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Writing $r^2=x^2+y^2+z^2$, the gradient is $\mathbf{v}=2e^{r^2}(x,y,z)$. Its divergence, equivalently the Laplacian of $\varphi$, isOn the unit sphere this scalar is the constant $10e$. The lower hemisphere has surface area $2\pi$, soThis is a scalar surface integral of the divergence, not the electric- or vector-field flux integral. There is no normal component to insert.

Write $c=\cos \varphi$ and $s=\sin \varphi$. The needed coefficient derivatives are $c_x=s^2/r$, $s_x=-sc/r$, $c_y=-sc/r$ and $s_y=c^2/r$. Apply the product rule to both terms of $f_x=c{g}_r-(s/r)g_{\varphi }$ and use the chain rule for the derivatives of $g$. Collecting terms givesSimilarly, differentiating $f_y=s{g}_r+(c/r)g_{\varphi }$ givesOn addition, both the mixed partial derivative terms and the first angular derivative terms cancel. Using $s^2+c^2=1$, the Laplacian in polar coordinates is

Write the scalar field as $u=R(r)Y(\theta ,\varphi )$, where $R=r+r^{-2}$ and $Y=\sin \theta \cos \varphi$. Its radial contribution to the Laplacian in spherical coordinates isFor the angular contribution, $Y_{\theta }=\cos \theta \cos \varphi$ and $Y_{\varphi \varphi }=-\sin \theta \cos \varphi$, soCombining the radial and angular terms provesThe calculation uses $0<\theta <\pi$ to avoid coordinate singularities. In Cartesian coordinates, $u=x(1+r^{-3})$, which extends smoothly across the polar axis for $r>0$. It therefore satisfies Laplace's equation there as well. The origin is a genuine singularity and is excluded.

In the Cartesian coordinate system, the Laplacian of a scalar field is the sum of its three unmixed second partial derivatives. Each squared coordinate here contributes one, soEquivalently $\nabla \psi =(x,y,z)$ and its divergence is three.

Put $q=x^{2}yz$. The chain rule and product rule giveSumming these partial derivatives gives the Laplacian

Apply the Laplacian in polar coordinates twice to the proposed separated function. For any real exponent $m$,so the biharmonic equation becomesIts four distinct roots are $m=-2,0,2,4$. The separated polar solutions of the biharmonic equation therefore give the complete radial family in this angular sector,Equivalently, its radial equation is a fourth-order Cauchy-Euler equation; four distinct powers supply four linearly independent solutions, so no logarithmic terms are required.

Interior. Continuity with value zero at the origin removes both $r^{-2}\cos 2\theta$ and $\cos 2\theta$. The latter depends on the approach direction even though it is bounded. The two boundary conditions at unit radius then give $C+D=0$ and $2C+4D=1$. ConsequentlyThis extends smoothly through the origin: $r^2\cos 2\theta =x^2-y^2$ and $r^4\cos 2\theta =(x^2+y^2)(x^2-y^2)$ are polynomials. Its radial derivative at $r=1$ is $(2r^3-r)\cos 2\theta =\cos 2\theta$.

Exterior. The condition on the radial derivative at infinity excludes the $r^2$ and $r^4$ terms. A nonzero $B\cos 2\theta$ is permitted, since the required decay concerns the radial derivative. Now $A+B=0$ and $-2A=1$, givingIndeed ${\partial }_r{\psi }_{\mathrm{out}}=r^{-3}\cos 2\theta \to 0$, and its value at unit radius is zero. These are particular interior and exterior solutions for the biharmonic streamfunction; no condition requiring $\psi$ itself to vanish at infinity was imposed.

For a separated harmonic function $R(r)\Theta (\theta )$ in polar coordinates, single-valuedness forces integer angular frequencies. The radial factors for frequency $m\ge 1$ are $r^m,r^{-m}$, while frequency zero has a constant and $\ln r$. Finiteness at the origin excludes $r^{-m}$ and $\ln r$.

The source's angular factor is $\cos \theta$ in every term, as printed, rather than $\cos (n\theta )$. The power-law angular mode for the planar Poisson equation givesHence, assuming the series permits termwise differentiation, the general finite, single-valued solution isThe arbitrary harmonic series must itself converge to a finite function on the prescribed disk. If ${\alpha }_0\ne 0$, the source has a direction-dependent limit at the origin; the displayed potential still extends finitely there, but the equation at that point is understood weakly rather than as a smooth classical source.

For the special source, a particular potential is $r^3\cos \theta /8$. Its boundary cosine term is canceled by $-a^{2}r\cos \theta /8$, and the boundary sine term is supplied by $(r/a{)}^2\sin 2\theta$. The resulting Dirichlet boundary conditions has solutionIt is finite at zero, has the required boundary values, and its Laplacian is $r\cos \theta$.

Away from the origin, $\hat{\mathbf{r}}=\nabla r$. Hence the curl of a gradient gives $\nabla \times \hat{\mathbf{r}}=0$. Its divergence is the Laplacian of the radial coordinate, which in spherical coordinates isEquivalently, differentiating $x_i/r$ in Cartesian coordinates gives $3/r-r^2/r^3=2/r$.

For $s=|\mathbf{x}-{\mathbf{x}}_0|>0$, the radial Laplacian gives ${\nabla }^2(1/s)=s^{-2}{\partial }_s(s^2{\partial }_s(1/s))=0$. The outward flux of $\nabla [-1/(4\pi s)]$ through a small sphere centred at the source is $[1/(4\pi s^2)]4\pi s^2=1$. Applying Green's second identity to a smooth test function outside that small sphere and shrinking its radius therefore gives the distributional identityThe kernel tends to zero at infinity. Thus it is the fundamental solution with the sign convention $\Delta G=+\delta$. The sign will reverse when converting it into the electrostatic potential of a positive point charge through Poisson's equation.

Away from its source, a radial two-dimensional Green's function satisfies $G^{\prime \prime }+G^{\prime }/r=0$, so $G=C\log r+C_0$. The divergence theorem on a disk of radius $\rho$ around the source requires $\int {\nabla }^{2}GdA=1$, hence $2\pi \rho G^{\prime }(\rho )=1$ and $C=1/(2\pi )$. Choosing the additive constant givesThe equality with the Dirac delta is distributional; away from the source the Laplacian vanishes.

For the wedge, use the finite image construction for a planar wedge. In complex coordinates $z_0=r_0{e}^{i{\theta }_0}$, place three positive sources at $z_0{e}^{2k\pi i/3}$ and three negative sources at $\overline{z_0}{e}^{2k\pi i/3}$, $k=0,1,2$. The positive source at $k=0$ is the actual charge. The five images are thereforeOn either grounded ray, reflection pairs each positive source with an equally distant negative source, so their logarithmic potentials cancel. All images lie outside the open physical wedge. Summing their contributions with coefficient $-e/(2\pi {\epsilon }_0)$ gives the required potential. The figure shows the bisecting-source configuration.

For $r_0=1$, ${\theta }_0=\pi /6$, the positive points are the three roots of $z^3=i$ and the negative points are the roots of $z^3=-i$. Thus the logarithmic sum isBoth boundaries have $\sin (3\theta )=0$, verifying the zero potential directly. An arbitrary reference length inside the individual logarithms cancels because the six signed charges sum to zero.

For a spherically symmetric number density, the radial Laplacian is $r^{-2}{\partial }_r(r^2{\partial }_{r}n)$. Applying separation of variables and choosing the separation constant $-k^2$ givesso the ordinary differential equations areThe radial problem is an eigenvalue problem for $-\Delta$ with Dirichlet boundary conditions, hence $k^2>0$: multiplying the radial equation by $r^2\overline{R}$ and integrating gives $k^2{\int }_0^a{r}^2|R{|}^{2}dr={\int }_0^a{r}^2|R^{\prime }{|}^{2}dr>0$ for a nonzero regular mode vanishing at $a$.

To solve it explicitly, set $R=f/r$, corresponding to the suggested power $p=-1$. The equation reduces to $f^{\prime \prime }+k^{2}f=0$, soFiniteness at the origin forces $B=0$, and the Dirichlet boundary condition at $r=a$ requires $k=m\pi /a$ for an integer $m\ge 1$. The absence of interior zeros excludes $m\ge 2$, selecting the fundamental normal mode. If $n_c>0$ denotes the initial central density, the result iswhere the sine ratio is defined as $1$ at the origin. It is strictly positive in the interior for every finite time. The arbitrary positive scale $n_c$ cannot be fixed without additional initial data.

The spatial factor is time independent, so the density and the total neutron population have the same exponential growth rate. Thus the spherical reaction-diffusion growth threshold isAt equality the fundamental mode is stationary; below it the mode decays. For fixed $\lambda >0$, the equivalent critical-radius condition is $a>\pi /\sqrt{\lambda }$. This is the competition between multiplication in the bulk and diffusion loss at the surface in the stated dimensionless model.

Since $\mathbf{V}=\nabla u$, its divergence is the Laplacian of the velocity potential. Using the Cauchy-Riemann equations and equality of mixed partial derivatives givesThus this is an incompressible flow, and $u$ is a harmonic function.

The time dependence gives $u_{tt}=-{\omega }^{2}u$, and the angular derivative vanishes for these radial normal modes. The Laplacian in polar coordinates therefore yieldsMultiplication by $r$ puts the Sturm-Liouville problem in the formThe relevant self-adjoint operator is $\mathcal{A}R=-r^{-1}(r{R}^{\prime }{)}^{\prime }$ in the weighted inner product with weight $r$. A physical displacement is regular at the origin, excluding the logarithmic singular solution. In particular $R$ is bounded and $r{R}^{\prime }\to 0$ there; regular eigenfunctions also have $R^{\prime }(0)=0$. This origin condition is needed in addition to the fixed outer boundary.

Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$, for which $\Delta =4{\partial }_z{\partial }_{\bar{z}}$. Since $H$ and $G$ are holomorphic,The Laplacian has real coefficients, so taking real and imaginary parts provesThis is the complex analytic representation of biharmonic functions; $F$ itself is generally not holomorphic because of its $\bar{z}$ factor.

Write $q=|r-r^{\prime }|$. Away from the source, the radial Laplacian gives ${\nabla }^2(-1/(4\pi q))=q^{-2}{\partial }_q[q^2/(4\pi q^2)]=0$. Through a small source-centred sphere, the outward flux of its gradient isThis identifies a unit Dirac delta at the source. More explicitly, for a smooth compactly supported test function $\psi$, apply Green's second identity outside the radius-$\epsilon$ ball. The inner outward normal of that punctured domain is $-e_q$. The term $H{\partial }_n\psi$ has integral $O(\epsilon )$, while $-\psi {\partial }_{n}H$ tends to $\psi (r^{\prime })$. Thus $\int H{\nabla }^2\psi dV=\psi (r^{\prime })$, proving the distributional identityWhenever the integral converges, convolution with this Green function of the Laplacian solves the Poisson equation for Newtonian gravity:The harmonic term represents the homogeneous freedom and is fixed by boundary conditions. For an isolated finite mass distribution, regularity away from the prescribed sources and the normalization $\Phi \to 0$ at infinity select $h=0$.

For a point mass at the origin, the delta integral gives the Newtonian potential of a point mass,

Only the quartic term contributes to the Laplacian. The mixed cubic has zero second derivative in each coordinate, while the exponential–trigonometric pairs are harmonic:Using $z=r\cos \alpha$ and the Jacobian determinant from part (a), the divergence theorem givesThe term containing $\sin \alpha {\cos }^2\alpha$ integrates to zero. Equivalently, the second vertical moment of a solid ring torus with major radius one is ${\int }_V{z}^{2}dV={\pi }^2{R}^4/2$.

Define the candidate potential by convolving the decaying Green function with the source. Acting with the Laplacian in the observation variable gives, in the sense of distributions,The final equality is the sifting property of the Dirac delta. The kernel found in part (ii) is locally integrable, and its convolution with a sufficiently regular rapidly decaying source also vanishes at spatial infinity. Thus it satisfies the Poisson equation and the required condition at infinity.

The difference from any other regular decaying solution is a harmonic function on all of space vanishing at infinity, hence is zero by the maximum principle. ThereforeOne can also obtain this expression from Green's second identity on larger and larger balls, with the singularity of the kernel treated distributionally. The surface term at infinity vanishes for the usual $1/r$ potential decay and $1/r^2$ gradient decay.

The derivative of an analytic function must be independent of the direction of approach. Taking a real increment gives $f^{\prime }(z)=u_x+i{v}_x$; taking an imaginary increment givesEquality of real and imaginary parts proves the Cauchy-Riemann equations:A real harmonic function is a twice continuously differentiable function whose Laplacian vanishes. For the given logarithmic function and $x^2+y^2>0$,It is therefore harmonic everywhere on the punctured complex plane.

On any domain admitting a branch of the complex logarithm, chooseIts derivatives are $v_x=-2y/(x^2+y^2)$ and $v_y=2x/(x^2+y^2)$, as required. For example, the principal complex logarithm gives such a harmonic conjugate on the plane cut along the negative real axis.

There is no single-valued harmonic conjugate on the whole punctured plane. Indeed its putative differential would satisfywhereas a globally defined function has zero integral of its differential around a closed curve. This is the log modulus has no global harmonic conjugate on the punctured plane obstruction. Thus the conjugate requested in the question must be interpreted locally or on a cut plane.

In Cartesian coordinates, write ${\partial }_i=\partial /\partial x_i$ and use the Levi-Civita symbol. The contraction of two Levi-Civita symbols givesHere the Kronecker delta selects the indicated indices, and continuous second partial derivatives allow their order to be interchanged. Identifying the gradient, divergence and componentwise Laplacian proves the curl of the curl identity:

Since $\mathbf{F}=\nabla \Omega$, its divergence is the Laplacian of $\Omega$:The divergence theorem changes the outward flux through the whole closed boundary into $I={\int }_V(6x+4z)dV$. Rotational symmetry makes the $x$ term vanish. With $z=(u^2-v^2)/2$ and the Jacobian determinant from part (c),Both the curved face and the top disk are included in this surface integral through the divergence theorem.