Graph Laplacians and effective conductance
Question 1
All edges of the six-node graph below have unit conductance.
1
2
3
5
6
4
(a) Write the 6 × 6 graph Laplacian. (4 marks)
(b) Set node 1 to unit voltage and node 4 to zero. Find all other voltages and the effective conductance. (8 marks)
(c) Instead feed a unit current into node 1 and remove it at node 4. Find the terminal voltage difference and the
current in edge 3–6. (8 marks) (Total: 20 marks)
13
MATH40007 Revised Historical-Coverage Practice Pack Questions
Worked solution and marking guidance
Question 1
The Laplacian is
𝐿 =
©
«
2 −1 −1 0 0 0
−1 3 0 0 −1 −1
−1 0 3 0 −1 −1
0 0 0 2 −1 −1
0 −1 −1 −1 4 −1
0 −1 −1 −1 −1 4
ª®®®®®®®
¬
.
With 𝑉1 = 1, 𝑉4 = 0, symmetry and KCL give
𝑉2 = 𝑉3 = 3
5 , 𝑉 5 = 𝑉6 = 2
5 ,
and 𝐶eff = 2(1 − 3/5) = 4/5. For unit current, scale all voltages by 1 /(4/5) = 5/4. Thus the terminal voltage difference is
5/4,
𝑉2 = 𝑉3 = 3
4 , 𝑉 5 = 𝑉6 = 1
2 ,
and the current from 3 to 6 is 3/4 − 1/2 = 1/4. Question difficulty allocation: 𝐴 = 8, 𝐵 = 4, 𝐶 = 4, 𝐷 = 4.
Green matrices and reciprocity
Question 2
Consider the path graph 0− 1 − 2 − 3 − 4, with nodes 0 and 4 grounded and unit conductances. Let𝐾 be the reduced
Laplacian on nodes 1,2,3.
(a) For each 𝑗 = 1, 2, 3, set node 𝑗 to unit voltage while the other internal nodes satisfy KCL. Find the internal
voltage vector. (6 marks)
(b) Find the current divergence at the unit-voltage node in each case. (4 marks)
(c) Prove the reciprocity identity 𝑥𝑇 b𝑓 = b𝑥𝑇 𝑓 for any two voltage/divergence pairs. (4 marks)
(d) Find the matrix 𝑀 such that 𝑥 = 𝑀 𝑓 for every divergence vector 𝑓 . (6 marks) (Total: 20 marks)
14
MATH40007 Revised Historical-Coverage Practice Pack Questions
Worked solution and marking guidance
Question 2
The reduced Laplacian is
𝐾 = ©
«
2 −1 0
−1 2 −1
0 −1 2
ª®
¬
.
The three unit-voltage responses are
(1, 2/3, 1/3)𝑇 , (1/2, 1, 1/2)𝑇 , (1/3, 2/3, 1)𝑇 .
The corresponding source divergences are 4/3, 1, 4/3. Reciprocity follows from 𝑓 = 𝐾𝑥, b𝑓 = 𝐾b𝑥 and symmetry:
𝑥𝑇 b𝑓 = 𝑥𝑇 𝐾b𝑥 = (𝐾𝑥 )𝑇b𝑥 = b𝑥𝑇 𝑓 .
The Green matrix is
𝑀 = 𝐾 −1 = 1
4
©
«
3 2 1
2 4 2
1 2 3
ª®
¬
.
Question difficulty allocation: 𝐴 = 8, 𝐵 = 4, 𝐶 = 4, 𝐷 = 4.
Prism spectra and diffusion
Question 3
A triangular prism graph consists of two triangles with corresponding vertices joined. In block form its Laplacian
is
𝐾 =
𝐴 −𝐼3
−𝐼3 𝐴
, 𝐴 = ©
«
3 −1 −1
−1 3 −1
−1 −1 3
ª®
¬
.
1
2 3
4
5 6
(a) Explain the block form of 𝐾. (4 marks)
(b) Use vectors (𝑒, 𝑒)𝑇 and (𝑒, −𝑒)𝑇 to find a complete orthonormal eigensystem. (8 marks)
(c) For diffusion 𝑥′ = −𝐾𝑥, describe the long-time behaviour and identify the slowest non-constant decay rate. (8
marks) (Total: 20 marks)
15
MATH40007 Revised Historical-Coverage Practice Pack Questions
Worked solution and marking guidance
Question 3
Let 𝐿3 be the triangle Laplacian. Since 𝐴 = 𝐿3 + 𝐼, symmetric vectors (𝑒, 𝑒)𝑇 have eigenvalues of 𝐿3, while antisymmetric
vectors (𝑒, −𝑒)𝑇 have eigenvalues of 𝐿3 + 2𝐼. The triangle eigenvalues are 0, 3, 3, hence the prism eigenvalues are
0, 3, 3, 2, 5, 5 .
Use any orthonormal triangle basis 𝑒0 = (1, 1, 1)/
√
3 and two orthonormal vectors perpendicular to it, then normalise (𝑒, 𝑒)
and (𝑒, −𝑒) by 1/
√
2. For 𝑥′ = −𝐾𝑥, the constant mode is conserved. Every other component decays exponentially. The
slowest non-constant rate is
2 .
If the initial mean is zero, 𝑥(𝑡) → 0. Question difficulty allocation: 𝐴 = 8, 𝐵 = 8, 𝐶 = 0, 𝐷 = 4.
7
MATH40007 Revised Historical-Coverage Practice Pack Course-method solutions
Image electrodes and current density
Question 4
The conductor is the region 𝑥 > 0 outside a circular electrode. Define
ℎ(𝑧) = 𝑉
log 𝜌 log
𝑎 − 𝑧
𝑎 + 𝑧
, 0 < 𝜌 < 1,
and take the electrode boundary to be
𝑎−𝑧
𝑎+𝑧
= 𝜌.
(a) Show that the wall 𝑥 = 0 is grounded. (3 marks)
(b) Show that the electrode is a circle, find its centre and radius, and show that it is at voltage𝑉. (5 marks)
(c) Find the complex current density. (4 marks)
(d) Find the wall-current density, its maximum magnitude and the total outward current through the wall. (4 marks)
(e) Show that the limit 𝜌 → 0, 𝑉/log 𝜌 → − 𝑚/(2𝜋) gives a point source of strength 𝑚 at 𝑧 = 𝑎 near a grounded
wall. (4 marks) (Total: 20 marks)
16
Worked solution and marking guidance
Question 4
For 𝑧 = 𝑖𝑦, the ratio (𝑎 − 𝑧)/( 𝑎 + 𝑧) has modulus 1, hence the wall is grounded. The circle equation is
(𝑥 − 𝑥0)2 + 𝑦2 = 𝑟2, 𝑥 0 = 𝑎(1 + 𝜌2)
1 − 𝜌2 , 𝑟 = 2𝑎𝜌
1 − 𝜌2 .
On it, ℜ log(( 𝑎 − 𝑧)/( 𝑎 + 𝑧)) = log 𝜌, so 𝜙 = 𝑉.
𝐽𝑥 − 𝑖𝐽 𝑦 = −ℎ′ (𝑧) = 2𝑎𝑉
log 𝜌 (𝑎2 − 𝑧2) .
On the wall,
𝐽𝑥 (0, 𝑦) = 2𝑎𝑉
log 𝜌 (𝑎2 + 𝑦2) , 𝐽 𝑦 = 0.
Its magnitude is largest at 𝑦 = 0. The total outward current is
−
∫ ∞
−∞
𝐽𝑥 𝑑𝑦 = − 2𝜋𝑉
log 𝜌 > 0.
If 𝑉/log 𝜌 → −𝑚/(2𝜋) and 𝜌 → 0, the circle shrinks to𝑧 = 𝑎 and the logarithmic singularity has source strength𝑚. Question
difficulty allocation: 𝐴 = 8, 𝐵 = 4, 𝐶 = 4, 𝐷 = 4.
8