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For $0<t<1$, split the Green's function integral where the absolute value changes sign:The two terms inside the bracket are respectivelyAdding them yieldsThe affine term gives $x^{\prime \prime }-4x=t$, while both exponential terms are homogeneous solutions. The values at the joins are $x(0)=(3e^{-2}-1)/16$ and $x(1)=-(1+e^{-2})/16$, agreeing with parts (i) and (ii). Their first derivatives also match: $x^{\prime }(0)=(3e^{-2}-1)/8$ and $x^{\prime }(1)=(1+e^{-2})/8$. Thus the pieces introduce no unwanted Dirac delta distributions.

Let $G(k)=\tilde{g}(k)$. Since $g^{\prime }=-xg$, differentiation under the Fourier transform integral and integration by parts giveThus $G(k)=G(0)e^{-k^2/2}$. The Gaussian integral gives $G(0)=\sqrt{2\pi }$, proving the Fourier transform of a Gaussian formulaFor the higher derivatives, start with $r_0=1$. If $g^{(m)}=r_{m}g$, the product rule givesIf $r_m$ has degree $m$ and leading coefficient $(-1{)}^m$, the recurrence gives degree $m+1$ and leading coefficient $(-1{)}^{m+1}$. Induction proves the assertion for every $m$. These polynomials are $r_m=(-1{)}^m{\mathrm{He}}_m$, where ${\mathrm{He}}_m$ is the Probabilists' Hermite polynomial.

Assume the necessary derivatives decay enough for all boundary terms to vanish. One integration by parts givesRepeating proves the Fourier transform of a derivative rule $\boxed{\widetilde{f^{(m)}}=(ik{)}^m\tilde{f}}$. On the other hand, differentiation under the integral givesand therefore $\boxed{\widetilde{x^{m}f}=i^m{\tilde{f}}^{(m)}}$.

Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$, with $a_n\ne 0$. Applying the second rule and part (a) giveswhereOnly the $j=n$ term can contribute degree $n$. Its leading coefficient is $\sqrt{2\pi }{a}_n(-i{)}^n\ne 0$, so $q_n$ has exactly the same degree $n$, with no possible leading-term cancellation. The coefficients of this transformed polynomial can be complex even when those of $p_n$ are real.

For an analytic function, the complex differentiability limit is independent of the direction of approach. A real increment gives $f^{\prime }(z)=u_x+i{v}_x$, whereas an imaginary increment givesEquality of real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if $u,v$ have continuous first partial derivatives near a point and obey these equations, their real differentiability gives, for a complex increment $h$,Dividing by $h$ proves the complex differentiability there. Continuous first partial derivatives are a sufficient regularity hypothesis for this converse; agreement along the two coordinate directions alone is only a necessary condition.

Using the complex cosine identity,we findIts real part is exactly the prescribed function. Thus a harmonic conjugate is $v=y\cos x\cosh y-x\sin x\sinh y$, and the general answer isTo check the only freedom, two analytic functions with this real part differ by a function with zero real part. The Cauchy-Riemann equations then force both partial derivatives of its imaginary part to vanish, so it is constant on the connected domain. No normalization was supplied to fix $C$.

Take the fixed source and receiver to be $A=(a,h_A)$ and $B=(b,h_B)$ on the same side of the mirror, with $h_A,h_B>0$. In the uniform medium, the Fermat principle minimizes the length through a reflection point $P=(s,0)$:Its stationary equation isThe two fractions are the signed tangential components of the ray directions. Equality of their magnitudes gives $\sin {\theta }_i=\sin {\theta }_r$, where both angles are measured from the normal vector to the mirror and lie in $[0,\pi /2)$. Hence the law of reflection isFor uniqueness, differentiate once more:Thus $L$ is a strictly convex function, its derivative increases from $-2$ to $2$, and it has exactly one stationary point, its global minimum. Explicitly,Substitution gives $(s_{\ast }-a)/h_A=(b-s_{\ast })/h_B$, verifying the reflection angles and the existence of that unique solution. Equivalently it is the point where the line from $A$ to the mirror image $(b,-h_B)$ crosses the mirror.

First expand the variational integrand:Its needed partial derivatives are ${\mathcal{L}}_f=3f^{\prime }+4f$ and ${\mathcal{L}}_{f^{\prime }}=2f^{\prime }+3f-x$. The Euler-Lagrange equation is thereforeThe general solution is $f=A\cosh (\sqrt{2}x)+B\sinh (\sqrt{2}x)-1/4$. Subtracting the two endpoint equations gives $B=0$; adding them gives $A=1/[4\cosh \sqrt{2}]$. HenceThis is the unique stationary solution. It is also the unique global minimizer: after discarding the fixed boundary terms, integration by parts rewrites $I$ as ${\int }_{-1}^1[f^{\prime 2}+2f^2+f-2x^2]dx$. For an endpoint-zero variation $\eta$, the linear part vanishes at the displayed solution andwith equality only for $\eta =0$.

Write the endpoint condition in its linear Robin boundary condition form $y^{\prime }(1)=y(1)/\alpha$. For a positive eigenvalue $\lambda =q^2$, the left Dirichlet boundary condition gives $y=C\sin (qx)$, and the right condition givesFor each positive root $q_n$, the squared norm isConsequently the sine coefficients in any complete expansion areWhen $\alpha >1$, these are all modes and the Sturm-Liouville eigenfunction expansion gives exactly the stated sine series, with the usual convergence interpretation. Positivity of all eigenvalues follows from integration by parts and the Cauchy-Schwarz inequality:for a nonzero admissible $y$, since $y(1)={\int }_0^1{y}^{\prime }dx$.

However, the original PDF states only $\alpha >0$, and with that range its sine-only completeness claim is false. The missing modes are as follows. At $\alpha =1$, $\lambda =0$ admits $y=x$, so the correct expansion isFor $0<\alpha <1$, a negative eigenvalue $\lambda =-{\kappa }^2$ admits $y=\sinh (\kappa x)$, withThere is exactly one positive root: $\tanh \kappa /\kappa$ decreases strictly from $1$ to zero. For example, its derivative has the sign of $\kappa {\mathrm{sech}}^2\kappa -\tanh \kappa <0$, because $\sinh \kappa \cosh \kappa >\kappa$. Thus the corrected expansion isThese are the nonpositive modes in a Dirichlet-Robin interval expansion. As an explicit counterexample to the printed claim, at $\alpha =1$ take $f(x)=x$: it satisfies both boundary conditions, and each positive root obeysbecause $\tan q_n=q_n$. Thus all $a_n$ vanish despite $f\ne 0$. Likewise $f=\sinh (\kappa x)$ is a counterexample for $\alpha <1$, by orthogonality of Sturm-Liouville eigenfunctions with distinct eigenvalues. The formulas above give the full solution without silently assuming away the omitted modes.

Choose the symmetric tensor fieldThe product rule and equality of mixed partial derivatives giveThis is the Euclidean scalar-gradient stress identity. Substitute it into part (e), and use ${\Sigma }_{ki}{\hat{n}}_k=({\partial }_n\psi ){\partial }_i\psi -\frac{1}{2}|\nabla \psi {|}^2{\hat{n}}_i$, where ${\partial }_n\psi =\hat{\mathbf{n}}\cdot \nabla \psi$. Reassembling the components gives the required pure surface integral:Sufficient smoothness of $\psi$ is understood for the displayed derivatives and applications of the divergence theorem.

For the zero-data initial value problem, choose a causal Green function satisfying $L_{x}G(x,\xi )=\delta (x-\xi )$ and $G=0$ for $x<\xi$. Continuity at $x=\xi$ prevents a derivative of the Dirac delta distribution, while integration across the source givesFor $x>\xi$, part (i) gives the shifted homogeneous solutions. The zero value at the join removes the cosine term, and the unit derivative jump fixes the sine coefficient. Therefore the causal Green function of a damped oscillator isEquivalently, multiply the first expression by the Heaviside step function $H(x-\xi )$. The required solution isIts initial value and first derivative vanish: the interval has zero length at $x=0$ and the kernel vanishes at its upper limit. Differentiating twice shows that the upper-limit contribution to $y^{\prime \prime }$ is precisely $g(x)$, while the remaining homogeneous kernel terms cancel in $Ly$.

For fixed initial and final coordinates, the two Euler-Lagrange equations areIndeed the first variation of the functional isIntegration by parts moves the time derivatives from the variations; their endpoint values are zero. The fundamental lemma of the calculus of variations then gives the two equations above.

The needed partial derivatives areSubstitution into the Euler-Lagrange equations givesThe ${\gamma }_1$ terms cancel in the first equation, leavingThese are accompanied by $x(0)=y(0)=0$ and $x(T)=y(T)=S$. The cancellation is consistent with part (b): the term proportional to ${\gamma }_1$ is the total derivative of $\alpha {\gamma }_1(x+\beta x^2/2)$.

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.

Use the bead angles ${\theta }_1,{\theta }_2,{\theta }_3$ as generalized coordinates. The straight-line spring length between two beads follows from the law of cosines:The hoop is horizontal, so the gravitational potential energy is constant and can be omitted. The kinetic energy is $(m/2){\sum }_i{\dot{\theta }}_i^2$, giving the LagrangianAt ${\theta }_i^0=\Theta +2\pi (i-1)/3$, every spring has its natural length, so the potential energy is zero and its first derivatives vanish. This is an equilibrium for every constant $\Theta$. It is stable against relative displacement because each spring energy is nonnegative; the overall rotation is a neutral direction.

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

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.

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.

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 $H(t)(e^{-2t}-e^{-3t})$.