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Let ${\varphi }_1,{\varphi }_2$ be two sufficiently regular solutions with the same source and boundary data, and set $w={\varphi }_1-{\varphi }_2$. Then ${\nabla }^{2}w=0$ and $\alpha {\partial }_{n}w+w=0$. Green's first identity givesThe left side is nonnegative, so it vanishes. Thus $w$ is constant on each connected component. Its normal derivative is zero, and the boundary condition then forces that constant to be zero. This proves at most one solution. The Poisson uniqueness with a nonnegative Robin normal coefficient proof never divides by $\alpha$, so it includes the points or boundary portions where $\alpha =0$.

For the exponential-sine modes, direct differentiation givesTheir sum is zero, so both modes are harmonic functions for every real $\ell$.

For the square, the horizontal Dirichlet data suggest separation of variables in the form $\varphi (x,y)=F(x)\sin (ky)$, where the positive integer $k$ makes the sine vanish at both horizontal edges. The Laplace equation gives $F^{\prime \prime }-k^{2}F=0$, hence $F=A\cosh (kx)+B\sinh (kx)$. At $x=0$, the outward normal points in the negative $x$ direction. The Robin boundary condition is therefore $F(0)-F^{\prime }(0)=0$, or $A=kB$. The right edge requires $F(\pi )=1$, fixing the remaining coefficient. The resulting separated Laplace mode with a Robin edge isThe denominator is positive for $k>0$. The function satisfies the Laplace equation and all four boundary conditions directly; for example its left-edge value and $x$ derivative agree, giving the correct outward-normal Robin sign. It extends regularly to the corners.

On the left edge the uniqueness coefficient is $\alpha =1$; on the other three edges it is $\alpha =0$ with the prescribed Dirichlet value. The same Green's first identity argument applies to the square's piecewise smooth boundary, whose corners have zero boundary measure. Thus this is the unique regular solution, or equivalently the unique finite-energy solution with these boundary traces. Unbounded corner singularities are outside that boundary-value class.

Use separation of variables in Laplace's equation: writing $\varphi =R(r)\Theta (\theta )$ givesSingle-valuedness requires $2\pi$-periodicity, so the nonconstant angular Fourier modes have integer $n\ge 1$ and are $\cos n\theta ,\sin n\theta$. Their radial solutions are $r^n,r^{-n}$. The zero angular mode has radial solutions $1,\log r$. These give the harmonic Fourier expansions in planar concentric domains used below.

For the annulus with $0<a<b$, constant boundary data suggest a radial harmonic function $A+B\log r$. Enforcing both boundary values givesThis has the required boundary values and satisfies $(r{\varphi }_r{)}_r=0$. It is unique: the difference of two continuous solutions is harmonic in the annulus and zero on both boundary circles, so the maximum principle for harmonic functions, applied to the difference and its negative, makes it zero.

For admissible initial data, for example a square-integrable function on the interval, expand in the sine eigenfunctions of the Dirichlet Laplacian:The expansion follows either from separation of variables or the complete sine orthogonal basis. If $\kappa <{\pi }^2$, every growth rate is negative and Parseval's identity givesThus all finite-energy data decay. The convergence is also uniform in $x$ after any fixed positive time: apply Cauchy-Schwarz inequality to the sine series, using the summable factors $e^{-2(j^2-1){\pi }^{2}t}$, to bound its supremum by a constant times $e^{-({\pi }^2-\kappa )t}$ for $t\ge t_0>0$.

Conversely choose $u(x,0)=\sin \pi x$. Then $u(x,t)=e^{(\kappa -{\pi }^2)t}\sin \pi x$, which is stationary at equality and grows when $\kappa >{\pi }^2$. All initial conditions decay if and only if $\kappa <{\pi }^2$. This is the continuous part of the reaction-diffusion spectral decay threshold.

Use separation of variables, $\varphi =R(r)\Theta (\theta )$. After multiplying the Laplace equation by $r^2/(R\Theta )$, the two separated equations areSingle-valuedness requires $2\pi$-periodicity. Thus the nonconstant angular modes have $\lambda =n^2$, $n=1,2,\dots$, and angular factors $\cos n\theta ,\sin n\theta$. Their radial factors are $r^n,r^{-n}$. The zero mode has constant angular factor and radial solution $A+B\log r$; a term proportional to $\theta$ would be multivalued.

Finiteness at the origin removes the logarithm and negative powers. In the disk the general regular harmonic expansion isFor the exterior, regularity at every finite radius allowsCoefficients must give convergent harmonic expansions on the domain and the intended boundary regularity at $r=a$. If “finite” is intended to mean bounded as $r\to \infty$, then $D_0=C_n=E_n=0$ and only the constant and decaying modes remain. Infinity is not included in the printed interval; moreover part (b)'s uniform background flow necessarily has an unbounded potential. Stating both interpretations avoids discarding that background mode.

Irrotational flow means $\nabla \times \mathbf{u}=0$; incompressible flow means $\nabla \cdot \mathbf{u}=0$. The given streamfunction automatically satisfies the second condition, since ${\partial }_x{\psi }_y-{\partial }_y{\psi }_x=0$. For an oriented path with unit tangent $(dx/ds,dy/ds)$, choose its right-hand unit normal $\mathbf{n}=(dy/ds,-dx/ds)$. ThenThe opposite normal reverses the sign; the source's unspecified normal orientation must be chosen consistently. The volume flux between two streamlines is thus their streamfunction difference. A closed contour has zero net flux, expressing mass conservation for constant-density flow without a source inside.

For the channel, a rigid boundary has zero normal velocity and hence constant streamfunction. Choose zero on the lower wall. Uniform inflow at infinity must have horizontal velocity $-m$, so there $\psi =-my$. The upper wall then has constant $-m$. The unbroken left wall connects to it and also has constant $-m$. The jump by $m$ between the walls at the corner represents the withdrawn flux. Finally the vorticity is $v_x-u_y=-\Delta \psi$, so irrotationality gives $\Delta \psi =0$.

Write $\psi =-my+w$. Then $w$ vanishes at $y=0,1$ and at infinity, while $w(0,y)=-m(1-y)$ for $0<y<1$. Separation of variables gives the harmonic sine mode in a half-strip $e^{-n\pi x}\sin n\pi y$. The Fourier sine series coefficients of the left boundary areTherefore the corner sink in a semi-infinite channel isFor $x>0$ the series and its differentiated series converge on compact subsets, making the sum harmonic. At $x=0$, ${\sum }_{n\ge 1}\sin (n\pi y)/n=\pi (1-y)/2$ for $0<y<1$, so the left-wall value is exactly $-m$. The other wall values and far-field limit follow directly.

The plus sign printed in the final series is inconsistent with the printed boundary conditions. It would give $m-2my$ on the left wall, not $-m$. The negative sign above is required. As a local flux check, near the corner the wall values give $\psi \sim -2m\theta /\pi$; the radial velocity is $-2m/(\pi r)$ and its outward flux across a quarter-circle is $-m$, precisely withdrawal.

For the single-frequency response, use separation of variables with $u=R(r)\sin \omega t$. Set $k=\omega /c$ and $V=rR$; then $V^{\prime \prime }+k^{2}V=0$, so $V=C\sin kr+D\cos kr$. A finite origin value forces $D=0$, since $\cos kr/r$ is singular. Since $\sin kr/r\to k$, the prescribed origin amplitude gives $C=A/k$. The regular time-harmonic spherical wave is thereforewith the quotient defined by its limit at zero.

This answers the literal finite-origin condition in the class of regular single-frequency responses. It is a standing combination of inward and outward waves: the numerator is proportional to $\cos (kr-\omega t)-\cos (kr+\omega t)$. A purely outgoing wave from an ideal point source would instead be proportional to $r^{-1}\sin [\omega (t-r/c)]$, which has no finite prescribed value $u(0,t)$. Such a source must be specified by a singularity strength, for example ${\lim }_{r\downarrow 0}ru$. The source's wording supplies neither an initial field nor an outgoing singular-source normalization; without restriction to the stated harmonic response, a general initial-value problem is not uniquely determined by the one origin condition.

In polar coordinates, the Laplace equation isFor a single-valued, $2\pi$-periodic harmonic function on an annulus, separation of variables and its angular Fourier series giveIndeed a nonzero angular mode has radial equation $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, with solutions $r^{\pm n}$; the zero mode has solutions 1 and $\log r$. This expression describes the periodic annular problem here; angular terms not periodic, such as $\theta$, are excluded.

Decay at the origin leaves only positive powers in the interior, and decay at infinity leaves only negative powers in the exterior. Continuity on the unit circle matches the corresponding Fourier coefficients. If the common cosine coefficient is $K_n$, the interior and exterior terms are $K_n{r}^n\cos n\theta$ and $K_n{r}^{-n}\cos n\theta$. Their exterior-minus-interior radial derivative at the circle is $-2n{K}_n\cos n\theta$. Thus harmonic matching across a circle with a derivative jump gives $K_2=-1/4$, $K_4=-1/8$, and zero for every other sine or cosine mode. ConsequentlyEach branch satisfies the Laplace equation and its respective decay condition. Both give $-\cos 2\theta /4-\cos 4\theta /8$ at $r=1$, and direct differentiation gives the required jump $\cos 2\theta +\cos 4\theta$. The same mode calculation leaves no nonzero homogeneous solution in this class.

The separated disk modes. Away from the plane of a razor-thin disk, there is no volume source, so the Newtonian gravitational potential obeys the Laplace equation in cylindrical coordinates. Separation of variables givesHere $k>0$ is a radial wavenumber, with units of inverse length. Regularity on the axis selects the Bessel function of the first kind $J_0(kR)$ rather than the singular second solution. Decay away from the disk, continuity across it and reflection symmetry select $Z=e^{-k|z|}$. These conditions apply to each nonzero-wavenumber mode; a uniform sheet is a separate zero-mode problem.

Integrate the Poisson equation for Newtonian gravity across the disk to obtain the jump conditionFor the unit-amplitude separated mode, the jump is $-2k{J}_0(kR)$. ThusAn individual basis density is signed; positive physical surface densities are built by superposition.

An arbitrary axisymmetric surface density. Define its order-zero Hankel transform byMatching the coefficients of the mode densities gives the Hankel representation of thin-disk gravity,The boundary conditions exclude added external fields. For sufficiently decaying surface density, this is the isolated-disk potential; for infinite scale-free disks, only potential differences or regularized forces need be finite.

The Mestel disk. Put $C={\Sigma }_0{R}_0$. The Bessel function integral ${\int }_0^{\infty }{J}_0(k{R}^{\prime })d{R}^{\prime }=1/k$, understood with a decaying regulator, gives $\hat{\Sigma }=C/k$. Differentiate the potential before removing the regulator and use $J_0^{\prime }=-J_1$:The last integral is $1/R$, since ${\int }_0^{\infty }{J}_1(u)du=[-J_0(u){]}_0^{\infty }=1$. Therefore the circular speed and enclosed mass areConsequently $v_c^2=GM(R)/R$. This equality follows from the specific Mestel disk calculation; the spherical shell theorem cannot be applied to general disks. Its total mass and absolute potential referenced to infinity diverge. A convenient finite midplane reference is $\Phi (R,0)=v_0^2\log (R/R_0)$.

The rotating distribution function. Use specific energy $E=(v_R^2+v_{\varphi }^2)/2+\Phi$ and $L_z=R{v}_{\varphi }$. For the printed positive exponential, take $\epsilon =-E$ and $s:=\sigma >0$; $s$ has units of squared velocity. If instead $\epsilon$ means ordinary total energy and $\sigma >0$, the velocity integral diverges, so that literal convention cannot define the requested galactic distribution function. Equivalently one can retain ordinary energy and choose the printed $\sigma$ negative.

With the binding-energy convention, write the amplitude as $A$ to distinguish it from the function:Integration over both radial velocities and positive azimuthal velocities givesThe half-line integral follows by setting $u=v_{\varphi }^2/(2s)$ in the defining Gamma function integral and requires $q>-1$. Matching its radial power and normalization to the Mestel disk yields the rotating Mestel disk distribution function parametersSince $v_0^2,s>0$, the required integrability condition is automatic. The amplitude uses the stated potential reference; changing the additive potential constant changes $A$ without changing the physical distribution.

The Fourier coefficients satisfy $a_0=a_n=0$ andThus the square wave has Fourier seriesIt equals $f$ away from its jump discontinuities, and converges to $0$ at the jumps. The assigned endpoint values do not affect its Fourier coefficients.

For the Poisson equation, use separation of variables $\varphi ={\sum }_{n\text{ odd}}{b}_n{R}_n(r)\sin (n\theta )$. The radial equation is $r^2{R}_n^{\prime \prime }+r{R}_n^{\prime }-n^2{R}_n=r^2$. Its bounded solution with zero Dirichlet boundary condition at $r=1$ isIndeed the negative power is excluded at the centre, and the boundary fixes the coefficient of $r^n$. No resonance at $n=2$ occurs because only odd modes are forced. ThereforeThis is a Poisson equation on a disk with angular forcing. The series converges uniformly on the closed disk: for $n\ge 3$ each term is bounded by a constant times $n^{-3}$. Its first spatial derivatives also converge uniformly. For instance the radial derivative and $r^{-1}$ times the angular derivative of each term are bounded by constants times $n^{-2}$; the $n=1$ harmonic term is smooth in Cartesian coordinates. It defines a function with zero boundary data and the required weak formulation. Applying the Laplacian to finite partial sums gives the Fourier partial sums of $f$, which converge in $L^2$; passing to the limit against smooth compactly supported test functions verifies the Poisson equation. It holds classically off the jump rays and the centre, where the forcing is locally constant. It cannot be interpreted as a global classical equation with continuous second derivatives for discontinuous forcing. Uniqueness follows from the zero-boundary Dirichlet energy identity for the difference of two solutions.

Finally orthogonality gives ${\int }_0^{2\pi }f(\theta )\sin (n\theta )d\theta =\pi b_n$. The radial integral isUniform convergence of $\varphi$ and boundedness of $f$ justify integration term by term, soThe negative sign also agrees with $\int \varphi \Delta \varphi =-\int |\nabla \varphi {|}^2$. The original PDF's radial denominator is $n^2-4$; the converted TeX's $n^2-1$ is a transcription error.

The logistic equation has the constant solutions $x=0$ and $x=1$. For a nonconstant solution, separation of variables givesSolving and imposing the initial value yieldsThis expression also includes the two equilibria when $x_0=0$ or $1$. For other initial values it is taken on the maximal interval containing $t=0$ on which its denominator does not vanish.

The phase portrait follows from the sign of $x(1-x)$: arrows point left for $x<0$, right for $0<x<1$, and left for $x>1$. Thus zero is an unstable equilibrium point and one is asymptotically stable. The requested sketch is shown below.

If $0<x_0<1$, the denominator is positive for every real $t$. The solution stays between the equilibria and increases monotonically, withMore precisely $x\sim [x_0/(1-x_0)]e^t$ in the remote past and $1-x\sim [(1-x_0)/x_0]e^{-t}$ in the remote future. Outside that interval there is a finite-time pole: for $x_0<0$ it occurs in the future, and for $x_0>1$ in the past, at $t_{\ast }=\log [(x_0-1)/x_0]$. This does not affect the globally defined trajectories requested between zero and one.

Seek a power series $y(x)={\sum }_{k\ge 0}{a}_k{x}^k$ in the Legendre differential equation. Equating coefficients of $x^k$ givessoThe even and odd coefficients form separate chains. Choose a nonzero seed of the same parity as $n$ and set the other seed to zero. In the chosen chain, none of the factors vanishes before $k=n$, and the factor $k-n$ vanishes at $k=n$. Therefore this chain terminates with a nonzero $a_n$: it gives a polynomial of degree exactly $n$.

To impose the normalization, its value at $x=1$ must be nonzero. Suppose a nonzero polynomial solution were $y=(x-1{)}^{m}q(x)$ with $m\ge 1$ and $q(1)\ne 0$. The lowest-order coefficient in $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }$ would be $-2m^{2}q(1)(x-1{)}^{m-1}$, whereas $n(n+1)y$ has order $m$. This cannot satisfy the equation. Thus divide by $y(1)$ to obtain the Legendre polynomial $P_n$ with $P_n(1)=1$. Substitution or the coefficient recurrence givesFor an axisymmetric solution of the Laplace equation regular at the two polar axes, the separated modes and their superposition in spherical polar coordinates areThis is the usual separation of variables expansion; the singular angular solutions are excluded by regularity at the axes.

Here ${\sin }^2\theta =\frac{2}{3}(P_0(\cos \theta )-P_2(\cos \theta ))$, so only degrees $0$ and $2$ are required. The degree-zero radial function $A_0+B_0/r$ takes the same value $2/3$ at $r=a,b$; since $a<b$, this forces $B_0=0$, $A_0=2/3$. Write the degree-two coefficient as $-(2/3)T(r)$, where $T=A{r}^2+B{r}^{-3}$ and $T(a)=T(b)=1$. Multiplying these boundary equations by $a^3,b^3$ and subtracting givesConsequentlyEach term is harmonic in the shell, and the radial factor is $1$ at both boundaries, so this satisfies both Dirichlet boundary conditions. The maximum principle for harmonic functions, applied to the difference of two solutions, proves uniqueness among solutions continuous on the closed shell.

During ejection, the remaining mass is $m(t)=m_o+m_w-Qt$, and the water is exhausted at $t_c=m_w/Q$. Consider the momentum of the octopus and the small portion of water expelled during $dt$. Initially it is $mu$. At the end of this interval, the remaining body has mass $m-Qdt$ and velocity $u+du$, while the expelled mass $Qdt$ has laboratory velocity $u-V$, to first order. Hence the total final momentum isThe external impulse from quadratic drag is $-\alpha u^{2}dt$. Equating this to the change in total momentum, and dividing by $dt$, proves the equation of motionThe outgoing water's momentum flux is essential: applying $d(mu)/dt$ to the remaining body alone would omit it.

For dimensional analysis, the independent physical inputs are $m_o,m_w,Q,V,\alpha$, with dimensionsThe scales $m_o$, $m_o/Q$ and $V{m}_o/Q$ fix units of mass, time and length. In these units, the only remaining independent dimensionless variables areFor example, with $s=Qt/m_o$ and $w=u/V$, the equation of motion becomes $(1+\lambda -s)dw/ds=1-\mu w^2$, with $w(0)=0$ and ejection ending at $s=\lambda$. Thus the dimensionless terminal value depends only on $\lambda ,\mu$, proving $u_c=Vf(\lambda ,\mu )$.

For its explicit value, put $m_0=m_o+m_w$ and $U=\sqrt{QV/\alpha }$. The initially stationary solution stays below $U$, since the acceleration vanishes at $U$ and is positive below it. Separation of variables givesThe two integrals yieldThis is constant mass-loss propulsion with quadratic drag. At $t_c$, the remaining mass is $m_o$, givingThus $f(\lambda ,\mu )={\mu }^{-1/2}\tanh (\sqrt{\mu }\log (1+\lambda ))$, verifying the required dimensional form. The limit as drag tends to zero is $V\log (1+\lambda )$, the ordinary rocket equation, while the positive-drag answer is below the terminal velocity $U$.

The axis-regular separation of variables expansion for an axisymmetric harmonic potential isThe radial powers solve $r^2{R}^{\prime \prime }+2r{R}^{\prime }-n(n+1)R=0$, while the regular angular solutions are Legendre polynomials. For an exterior solution tending to zero at infinity, retain only the negative radial powers. The boundary polynomial decomposes as $1+x^2=\frac{4}{3}{P}_0(x)+\frac{2}{3}{P}_2(x)$. ThereforeIt has the required boundary value and decay. Uniqueness follows from the maximum principle for harmonic functions for a harmonic difference, using its zero boundary values and vanishing limit at infinity.

The overdamped filament bending equation is linear, so sinusoidal forcing has a sinusoidal steady response at the same frequency. Write $h=h_0\mathrm{Re}[e^{i\omega t}F(\eta )]$ with $\eta =x/\ell$. Substitution givesUsing the elastohydrodynamic penetration length from the previous part reduces this ordinary differential equation toThe boundary conditions become $F(0)=1$, $F^{\prime \prime }(0)=0$, and decay of $F$ and its derivatives at infinity. The derivative condition expresses zero bending moment at the driven end, rather than a clamped slope. This separation of variables constructs the periodic response; an unspecified initial shape can add a transient solution of the homogeneous evolution problem.

For small transverse displacement, the wave equation on a string iswhere $T$ is the tension and $\rho$ is mass per unit length. Using separation of variables, put $y=X(x)F(t)$, the ratios $X^{\prime \prime }/X=F^{\prime \prime }/(c^{2}F)$ equal a constant. The fixed-end conditions $X(0)=X(L)=0$ exclude nontrivial solutions for nonnegative separation constants; for $-{\kappa }^2$ they give $X=\sin (\kappa x)$ and $\kappa =n\pi /L$, $n\ge 1$. The associated angular frequencies are ${\omega }_n=n\pi c/L$. Thus the general superposition of separated normal modes isThe string is initially straight and at rest before the impulse, so $A_n=0$. Assume the hammer interval lies inside $[0,L]$. The imposed velocity $g(x)=y_t(x,0)$ has Fourier sine series coefficientTherefore $B_n=g_n/{\omega }_n$, giving the amplitude of every excited normal mode.

The string energy is $\frac{\rho }{2}{\int }_0^L(y_t^2+c^2{y}_x^2)dx$. At the impulse it is $E_{\mathrm{total}}=\rho a{v}^2/2$. Orthogonality of the sine modes gives $E_n=\rho L{g}_n^2/4$ initially; each mode subsequently exchanges kinetic and elastic energy while retaining that total. Thus the modal energy of a localized velocity impulse on a string has fractionsFor $l=L/3$ and $a=L/10$, a mode is absent exactly when $\sin (n\pi /3)=0$ or $\sin (n\pi /20)=0$. Hence the unexcited modes are precisely

For the disc, put $u(r,\phi )=R_0(r)\Theta (\phi )$. The polar Laplace equation separates aswhere periodicity requires an integer $n\ge 0$. The radial ansatz $R_0=r^{\alpha }$ gives $\alpha =\pm n$; for $n=0$ the solutions are a constant and $\log r$. Regularity at the origin excludes negative powers and the logarithm.

Writing the boundary data as a Fourier serieswith $a_0=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_D$, $a_n={\pi }^{-1}{\int }_0^{2\pi }{u}_D\cos n\phi$, and the analogous formula for $b_n$, givesEquivalently it is the Poisson integral on the unit disk with kernelFor continuous boundary data this integral converges to the prescribed value at every boundary point, is smooth and harmonic inside, and uniqueness follows from the maximum principle for harmonic functions.

For the square, superpose four solutions, each carrying one side's data and zero data on the other three sides. Set $k_n=n\pi /a$ andSeparation of variables gives the explicit sumEach summand has zero Laplacian and the required zero values on three open sides. The Fourier sine series supplies the fourth side. For regular continuous data with matching values at the corners, the harmonic extension also takes the common corner values; termwise boundary evaluation of the sine series at the corners is not valid in general. For incompatible corner data there is no continuous solution on the closed square, but the displayed harmonic solution still attains the prescribed data on each open side.

For an axisymmetric Newtonian gravitational potential, the vacuum Laplace equation in cylindrical coordinates is $R^{-1}{\partial }_R(R{\partial }_R\Phi )+{\partial }_z^2\Phi =0$. Use separation of variables, $\Phi =X(R)Z(z)$. Dividing by $XZ$ and setting the separated constant to $k^2$, with $k>0$, givesThe first equation has solutions $e^{\pm kz}$; with $u=kR$, the second is the order-zero Bessel differential equation. Regularity at the axis selects the Bessel function of the first kind $J_0(kR)$, rather than the singular Bessel function of the second kind $Y_0$. ThusThere are also zero-separation-constant modes, and other boundary conditions can select other separated families. For a localized reflection-symmetric thin disk, decay away from the disk selects $e^{-k|z|}$ on the respective half-spaces.

Apply the divergence theorem to the Poisson equation for Newtonian gravity across a narrow pillbox enclosing disk area. Writing the volume mass density as $\Sigma (R)\delta (z)$ givesReflection symmetry makes ${\Phi }_z(R,0^{+})=2\pi G\Sigma (R)$. A superposition of the regular vacuum modes therefore satisfiesThe order-zero Hankel transform and its inversion now determine the Hankel representation of thin-disk gravity:The sign is fixed by the positive derivative of an attractive Newtonian gravitational potential above the disk. For ordinary finite disks the representation uses the isolated boundary condition, fixing any extra vacuum field or additive constant. Infinite scale-free disks require suitable regularization of the potential itself.