The Fourier transform of a derivative is $i\omega$ times the transform under the stated convention. Transforming the linear ordinary differential equation givesThe applied force is $F(t)=H(t)e^{-t}$, with $H$ the Heaviside step function, so $\tilde{F}=(1+i\omega {)}^{-1}$. Partial fractions giveUsing the transform pair $H(t)e^{-ct}\leftrightarrow (c+i\omega {)}^{-1}$ yieldsIts value and first derivative both vanish at $t=0^{+}$, so the joining introduces no spurious Dirac delta terms. Equivalently, this is the convolution of the force with the causal impulse response $H(t)(e^{-2t}-e^{-3t})$.

Let $B=|b|>0$. For $a>0$, integrate $e^{iaz}/(z^2+B^2)$ over the upper semicircle. Jordan's lemma removes the arc, and the residue theorem at $z=iB$ givesTaking the real part and using evenness in both $x$ and $a$ yieldsThe case $a=0$ follows directly from an inverse tangent primitive. The complex exponential is essential on the contour: the cosine alone has a growing exponential component.

For $a>0$, differentiation under the integral sign gives $-I_1^{\prime }(a)={\int }_0^{\infty }x\sin (ax)/(x^2+B^2)dx$. This derivative is a conditionally convergent improper integral; the Dirichlet test makes its tails uniform on compact subintervals of $a>0$, justifying the differentiation. Setting $a=1$ givesFinally let $J(a)={\int }_0^{\infty }\sin (ax)/[x(x^2+B^2)]dx$. Its differentiated integrand is absolutely integrable, so $J^{\prime }(a)=I_1(a)$ and $J(0)=0$. HenceIf $b=0$, $I_1$ and $I_3$ diverge at the origin, whereas $I_2={\int }_0^{\infty }\sin x/xdx=\pi /2$ is the corresponding Dirichlet integral.

The function $e^{-z^2}$ is entire, so Cauchy's integral theorem makes its contour integral around the rectangle vanish. The lower horizontal integral tends to $\sqrt{\pi }$ by the Gaussian integral. On a vertical side $z=\pm R+it$, with $t$ between $0$ and $b$, its modulus is $e^{-R^2+t^2}\le e^{-R^2+b^2}$; hence both vertical integrals tend to zero. The upper side has the opposite orientation and contributesConsequently the integral in this expression is $\sqrt{\pi }{e}^{-b^2}$. Its imaginary part vanishes because $e^{-x^2}\sin (2bx)$ is odd, and its real part is twice the positive-half-line integral because the integrand there is even. ThereforeFor negative $b$ the rectangle lies below the real axis and the same cancellation holds; alternatively use the even dependence on $b$.

Differentiate the Green-function representation, valid away from the charged sheet:For $0<x<L$, the two inverse tangent limits at $y\downarrow 0$ are $\pi /2$ and $-\pi /2$; approaching from below reverses both signs. ThusThis agrees with the normal electric field jump obtained by applying Gauss's law to a thin pillbox enclosing unit surface charge density.

Use the Green-function representation to integrate over the charged segment:For $y\ne 0$, a primitive with respect to $u$ isas direct differentiation verifies using the derivative of inverse tangent. ThereforeThis expression is even in $y$, since $y\arctan (u/y)$ is even. On $y=0$, take the continuous limit $F(u,0)=u\log u^2-2u$, with $u\log u^2=0$ at $u=0$. The finite segment has a logarithmically growing potential at large distance, but its electric field decays; the boundary condition constrains the field rather than the potential.

For the rotationally invariant free-space fundamental solution of the Laplace equation, put $s=\sqrt{(x-x^{\prime }{)}^2+(y-y^{\prime }{)}^2}$. Away from the source, the radial Laplacian gives $(s{G}^{\prime }{)}^{\prime }=0$, hence $G=A\log s^2+C$. Integrate its Poisson equation over a disk of radius $r$ about the source. The divergence theorem gives $2\pi r(2A/r)=-1/{ϵ}_0$, soA fixed reference length can be inserted inside the logarithm and absorbed into $C$. The resulting electric field decays like $1/s$.

Strictly, the stated pointwise decay as $y\to \pm \infty$ alone does not justify “most general”: one can also add the harmonic function $\mathrm{Re}{e}^{(x+iy{)}^2}$, whose gradient has that decay for every fixed $x$. This is the same growth issue as a zero-boundary harmonic function with pointwise vertical decay. The displayed answer is the usual isotropic free-space choice; requiring the field to vanish uniformly in every direction at spatial infinity removes these extra harmonic additions. An additive constant remains undetermined.

Since the electric field is irrotational, introduce an electrostatic potential with $\mathbf{E}=-\nabla \phi$. Gauss's law becomes the Poisson equationThe curl equation is automatic because the curl of a gradient vanishes. Independence of $z$ makes $E_z$ constant, and its required decay forces it to be zero, so a $z$-independent potential suffices.

The plane separates the exterior into two half-domains. Use an odd Legendre extension for an equatorial Dirichlet plane in each. On the upper hemisphere, extend $u+2u^2$ to the odd function $u+2u|u|$ on the whole sphere. On the lower hemisphere, the appropriate odd extension is $u-2u|u|$. Define, for odd $\ell$,The orthogonality and parity of the Legendre polynomials give the solution in both half-domains asSet its value on $u=0$ to zero. The first coefficients are $b_1=3/2$ and $b_3=7/12$. All odd Legendre polynomials vanish at zero, so the plane condition holds. On the sphere the sum is $2u|u|$, and multiplication by $\mathrm{sgn}(u)$ recovers $2u^2$ on either side. Thus the original sphere data and decay at infinity both hold. A jump of the normal derivative at the sheet is permitted; the sheet can support induced surface charge density.

Set $u=\cos \theta$. The boundary data simplify to $u+2u^2=\frac{2}{3}{P}_0(u)+P_1(u)+\frac{4}{3}{P}_2(u)$. The axisymmetric harmonic extension from a sphere keeps only decaying radial powers outside and regular powers inside. HenceandEach Legendre polynomial coefficient agrees at $r=a$. The exterior solution tends to zero, and the interior solution is finite at $r=0$. The maximum principle for harmonic functions proves uniqueness in these respective classes.

For an axisymmetric harmonic function, put $\Psi =R(r)\Theta (\theta )$ in the Laplacian in spherical coordinates and multiply by $r^2/(R\Theta )$. Separation of variables givesWith $u=\cos \theta$, the angular equation becomes the Legendre differential equation. Regularity at both poles selects the Legendre polynomials $P_{\ell }(u)$ with integer $\ell \ge 0$; the singular second solutions are excluded. The radial Euler-Cauchy equation has solutions $r^{\ell }$ and $r^{-\ell -1}$. Superposition therefore giveswith coefficients chosen to ensure convergence and the required boundary regularity. The axisymmetric harmonic extension from a sphere gives these coefficients directly from boundary data.

For the symmetric group representation in part (a), the identity and three transpositions contribute $4I$ to $A={\sum }_{g}D(g^2)$. The two 3-cycles contribute $D(r)+D(r^2)=-I$. ThusMore generally, the central average of representation matrices at squares commutes with the representation, because conjugation permutes the summands. For a complex irreducible representation, Schur's lemma therefore gives $A=cI$. Taking the trace yields $c\dim D=|G|\nu$, proving the claimed formula under its correct hypotheses.

For the Klein four-group there is no two-dimensional irreducible representation, as established in part (b). Its genuine one-dimensional irreducibles have $A=4$ and $\nu =1$, so $A=(4\cdot 1/1)I$.

If the intended alternative is the real quarter-turn representation of the cyclic group of order four, its squared matrices sum to $I-I+I-I=0$. Thus $A=0=(4\cdot 0/2)I$ in that example. This equality is a direct calculation; the complex version of Schur's lemma cannot be applied to its reducible complexification.

The requested two-dimensional irreducible representation does not exist for the Vierergruppe. The Klein four-group $V_4$ is an abelian group in which every nonidentity element has order two. Its complex irreducible representations are all one-dimensional. The same holds over the real numbers here: the commuting representing involutions are simultaneously diagonalizable with real eigenvalues $\pm 1$, so every two-dimensional representation has an invariant line. Each genuine irreducible character satisfies $\chi (g^2)=\chi (1)=1$, and hence has Frobenius-Schur indicator $\nu =1$.

The printed notation $C_4$ suggests a different possible interpretation, which must not be conflated with the Klein four-group. For the cyclic group of order four, a real two-dimensional irreducible representation is generated byIts character on $1,J,-I,-J$ is $(2,0,-2,0)$, giving $\nu =(2-2+2-2)/4=0$. Over the complex numbers this representation splits into two conjugate one-dimensional characters, so it is not a two-dimensional complex irreducible representation. These alternatives explain what can be computed without silently changing the printed group or the ground field.

The printed wording calls ${\Sigma }_3$ cyclic, but its meaning in question 9 is the symmetric group $S_3$. Under that interpretation, take its planar dihedral group representation, withHere $r^3=s^2=1$ and $srs=r^{-1}$. This representation is irreducible over the complex numbers: the two distinct rotation eigenspaces are exchanged by the reflection, so neither is invariant under the whole group. Its character is $2$ on the identity, $-1$ on the two 3-cycles and $0$ on the three transpositions. Squaring maps each transposition to the identity and permutes the 3-cycles. Its Frobenius-Schur indicator is thereforeIf “cyclic” were intended literally, no two-dimensional complex irreducible representation would exist, since an irreducible representation of a finite abelian group is one-dimensional.

Write the equation as $(L+\beta )y=-\cos x/x$ with the positive Sturm-Liouville operator from part (a). A Sturm-Liouville eigenfunction expansion therefore gives $a_n=-f_n/(4n^2+\beta )$, where weighted orthogonal projection givesHere the product-to-sum formulas reduce the integral to half the sum of the integrals of $\sin ((2n+1)x)$ and $\sin ((2n-1)x)$. ConsequentlyAt $x=0$, use the continuous limits of the terms. The coefficients are $O(n^{-3})$, so the resulting endpoint series converges. Completeness is the ordinary Fourier sine series completeness after $v=xy$. For an independent check, solving $v^{\prime \prime }-\beta v=\cos x$ with zero endpoint values givesIts numerator vanishes at zero, and it has the same orthogonal projection coefficients. Positivity of $L+\beta$ proves uniqueness.

Set $v=xu$. Direct differentiation gives $u^{\prime \prime }+2u^{\prime }/x=v^{\prime \prime }/x$, so $-v^{\prime \prime }=\lambda v$. Finiteness of $u$ at zero requires $v(0)=0$, and the other endpoint gives $v(\pi /2)=0$. The Dirichlet eigenvalues are positive: integration by parts gives $\lambda \int |v{|}^2=\int |v^{\prime }{|}^2$, and a nonzero $v$ cannot have zero right-hand side. Thus $v=C\sin (\sqrt{\lambda }x)$, and the second endpoint givesDefine $u_n(0)=2n$ by its limit. This is the radial Dirichlet spectrum in three dimensions with radius $\pi /2$. Its weighted squared norm is ${\int }_0^{\pi /2}{x}^2{u}_n^{2}dx=\pi /4$.

Multiplication by $x^2$ gives the Sturm-Liouville formso the weight is $w(x)=x^2$. Use the weighted inner product $⟨f,g{⟩}_w={\int }_0^{\pi /2}{x}^2\overline{f}gdx$ and the Sturm-Liouville operator $L=-x^{-2}(x^{2}d/dx{)}^{\prime }$. Integration by parts gives its boundary form:The upper endpoint term vanishes by the Dirichlet boundary conditions, and the lower one vanishes for the regular solutions. More precisely, the radial Dirichlet spectrum in three dimensions identifies $L$ under the unitary transformation $v=xu$ with $-v^{\prime \prime }$ on $(0,\pi /2)$ with zero endpoint values; this specifies the self-adjoint realization at the singular endpoint as well as proving the formal identity.

For eigenfunctions $u_a,u_b$ with different eigenvalues, the same identity gives $({\lambda }_b-{\lambda }_a)⟨u_a,u_b{⟩}_w=0$. Taking real eigenfunctions proves the required orthogonality of Sturm-Liouville eigenfunctions.

Write $\Delta V=V_2-V_1>0$ and ${\gamma }_i=(1-v_i^2{)}^{-1/2}$, the initial Lorentz factor. The conserved energy from part (a) is $E=m{\gamma }_i+V_1$. To emerge on the higher plateau with positive speed, the particle needs $E>m+V_2$. HenceAbove this threshold, conservation of energy gives ${\gamma }_f={\gamma }_i-\Delta V/m>1$. The transmitted velocity is positive, soBelow threshold there is a turning point within the steep transition, where $\dot{x}=0$ and $V=E-m<V_2$. The particle returns to the lower plateau. Its final potential energy equals its initial value, so its final Lorentz factor equals ${\gamma }_i$ but its direction is reversed:At exact threshold the limiting speed on the upper plateau is zero. This is the marginal case, not passage with positive outgoing speed. The ideal discontinuous potential does not specify the detailed trajectory at equality; a smooth transition can make the particle approach the top asymptotically. The final-speed formulas above cover the strictly transmitted and reflected regimes requested.

The generalized momentum of this Lagrangian isThe Euler-Lagrange equation therefore givesBecause $\mathcal{L}$ has no explicit time dependence, the energy form of the Beltrami identity gives a conserved quantity $E=\dot{x}\partial \mathcal{L}/\partial \dot{x}-\mathcal{L}$. HereFor a direct check of conservation of energy, differentiate:This is the rest-plus-kinetic energy of a relativistic particle, in units with light speed one, plus its potential energy.

In the shifted variable $t=y-1$, substitute the Frobenius solution $x={\sum }_{j\ge 0}{a}_j{t}^{j+\sigma }$ into $t(t+2)x^{\prime \prime }+x^{\prime }+x=0$. Equating the coefficient of $t^{j+\sigma }$ givesThus the linear recurrence relation, valid for $j=0,1,\dots$, isFor the indicial root $\sigma =0$, this becomesFor the indicial root $\sigma =1/2$, rename the coefficients $b_j$; thenWith $a_0=b_0=1$, the first terms of the two Frobenius solutions are

Set $t=y-1$. Then $y^2-1=t(t+2)$, and a Frobenius solution has the formIn the differential equation, the lowest power is $t^{\sigma -1}$, whose coefficient isThe indicial equation and its indicial roots are consequentlyTheir difference is not an integer, so the Frobenius method produces two independent series without a logarithmic term:Choose a branch of $(y-1{)}^{1/2}$, or take $y>1$ for real solutions. A general local solution is $C_1{x}_1+C_2{x}_2$ by linear independence and the superposition principle. The analytic factors converge at least for $|y-1|<2$, up to the next singular point at $y=-1$.