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Use the same Fourier transform convention, now in the sense of tempered distributions, since the homogeneous oscillations are not integrable. Transforming givesSince $q>0$, the two roots are distinct. A distribution annihilated by this polynomial is a linear combination of Dirac delta distributions at its roots:Fourier inversion therefore givesThe constants may be complex. Differentiating the factor $e^{-irt}$ also verifies that the original ordinary differential equation reduces to $\ddot{\psi }+q^2\psi =0$ for $\varphi =e^{-irt}\psi$.

The convolution theorem gives the causal responseDefine the integrated impulse response, for $s\ge 0$, byDirect differentiation gives $F^{\prime }(s)=e^{-ps}\sin (qs)/q$ and $F(0)=F^{\prime }(0)=0$. ThusIn particular, after switch-off the answer isThe response and its first derivative are continuous at both switching times, and the post-pulse motion is a freely decaying damped harmonic oscillator. The expression described as valid for $t<1$ in the paper is understood for $0<t<1$; before the pulse the causal output is zero.

Take the causal solution, extending the input and output by zero before the device starts at $t=0$. The two zero initial conditions ensure that this extension adds no impulsive terms to the first two distributional derivatives. The Fourier transform of a derivative is multiplication by $i\omega$, soFactoring the quadratic and taking partial fraction decomposition yieldsThe inverse Fourier transform follows directly by integrating decaying exponentials:where $H$ is the Heaviside step function. This damped-oscillator impulse response has $R(0^{+})=0$ and $R^{\prime }(0^{+})=1$, as required for a unit impulse.

Use the Fourier transform conventionThe convolution is $(f\ast g)(t)={\int }_{-\infty }^{\infty }f(\tau )g(t-\tau )d\tau$. For integrable $f,g$, Fubini's theorem and the substitution $s=t-\tau$ giveHence the convolution theorem is

Use the counterclockwise rectangular contour with vertices $-R,R,R+i\pi ,-R+i\pi$. It encloses just the simple pole $i\pi /2$, whose residue is $e^{k\pi /2}/i$. Let $I_R={\int }_{-R}^R{e}^{-ikx}/\cosh xdx$. The identitymakes the top edge, traversed right to left, contribute $e^{k\pi }{I}_R$. On either vertical edge,Thus the sum of the vertical integrals tends to zero, with absolute value at most $2\pi e^{|k|\pi }/\sinh R$. The residue theorem and $R\to \infty$ now giveThe argument works for every fixed real $k$, including zero and negative values, and gives the Fourier transform of the hyperbolic secant with the convention $\tilde{f}(k)=\int e^{-ikx}f(x)dx$.

The exponential numerator is entire and never zero. The denominator vanishes precisely when $e^{2z}=-1$, givingAt these points $\sinh z_n=i(-1{)}^n\ne 0$, so every zero of the denominator is simple and hence every singularity is a simple pole. Using the residue formula in (b),

A simple pole at $z_0$ means that $f$ is holomorphic in a punctured neighbourhood and $(z-z_0)f(z)$ extends holomorphically to a nonzero value at $z_0$. Equivalently its Laurent series begins with $a_{-1}/(z-z_0)$ and has no higher negative powers. The residue isIn particular, if $f=g/h$ with $g(z_0)\ne 0$, $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, this becomes $\mathrm{Res}(f,z_0)=g(z_0)/h^{\prime }(z_0)$. The associated residue theorem isfor a positively oriented simple contour with no poles on it.

Write $f=u+iv$ and $z=x+iy$. For a positively oriented, piecewise smooth simple closed contour bounding a region where $u,v$ have continuous first partial derivatives, Green's theorem givesThe Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$, so both integrands vanish:This proves Cauchy's integral theorem under the standard regularity assumptions for Green's theorem. More general admissible contours follow by subdivision; every enclosed region must be free of singularities.

The outward normal of the exterior region points into the excluded disc, so it is $-\mathbf{r}$ on the unit circle. Differentiating the Dirichlet Green function from (e), and writing $s=|{\mathbf{r}}^{\prime }|$ and $d=|\mathbf{r}-{\mathbf{r}}^{\prime }|$, givesHere $|\mathbf{r}-\mathbf{p}{|}^2=d^2/s^2$ was used. The concentrated boundary data have unit integral, so the boundary representation evaluates this exterior Poisson kernel at $\mathbf{r}=(-1,0)$. Thus the bounded solution on either part of the exterior $x$-axis isIt is positive on both intervals, diverges as $x^{\prime }\to -1^{-}$ at the boundary source, and tends to $1/(2\pi )$ as $|x^{\prime }|\to \infty$. Boundedness at infinity selects this solution; otherwise one could add an arbitrary multiple of $\log |{\mathbf{r}}^{\prime }|$ without changing the circle's boundary data.

For a source in the exterior, its inverse point $\mathbf{p}$ lies strictly inside the excluded disc. The method of images construction of the bounded exterior disk Green function therefore allows an opposite logarithmic source there without adding a singularity to the domain. The candidate Dirichlet Green function isIts Laplacian in the exterior is exactly $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$, since the image singularity is outside that domain. The identity in (d) makes $G=0$ on the unit circle. Moreoverso it obeys the required boundedness condition. These checks establish the stated logarithm-of-a-ratio expression with $A=1/(2\pi )$. Bounded exterior Dirichlet uniqueness for the Poisson equation follows by inversion in a circle and removal of the bounded singularity at the origin.

Let $s=|{\mathbf{r}}^{\prime }|>0$ and $\mathbf{p}={\mathbf{r}}^{\prime }/s^2$. For $|\mathbf{r}|=1$, expanding both squared Euclidean distances givesTaking nonnegative square roots provesThis is the distance identity behind inversion in a circle and the method of images for a circular boundary.

Green's second identity, with the outward normal derivative, isSet $u=\Phi$ and $v=G(\cdot ;{\mathbf{r}}^{\prime })$, interpreting the logarithmic singularity by a small excised disc or distribution theory. The volume integral is $\Phi ({\mathbf{r}}^{\prime })$ because $\Delta \Phi =0$ and $\Delta G=\delta$. On the boundary $G=0$ and $\Phi =f$, leavingFor the exterior problem below, apply the identity first in a large truncated region. For bounded harmonic functions the extra circle at infinity contributes zero: $G$ tends to a constant, its radial derivative is $O(r^{-2})$, and a bounded exterior harmonic function also has radial derivative $O(r^{-2})$. This is the condition under which the same boundary representation applies there.

The Dirichlet Green function is the solution, with the source point held fixed, ofIts singularity is the logarithmic one found in (a); subtracting that fundamental solution leaves a harmonic function. An unbounded region also requires a condition at infinity, here boundedness.

A fundamental solution of the Laplace equation satisfies ${\Delta }_{\mathbf{r}}{G}_0=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ in the sense of distributions, without a boundary condition. Choose it rotationally symmetric about the source. Away from the source, with $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$, the radial Laplace equation is $(s{G}_0^{\prime }{)}^{\prime }=0$, giving $G_0=A\log s+B$.

Integrating over a small disc around the source and using the divergence theorem fixes the normalization:ThereforeThe additive constant is arbitrary; adding other entire harmonic functions is also possible if rotational symmetry is not imposed. The positive sign here follows from the convention $\Delta G_0=+\delta$.

Reflecting the solution across $\theta =\pi \alpha$ preserves its Dirichlet boundary conditions and outer boundary data, so uniqueness makes the solution reflection-symmetric. Sinceonly odd angular Fourier modes survive. Put $\rho =(r/R{)}^{1/(2\alpha )}$ and $\psi =\theta /(2\alpha )$. On the symmetry axis $\psi =\pi /2$, comparison of the convergent power series givesThusThis is the imaginary part of $\sin (\rho e^{i\psi })$, so a useful closed form isAt either straight edge $\sin \psi =0$, while on the symmetry axis the formula is $\sinh \rho$. Every term is harmonic in the sector, and the factorial denominator gives convergence up to $r=R$, verifying the full solution.

The Fourier orthogonality relation on this angular interval isProjecting the outer boundary condition onto each angular eigenfunction givesThe solution is unique in the class of bounded harmonic functions with the prescribed boundary trace. For continuous boundary data, the difference of two solutions is zero on the boundary and bounded at the vertex, so the maximum principle for harmonic functions makes it zero throughout the sector. Equivalently all its Fourier sine series coefficients vanish. The usual convergence interpretation applies if the data are only piecewise smooth.

For the harmonic extension in a circular sector, write a separated harmonic function as $F(r)H(\theta )$. The Laplace equation in polar coordinates yieldsThe angular Dirichlet boundary conditions select the Fourier sine basisIndeed, the angular equation has no nonzero solution with nonpositive ${\nu }^2$: multiplying $-H^{\prime \prime }={\nu }^{2}H$ by $H$ and integrating gives $\int (H^{\prime }{)}^2={\nu }^2\int H^2$. The radial Cauchy-Euler equation has solutions $r^{{\nu }_n}$ and $r^{-{\nu }_n}$. Boundedness at the vertex excludes the latter, so superposition gives

Distinguish the quaternion group generators $\mathbf{i},\mathbf{j},\mathbf{k}$ from the complex scalar $i$. Denote their proposed image matrices byDirect multiplication givesFor example $AB=\mathrm{diag}(i,-i)=C$ and $ABC=C^2=-I_2$. Sending $I$ to $I_2$ and each negative quaternion element to the negative of its positive image now reproduces every quaternion product. The matrices all have determinant one and are invertible, so this is a group representation in $GL(2,\mathbb{C})$.

The eight matrices $\pm I_2,\pm A,\pm B,\pm C$ are distinct. Therefore the group homomorphism is injective, and the representation is faithful; equivalently its kernel is $\{I\}$.

The character of a representation is the trace of its representing matrix. ConsequentlyThese are constant on the five conjugacy classes $\{I\}$, $\{-I\}$, $\{\pm \mathbf{i}\}$, $\{\pm \mathbf{j}\}$, $\{\pm \mathbf{k}\}$.

The complete list of subgroups isTo see completeness, Lagrange's theorem allows only orders $1,2,4,8$. The only element of order two is $-I$, which fixes the order-two subgroup. Every noncentral element has order four and generates the corresponding listed cyclic subgroup. Any order-four subgroup contains such an element, because there are only two central elements; it must equal the cyclic subgroup generated by that element. This accounts for all possibilities.

Here $a^2=-I$, $a^3=-a$ and $a^4=I$, so $K$ is a cyclic group of order four. Its multiplication table isWrite the specified image of $a$ as $A$. Direct matrix multiplication givesA group representation is a group homomorphism to a general linear group, so the images of all powers of $a$ are forced:All four images are invertible. Defining $\Phi (a^j)=A^j$ is well-defined modulo four because $A^4=I_2$, and $A^{j+\ell }=A^j{A}^{\ell }$ verifies every product in the table. These observations prove both existence and uniqueness of this representation with the prescribed image.

A group homomorphism preserves multiplication:It automatically sends the identity element to the identity and sends inverse elements to inverses. A group isomorphism is a bijective group homomorphism. For finite groups this requires equal group orders, and, once those orders agree, either injectivity or surjectivity of the group homomorphism is sufficient. Equal orders alone do not suffice.