past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperia_3_2022.pdf

= 1E
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=1E}

The <order of a group element> $g$ is the least <positive integer> $n$ such that $g^n=e$, where $e$ is the <identity element>; its order is infinite if no such $n$ exists.

Let $g$ have finite order $n$. A <group homomorphism> preserves the <group operation> and the identity, so
$$
\phi(g)^n=\phi(g^n)=\phi(e)=e.
$$
The order of an element divides every positive exponent that gives the identity. Hence
$$
\boxed{\operatorname{ord}(\phi(g))\mid\operatorname{ord}(g)}.
$$

If $\phi$ is a <surjective function> and $h\in H$ has order $m$, choose $g\in G$ with $\phi(g)=h$. The first result gives $m\mid n$, where $n=\operatorname{ord}(g)$. The element
$$
g^{n/m}
$$
then has order $m$, because the order of $g^k$ is $n/\gcd(n,k)$.

A <group homomorphism> $C_9\to S_4$ is determined by the image of a <generator of a group>[generator] of the <cyclic group> $C_9$, and that image must have order dividing $9$. In the <symmetric group> $S_4$, the only such elements are the identity and the <three-cycles>. There are
$$
\binom43(3-1)!=4\cdot2=8
$$
three-cycles. Therefore the number of homomorphisms is
$$
\boxed{1+8=9}.
$$

Solved by gpt-5.6-sol high.

= 2E
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=2E}

A <group> $G$ is <abelian group>[abelian] when $xy=yx$ for every $x,y\in G$. It is <cyclic group>[cyclic] when some <generator of a group> $g$ satisfies
$$
G=\langle g\rangle=\{g^n:n\in\mathbb Z\}.
$$
If $x=g^r$ and $y=g^s$ belong to a cyclic group, then
$$
xy=g^{r+s}=g^{s+r}=yx,
$$
so every cyclic group is abelian.

The <Klein four-group> $C_2\times C_2$ is abelian, but every nonidentity element has order two, so no element generates all four elements. Thus an abelian group need not be cyclic.

The condition on proper <subgroups> does not force $G$ to be abelian. The <quaternion group>
$$
Q_8=\{\pm1,\pm i,\pm j,\pm k\}
$$
is nonabelian because $ij=k$ whereas $ji=-k$. Its proper subgroups are the trivial subgroup, $\{\pm1\}$, and the three cyclic subgroups
$$
\langle i\rangle,\qquad\langle j\rangle,\qquad\langle k\rangle,
$$
each of order four. They are all cyclic, so
$$
\boxed{\text{the answer is no}}.
$$

Solved by gpt-5.6-sol high.

= 3A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= Solution
{parent=3A}

Set
$$
u=\frac{x}{y},\qquad v=xy.
$$
Because the region lies in the <positive quadrant>, its four inequalities become
$$
1\leq u\leq\alpha,\qquad 1\leq v\leq\alpha.
$$
Thus this <change of variables formula>[change of variables] sends $D$ to a rectangle in the $uv$-plane. Its <Jacobian determinant> is
$$
\frac{\partial(u,v)}{\partial(x,y)}
=
\begin{vmatrix}
1/y&-x/y^2\\
y&x
\end{vmatrix}
=\frac{2x}{y}=2u,
$$
so
$$
dx\,dy=\frac{du\,dv}{2u}.
$$
Since $x^2=uv$, the <double integral> is
$$
\begin{aligned}
\iint_Dx^2\,dx\,dy
&=\int_1^\alpha\int_1^\alpha
uv\,\frac{du\,dv}{2u}\\
&=\frac12\left(\int_1^\alpha du\right)
\left(\int_1^\alpha v\,dv\right)\\
&=\boxed{\frac{(\alpha-1)(\alpha^2-1)}4}.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

= 4A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= Solution
{parent=4A}

Use $x>0$ as the parameter of the <smooth curve>:
$$
\mathbf r(x)=(x,\log x,0).
$$
Its <velocity vector> and speed are
$$
\mathbf r'(x)=\left(1,\frac1x,0\right),
\qquad
|\mathbf r'(x)|=\frac{\sqrt{x^2+1}}x.
$$
The <unit tangent vector> is therefore
$$
\boxed{\mathbf t(x)=\frac{(x,1,0)}{\sqrt{x^2+1}}}.
$$

The parameter-independent formula for the <curvature of a space curve> gives
$$
\kappa(x)
=\frac{|\mathbf r'(x)\times\mathbf r''(x)|}
{|\mathbf r'(x)|^3}
=\boxed{\frac{x}{(x^2+1)^{3/2}}}.
$$
Differentiating,
$$
\kappa'(x)=\frac{1-2x^2}{(x^2+1)^{5/2}}.
$$
Hence $\kappa$ increases for $0<x<1/\sqrt2$ and decreases for $x>1/\sqrt2$. Its <global maximum> occurs at $x=1/\sqrt2$ and equals
$$
\boxed{\kappa_{\max}=\frac{2}{3\sqrt3}}.
$$
The corresponding point is
$$
\boxed{\left(\frac1{\sqrt2},-\frac12\log2,0\right)}.
$$

Solved by gpt-5.6-sol high.

= 5E
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=5E}

A <group action> of $G$ on $X$ is a map $(g,x)\mapsto gx$ such that $ex=x$ and $(gh)x=g(hx)$. The <orbit of a group action>[orbit] and <stabilizer subgroup>[stabilizer] of $x$ are
$$
Gx=\{gx:g\in G\},
\qquad
G_x=\{g\in G:gx=x\}.
$$
For a <finite group>, the map
$$
gG_x\longmapsto gx
$$
is a well-defined <bijection> from the left cosets of $G_x$ to $Gx$. Each coset has $|G_x|$ elements, so the <orbit-stabilizer theorem> is
$$
\boxed{|G|=|Gx|\,|G_x|}.
$$

The <Cauchy theorem for groups> says that if a <prime number> $p$ divides $|G|$, then $G$ has an element of order $p$. To prove it, let
$$
X=\{(g_1,\ldots,g_p)\in G^p:g_1\cdots g_p=e\}.
$$
The first $p-1$ entries determine the last, so $|X|=|G|^{p-1}$, which is divisible by $p$. The <cyclic group> $C_p$ acts on $X$ by cyclically rotating the entries; rotation preserves the product condition because
$$
g_2\cdots g_pg_1=g_1^{-1}(g_1\cdots g_p)g_1=e.
$$
Every orbit has size one or $p$. The fixed points are exactly the tuples $(g,\ldots,g)$ with $g^p=e$. Their number is therefore divisible by $p$. Since the identity gives one fixed point, there is another, and its entry has order $p$.

Now let $|G|=33$. Cauchy's theorem gives a subgroup $H$ of order $11$. Let $H$ act by the <conjugation action> on the set $\mathcal X$ of subgroups of order $11$. It fixes $H$. If it also fixed $K\ne H$, then $H$ would normalize $K$; since $H\cap K=\{e\}$, the product $HK$ would be a subgroup of order $121$, which is impossible. Every other $H$-orbit in $\mathcal X$ therefore has size $11$, so
$$
|\mathcal X|\equiv1\pmod{11}.
$$
Distinct members of $\mathcal X$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal X|\leq33$, forcing $|\mathcal X|=1$. Thus $H$ is a <normal subgroup>.

Conjugation now defines a <group homomorphism>
$$
G\longrightarrow\operatorname{Aut}(H).
$$
Because $H\cong C_{11}$, its <automorphism group> has order $10$. By the <Lagrange theorem>, the image has order dividing both $33$ and $10$, so the image is trivial and $H$ lies in the <center of a group>[center] of $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\operatorname{lcm}(11,3)=33$. Therefore
$$
\boxed{G=\langle hx\rangle\cong C_{33}}.
$$

Solved by gpt-5.6-sol high.

= 6E
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=6E}

A <Möbius transformation> of the <Riemann sphere> is a map
$$
z\longmapsto\frac{az+b}{cz+d},
\qquad ad-bc\ne0,
$$
with its natural values at the pole and at infinity.

For three distinct points $z_1,z_2,z_3$, define
$$
T_z(\zeta)=
\frac{(\zeta-z_2)(z_1-z_3)}
     {(\zeta-z_3)(z_1-z_2)}.
$$
The usual limiting conventions cover an infinite $z_i$. This Möbius transformation sends $(z_1,z_2,z_3)$ to $(1,0,\infty)$. Defining $T_w$ similarly, the map
$$
\boxed{f=T_w^{-1}\circ T_z}
$$
sends $z_i$ to $w_i$. If two Möbius transformations do so, their quotient fixes $0,1,\infty$. A Möbius transformation fixing infinity is affine, and fixing zero and one then makes it the identity. This proves uniqueness.

With this convention, the <cross-ratio> is
$$
[z_1,z_2,z_3,z_4]
=T_z(z_4)
=\frac{(z_4-z_2)(z_1-z_3)}
{(z_4-z_3)(z_1-z_2)}.
$$
Substitution shows that translations, nonzero scalings, and inversion preserve it; since these generate the Möbius group, every Möbius transformation preserves cross-ratios.

Conversely, suppose a <bijection> $f$ of the Riemann sphere preserves every cross-ratio. Let $m$ be the unique Möbius transformation agreeing with $f$ at three chosen points $z_1,z_2,z_3$. For any other $z$, preservation by $f$ and $m$ gives
$$
[z_1,z_2,z_3,z]
=[f(z_1),f(z_2),f(z_3),f(z)]
=[f(z_1),f(z_2),f(z_3),m(z)].
$$
The last coordinate in a cross-ratio with three fixed distinct entries is injective, so $f(z)=m(z)$. The equality already holds at the three base points, hence $f=m$ everywhere and $f$ is Möbius.

Finally, the map $z\mapsto a\overline z+b$ is constant when $a=0$ and therefore is not Möbius. If $a\ne0$, it is bijective and fixes infinity. Were it Möbius, it would have the affine form $\alpha z+\beta$. Equality on real $z$ forces $\alpha=a$ and $\beta=b$, whereas equality at $z=i$ would require $ai=-ai$, contradicting $a\ne0$. Thus
$$
\boxed{\text{there are no such }a,b}.
$$

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=7E}

A <subgroup> $N\leq G$ is <normal subgroup>[normal] when
$$
gNg^{-1}=N\qquad(g\in G).
$$
Its left and right cosets then agree, and multiplication
$$
(gN)(hN)=ghN
$$
is well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the <quotient group> $G/N$.

For a <group homomorphism> $\theta:G\to H$, its <kernel of a group homomorphism>[kernel] and <image of a group homomorphism>[image] are
$$
\ker\theta=\{g:\theta(g)=e_H\},
\qquad
\operatorname{im}\theta=\{\theta(g):g\in G\}.
$$
The kernel is normal because
$$
\theta(gkg^{-1})
=\theta(g)\theta(k)\theta(g)^{-1}
=e_H
$$
whenever $k\in\ker\theta$. Conversely, if $K\trianglelefteq G$, the quotient map
$$
q:G\longrightarrow G/K,\qquad q(g)=gK,
$$
is a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the map
$$
G/\ker\theta\longrightarrow\operatorname{im}\theta,
\qquad
g\ker\theta\longmapsto\theta(g)
$$
is well defined and bijective and preserves multiplication. This is the <first isomorphism theorem>.

Define
$$
\Phi:(\mathbb R,+)\longrightarrow(\mathbb C\setminus\{0\},\cdot),
\qquad
\Phi(t)=e^{2\pi it}.
$$
<Euler's formula> shows that $\Phi$ is a homomorphism, its image is the <complex unit circle>, and its kernel is $\mathbb Z$. The first isomorphism theorem therefore gives
$$
\boxed{\mathbb R/\mathbb Z\cong S^1}.
$$
The image of $\mathbb Q/\mathbb Z$ consists exactly of the <roots of unity>: if $t=p/q$, then $\Phi(t)^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.

Solved by gpt-5.6-sol high.

= 8E
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=8E}

The set $S(\mathbb N)$ consists of all <bijections> $\mathbb N\to\mathbb N$. The composition of two bijections is a bijection, composition is associative, the identity map is an identity, and every bijection has an inverse, so $S(\mathbb N)$ is a <group>.

An element of $S_{\mathrm{fin}}(\mathbb N)$ has finite <support of a permutation>[support]. The support of a composition is contained in the union of the two supports, and a permutation and its inverse have the same support. The identity has empty support. Hence $S_{\mathrm{fin}}(\mathbb N)$ is a <subgroup>.

Let $\sigma$ be a cycle and choose $n$ that it moves. Because $\sigma$ has finite support, the sequence
$$
n,\sigma(n),\sigma^2(n),\ldots
$$
must repeat. Since $\sigma$ is invertible, its first repetition returns to $n$; let the least positive return time be $l$. The cycle condition says that every moved point occurs in this orbit, so $\sigma^l$ fixes every point. No smaller positive power fixes $n$. Therefore
$$
\boxed{\operatorname{ord}(\sigma)=l},
$$
which is finite.

For $\tau\in S_{\mathrm{fin}}(\mathbb N)$, partition its finite support into the orbits of the <cyclic group> $\langle\tau\rangle$. On each orbit, let $\sigma_i$ agree with $\tau$ and fix every point outside that orbit. Then each $\sigma_i$ is a <permutation cycle>, their supports are pairwise disjoint, and
$$
\tau=\sigma_1\cdots\sigma_k.
$$
Disjoint cycles commute because at every point at most one of them acts nontrivially.

Writing $l_i=\operatorname{ord}(\sigma_i)$, a power $\tau^m$ is the identity exactly when every $\sigma_i^m$ is the identity, equivalently when every $l_i$ divides $m$. Thus
$$
\boxed{\operatorname{ord}(\tau)
=\operatorname{lcm}(l_1,\ldots,l_k)}.
$$

Solved by gpt-5.6-sol high.

= 9A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=9a}
{scope}

= Solution
{parent=a}

Since $r=(x_jx_j)^{1/2}$, the <chain rule> gives
$$
\boxed{\frac{\partial r}{\partial x_i}=\frac{x_i}{r}}.
$$
For the radial <vector field> $g(r)\mathbf x$, the <product rule> and <Einstein summation convention> give
$$
\begin{aligned}
\nabla\cdot\bigl(g(r)\mathbf x\bigr)
&=\frac{\partial}{\partial x_i}\bigl(g(r)x_i\bigr)\\
&=g'(r)\frac{x_i}{r}x_i+3g(r)\\
&=\boxed{rg'(r)+3g(r)}.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

= b
{parent=9a}
{scope}

= Solution
{parent=b}

Differentiating the <spherical coordinate system>
$$
\mathbf x=(r\sin\theta\cos\phi,\,
r\sin\theta\sin\phi,\,
r\cos\theta)
$$
gives the orthogonal decomposition
$$
d\mathbf x
=\mathbf e_r\,dr+r\mathbf e_\theta\,d\theta
+r\sin\theta\,\mathbf e_\phi\,d\phi,
$$
where
$$
\begin{aligned}
\mathbf e_r&=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta),\\
\mathbf e_\theta&=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta),\\
\mathbf e_\phi&=(-\sin\phi,\cos\phi,0).
\end{aligned}
$$
Thus the <scale factors of orthogonal coordinates> are
$$
\boxed{h_r=1,\qquad h_\theta=r,\qquad h_\phi=r\sin\theta}.
$$

Since the <total differential> satisfies
$$
df=f_r\,dr+f_\theta\,d\theta+f_\phi\,d\phi
=d\mathbf x\cdot\nabla f,
$$
comparison of coefficients gives the <gradient>
$$
\boxed{
\nabla f
=\mathbf e_r\frac{\partial f}{\partial r}
+\mathbf e_\theta\frac1r\frac{\partial f}{\partial\theta}
+\mathbf e_\phi\frac1{r\sin\theta}\frac{\partial f}{\partial\phi}}.
$$

Solved by gpt-5.6-sol high.

= c
{parent=9a}
{scope}

= Solution
{parent=c}

Each field has only an azimuthal component and is independent of $r$ and $\phi$. The <curl in spherical coordinates> therefore reduces to
$$
(\nabla\times\mathbf A)_r
=\frac1{r\sin\theta}
\frac{\partial}{\partial\theta}
\bigl(\sin\theta A_\phi\bigr),
\qquad
(\nabla\times\mathbf A)_\theta
=-\frac1r\frac{\partial}{\partial r}(rA_\phi),
$$
and its $\phi$ component vanishes. The <half-angle identities>
$$
\sin\theta\tan\frac\theta2=1-\cos\theta,
\qquad
\sin\theta\left(-\cot\frac\theta2\right)=-(1+\cos\theta)
$$
show in both cases that
$$
\boxed{\nabla\times\mathbf A_+
=\nabla\times\mathbf A_-
=\frac{\mathbf e_r}{r^2}
=\frac{\mathbf x}{r^3}}.
$$

Apply part (a) with $g(r)=r^{-3}$. Then
$$
\nabla\cdot\frac{\mathbf x}{r^3}
=r(-3r^{-4})+3r^{-3}=0
$$
away from the origin, explicitly confirming that each resulting <curl> has zero <divergence>.

Solved by gpt-5.6-sol high.

= d
{parent=9a}
{scope}

= Solution
{parent=d}

Part (c) gives
$$
\nabla\times(\mathbf A_+-\mathbf A_-)=0.
$$
The half-space $x_1>0$ is a <simply connected domain>, so the <Poincare lemma> makes this curl-free field a <conservative vector field>: it is the gradient of a single-valued scalar potential.

Using $\tan u+\cot u=2/\sin(2u)$,
$$
\mathbf A_+-\mathbf A_-
=\frac2{r\sin\theta}\mathbf e_\phi.
$$
The spherical-coordinate formula from part (b) shows that a potential must satisfy $f_\phi=2$ and may be independent of $r,\theta$. On $x_1>0$ the azimuthal angle has the single-valued branch
$$
\phi=\arctan\frac{x_2}{x_1}.
$$
Hence one solution, up to an <additive constant>, is
$$
\boxed{f(\mathbf x)=2\phi
=2\arctan\frac{x_2}{x_1}}.
$$

Solved by gpt-5.6-sol high.

= 10A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= Solution
{parent=10A}

On the part with $1\leq z\leq2$, write
$$
\mathbf r(\rho,\phi)
=\left(\rho\cos\phi,\rho\sin\phi,
\sqrt{\rho^2-1}\right),
\qquad
\sqrt2\leq\rho\leq\sqrt5.
$$
This is the upper portion of a <one-sheet hyperboloid>[hyperboloid of one sheet]. Choosing the <orientation of a surface>[outward orientation], whose normal points radially away from the $z$-axis, the <vector area element> is
$$
\begin{aligned}
d\mathbf S
&=(\mathbf r_\phi\times\mathbf r_\rho)\,d\rho\,d\phi\\
&=\boxed{\left(
\frac{\rho^2\cos\phi}{\sqrt{\rho^2-1}},
\frac{\rho^2\sin\phi}{\sqrt{\rho^2-1}},
-\rho\right)d\rho\,d\phi}.
\end{aligned}
$$
Its radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.

For
$$
\mathbf A=(-yz^2,xz^2,0),
$$
the <curl> is
$$
\nabla\times\mathbf A=(-2xz,-2yz,2z^2).
$$
On the surface, $z^2=\rho^2-1$, and therefore
$$
(\nabla\times\mathbf A)\cdot d\mathbf S
=-2\rho(2\rho^2-1)\,d\rho\,d\phi.
$$
Direct integration gives
$$
\begin{aligned}
\int_S\nabla\times\mathbf A\cdot d\mathbf S
&=-2\int_0^{2\pi}\int_{\sqrt2}^{\sqrt5}
\rho(2\rho^2-1)\,d\rho\,d\phi\\
&=-2\pi[\rho^4-\rho^2]_{\sqrt2}^{\sqrt5}\\
&=\boxed{-36\pi}.
\end{aligned}
$$

To verify this with the <Stokes theorem>, note that on a circle of fixed $\rho,z$,
$$
\mathbf A=z^2\rho\,\mathbf e_\phi,
\qquad
d\mathbf r=\rho\mathbf e_\phi\,d\phi.
$$
The induced <boundary orientation> runs in the negative $\phi$ direction on the top circle and the positive $\phi$ direction on the bottom circle. Hence
$$
\oint_{\partial S}\mathbf A\cdot d\mathbf r
=2\pi(1^2)(\sqrt2)^2
-2\pi(2^2)(\sqrt5)^2
=4\pi-40\pi
=-36\pi.
$$

For $S'$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\phi$ direction at $z=-1$, where $\rho^2=2$, and negative $\phi$ direction at $z=\sqrt2$, where $\rho^2=3$. Thus
$$
\boxed{
\int_{S'}\nabla\times\mathbf A\cdot d\mathbf S
=2\pi(1)(2)-2\pi(2)(3)
=-8\pi}.
$$

Solved by gpt-5.6-sol high.

= 11A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= i
{parent=11a}
{scope}

= Solution
{parent=i}

Let $\phi_1,\phi_2$ be two solutions with the same <Neumann boundary condition>, and set $u=\phi_1-\phi_2$. Then
$$
\nabla^2u-m^2u=0\quad\hbox{in }V,
\qquad
\frac{\partial u}{\partial n}=0\quad\hbox{on }S.
$$
Multiplying by $u$ and applying <Green's first identity> gives
$$
\int_V\left(|\nabla u|^2+m^2u^2\right)dV
=\int_Su\frac{\partial u}{\partial n}\,dS
=0.
$$
The integrand is nonnegative. If $m>0$, both terms can vanish only when $u=0$, so the solution of the <modified Helmholtz equation> is unique.

If $m=0$, the identity only forces $\nabla u=0$. On a connected region, $u$ may be any constant. Thus solutions of the <Laplace equation> with prescribed normal derivative are unique only up to an additive constant, so
$$
\boxed{\text{uniqueness fails when }m=0}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=11a}
{scope}

= Solution
{parent=ii}

Put $\eta=\psi-\phi$. The common <Dirichlet boundary condition> gives $\eta=0$ on $S$. Expanding the <energy functional>,
$$
\begin{aligned}
E[\psi]-E[\phi]
={}&\int_V\left(|\nabla\eta|^2+m^2\eta^2\right)dV\\
&+2\int_V\left(\nabla\phi\cdot\nabla\eta
+m^2\phi\eta\right)dV.
\end{aligned}
$$
<Integration by parts> and the field equation make the cross term zero:
$$
\int_V\left(\nabla\phi\cdot\nabla\eta+m^2\phi\eta\right)dV
=\int_S\eta\frac{\partial\phi}{\partial n}\,dS
-\int_V\eta(\nabla^2\phi-m^2\phi)\,dV
=0.
$$
Consequently
$$
\boxed{
E[\psi]-E[\phi]
=\int_V\left(|\nabla\eta|^2+m^2\eta^2\right)dV\geq0}.
$$

For $m>0$, equality holds exactly when $\eta=0$. For $m=0$, equality holds exactly when $\nabla\eta=0$, so $\eta$ is constant on the connected region; its zero boundary value then again forces $\eta=0$. The minimizing property therefore proves uniqueness for the Dirichlet problem in both cases:
$$
\boxed{\psi=\phi\text{ is the equality condition for every }m\geq0}.
$$

Solved by gpt-5.6-sol high.

= 12A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=12a}
{scope}

= Solution
{parent=a}

Let a change of orthonormal coordinates be represented by a <rotation matrix> $R$. Since both <angular momentum> and <angular velocity> are vectors,
$$
L'_i=R_{ip}L_p,
\qquad
\omega'_j=R_{jq}\omega_q.
$$
Using $L_i=I_{ij}\omega_j$ in both frames gives
$$
I'_{ij}R_{jq}\omega_q=R_{ip}I_{pq}\omega_q
$$
for every vector $\boldsymbol\omega$. Multiplication by $R^{-1}=R^T$ yields
$$
\boxed{I'_{ij}=R_{ip}R_{jq}I_{pq}},
$$
which is exactly the transformation law for a rank-two <tensor>. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf r=\mathbf x-\mathbf a$. The <vector triple product> gives
$$
\mathbf r\times(\boldsymbol\omega\times\mathbf r)
=r^2\boldsymbol\omega
-\mathbf r(\mathbf r\cdot\boldsymbol\omega).
$$
Therefore
$$
\boxed{
I_{ij}(\mathbf a)
=\rho\int_B
\left[(x_k-a_k)(x_k-a_k)\delta_{ij}
-(x_i-a_i)(x_j-a_j)\right]dV}.
$$
At the centre of mass,
$$
\boxed{
I_{ij}(\mathbf0)
=\rho\int_B(x_kx_k\delta_{ij}-x_ix_j)\,dV}.
$$
Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho\int_BdV=M$. Hence the <parallel axis theorem> in tensor form is
$$
\boxed{
I_{ij}(\mathbf a)
=I_{ij}(\mathbf0)
+M(a_ka_k\delta_{ij}-a_ia_j)}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=12a}
{scope}

= Solution
{parent=b}

Reflection symmetry of the cube in each coordinate plane makes every off-diagonal integral
$$
-\rho\int_Bx_ix_j\,dV\qquad(i\ne j)
$$
vanish because its integrand is <odd function>[odd] in one coordinate. Permuting the three coordinate axes leaves the cube unchanged, so its three diagonal moments are equal. Thus its <inertia tensor> at the centre has the isotropic form
$$
I_{ij}(\mathbf0)=\lambda\delta_{ij}.
$$

For
$$
\mathbf a=\frac\ell2(1,1,0),
\qquad
|\mathbf a|^2=\frac{\ell^2}{2},
$$
part (a) and $\lambda=M\ell^2/6$ give
$$
\boxed{
I(\mathbf a)
=\frac{M\ell^2}{12}
\begin{pmatrix}
5&-3&0\\
-3&5&0\\
0&0&8
\end{pmatrix}}.
$$
The displayed symmetric matrix has orthogonal <eigenvectors>
$$
(1,1,0),\qquad(1,-1,0),\qquad(0,0,1)
$$
with corresponding <eigenvalues>
$$
\frac{M\ell^2}{6},
\qquad
\frac{2M\ell^2}{3},
\qquad
\frac{2M\ell^2}{3}.
$$
For a unit angular velocity, $|\mathbf L|=|I\boldsymbol\omega|$ is minimized and maximized along eigenvectors for the smallest and largest eigenvalues. Their ratio is
$$
\boxed{
\frac{|\mathbf L|_{\max}}{|\mathbf L|_{\min}}
=\frac{2M\ell^2/3}{M\ell^2/6}=4}.
$$

Solved by gpt-5.6-sol high.