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A Givens rotation ${\Omega }^{[p,q]}$ is the identity except in rows and columns $p,q$, where it has the blockFor $a=A_{pj}$ and $b=A_{qj}$, choosewhen $(a,b)\ne (0,0)$. ThenIf both entries vanish, any angle works.

For the given matrix, first useIt givesThen useThus $R={\Omega }^{[2,3]}{\Omega }^{[1,3]}A$ isAll leading row entries are positive. The resulting QR decomposition by Givens rotations iswhereand $R$ is the matrix above. Since it is a product of transposed rotations, $Q$ is orthogonal.

Let $x$ point downslope and $z$ point normally away from the plane, with $0\le z\le h$. Write the parallel velocity as $u(z)e_x$ and take $\tau >0$ to be the magnitude of the air's upslope stress. The steady equations arewith boundary conditionsThe free-surface condition makes $p_x=0$, and integration gives

The surface velocity, downslope shear exerted by the fluid on the plane, and volume flux per unit width areThus the inclined viscous film with opposing surface shear reverses in the three senses whenThe order of increasing required air stress is therefore

Differentiate under the integral and use ${\nabla }_x|x-x^{\prime }{|}^{-1}=-{\nabla }_{x^{\prime }}|x-x^{\prime }{|}^{-1}$:The omitted boundary term vanishes if the current is localized sufficiently rapidly. A steady current obeys charge conservation ${\nabla }^{\prime }\cdot J=0$, hence

For $r=|x|$ much larger than the source size,Localization and ${\nabla }^{\prime }\cdot J=0$ imply $\int J{d}^3{x}^{\prime }=0$. They also implyby integrating ${\partial }_k^{\prime }(x_i^{\prime }{x}_j^{\prime }{J}_k)$. Thus the first nonzero moment is antisymmetric and can be written using the magnetic dipole momentConsequently the Coulomb-gauge vector potential of a localized steady current has far fieldThe dimensions are

Continuity at $x=\xi$ is automatic in the proposed expression. Integrating the differential equation through $x=\xi$ gives the derivative jumpThusIn terms of the Wronskianone hasThe boundary conditions hold because $y_1^{\prime }(0)=0$ and $y_2^{\prime }(1)=0$.

The Abel identity gives $W^{\prime }=-\alpha W$. If $\alpha =0$, the Wronskian and hence $c$ are constant. The two branches of the displayed formula are then interchanged by $x\leftrightarrow \xi$, provingThis is the symmetry of the Neumann Green function for a second-order ordinary differential equation in the self-adjoint case.

For $G^{\prime \prime }-G=\delta (x-\xi )$, chooseTheir Wronskian isWriting $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$ givesThe solution of the inhomogeneous problem is $y(x)={\int }_0^{1}G(x,\xi )\xi d\xi$. Equivalently, solving directly givesThe two Neumann conditions yield $B=1$ andTherefore