Fundamental subspaces and an electrical network
**(a)** Let
\[
M=\begin{pmatrix}
1&2&0\\
2&4&1\\
3&6&1\\
4&8&2
\end{pmatrix}.
\]
Find a basis, the dimension, and the ambient space for each of
\(\mathcal C(M)\), \(\mathcal N(M)\), \(\mathcal C(M^T)\), and
\(\mathcal N(M^T)\). **[8 marks]**
**(b)** In the conductance network below, the number beside each edge is its
conductance. Vertices 1 and 2 are held at potentials 1 and 0 respectively.
**(i)** Write Kirchhoff equations for the potentials \(V_3,V_4\).
**[2 marks]**
**(ii)** Find \(V_3,V_4\). **[4 marks]**
**(iii)** Find the magnitude and direction of the current in edge \(3\!-\!4\).
**[3 marks]**
**(iv)** Find the effective conductance between vertices 1 and 2.
**[3 marks]**
Worked solution and marking guidance
(a) Since column 2 is twice column 1 and columns 1 and 3 are independent, rank(M)=2.
\[
\mathcal C(M)=\operatorname{span}\{(1,2,3,4)^T,(0,1,1,2)^T\}\subset\mathbb R^4,\quad \dim=2.
\]
\[
\mathcal N(M)=\operatorname{span}\{(-2,1,0)^T\}\subset\mathbb R^3,\quad \dim=1.
\]
\[
\mathcal C(M^T)=\operatorname{span}\{(1,2,0)^T,(2,4,1)^T\}\subset\mathbb R^3,\quad \dim=2.
\]
\[
\mathcal N(M^T)=\operatorname{span}\{(-1,-1,1,0)^T,(0,-2,0,1)^T\}\subset\mathbb R^4,\quad \dim=2.
\]
(b) KCL gives \(4V_3-V_4=2\) and \(4V_4-V_3=1\), so
\[
V_3=\frac35,\qquad V_4=\frac25.
\]
Hence \(I_{34}=V_3-V_4=\frac15\), directed from 3 to 4, and
\[
C_{\rm eff}=2(1-V_3)+(1-V_4)=\frac75.
\]
Networks and random walks
All edges in the three diagrams below have unit conductance. Treat the three parts separately.
**(a)** In network A, \(V_s=1\) and \(V_t=0\).
Find \(V_a,V_b,V_c\) and the effective conductance between \(s\) and \(t\).
**[8 marks]**
**(b)** In network B, \(V_s=1\) and \(V_t=0\).
Use symmetry where appropriate to find \(V_p,V_q,V_r\) and the effective
conductance between \(s\) and \(t\). **[6 marks]**
**(c)** A simple random walker on the final graph chooses uniformly
among neighbouring vertices. Starting at \(b\), find the probability that it
hits \(R\) before \(L\). State the boundary-value problem used. **[6 marks]**
Worked solution and marking guidance
(a) Harmonicity gives
\[
4a-b-c=1,\qquad 3b-a-c=1,\qquad 3c-a-b=0.
\]
Thus \(a=\frac12,\ b=\frac58,\ c=\frac38\), and
\[
C_{\rm eff}=(1-a)+(1-b)=\frac78.
\]
(b) Symmetry gives \(p=q=x\). KCL at \(p\) and \(r\) gives
\[
2x-r=1,\qquad 3r-2x=0.
\]
Hence \(p=q=\frac34,\ r=\frac12\), and
\[
C_{\rm eff}=2\left(1-\frac34\right)=\frac12.
\]
(c) Let \(h(v)=\Pr_v(\tau_R<\tau_L)\). Then \(h(L)=0,\ h(R)=1\), and
\(h\) is harmonic at \(a,b,c\). The equations imply
\[
h(a)=h(b)=\frac14,\qquad h(c)=\frac12.
\]
Therefore the required probability is \(\boxed{\frac14}\).
Diffusion on a five-node path
Diffusion takes place on the unit-weight path shown below.
The boundary concentrations are fixed at \(x_1(t)=1\) and \(x_5(t)=0\).
Let \(y(t)=(x_2(t),x_3(t),x_4(t))^T\), and suppose
\(y(0)=(0,1,0)^T\).
**(a)** Derive the reduced system \(\dot y=-Ky+f\).
**[4 marks]**
**(b)** Find the steady state \(y_\infty\). **[4 marks]**
**(c)** Find an orthonormal set of eigenvectors of \(K\) and the corresponding
eigenvalues. **[4 marks]**
**(d)** Hence obtain an explicit modal formula for \(y(t)\). **[6 marks]**
**(e)** State the initial value and the long-time limit as checks on your
formula. **[2 marks]**
Worked solution and marking guidance
The reduced system is
\[
\dot y=-Ky+f,\qquad
K=\begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix},
\qquad f=\begin{pmatrix}1\\0\\0\end{pmatrix}.
\]
Solving \(Ky_\infty=f\) gives
\[
y_\infty=\begin{pmatrix}3/4\\1/2\\1/4\end{pmatrix}.
\]
An orthonormal eigenbasis is
\[
\phi_1=\begin{pmatrix}1/2\\1/\sqrt2\\1/2\end{pmatrix},\quad
\phi_2=\begin{pmatrix}1/\sqrt2\\0\\-1/\sqrt2\end{pmatrix},\quad
\phi_3=\begin{pmatrix}1/2\\-1/\sqrt2\\1/2\end{pmatrix},
\]
with eigenvalues \(2-\sqrt2,\ 2,\ 2+\sqrt2\), respectively. Since
\(z(0)=y(0)-y_\infty=(-3/4,1/2,-1/4)^T\), the modal coefficients are
\[
b_1=\frac{1/\sqrt2-1}{2},\qquad
b_2=-\frac{1}{2\sqrt2},\qquad
b_3=-\frac{1+1/\sqrt2}{2}.
\]
Therefore
\[
y(t)=y_\infty+
b_1e^{-(2-\sqrt2)t}\phi_1+
b_2e^{-2t}\phi_2+
b_3e^{-(2+\sqrt2)t}\phi_3.
\]
At \(t=0\) this gives \((0,1,0)^T\), and as \(t\to\infty\) it tends to
\((3/4,1/2,1/4)^T\).
Electrically heated conducting rod
A uniform rod occupies \(0\le x\le2\). Its electrical conductivity
is \(C=3\), its thermal conductivity is \(k=2\), and there are no other
heat sources. The electrical potential satisfies
\(\phi(0)=0,\ \phi(2)=4\). Initially, both ends are maintained at
\(T(0)=T(2)=10\). All electrical power is converted into heat.
**(a)** State the electrical boundary-value problem and find
\(\phi(x)\) and the current density \(J\). **[4 marks]**
**(b)** Find the volumetric Joule-heating rate. **[3 marks]**
**(c)** State and solve the steady thermal boundary-value problem.
**[5 marks]**
**(d)** Find the outward heat flux at each end. **[4 marks]**
**(e)** Verify global energy conservation. **[2 marks]**
**(f)** The left end is now thermally insulated while \(T(2)=10\) remains
fixed. Find the new steady temperature and the outward heat flux at \(x=2\).
**[2 marks]**
Worked solution and marking guidance
The electrical equation is \((C\phi')'=0\), with
\(\phi(0)=0,\phi(2)=4\). Thus
\[
\phi(x)=2x,\qquad J=-C\phi'=-6.
\]
(The opposite current sign is acceptable if the positive direction is stated.)
The Joule-heating rate is
\[
q=C(\phi')^2=3(2)^2=12.
\]
The steady heat equation and boundary conditions are
\[
-kT''=q,\qquad T(0)=T(2)=10.
\]
Hence
\[
T(x)=10+6x-3x^2.
\]
The conductive heat flux in the positive \(x\)-direction is
\(q_h=-kT'=-12+12x\). Therefore the outward heat flux is 12 at
both \(x=0\) and \(x=2\). Their sum is 24, equal to
\(\int_0^2 12\,dx=24\).
With \(T'(0)=0,\ T(2)=10\), the new solution is
\[
T(x)=22-3x^2,
\]
and all generated heat leaves at the right end, where the outward flux is 24.