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Two formal grammars over the same terminal alphabet are isomorphic when there is a bijection $\varphi :V_0\to V_1$ between their nonterminals such that $\varphi (S_0)=S_1$ andwhere $\varphi$ is extended to words by fixing each terminal and acting symbol by symbol.

Two formal grammars over the same terminal alphabet are equivalent when they generate the same language:

For the formal grammar $G=(\Sigma ,V,P,S)$,Thus $L(G)$ contains exactly the finite terminal words reachable from the start symbol by a finite grammar derivation.

Put $s={\gamma }^{-1/2}$. ThenFor an eigenvalue $\lambda$, the corresponding $2\times 2$ block has characteristic polynomialFor $\lambda =1$,so the polynomial isFor $\lambda =\gamma$, it isIts discriminant is $\beta (\beta -4)<0$, so its conjugate roots have product $\beta$ and modulus $\sqrt{\beta }$. Both blocks therefore have the same spectral radius, andThis is the heavy-ball rate for a two-eigenvalue diagonal quadratic.

By contrast, the steepest-descent factor isThe heavy-ball factor is $1-{\gamma }^{-1/2}$, which is substantially smaller for $\gamma \gg 1$: it gives an iteration scale of order $\sqrt{\gamma }$ rather than order $\gamma$.

Because $A{x}^{\ast }=b$, the error $e^{(k)}=x^{\ast }-x^{(k)}$ obeysThus the heavy-ball error propagation matrix is

If $A=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$, permute the coordinates fromtoFor the corresponding permutation matrix $P$,

Since $\nabla F(x^{(k)})=-r^{(k)}$, the Heavy-ball method isMultiplication by $A$ and use of $A{x}^{(j)}=b-r^{(j)}$ give the heavy-ball residual recurrenceWith the usual initialization $x^{(-1)}=x^{(0)}=0$, both initial residuals equal $b$. Induction in the recurrence shows thatfor a polynomial $p_k$ of degree at most $k$. ThereforeThis is ${\mathcal{K}}_k(A,b)$ under the zero-indexed Krylov subspace convention in the question; under the convention whose order-$j$ space ends at $A^{j-1}b$, it is ${\mathcal{K}}_{k+1}(A,b)$.

Forthe gradient is $\nabla F(x)=Ax-b=-r$. The steepest descent method with exact line search for a positive-definite quadratic therefore uses

For $A=\mathrm{diag}(1,\gamma )$ and $b=0$, the minimizer is $x^{\ast }=0$. PutStarting from $x^{(0)}=(\gamma ,1{)}^T$, direct substitution gives ${\alpha }_k=2/(\gamma +1)$ and, inductively,The Euclidean norm is unchanged by the alternating sign, soFor a positive-definite matrix, the spectral condition number of a positive-definite matrix isHere ${\kappa }_2(A)=\gamma$, and hence

The Conjugate gradient method starts with $p_0=r_0$ and updatesBy finite termination of the conjugate gradient method, it reaches the exact solution in at most the number of distinct eigenvalues. This matrix has two, so at mostare required in exact arithmetic.

The leading shear stress exerted by the fluid on the lower surface isThe horizontal force per unit span is ${\int }_0^L\tau dx$. Substituting the pressure gradient from part (c) givesUsing the value of $C$ and setting this force to zero yieldsConsequently the zero-shear lower wall in a translating and squeezing finite gap condition is

When $U=0$, putThe pressure gradient from part (c) reduces toBecause $h^{-3}>0$, $x_{\ast }$ is a weighted mean of points in $[0,L]$ and therefore belongs to $[0,L]$; it lies strictly inside unless the weight is singularly concentrated at an endpoint. Thus $p_x(x_{\ast })=0$, proving that the pressure gradient vanishes somewhere between the ends.

Integrating the incompressible continuity equation $u_x+v_y=0$ across the moving gap, with $v(x,0)=0$ and $v(x,h)=-V$, gives the Reynolds lubrication equationHencefor a constant $C$, and the flux formula from part (b) givesDefine the spatial meanBecause both ends meet fluid at pressure $p_0$, the pressure recovery condition in lubrication flow isIt determinesTherefore the requested pressure gradient in a translating and squeezing finite gap is

The leading streamwise momentum equation in lubrication theory iswith $p=p(x)$. The no-slip boundary condition isTwice integrating in $y$ gives the local Couette-Poiseuille flow in a thin gapIts volume flux per unit span is