

= AT,B (L−B),

where we have used that the unit normal vector is NNNN (r) = uz for all r ∈ T,B.

20

Problem 4: Second derivatives with ∇

(a) ∇ · [∇φ] =∑

i ∂i[∂iφ] =[ ∑

i ∂2i

]φ = ∇2φ

(b) It is sufficient, to consider the 1-component of ∇×[∇× F

],

∂2(∂1F2 − ∂2F1) − ∂3(∂3F1 − ∂1F3)

= ∂1(∂2F2 + ∂3F3) − (∂22 + ∂2

3)F1

= ∂1(∂1F1 + ∂2F2 + ∂3F3) − (∂21 + ∂2

2 + ∂23)F1 ≡

[∇[∇ · F

]−∇2F

]1.

21

Worksheet 8

Problem 0: Linear ODEs (not on the worksheet)

Problem: Find the general real-valued solutions f(x) of the following ODEs.1

(a) f ′′′(x) + 9f ′′(x) + 26f ′(x) + 24f(x) = 0,

(b) f ′′′(x)− 11f ′′(x) + 55f ′(x)− 125f(x) = 0,

(c) f ′′′(x)− 2f ′′(x)− 5f ′(x) + 10f(x) = 0,

(d) f ′′(x)− f(x) = 0,

(e) f ′′(x) + f(x) = 0.

Solution: In all cases (a–e), the exponential ansatz f(x) = eλx leads to an algebraicequation for λ ∈ C. In cases (a–c), this equation is third-order: one of its three zeros canbe guessed to be λ1 = −2 (a), λ1 = 5 (b) and λ1 = 2 (c); the remaining two zeros λ2 andλ3 are then obtained via polynomial division. Cases (d) and (e) are (almost) trivial.Eventually, the wanted real-valued solutions to the ODEs are:

(a) f(x) = Ae−2x +Be−3x + Ce−4x,

(b) f(x) = Ae5x + (B cos 4x+ C sin 4x)e3x,

(c) f(x) = Ae2x +Be√

5x + Ce−√

5x,

(d) f(x) = Aex +Be−x,

(e) f(x) = A cosx+B sinx,

where A,B,C ∈ R are constants with arbitrary values.

1Theorem of Vieta: A polynomial f(x) with given zeros x1, x2, x3 is

f(x) = x3 +Ax2 +Bx+ C,

when A = −(x1 + x2 + x3), B = x1x2 + (x1 + x2)x3 and C = −x1x2x3.We note that, for integer values A,B,C ∈ Z, it is sufficient that either x1, x2, x3 ∈ Z or

x1,2 = a±√b, x3 = c (a, b, c ∈ Z)

(where, for b < 0, it is understood that√b = i

√−b ).

In the latter case, A = −(2a+ c), B = a2 − b+ 2ac, C = (b− a2)c.In part (a), (x1, x2, x3) = (−2,−3,−4). In part (b), (a, b, c) = (3,−16, 5). In part (c), (a, b, c) = (0, 5, 2).

22

Problem 3:

(a) Writing r ≡ |r| = [x2 + y2 + z2]1/2 and applying the chain rule, we find

∂

∂xφ(r) ≡ ∂

∂xf(r) ≡ ∂

∂xf([...]1/2

)= f ′

([...]1/2

)· 1

2[...]−1/2 · 2x = f ′(r)

x

r.

In the same way, we find ∂∂yφ(r) = f ′(r)y

rand ∂

∂zφ(r) = f ′(r) z

r. Consequently,

∇φ(r) =

∂∂xφ(r)

∂∂yφ(r)

∂∂zφ(r)

= f ′(r)1

r

xyz

= f ′(r)r

r.

(b) Now, since ∂∂xf ′(r) = f ′′(r)x

ras we know from part (a), the product rule yields

∂2

∂x2φ(r) ≡ ∂

∂x

[f ′(r)

x

r

]= f ′′(r)

x2

r2+ f ′(r)

∂

∂x

(xr

).

Writing again r = [x2 + y2 + z2]1/2, we have

∂

∂x

(xr

)≡ ∂

∂xx[...]−1/2 = [...]−1/2 − x

2[...]−3/2 · 2x =

1

r− x2

r3.

In summary,

∂2

∂x2φ(r) = f ′′(r)

x2

r2+ f ′(r)

(1

r− x2

r3

).

With the corresponding expressions for ∂2

∂y2φ(r) and ∂2

∂z2φ(r), we eventually obtain

∇2φ(r) ≡(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)φ(r) = f ′′(r) +

2

rf ′(r),

where we have used x2 + y2 + z2 = r2.

(c) With the spherical surface area A∂Ω = 4πR2, the given formula here reads∮∂Ω

dA ·[∇φ(r)

]= 4πR2

⟨NNNN (r) · ∇φ(r)

⟩r∈∂Ω

.

For r ∈ ∂Ω , we have NNNN (r) = rR

and, due to part (a), ∇φ(r) = f ′(R) rR

.Therefore, NNNN (r) · ∇φ(r) = f ′(R) r

R· rR

= f ′(R) is a constant for r ∈ ∂Ω ,∮∂Ω

dA ·[∇φ(r)

]= 4πR2f ′(R). (3)

On the other hand, using ∇ · ∇φ(r) ≡ ∇2φ(r) and the result of part (b), we have∫Ω

d3r[∇ · ∇φ(r)

]≡

∫Ω

d3r[∇2φ(r)

]=

∫ R

0

dr

∫ π

0

dθ

∫ 2π

0

dφ (r2 sin θ)

[f ′′(r) +

2

rf ′(r)

]= 4π

∫ R

0

dr[r2f ′′(r) + 2rf ′(r)

].

Since the integrand is the derivative g′(r) of the function g(r) = r2f ′(r), we have∫Ω

d3r[∇ · ∇φ(r)

]= 4π

[g(r)

]R0

= 4πR2 f ′(R).

This agrees with the value of the integral (3), thus confirming Gauss’ theorem.

23

Worksheet 9

Problem 1: Linear ODEs

Problem 2: Harmonic Oscillator

Problem 3: Divergence in Spherical Coordinates

Problem 4: Diffusion equation in 3D

Due to the continuity equation ∂ρ∂t

+∇ · J = 0, the divergence of the wanted vector fieldJ(r, t) can simply be obtained by taking a derivative of the given function ρ(r, t),

∇ · J(r, t) ≡ − ∂

∂tρ(r, t) =

( 3

2t− r2

4Dt2

)ρ(r, t).

The spherical symmetry suggests a solution J(r, t) with the spherical components

Jr(r, θ, φ, t) = Jr(r, t), Jθ(...) = Jφ(...) = 0.

In this case, the formula for divergence in spherical coordinates (problem 3) yields

1

r2

∂

∂r

[r2Jr(r, t)

]≡ ∇ · J(r, t) =

( 3

2t− r2

4Dt2

)ρ(r, t).

Multiplying with r2 and using the given expression for ρ(r, t), we get

∂

∂r

[r2Jr(r, t)

]=

M

(4πDt)3/2

(3r2

2t− r4

4Dt2

)e−r

2/4Dt ≡ M

(4πDt)3/2

∂

∂r

r3

2te−r

2/4Dt,

where we have identified a derivative in the second step. Consequently,

Jr(r, t) =M

(4πDt)3/2

r

2te−r

2/4Dt, J(r, t) = Jr(r, t)ur.

This vector field describes a mass current, flowing radially outward at any position r.Of course, any additional field J1(r, t) with zero divergence, ∇ · J1(r, t) = 0, may beadded to J(r, t), still yielding a possible current density compatible with the given densitydistribution ρ(r, t).

An alternative (but equivalent) reasoning starts from the equation

4πr2Jr(r, t) ≡∮∂Ωr

dA · J(r, t) = − d

dt

∫Ωr

d3r ρ(r, t) ≡ −∫Ωr

d3r∂

∂tρ(r, t),

where Ωr is a sphere with radius r around the origin r = 0. Using the above expressionfor − ∂

∂tρ(r, t), we find

Jr(r, t) =1

4πr2

M

(4πDt)3/2

∫ r

0

dr 4πr2( 3

2t− r2

4Dt2

)e−r

2/4Dt,

which leads to the same result.

24

Problem 5: Fourier Analysis

For f(x) = x, let us evaluate the integral

I5 =

∫ a

−adx f(x) sin

(5π

ax)

≡∫ a

−adx x sin

(5π

ax)

=

[( a5π

)2

sin(5π

ax)− a

5πx cos

(5π

ax)]a−a

= −2a2

5πcos(5π) =

2a2

5π. (4)

Now, we evaluate the same integral, but use instead of the explicit expression f(x) = x theFourier series f(x) =

∑∞n=1An sin(nπ

ax) (with yet unknown coefficients A1, A2, A3, ...),

I5 =

∫ a

−adx

∞∑n=1

An sin(nπ

ax)

sin(

5π

ax)

=∞∑n=1

An

∫ a

−adx sin

(nπax)

sin(5π

ax). (5)

It is easy to see that2∫ a

−adx sin

(nπax)

sin(5π

ax)

=

0 (for n 6= 5),a (for n = 5).

Consequently, in the infinite sum of Eq. (5), all summands with n 6= 5 are zero and onlythe summand with n = 5 contributes: I5 = A5a or, equivalently,

A5 =I5

a≡ 2a

5π,

where we have inserted the earlier result (4) in the second step.Obviously, we generally have

An =Ina,

where In can be calculated directly for any n ∈ N, like I5 in Eq. (4), with the generalresult

In = −2a2

nπcos(nπ) =

2a2

nπ(−1)n−1 (n = 1, 2, 3, ...).

Individually, these numbers are I1 = 2a2

π, I2 = −2a2

2π, I3 = 2a2

3π, I4 = −2a2

4π, I5 = 2a2

5π, ...

Consequently, the Fourier series f(x) =∑∞

n=1 An sin(nπax) for f(x) = x reads

f(x) ≡ x =∞∑n=1

2a

nπ(−1)n−1 sin

(nπ

ax)

=2a

π

[sin(πax)− 1

2sin

(2π

ax

)+

1

3sin

(3π

ax

)−+...

].

2Using the formula sinα sinβ = 12 [cos(α− β)− cos(α+ β)], we obtain∫ a

−a

dx sin(nπax)

sin(5π

ax)

=1

2

∫ a

−a

dx cos

((n− 5)π

ax

)− 1

2

∫ a

−a

dx cos

((n+ 5)π

ax

).

For n ∈ N, but n 6= 5, both integrals on the RHS are zero. For n = 5, the first one is 12

∫ a

−adx 1 = a,

while the second one is still zero.

25

Choosing the particular value a = π, this Fourier series becomes (see the figure)

f(x) ≡ x = 2[

sinx − 1

2sin 2x +

1

3sin 3x − 1

4sin 4x +

1

5sin 5x −+ ...

]. (6)

-5 5

-3

-2

-1

1

2

3

-5 5

-3

-2

-1

1

2

3

-5 5

-3

-2

-1

1

2

3

-5 5

-3

-2

-1

1

2

3

Figure 1: The blue curves represent the sum of the first N terms of the series Eq. (6),for (from top to bottom): N = 1, N = 2, N = 5, and N = 10. In each panel, thecorresponding periodic function f(x) of Eq. (7) is shown in red.

Remark: The function f(x) = x is represented by this Fourier series only for x ∈ (−a, a).For general x ∈ R, this series yields a periodic function f(x), with period 2a:

f(x) = x− νxa, (7)

where, depending on x, νx ∈ 0,±2,±4, ... is the even integer that comes closest to xa,

x ∈ (−a, a) : νx = 0, x ∈ (a, 3a) : νx = 2, etc.

[For x ∈ ±a,±3a,±5a, ..., when νx is ill-defined, we have f(x) = 0.]

26

Problem 6: Temperature Distribution

With the ansatz T (x, y) = X(x)Y (y), the Laplace equation yields

∇2T (x, y) ≡ X ′′(x)Y (y) + X(x)Y ′′(y) = 0.

Dividing both sides by T (x, y) = X(x)Y (y), we obtain

X ′′(x)

X(x)+Y ′′(y)

Y (y)= 0,

which is possible only when each summand on the LHS is a constant,

X ′′(x) = −k2X(x), Y ′′(y) = k2Y (y).

The general solutions of these ODEs (for any value of the yet unknown k) are

X(x) = A sin(kx) + B cos(kx), Y (y) = Ceky + De−ky (A,B,C,D ∈ R).

• The condition T (0, y) = 0 requires that B = 0.• Then, the condition T (a, y) = 0 requires that ka = nπ, with n ∈ Z,

k = nπ

a≡ kn (n ∈ Z). (8)

• The condition T (x, 0) = 0 requires that D = −C, and we have

T (x, y) = A sin(knx)[ekny − e−kny

]≡ A sin(knx) 2 sinh(kny) (n ∈ N).

Here, the integer n ∈ N is not fixed. In fact, any linear combination of functions of thistype satisfies all boundary conditions, and the most general solution to our problem reads

T (x, y) =∞∑n=1

An sin(knx) 2 sinh(kny), (9)

with yet unknown coefficients A1, A2, A3, ...• The An are fixed by the remaining condition T (x, b) = T0,

T (x, b) =∞∑n=1

An 2 sinh(knb)︸ ︷︷ ︸αn

sin(knx)

≡∞∑n=1

αn sin(knx) = T0.

This is the Fourier series of the constant function f(x) = T0 for 0 ≤ x ≤ a.Fourier analysis (see problem 5) yields here the coefficients

αn ≡ An 2 sinh(knb) =

4T0nπ

(n = 1, 3, 5, ...),0 (n = 2, 4, 6, ...).

With these coefficients, Eq. (9) reads

T (x, y) =∑

n=1,3,5,...

4T0nπ

sinh(knb)sin(knx) sinh(kny)

=4T0

π

∞∑ν=1

sin(k2ν−1x)

2ν − 1

sinh(k2ν−1y)

sinh(k2ν−1b). (10)

Using Eq. (8) for k1, k3, k5, ..., this expression can be evaluated for any given T0, a and b.In practice, it will be sufficient to truncate the infinite sum at some maximum value of ν.

27

Figure 2: Temperature distribution on the plate (with width a = 3 and height b = 1),obtained from Eq. (10) by truncating the infinite sum after the term with ν = 100.The six small colored dots on the right give the temperature key (from top to bottom):T0 (red), 0.8T0 (orange), 0.6T0 (yellow-green), 0.4T0 (green), 0.2T0 (blue), 0 (dark blue).

Table: Numerical values of T (x, y) in units of T0, obtained from Eq. (10) by truncatingthe infinite sum after the term with ν = 100.

x : 0.2 0.4 0.6 0.8 1.0 1.2 1.4

y = 0.8 0.479 0.664 0.735 0.767 0.783 0.790 0.793y = 0.6 0.252 0.416 0.504 0.549 0.572 0.584 0.589y = 0.4 0.138 0.245 0.312 0.352 0.373 0.384 0.389y = 0.2 0.063 0.114 0.149 0.171 0.184 0.190 0.193

Figure 3: 3D plot of the function T (x, y), in units of T0.

28