With $d=\Delta x$, define the semidiscrete Fourier transform byFor finite-support sequences, expand the squared modulus and use orthogonality:ConsequentlyFor a general square-summable sequence, truncate it to $|j|\le N$. The identity makes the transformed truncations Cauchy in $L^2$, and completeness defines the transform as their $L^2$ limit. Passing to the limit proves the Parseval identityThe infinite series need not converge absolutely pointwise; the $L^2$ interpretation is sufficient for the stability analysis.

A method has numerical convergence if it approximates the exact solution uniformly at mesh points on every fixed finite interval as $h\to 0$, whenever its starting approximations converge to the corresponding exact starting values, for sufficiently smooth solutions of a well-posed initial-value problem.

The Dahlquist equivalence theorem states that a fixed-coefficient linear multistep method is convergent if and only if it is consistent and zero-stable, under the usual Lipschitz condition on the differential equation and convergent starting data. In this family numerical consistency forces the parameters found in part (a), and part (b) proves that choice is not zero-stable. All other parameter choices are inconsistent. Therefore no member of the family is convergent.

A zero-stable has a uniformly bounded response to small perturbations of its starting values and accumulated residuals on a fixed time interval as $h\to 0$. Equivalently, its first characteristic polynomial satisfies the root condition for a multistep method: every root has modulus at most one, and each root on the unit circle is simple.

For the consistent parameters, the recurrence for $y^{\prime }=0$ is $y_{n+3}-2y_{n+1}+y_n=0$, with characteristic polynomialLet $r=(-1-\sqrt{5})/2$. Choose $y_1=r$ and $y_2=r^2$; together with $y_0=1$ these generate exactly $y_n=r^n$. Since $|r|=(1+\sqrt{5})/2>3/2$,This parasitic mode violates the root condition and grows without a bound independent of the number of steps. Scaling its starting amplitudes gives the same unbounded amplification for arbitrarily small initial perturbations. Thus the only consistent method in the family is not zero-stable.

Substitute a smooth exact solution into the linear multistep method and expand about $x_n$. Its residual isConsistency of a numerical method requires the coefficients of $y$ and $h{y}^{\prime }$ to vanish. Therefore $1+a+b=0$ and $3+a=1$, giving the unique choiceFor this choice the next coefficient is $9/2-1-2=3/2\ne 0$, so the residual is $\frac{3}{2}{h}^2{y}^{\prime \prime }+O(h^3)$. The normalized local truncation error, obtained by dividing the residual by $h$, is $O(h)$. Hence the method has order one.

At the current exact path point $v(\mu )$, stationarity is already zero and $Cy=\mu e$. Therefore $F_{\bar{\mu }}(v(\mu ))=(0,(\mu -\bar{\mu })e{)}^T$. Its Newton correction solvesThe matrix has no explicit dependence on the barrier parameter, so comparison with part (iv) giveswhen $J$ is nonsingular. If it is singular, the displayed vector is still a solution of the correction equations, but uniqueness of the correction cannot be inferred from differentiability of the path alone.

This step is exactly the tangent predictor in the stated first-order Taylor expansion. For nearby positive $\bar{\mu }$, it approximates the new path point with error $o(|\bar{\mu }-\mu |)$; a bounded second path derivative improves this to $O((\bar{\mu }-\mu {)}^2)$. Continuity also preserves strict positivity for sufficiently small parameter changes. Larger changes may require damping to keep both the constraint values and multiplier estimates positive.

The Jacobian matrix of $F_{\mu }$ in $(x,y)$ isThe top-left block is the Hessian matrix of $f-y^{T}c$ with $y$ held fixed. In the second block, differentiating $C(x)y$ in $x$ gives $YA$ and in $y$ gives $C$. The explicit derivative with respect to $\mu$ is $(0,-e{)}^T$. Differentiate $F_{\mu }(v(\mu ))=0$ by the chain rule:All blocks are evaluated on the current central path point.

The equation $F_{\mu }(x,y)=0$, together with $c(x)>0$ and $y>0$, is the perturbed Karush-Kuhn-Tucker conditions: stationarity is unchanged, while exact complementary slackness becomes $c_i(x)y_i=\mu$ for each $i$. A finite limit of such points as $\mu \downarrow 0$ satisfies the original conditions by continuity.

For a strictly feasible $x$, the second block uniquely specifies $y_i=\mu /c_i(x)$. Substitution into the first block gives precisely $\nabla B_{\mu }(x)=0$. Thuswithin the stated strictly positive domain. These are stationary barrier points; they need not all be local minimizers in a nonconvex problem.

On the strictly feasible region $c_i(x)>0$, the logarithmic barrier objective isWith $y_i(x)=\mu /c_i(x)$, its gradient isThe positive coefficients $y_i$ are the barrier's multiplier estimates.

For $\mathcal{L}=f-y^{T}c$, the inequality-constrained Karush-Kuhn-Tucker conditions areUnder an appropriate constraint qualification, the second-order necessary condition is $d^{T}H(x_{\ast },y_{\ast })d\ge 0$ on the critical coneEquivalently active constraints with positive multiplier require $A_{i}d=0$, while active zero-multiplier constraints retain $A_{i}d\ge 0$. Inactive constraints impose no first-order tangent restriction. This distinction is why the relevant cone need not be the nullspace of every active row.

Stationarity of the equality-constrained quadratic model gives $g_k+H_k{s}_k-A_k^T{y}_{k+1}=0$, while feasibility gives $A_k{s}_k=-c_k$. A symmetric system linking the step and new multiplier isLet $Z$ have columns forming a basis of $\ker A_k$. The saddle-point matrix is nonsingular exactly when $A_k$ has full row rank and $Z^T{H}_{k}Z$ is nonsingular. Full row rank is necessary because a nonzero vector in $\ker A_k^T$ would give a null vector of the block matrix. Under full row rank, solving a homogeneous system forces $s=Zq$ and then $Z^T{H}_{k}Zq=0$; this proves sufficiency and the converse. Positive definiteness of the reduced Hessian is sufficient for a constrained minimum but is stronger than the nonsingularity condition asked here.

A null-space method first finds $s_p$ with $A_k{s}_p=-c_k$, then solvesand recovers $y_{k+1}$ from $A_k^T{y}_{k+1}=g_k+H_k{s}_k$. It needs the stated full row rank and nonsingular reduced matrix; Cholesky decomposition is available when the latter is positive definite.

A range-space method additionally assumes $H_k$ is invertible. It solves the Schur complement equationsIf $H_k$ is positive definite and $A_k$ has full row rank, the Schur matrix is positive definite and both solves can use Cholesky decomposition. An indefinite but invertible $H_k$ instead requires a suitable indefinite factorization; the Schur matrix must remain nonsingular.

The linearized constraints may be inconsistent, for example when $-c(x_k)$ is not in the image of a linear map defined by $A(x_k)$. Even if feasible, the quadratic objective may have negative curvature along its feasible tangent space, making it unbounded below and leaving no local minimizer.

Fletcher's Sℓ1QP replaces the hard linearized constraints by an L1 exact penalty and bounds the step inside a trust region. One usual form isThe zero step is always admissible. The trust region is compact, so a minimum exists even when the model Hessian matrix is indefinite; the constraint penalty measures infeasibility rather than making the subproblem impossible to solve. Adjusting its weight promotes feasibility, while the radius keeps the local model within a controlled neighbourhood. Introducing nonnegative positive- and negative-part slack variables converts the absolute values to an equivalent smooth quadratic program with box constraints.^[ref]

Complete the square in the augmented Lagrangian method objective:For fixed $u,\mu$, the last term is constant in $x$. Thus its minimization is the quadratic penalty method applied to the shifted constraint $c(x)=\mu u$.

Its practical advantage is that multiplier updates, typically $u^{+}=u-c(x)/\mu$, can drive the original constraint residual to zero while $\mu$ stays moderately sized. The plain quadratic penalty method generally needs $\mu$ very small to produce the same feasibility accuracy, exposing the conditioning problem in part (ii).

With $y=-c/\mu$, differentiation of the quadratic penalty method objective givesThe primal Newton system is thereforeFor bounded multipliers and regular derivatives, $H$ remains of ordinary size while ${\mu }^{-1}{A}^{T}A$ becomes large in directions normal to the constraints. If the feasible tangent space is nontrivial and retains ordinary-size curvature, this produces a large condition number as $\mu \downarrow 0$.

Introduce $w=Ap/\mu$. The same step is obtained from the symmetric augmented systemEliminating $w$ recovers the primal system exactly. This matrix contains no explicit ${\mu }^{-1}$ entries and tends to a saddle-point matrix. Under full row rank of the limiting constraint Jacobian matrix and nonsingularity of the limiting reduced Hessian, that limiting matrix is nonsingular, so this formulation avoids the artificial penalty-induced separation of scales.

These conditioning statements require regularity beyond bounded $y$ alone. A rank-deficient constraint Jacobian matrix or singular reduced Hessian can still make the augmented system ill-conditioned. If $m=n$ and $A$ is nonsingular, all directions may scale together in the primal system, so a large condition number is not automatic there either.

Use the optimization Lagrangian $\mathcal{L}(x,y)=f(x)-y^{T}c(x)$, with $A(x)=Dc(x)$ and $H(x,y)={\nabla }_{xx}^2\mathcal{L}(x,y)$. Under the usual linear independence constraint qualification, a local minimizer has multipliers $y_{\ast }$ satisfying the equality-constrained Karush-Kuhn-Tucker conditionsThe second-order necessary condition isOnly feasible tangent directions enter this last test; the Hessian matrix need not be positive semidefinite on the entire ambient space.

From $B_k{p}_k=-g_k$ and the matrix 2-norm bound, $\Vert g_k{\Vert }_2\le {\kappa }_b\Vert p_k{\Vert }_2$, so $\Vert p_k{\Vert }_2\ge \Vert g_k{\Vert }_2/{\kappa }_b$. Let $C=2\tau \epsilon (1-\beta )/\gamma >0$. The summand in the supplied decrease bound satisfiesThus, with the fixed positive constant $c=\beta \epsilon \min ({\alpha }_{\mathrm{init}}/{\kappa }_b,C)$,The nonnegative series therefore converges, forcing its terms to zero:The argument uses the given angle condition and upper bound on $\Vert B_k{\Vert }_2$; it does not need an additional uniform lower eigenvalue bound on $B_k$.

If the first trial is accepted, ${\alpha }_k={\alpha }_{\mathrm{init}}$. Otherwise the preceding trial was rejected and so exceeded ${\alpha }_{\max }$, since part (ii) guarantees acceptance below that threshold. The accepted step, reduced by the factor $\tau$, consequently exceeds $\tau {\alpha }_{\max }$. In either case,The angle condition $-g_k^T{p}_k\ge \epsilon \Vert g_k{\Vert }_2\Vert p_k{\Vert }_2$ gives ${\alpha }_{\max }\ge 2\epsilon (1-\beta )\Vert g_k{\Vert }_2/(\gamma \Vert p_k{\Vert }_2)$. Hence the backtracking line search has the lower bound

The quadratic model has gradient $g(x_k)+B_{k}p$ and positive-definite Hessian matrix $B_k$. Its unique minimizer is thereforeIf $g(x_k)\ne 0$, then $p_k\ne 0$ and $g(x_k{)}^T{p}_k=-p_k^T{B}_k{p}_k<0$, proving that it is a descent direction.

Backtracking line search starts with a trial step and repeatedly reduces it until the Armijo condition holds. Part (ii) guarantees finite termination because sufficiently small steps are acceptable. The test guarantees a specified positive fraction of the decrease predicted by the linear model, namely $f(x_k+\alpha p_k)\le f(x_k)+\alpha \beta g(x_k{)}^T{p}_k<f(x_k)$ for an accepted positive step. It does not require solving the line-minimization problem exactly.

The descent lemma follows by integrating the gradient along the line and using its Lipschitz continuity:This upper bound is at most the right side of the Armijo condition when $\alpha [(1-\beta )g^{T}p+\gamma \alpha \Vert p{\Vert }_2^2/2]\le 0$. Since $g^{T}p<0$, it holds forThe zero step gives equality; every positive step in this interval satisfies the prescribed sufficient-decrease test.

At an unconstrained local minimizer of a twice continuously differentiable objective function, the first-order optimality condition is $\boxed{\nabla f(x_{\ast })=0}$. The second-order necessary condition is $d^T{\nabla }^{2}f(x_{\ast })d\ge 0$ for every direction $d$, so the Hessian matrix is positive semidefinite.

For the Euclidean norm, the unit steepest descent direction is $-g/\Vert g{\Vert }_2$ when $g=\nabla f(x)\ne 0$; the algorithm usually uses the unnormalized direction $p=-g$. It is a descent direction because $\boxed{g^{T}p=-\Vert g{\Vert }_2^2<0}$.

Two limitations of unpreconditioned steepest descent method in high-accuracy smooth optimization are its slow zigzagging on ill-conditioned valleys and its sensitivity to scaling of the coordinates. It uses no curvature information to correct those directions, so it can require many function and gradient evaluations compared with a suitable Newton or quasi-Newton method. These are reasons to prefer curvature-aware methods in that setting, rather than a claim that gradient-based methods are never useful.

The standard conjugate gradient convergence bound isHere $\kappa \le 5$, so the factor in parentheses is at most $q=(3-\sqrt{5})/2\simeq 0.381966$. Since $2q^{25}<10^{-10}$, 25 iterations suffice to reduce the energy-norm error by a factor exceeding $10^{10}$.

There are at most $m/20$ nonzero entries of $t$, and each contributes to at most two off-diagonals. Thus $A-I$ has at most $m^2/10$ nonzero entries. Initialize a matrix-vector product with the diagonal contribution $p$, which costs no arithmetic because the diagonal is one. Each off-diagonal term needs one multiplication and one addition, giving at most $m^2/5$ floating-point operations per product.

A standard conjugate gradient step adds two inner products, three vector updates and two scalar divisions, at most $10m$ further operations. Starting from $x_0=0$ makes $r_0=b$ and avoids an initial matrix-vector product; its squared norm needs $2m-1$ operations. The total for 25 steps is at mostThis supplies an explicit meaning of the paper's large-$m$ qualification. The quoted conjugate gradient convergence bound is the usual exact-arithmetic result.