Uniform <expectations> require nonempty <finite sets>; assume this throughout the analytic argument. For <real-valued functions> on $X\times Y$, define the <box norm> by
$$
\boxed{\|f\|_{\square}=
\left(\mathbb E_{x_0,x_1\in X\atop y_0,y_1\in Y}
\prod_{i,j=0}^1f(x_i,y_j)\right)^{1/4}.}
$$
All four variables are sampled as <independent random variables> with replacement, so repeated coordinates are included. The defining fourth power equals
$$
\mathbb E_{y_0,y_1}\left(\mathbb E_xf(x,y_0)f(x,y_1)\right)^2\geq0.
$$
If it vanishes, every summand vanishes. In particular, the terms $y_0=y_1=y$ give $\mathbb E_x f(x,y)^2=0$ for each $y$, so $f=0$. <Absolute homogeneity of a norm> follows directly: $\|af\|_{\square}=|a|\|f\|_{\square}$.

For the <triangle inequality>, prove the mixed <box Cauchy-Schwarz inequality>. Write
$$
\Lambda(f_{00},f_{01},f_{10},f_{11})=
\mathbb E_{x_0,x_1,y_0,y_1}\prod_{i,j=0}^1f_{ij}(x_i,y_j).
$$
Separating the two $y$ averages and applying the <Cauchy-Schwarz inequality> in $(x_0,x_1)$ gives
$$
|\Lambda|\leq A(f_{00},f_{10})^{1/2}A(f_{01},f_{11})^{1/2},
\quad A(f,g)=\mathbb E_{x,x'}\left(\mathbb E_y f(x,y)g(x',y)\right)^2.
$$
Expanding the square and reversing the order of finite sums,
$$
A(f,g)=\mathbb E_{y,y'}
\left(\mathbb E_xf(x,y)f(x,y')\right)
\left(\mathbb E_xg(x,y)g(x,y')\right)
\leq\|f\|_{\square}^2\|g\|_{\square}^2,
$$
by a second <Cauchy-Schwarz inequality>. Thus $|\Lambda|\leq\prod_{i,j}\|f_{ij}\|_{\square}$. Expand $\Lambda(f+g,f+g,f+g,f+g)$ into its sixteen multilinear terms and apply this bound to each term. Their upper bounds sum to $(\|f\|_{\square}+\|g\|_{\square})^4$. Taking fourth roots yields the <triangle inequality>. Together with <definiteness of a norm> and <absolute homogeneity of a norm>, this proves \b[the <box norm> is a <norm>.] Complex <functions> require conjugates and are outside the printed real-valued convention.

For the <bilinear correlation bound for the box norm>, use the normalized <L2 norm>: let $B=\mathbb E_{x,y}f(x,y)u(x)v(y)$ and $\|u\|_2^2=\mathbb E_xu(x)^2$, $\|v\|_2^2=\mathbb E_yv(y)^2$. The <Cauchy-Schwarz inequality> in $y$ gives
$$
|B|^2\leq\|v\|_2^2\mathbb E_y\left(\mathbb E_x f(x,y)u(x)\right)^2.
$$
The remaining factor is $\mathbb E_{x,x'}u(x)u(x')\mathbb E_y f(x,y)f(x',y)$. By the <Cauchy-Schwarz inequality> in $(x,x')$, its magnitude is at most
$$
\left(\mathbb E_{x,x'}u(x)^2u(x')^2\right)^{1/2}
\left(\mathbb E_{x,x'}\left(\mathbb E_yf(x,y)f(x',y)\right)^2\right)^{1/2}
=\|u\|_2^2\|f\|_{\square}^2.
$$
Taking square roots proves
$$
\boxed{|\mathbb E_{x,y}f(x,y)u(x)v(y)|\leq\|f\|_{\square}\|u\|_2\|v\|_2.}
$$

For the <tripartite graph>, assume its parts are nonempty and write $g_{XY},g_{YZ},g_{XZ}$ for its adjacency <indicator functions>. Let $f=g_{XY}-\alpha$, so $\|f\|_{\square}\leq c$. The normalized <triangle count> is
$$
\tau=\mathbb E_{x,y,z}(\alpha+f(x,y))g_{YZ}(y,z)g_{XZ}(x,z).
$$
The constant-degree hypothesis is exactly $\mathbb E_y g_{YZ}(y,z)=\beta$ for each $z$. Therefore the contribution of the constant $\alpha$ is
$$
\alpha\mathbb E_z\left(\mathbb E_y g_{YZ}(y,z)\right)
\left(\mathbb E_x g_{XZ}(x,z)\right)=\alpha\beta\gamma.
$$
For fixed $z$, apply the <bilinear correlation bound for the box norm> with $u_z(x)=g_{XZ}(x,z)$ and $v_z(y)=g_{YZ}(y,z)$. Since these are <indicator functions>, $\|v_z\|_2^2=\beta$ and $\|u_z\|_2^2=d_X(z)$, the fraction of $X$ adjacent to $z$. Averaging and applying the <Cauchy-Schwarz inequality> in $z$ gives the sharper <triangle counting with one box-uniform pair and constant opposite degree> estimate
$$
|\tau-\alpha\beta\gamma|
\leq c\sqrt\beta\,\mathbb E_z\sqrt{d_X(z)}
\leq c\sqrt{\beta\,\mathbb E_zd_X(z)}
=c\sqrt{\beta\gamma}\leq c.
$$
Multiplying by $|X||Y||Z|$ yields
$$
\boxed{\bigl|T(G)-\alpha\beta\gamma\,|X|\,|Y|\,|Z|\bigr|
\leq c\sqrt{\beta\gamma}\,|X||Y||Z|
\leq3c|X||Y||Z|.}
$$
The middle bound also handles $\beta=0$ or $\gamma=0$, when the <triangle count> is zero.

To connect explicitly with the suggested expansion, put $f_2=g_{YZ}-\beta$, $f_3=g_{XZ}-\gamma$. Both have overall mean zero, and $\mathbb E_y f_2(y,z)=0$ for every $z$. Consequently $\mathbb E_{x,y,z}f_2(y,z)f_3(x,z)=0$; all terms with constant $\alpha$ reduce to $\alpha\beta\gamma$. The terms containing $f$ combine into the single error just estimated. This is where the exact degree assumption is used. If any part is empty the <triangle count> itself is zero, but the printed uniform <expectations> and <edge density of a bipartite graph> values would be undefined; the nonempty-parts convention must therefore be stated rather than dividing by zero.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 129 / 4 / Solution

A <metric space> metric $d:X\times X\to[0,\infty)$ satisfies positivity, $d(x,y)=0$ exactly when $x=y$, symmetry, and the <triangle inequality>. A subset $U\subseteq X$ is an <open set> when every $x\in U$ has some $\varepsilon>0$ with the open ball $B_d(x,\varepsilon)\subseteq U$.

Both $\varnothing$ and $X$ are open. An arbitrary union of open sets is open because a point lies in one member supplying a ball. A finite intersection is open because the minimum of the finitely many available radii supplies a ball. Thus the metric-open sets satisfy the <topology axioms>.

For $C[0,1]$, the <Cauchy-Schwarz inequality> gives $d_1(f,g)\leq d_2(f,g)$, so the $d_2$ topology is at least as fine as the $d_1$ topology. To see strictness, define the continuous triangular spikes
$$
f_n(x)=\sqrt n\max\{1-nx,0\}.
$$
Then
$$
d_1(f_n,0)=\frac1{2\sqrt n}\longrightarrow0,
\qquad
d_2(f_n,0)^2=\int_0^{1/n}n(1-nx)^2\,dx=\frac13.
$$
Thus $f_n$ converges to zero in the <L1 norm> but not in the <L2 norm>. \b[The two metrics induce different topologies.]

Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2 / 4E / Solution

For a sufficiently regular solution with zero <Dirichlet boundary conditions>, differentiate its squared <L2 norm>. Since $u_t=-i(\Delta u-Vu)$,
$$
\frac{d}{dt}\|u(t)\|_{L^2}^2
=2\operatorname{Re}\int_\Omega\overline u\,u_t
=2\operatorname{Re}\left[-i\left(\int_\Omega\overline u\,\Delta u
-\int_\Omega V|u|^2\right)\right].
$$
<Integration by parts> and the zero boundary trace give $\int\overline u\Delta u=-\int|\nabla u|^2$. The potential contribution is real because $V$ is real. The quantity in parentheses is consequently real, and its product with $-i$ has zero real part. Hence
$$
\boxed{\|u(t)\|_{L^2(\Omega)}=\|u(0)\|_{L^2(\Omega)}.}
$$
This is the <energy method> for a <skew-adjoint> evolution. For general <L2 space> data, the same result follows from the <strongly continuous unitary group> generated by a self-adjoint realization of $\Delta-V$. A sufficient assumption is a bounded real potential with the Dirichlet Laplacian; arbitrary real potentials require appropriate domain and self-adjointness assumptions. The differentiated proof presumes the regularity needed for the displayed integrals, rather than asserting it for every pointwise real function $V$.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 341 / 4 / a / Solution

The starting point of a conforming <finite element method> is a <weak formulation> in a <Hilbert space> $X$. For example, a homogeneous <Dirichlet boundary condition> on a bounded <Lipschitz domain> gives $X=H_0^1(\Omega)$. One seeks
$$
a(u,v)=\ell(v)\qquad(v\in X),
$$
where $a$ is a <bounded bilinear form>, $|a(w,v)|\leq L\|w\|_X\|v\|_X$, and $\ell\in X^*$ is a bounded <linear functional>. A <conforming finite element space> $X_h\subset X$ is finite dimensional and is typically built from functions that are polynomial on each element of a <finite element mesh>. With a basis $\varphi_j$, the <Galerkin method> imposes the same identity only for $v_h\in X_h$, giving
$$
\boxed{Kc=F,\qquad K_{ij}=a(\varphi_j,\varphi_i),\quad F_i=\ell(\varphi_i).}
$$
The <stiffness matrix> is sparse when basis supports overlap only locally. Local element integrals assemble its entries; the <mass matrix> represents an $L^2$ term. This separates the choice of trial functions from the variational principle determining their coefficients.

The fundamental well-posedness theorem is the <Lax-Milgram theorem>: if $a(v,v)\geq\gamma\|v\|_X^2$ for some $\gamma>0$, there is a unique solution for each $\ell\in X^*$, with $\|u\|_X\leq\|\ell\|_{X^*}/\gamma$. The same theorem on $X_h$ gives a unique discrete solution. The constants must be independent of the mesh for uniform <stability of a numerical method>. Boundary conditions are built into the trial space when they are essential constraints; flux conditions arise through boundary terms in <integration by parts> and are natural constraints in the <weak formulation>.

When $a$ is symmetric and a <coercive bilinear form>, the <Ritz method> minimizes
$$
\mathcal E(v)=\tfrac12a(v,v)-\ell(v)
$$
over $X_h$. Its first variation is exactly the <Galerkin method>, so \b[Ritz minimization and Galerkin testing coincide for symmetric coercive problems]. The <energy norm> $\|v\|_a=\sqrt{a(v,v)}$ turns the approximation into an <orthogonal projection>. Indeed, subtracting the exact and discrete weak equations gives <Galerkin orthogonality>, $a(u-u_h,v_h)=0$. For every $v_h\in X_h$ the <Pythagorean identity> becomes
$$
\|u-v_h\|_a^2=\|u-u_h\|_a^2+\|u_h-v_h\|_a^2,
$$
so
$$
\boxed{\|u-u_h\|_a=\inf_{v_h\in X_h}\|u-v_h\|_a.}
$$
This exact best-approximation property is the central advantage of the <Ritz method>.

Symmetry is unnecessary for the <Galerkin method>. For any bounded, coercive form, the <Céa lemma> gives
$$
\boxed{\|u-u_h\|_X\leq\frac L\gamma
\inf_{v_h\in X_h}\|u-v_h\|_X.}
$$
To prove it, put $e=u-u_h$. <Galerkin orthogonality> yields $a(e,e)=a(e,u-v_h)$, so coercivity and boundedness give $\gamma\|e\|_X^2\leq L\|e\|_X\|u-v_h\|_X$. Divide when $e\neq0$ and take the <infimum>. Thus approximation capability plus uniform coercivity gives convergence. In contrast, minimizing $\tfrac12a(v,v)-\ell(v)$ for a nonsymmetric form differentiates its symmetric part and generally does not solve the original weak equation.

A concrete symmetric example is the one-dimensional reaction-diffusion problem from Question 5. Continuous <piecewise-linear hat functions> give the diffusion <stiffness matrix> $S$ with diagonal $2/h$ and neighbouring entries $-1/h$, and the <mass matrix> $M$ with diagonal $2h/3$ and neighbouring entries $h/6$. Therefore $K=S+M$ and $F_i=ih^2$. Its <positive-definite matrix> is the coordinate form of the continuous energy. The exact smooth solution $u(x)=x-\sinh(x)/\sinh1$ makes this an explicit test of the method, not just an abstract existence result.

For a nonsymmetric example, take positive $\varepsilon$, constant real $\beta$, and
$$
-\varepsilon u''+\beta u'+u=f,\qquad u(0)=u(1)=0.
$$
Its form is $a(u,v)=\varepsilon\int u'v'+\beta\int u'v+\int uv$. Since $\int v'v=0$, it has coercivity constant $\min(\varepsilon,1)$ in the full $H^1$ norm, although it is not symmetric when $\beta\neq0$. The <Galerkin method> remains well posed. On the uniform hat basis, $K=\varepsilon S+\beta C+M$, where $C_{i,i+1}=1/2$, $C_{i,i-1}=-1/2$ and $C_{ii}=0$. The <skew-symmetric matrix> $C$ cancels from $c^TKc$, proving nonsingularity. A Ritz energy would omit precisely this advection contribution. Small $\varepsilon$ may nevertheless make the approximation constant large and produce poorly resolved layers; algebraic solvability is not an accuracy guarantee.

Quantitative convergence uses a <finite element interpolation estimate>. On a <shape-regular mesh>, for piecewise-linear conforming elements and $u\in H^2(\Omega)$ in the usual low-dimensional setting,
$$
\|u-I_hu\|_{H^1}\leq Ch\|u\|_{H^2},\qquad
\|u-I_hu\|_{L^2}\leq Ch^2\|u\|_{H^2}.
$$
The <Céa lemma> therefore gives an $O(h)$ energy error. The <Aubin–Nitsche duality argument> improves the <L2 norm> error when the dual elliptic problem has $H^2$ regularity: solve $a(v,z)=(e,v)_{L^2}$ and use
$$
\|e\|_{L^2}^2=a(e,z-I_hz)
\leq Ch\|e\|_{H^1}\|z\|_{H^2}
\leq Ch\|e\|_{H^1}\|e\|_{L^2}.
$$
Consequently
$$
\boxed{\|u-u_h\|_{H^1}=O(h),\qquad
\|u-u_h\|_{L^2}=O(h^2),}
$$
with constants depending on the solution and regularity bounds. The second estimate is conditional on dual regularity; corners or insufficient data regularity can reduce the rate.

The same <Galerkin method> handles evolution equations. For the homogeneous <heat equation>, the <semidiscrete finite element heat equation> is $M\dot c+Sc=0$. Because $M$ is a <positive-definite matrix> and $S$ is a <positive semidefinite matrix>,
$$
\frac{d}{dt}(c^TMc)=-2c^TSc\leq0.
$$
The decreasing <quadratic form> is exactly the <L2 norm> squared of the finite-element function. This separates spatial <stability of a numerical method> from the subsequent time integrator, just as in the <Schrodinger equation> example. \b[Conformity, coercivity and approximation estimates together explain convergence.]

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 341 / 6 / Solution

Write $h=\Delta x$ and set $u_0=u_{M+1}=0$. In the discrete <L2 norm> $\|u\|_h^2=h\sum_{m=1}^M|u_m|^2$, the <energy method> gives
$$
\frac12\frac{d}{dt}\|u\|_h^2
=-\frac1h\sum_{m=0}^M|u_{m+1}-u_m|^2\leq0.
$$
To obtain this identity, shift the indices in the centered second-difference sum. The centered first-difference <matrix> is <skew-symmetric>, so its contribution has zero real part even for complex grid values. The homogeneous <Dirichlet boundary conditions> remove the endpoint terms.

Therefore the semidiscretization is \b[stable for every real $\alpha$ and every positive $h$], with the mesh-independent bound $\|u(t)\|_h\leq\|u(0)\|_h$. No restriction on $|\alpha|h$ is needed for this energy estimate.

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 3 / a / Solution

With $h=\Delta x$, $\tau=\Delta t$, $r=\tau/h^2$ and $c=\alpha\tau/h$, the <Forward Euler method> gives
$$
\boxed{u_m^{n+1}=(r-c/2)u_{m-1}^n+(1-2r)u_m^n+(r+c/2)u_{m+1}^n.}
$$
For a <Fourier mode> $e^{im\theta}$, the amplification factor is $G(\theta)=1-4r\sin^2(\theta/2)+ic\sin\theta$. Set $s=\sin^2(\theta/2)$. Then
$$
|G|^2-1=4s\{c^2-2r+(4r^2-c^2)s\}.
$$
The expression in braces is an <affine function> of $s$, so it is nonpositive for all $s\in[0,1]$ precisely when its two endpoint values are nonpositive. For $\tau>0$, these conditions are $c^2\leq2r$ and $r\leq1/2$. Thus <Forward Euler stability for centered advection-diffusion> requires exactly
$$
\boxed{\Delta t\leq\frac{(\Delta x)^2}{2},\qquad\alpha^2\Delta t\leq2.}
$$
For $\alpha=0$, only the first condition remains. If either bound fails, a band of <Fourier modes> has $|G|>1$, giving instability under <Von Neumann stability analysis>. At equality the amplification bound still holds, so the endpoints are included. This is discrete <L2 norm> stability; nonnegative stencil weights would impose the additional spatial restriction $|\alpha|\Delta x\leq2$.

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 3 / b / Solution

The <mass matrix> is a <positive-definite matrix>, and <integration by parts> with periodic boundary conditions gives $C^T=-C$. Hence
$$
\frac{d}{dt}(U^TMU)=2U^TCU=0.
$$
This is <energy conservation for semidiscrete Galerkin advection>: $U^TMU=\|u_h\|_{L^2(0,1)}^2$. Thus the semidiscretization is \b[stable and conserves its finite element <L2 norm>].

For a mesh-independent comparison with the grid norm, the <Fourier symbol> of $M$ is $h(2+\cos\theta)/3$, between $h/3$ and $h$. Consequently
$$
\frac h3\sum_m|U_m|^2\leq U^*MU\leq h\sum_m|U_m|^2.
$$
The conserved energy therefore gives a uniform bound in $\sqrt{h\sum_m|U_m|^2}$ as well. Equivalently, each <Fourier mode> evolves with the purely imaginary exponent $\sigma(\theta)=3i\sin\theta/[h(2+\cos\theta)]$. This concerns the semidiscrete method; a chosen time integrator must separately control those imaginary <eigenvalues>.

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 5 / b / Solution

Consider a bounded <Lipschitz domain> $\Omega$ and homogeneous <Dirichlet boundary conditions> for
$$
-\nabla\cdot(A(x)\nabla u)+c(x)u=f.
$$
Assume $A$ is a bounded real <symmetric matrix> field with $\xi^TA(x)\xi\geq a_0|\xi|^2$ uniformly, $c\geq0$ is bounded, and $f$ defines a bounded linear functional on $V=H_0^1(\Omega)$, the <zero-boundary Sobolev space>. The <weak formulation> is
$$
a(u,v)=\ell(v)\quad(v\in V),\qquad
 a(u,v)=\int_\Omega\nabla v^TA\nabla u+cuv,\quad\ell(v)=\langle f,v\rangle.
$$
The <Cauchy-Schwarz inequality> gives continuity $|a(u,v)|\leq M\|u\|_V\|v\|_V$. Uniform ellipticity and the <Poincare inequality> give a <coercive bilinear form>, $a(v,v)\geq\alpha\|v\|_V^2$. The <Lax-Milgram theorem> therefore supplies a unique <weak solution>, with $\|u\|_V\leq\|\ell\|_{V'}/\alpha$. Its proof represents $a$ by a <bounded linear operator> $T$ using the <Riesz representation theorem>; coercivity gives an injective operator with closed image, and a zero <orthogonal complement> makes that image all of $V$. This yields existence, uniqueness and the displayed estimate.

For a <conforming finite element space> $V_h\subseteq V$, the <Galerkin method> finds $u_h\in V_h$ such that $a(u_h,v_h)=\ell(v_h)$ for every $v_h\in V_h$. With a basis $\phi_1,\ldots,\phi_N$, this is the linear system $KU=F$, where the <stiffness matrix> is $K_{ij}=a(\phi_j,\phi_i)$ and $F_i=\ell(\phi_i)$. Its <positive-definite matrix> property proves unique discrete solvability. Local support gives a <sparse matrix> through elementwise assembly.

For the symmetric problem, the <Ritz method> minimizes $J(v)=a(v,v)/2-\ell(v)$ over $V_h$. Differentiating in every direction $v_h$ gives precisely the <Galerkin method> equations. Conversely, if $u_h$ solves them, $J(u_h+w_h)-J(u_h)=a(w_h,w_h)/2$, proving it is the unique minimum. The same argument over $V$ gives $J(v)-J(u)=\|v-u\|_a^2/2$, where $\|v\|_a=\sqrt{a(v,v)}$ is the <energy norm>. Thus the <Ritz method> chooses the best trial function in that norm.

Subtracting the continuous and discrete equations gives <Galerkin orthogonality>, $a(u-u_h,v_h)=0$. For arbitrary $v_h\in V_h$ and $e=u-u_h$, it follows that $a(e,e)=a(e,u-v_h)$. Coercivity and continuity prove the <Céa lemma>:
$$
\alpha\|e\|_V^2\leq M\|e\|_V\|u-v_h\|_V,
\qquad
\|u-u_h\|_V\leq\frac M\alpha\inf_{v_h\in V_h}\|u-v_h\|_V.
$$
In the symmetric <energy norm>, the <Cauchy-Schwarz inequality> improves the constant to one. Equivalently, <Galerkin orthogonality> gives the <Pythagorean identity> $\|u-v_h\|_a^2=\|u-u_h\|_a^2+\|u_h-v_h\|_a^2$. These estimates establish stability and show that approximation properties of the trial space determine convergence.

For continuous piecewise-linear elements on a <shape-regular mesh> in dimensions at most three, a <finite element interpolation estimate> gives $\|u-I_hu\|_{H^1}\leq Ch\|u\|_{H^2}$ when $u\in H^2(\Omega)$. Combining it with the <Céa lemma> yields $O(h)$ <energy norm> error. More generally, degree-$p$ elements give $O(h^p)$ error in $H^1$ under the corresponding regularity and approximation assumptions. Mere mesh refinement cannot supply an order whose required solution regularity is absent.

An additional order in the <L2 norm> follows under a dual <elliptic regularity> assumption. Solve $a(v,z)=(e,v)_{L^2}$ and assume $\|z\|_{H^2}\leq C\|e\|_{L^2}$. The <Aubin–Nitsche duality argument> uses <Galerkin orthogonality> to obtain
$$
\|e\|_{L^2}^2=a(e,z-I_hz)\leq Ch\|e\|_{H^1}\|z\|_{H^2}
\leq Ch\|e\|_{H^1}\|e\|_{L^2}.
$$
Therefore $\|e\|_{L^2}\leq Ch\|e\|_{H^1}$, giving $O(h^2)$ for the piecewise-linear case with $u\in H^2$. This improvement explicitly requires regularity of the dual problem.

As a concrete example, solve $-u''=1$ on $(0,1)$ with zero endpoint values. On a uniform mesh of spacing $h$, the <piecewise-linear hat functions> give $K_{jj}=2/h$, $K_{j,j\pm1}=-1/h$ and $F_j=h$. The discrete equations are $2U_j-U_{j-1}-U_{j+1}=h^2$, whose solution is $U_j=x_j(1-x_j)/2$. The exact solution is $u(x)=x(1-x)/2$, so $u_h$ is its piecewise-linear nodal interpolant. On each interval $[x_j,x_{j+1}]$,
$$
u-u_h=\frac12(x-x_j)(x_{j+1}-x).
$$
Integrating the squared error and squared derivative error gives the explicit rates
$$
\boxed{\|u-u_h\|_{H_0^1}=\frac h{\sqrt{12}},\qquad\|u-u_h\|_{L^2}=\frac{h^2}{\sqrt{120}},}
$$
where the $H_0^1$ norm here is $\|v'\|_{L^2}$.

For nonsymmetric coercive problems, the <Galerkin method>, <Galerkin orthogonality> and the <Céa lemma> still apply, but the symmetric quadratic minimization interpretation of the <Ritz method> generally does not. Nonconforming trial spaces, inexact integration and noncoercive equations need additional arguments beyond the conforming coercive theory proved here.

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 7 / Solution

The equation can be written
$$
u_t=iu_{xx}-iVu.
$$
For the squared <L2 norm>, periodic <integration by parts> and the reality of $V$ give
$$
\frac d{dt}\int_{-1}^1|u|^2\,dx
=2\operatorname{Re}\int_{-1}^1
\left(iu_{xx}-iVu\right)\overline u\,dx
=2\operatorname{Re}\left[
-i\int_{-1}^1|u_x|^2\,dx
-i\int_{-1}^1V|u|^2\,dx
\right]=0.
$$
\b[Thus the <Schrodinger equation> generates a unitary flow and $\|u(\cdot,t)\|_2$ is constant.]

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 341 / 3 / a / Solution

{c}
{title2=$\lVert\mathord\cdot\rVert_2$}

The $L^2$ norm is
$$
\lVert f\rVert_2=\left(\int|f|^2\right)^{1/2}.
$$
It is induced by the $L^2$ inner product and is the continuous analogue of the <Euclidean norm>.

Codex @codex  0

 Incoming links: L2 norm