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For a homogeneous isotropic solid with constant mass density $\rho >0$ and Lamé parameters $\lambda ,\mu$, the Navier-Cauchy equation isTaking the divergence, $d=\nabla \cdot u$ obeys $d_{tt}=c_P^2\Delta d$ with $c_P^2=(\lambda +2\mu )/\rho$. Taking the curl, $w=\nabla \times u$ obeys $w_{tt}=c_S^2\Delta w$ with $c_S^2=\mu /\rho$, because the curl of a gradient vanishes. These are the longitudinal P wave and transverse S wave speeds.

A suitable Helmholtz decomposition into a scalar-potential gradient and a divergence-free part separates the corresponding motions. In two-dimensional in-plane propagation, the P-SV displacement potentials implement this separation with two scalar wave equations. The constant-coefficient and isotropy assumptions matter: interfaces can couple P and SV modes, and arbitrary heterogeneous coefficients cannot be passed through the divergence and curl as if they were constants.

For a twice differentiable vector potential $\mathbf{B}$, commuting partial derivatives in $\mathbf{J}=\nabla \times \mathbf{B}$ gives the necessary conditionThe potential is not unique: for any sufficiently smooth scalar $\psi$, $\mathbf{B}+\nabla \psi$ has the same curl, since the curl of a gradient is zero. This is the usual gradient freedom of a vector potential. A divergence-free condition alone need not suffice globally on every domain; here only its necessity is requested.

Stokes theorem states that for an oriented surface with compatible boundary orientation,For the integrand in this problem, a convenient globally smooth vector potential isThe boundary of the upper ellipsoid is the ellipse $z=0$. The outward normal on the upper surface induces counterclockwise orientation viewed from above, so parameterize it as $\mathbf{r}(\theta )=(a\cos \theta ,b\sin \theta ,0)$, $0\le \theta \le 2\pi$. Along this curve the first component of $\mathbf{B}$ is zero and $dz=0$, leavingSince the last angular integral is $3\pi /4$,The result is independent of the vertical semiaxis $c$. As a sign check, the field is divergence-free, so closing the cap with the downward-oriented base disc makes its flux the positive integral of $x^2$ over that disc, which has the same value.

Integrate $f_x=y{e}^x$ with respect to $x$, obtaining $f=y{e}^x+g(y,z)$. Matching $f_y=e^x+2y{z}^3$ gives $g_y=2y{z}^3$, so $g=y^2{z}^3+h(z)$. Finally, $f_z=3y^2{z}^2+h^{\prime }(z)$ must equal the third component, so $h^{\prime }=0$. Thus a scalar potential isThe fundamental theorem for line integrals now gives ${\int }_{C}v\cdot dr=f(1,1,1)-f(0,0,0)=e+1$ for either path, confirming both direct computations. Since the curl of a gradient is zero for this smooth function, it also confirms the result in part (b). The potential is defined throughout ${\mathbb{R}}^3$, so no domain obstruction to path independence occurs.

Each component depends only on its own coordinate, so all cross partial derivatives vanish. Therefore the curl isThis also follows from the identity that the curl of a gradient vanishes for a twice continuously differentiable potential energy.

The curl of a gradient vanishes because the mixed second partial derivatives of the smooth function $\varphi$ agree. Therefore $\nabla \times \mathbf{v}=\nabla \times \nabla \varphi =0$ everywhere, andUnlike part (i), this integrand does not involve the problematic auxiliary field.

The curl of the first vector field iswhich is not identically zero. Thus $\mathbf{F}$ is not conservative on ${\mathbb{R}}^3$ and cannot equal a negative gradient there.

For the second field, direct differentiation showsTherefore $\mathbf{G}$ is conservative, and the requested negative-gradient potential isIts curl vanishes, consistently with the curl of a gradient identity.

Taking each partial derivative of the scalar potential givesThe factor $5y^4$ in the middle component comes from the chain rule. The potential is smooth, so the curl of a gradient identity gives $\boxed{\nabla \times \mathbf{F}=0}$.

Direct calculation of the curl givesSince $\mathbf{a}\ne 0$, the first vector field cannot be a gradient field: the curl of a gradient is zero. The second is a smooth polynomial field on all of the simply connected space ${\mathbb{R}}^3$, so its vanishing curl makes it a conservative vector field. A potential is also constructed explicitly in the next part. Thus ${\mathbf{v}}_2$ is a conservative vector field and ${\mathbf{v}}_1$ is not. A vanishing divergence alone, as for ${\mathbf{v}}_1$, is a different property and does not establish conservativity.

The potential is smooth everywhere, so its mixed partial derivatives commute. The curl of a gradient therefore vanishes:

For a twice continuously differentiable scalar potential, the curl of its gradient isEquality of mixed partial derivatives makes every component zero. ThusThis is the curl of a gradient identity.

The curl of a gradient is zero pointwise, so its surface integral over this sphere isFor this particular polynomial, $\nabla \Phi =2(x,y,z{)}^T$, whose curl vanishes directly as well.

The curl of a gradient is zero. Consequently the dot product in the exponent vanishes and

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 $3/r-r^2/r^3=2/r$.

In Cartesian coordinates, use the Levi-Civita symbol and the Einstein summation convention to expandThe contraction of the two symbols uses the Kronecker delta, and the final step is the product rule. Thus the convective acceleration identity isPut $\mathbf{\omega }=\nabla \times \mathbf{u}$. The curl of a gradient vanishes, and the curl of a cross product identity givesSince the divergence of a curl is zero, taking the curl of the first identity yieldsTwice continuous differentiability suffices for the mixed-derivative identities used here.

The divergence theorem states that a continuously differentiable vector field $\mathbf{F}$ on a volume $V$ with piecewise smooth closed boundary $S$ satisfieswhere $\hat{\mathbf{n}}$ is the outward unit normal. Take $\mathbf{F}=\nabla a\times \mathbf{B}$. Since the curl of a gradient is zero for a twice continuously differentiable scalar field, part (i) givesIntegrating this identity and applying the divergence theorem proves

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.

Use a velocity potential to reconstruct an irrotational flow from one scalar field. A velocity potential $\varphi (\mathbf{x},t)$ satisfiesIt has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is unchanged as a description of the velocity field when any function of time alone is added. For twice continuously differentiable $\varphi$, the identity curl of a gradient proves zero vorticity. In incompressible flow, taking the divergence also gives ${\nabla }^2\varphi =0$, the Laplace equation. This turns a three-component velocity field into a single harmonic function with appropriate boundary conditions. ^[ref]

For uniform flow $\mathbf{u}=(U,0,0)$, take $\varphi =Ux$. With $U=2\mathrm{m}{\mathrm{s}}^{-1}$, the change from $x=0$ to $x=3\mathrm{m}$ is $\Delta \varphi =6{\mathrm{m}}^2{\mathrm{s}}^{-1}$. By the fundamental theorem for line integrals, the line integral of velocity between these points equals that change along any path in the domain; a closed curve gives zero circulation.

For the potential line vortex, choose a local angle branch in polar coordinates and set $\varphi =K\theta$. Then $u_r={\partial }_r\varphi =0$ and $u_{\theta }=r^{-1}{\partial }_{\theta }\varphi =K/r$. One counterclockwise turn changes the continued velocity potential by $2\pi K$, although the velocity field returns to its original value. With $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$, this change is $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$, exactly the circulation. A full annulus therefore admits local angle branches and a multivalued velocity potential, but no global single-valued velocity potential when $K\ne 0$. Cutting the annulus along a radial line allows a single-valued angle branch on the remaining domain. The obstruction is path independence, not a failure of the local gradient formula.