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Differentiating the spherical coordinate systemgives the orthogonal decompositionwhereThus the scale factors of orthogonal coordinates are

Since the total differential satisfiescomparison of coefficients gives the gradient

A function $f:D\to \mathbb{R}$ on a convex set $D$ is convex whenfor all $x,y\in D$ and $0\le t\le 1$.

If $f$ is once differentiable, this is equivalent to monotonicity of its gradient:Equivalently, every tangent hyperplane supports the graph:If $f$ is twice differentiable, convexity is equivalent to the Hessian matrix being positive semidefinite throughout $D$.

Forthe Hessian isA real symmetric two-by-two matrix is positive semidefinite exactly when its two diagonal entries and its determinant are nonnegative. ThusThe largest convexity domain is thereforeFor $A\ne 0$, its boundary is the hyperbola $y=A^2/(36x)$ in the first quadrant and the domain lies above it. For $A=0$, it is the closed first quadrant. This is the convexity domain of x cubed plus y cubed plus Axy.

Let the displayed integrand be $F$. Since it has no explicit dependence on $u$ or $v$, the Euler-Lagrange equations for two fields areIts derivatives areThusUsing the Laplacian, gradient, and divergence, these combine into the static Navier-Cauchy equation

Define the softmax weightsThey obey $p_i\ge 0$ and ${\sum }_i{p}_i=1$. The gradient is their weighted mean,Differentiating once more gives the covariance-form Hessian matrixFor every unit vector $u$,where $G={\max }_i\Vert a_i{\Vert }_2$. The Hessian is a covariance matrix, so it is positive semidefinite; the displayed upper bound also gives ${\nabla }^2{f}_{\beta }⪯\beta G^{2}I$ in the Loewner order. Consequently $f_{\beta }$ has a Lipschitz gradient with

The first-order subgradient inequality for a differentiable convex function givesand, after interchanging $v$ and $w$,Adding and rearranging yieldsThus the gradient $F=\nabla f$ is a monotone operator.

Using the harmless normalization $1/(2n)$, the squared-loss empirical risk for linear prediction iswith explicit gradientStarting from any ${\beta }_0\in C$, projected gradient descent with step sizes ${\eta }_k>0$ is

For a fully explicit update, write ${\tilde{\beta }}_{k+1}=(u_k,t_k)$ and ${\rho }_k=\Vert u_k{\Vert }_2$. The projection from part (e) isEvery iterate therefore lies in the prescribed hypothesis class, and any convergent run under the standard convex-optimization step-size conditions targets its empirical-risk minimizer.

The Stokes theorem states that for an oriented smooth surface $S$ with positively oriented boundary $\partial S$,A field is conservative vector field if its line integral is path-independent, equivalently if every closed line integral vanishes. If $F=\nabla \varphi$, the fundamental theorem for line integrals givesThis is zero for every closed curve, so every globally defined gradient field is conservative.

The first field isso its integral around $C$ is zero and it is conservative with potential $e^{x^2}\cos (yz)$.

Along $C$, $dr=(-\sin s,\cos s,0)ds$ andsoThus $B_2$ is not conservative and cannot be a gradient on the stated punctured region.

Finally,Its integral around $C$ is zero, and it is conservative with the displayed potential. Therefore the three integrals are respectivelyand exactly $B_1,B_3$ are conservative gradient fields.

Vanishing net charge density does not require vanishing current. Positive and negative charge carriers can cancel in charge density while their oppositely directed motions add to a nonzero current. Charge conservation only requiresfor magnetostatics this becomes $\nabla \cdot J=0$.

Because $\nabla \cdot B=0$, one may introduce a magnetic vector potential withIt is not unique: $A+\nabla \chi$ gives the same field for any scalar $\chi$.

For the stated current,so it is consistent with stationary charge conservation. Direct calculation givesThus $J$ is a Beltrami field. For $\lambda \ne 0$, chooseThen $\nabla \cdot B=0$ and $\nabla \times B={\mu }_{0}J$. A convenient Coulomb-gauge potential issince $\nabla \times A=B$ and $\nabla \cdot A=0$. Gradient gauge terms may of course be added.

If $\lambda =0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is

Newton's method for minimization usesFor a quantitative local bound, suppose on a convex neighbourhood containing the iterates thatwith $m>0$, and let $x^{\ast }$ be the minimizer. The integral form of the gradient and the Hessian Lipschitz bound give the quadratic convergence bound for Newton's methodIf $q=M\Vert x_0-x^{\ast }\Vert /(2m)<1$, induction yieldsSince $f(x)-f(x^{\ast })\le L\Vert x-x^{\ast }{\Vert }^2/2$,

Forthe Fréchet derivative in direction $h$ isThus the Hilbert-space gradient isThe gradient descent update with step size $\gamma$ is consequentlywhich is exactly Landweber iteration. Its stationary points solve the normal equation and minimize the convex quadratic functional $J$.

Take downward velocity as positive. Lubrication theory reduces the axial momentum equation to a constant pressure gradient, and the no-slip conditions are $u(0)=0$, $u(h)=U$. Twice integrating givesThe flux per unit circumference and the total gap flux areThe shear at the cylinder iswith the fluid traction on the falling cylinder opposing its motion.

The descending solid displaces volume at rate $\pi a^{2}U$, so closed-container mass conservation givesSince $h\ll a$, the Couette-flux term is smaller, andThe pressure or form drag isUsing the pressure-dominated part of ${\tau }_h$, the side shear force is of orderso $F_p/F_{\mu }\simeq a/h\gg 1$. Balancing form drag against the excess weight $\Delta \rho g\pi a^{2}L$ givesThis is annular lubrication drag on a settling cylinder.

The diffusive flux is $j=-D{C}_x$. The total outward flux through both ends is thereforeIt diverges as $t\downarrow 0$: the initially uniform profile is incompatible with the suddenly imposed zero boundary values, creating an idealized infinite initial gradient.

With streamwise and normal velocities $u,v$, zero exterior pressure gradient gives the steady boundary-layer equationsAt the organism,and matching to the quiescent exterior requires $u(x,y)\to 0$ as $y\to \infty$.

The gradient is $\nabla f(x)=Ax-b=A(x-x^{\ast })$. Completing the square givesfor $x\ne x^{\ast }$, so $x^{\ast }$ is the unique global minimizer. Since $x-x^{\ast }=A^{-1}\nabla f(x)$,

For $X={\theta }^{T}Z$,The objective has gradient $b-V\theta$ and Hessian $-V$, which is negative definite. Its unique maximizer is therefore

Pressure acts normally to the circular cylinder and so has no moment about its axis. To leading lubrication order, the tangential shear at the cylinder isSubstitution of the pressure gradient givesA torque-free cylinder requires ${\int }_{-\infty }^{\infty }{\tau }_{c}dx=0$. Also,Using $Q=2h_0(V+a\Omega )/3$, the torque condition reduces toHence

Every term in part (d) is a spatial gradient. Extracting the longitudinal scalar coefficient, up to a spatially homogeneous function that can be absorbed into the zero mode, yields the Cosmological Euler equation in synchronous gauge:

Yes. Synchronous gauge in cosmology uses freely falling time lines and sets the lapse and shift perturbations to zero, so no gravitational-potential force appears explicitly in this component of momentum conservation. Metric perturbations still affect the other field equations, the evolution of matter variables, and the relation between coordinate-dependent variables and gauge-invariant observables; synchronous gauge also retains residual gauge modes.

In Newtonian gauge in cosmology, the time-time metric perturbation is a Newtonian gravitational potential. Its spatial gradient therefore appears explicitly as a force term in the Euler equation. The difference is a coordinate representation of the same covariant conservation law.

Boundary-layer theory assumes steady incompressible high-Reynolds-number flow, $\delta /x\ll 1$, weak streamwise viscous diffusion, nearly constant pressure across the layer, and outer pressure gradient $-{\rho }^{-1}{p}_x=U{U}^{\prime }$.

For $\varphi =-A{r}^k\cos (k\theta )/k$,At $\theta =\pm \pi /k$ the normal velocity vanishes; along the upper wall $U(x)=A{x}^{k-1}$. Boundary-layer scaling givesSet $m=k-1$. The Falkner-Skan equation becomesso $\alpha =k-1$, $\beta =-k/2$, $\gamma =1-k$, with
$f(0)=f^{\prime }(0)=0$ and $f^{\prime }(\infty )=1$. Finallyrequires $x\gg (\nu /A{)}^{1/k}$; thus $a=1/k$ and the tip region violates the approximation.