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Chapter 7 • Theory & Derivations

Unit 7: Absolute Stability Analysis, Stiff Differential Equations & Linear Multi-Step Methods

Dahlquist test equation, stability regions in the complex plane, stiff differential equations, A-stability, L-stability, Padé rational approximations, explicit and implicit Linear Multi-Step Methods (Adams-Bashforth, Adams-Moulton, Milne-Simpson, BDF), characteristic polynomials rho(r), sigma(r), root condition, and Dahlquist's First and Second Stability Barriers.

§7.1 The Dahlquist Test Equation, Amplification Factors & Stability Regions in C

1. Germund Dahlquist's Test Problem

To analyze whether numerical solutions remain bounded as $t \to \infty$ for fixed step size $h > 0$, Germund Dahlquist (1956) introduced the scalar complex test equation:

$$\frac{dy}{dt} = \lambda y, \quad \lambda \in \mathbb{C}, \quad \text{Re}(\lambda) < 0$$

The exact solution is $y(t) = y_0 e^{\lambda t}$. Since $\text{Re}(\lambda) < 0$, the exact solution asymptotically decays to zero:

$$\lim_{t \to \infty} |y(t)| = 0$$

Applying a numerical method with step size $h$ yields a discrete recurrence relation:

$$y_{n+1} = R(z) y_n, \quad z = h\lambda \in \mathbb{C}$$

where $R(z): \mathbb{C} \to \mathbb{C}$ is the stability function (or amplification factor). After $n$ steps:

$$y_n = [R(z)]^n y_0$$
Definition: Region of Absolute Stability

The region of absolute stability $\mathcal{S} \subset \mathbb{C}$ is the set of complex parameters $z = h\lambda$ for which the numerical solution decays or remains bounded:

$$\mathcal{S} = \left\{ z \in \mathbb{C} : |R(z)| < 1 \right\}$$

Its boundary is the locus $\partial \mathcal{S} = \{ z \in \mathbb{C} : |R(z)| = 1 \}$.


2. Stability Functions for Classic Methods

1. Forward Euler:
$$y_{n+1} = y_n + h(\lambda y_n) = (1 + z) y_n \implies R(z) = 1 + z$$

Stability condition: $|1 + z| < 1$, which is an open disk of radius 1 centered at $z = -1$. On the negative real axis ($\lambda \in \mathbb{R}^-$), stable step sizes require:

$$-2 < h\lambda < 0 \implies h < \frac{2}{|\lambda|}$$
2. Backward Euler (Implicit):
$$y_{n+1} = y_n + h\lambda y_{n+1} \implies (1 - z)y_{n+1} = y_n \implies R(z) = \frac{1}{1 - z}$$

Stability condition: $|1 - z| > 1$, which is the exterior of the disk of radius 1 centered at $z = 1$. Notice that $\mathcal{S}$ includes the entire left half-plane $\mathbb{C}^- = \{z : \text{Re}(z) < 0\}$.

3. Implicit Trapezoidal Rule (Crank-Nicolson):
$$y_{n+1} = y_n + \frac{h\lambda}{2}(y_n + y_{n+1}) \implies R(z) = \frac{1 + z/2}{1 - z/2}$$

For $\text{Re}(z) < 0$, $|1 + z/2| < |1 - z/2|$, so $|R(z)| < 1$ holds for all $z \in \mathbb{C}^-$.

4. Explicit Runge-Kutta Methods of Order $p \le 4$:

For any explicit $p$-stage, $p$-th order RK method ($p = 1, 2, 3, 4$), the stability function is the truncated Taylor polynomial of $e^z$:

$$R(z) = \sum_{k=0}^p \frac{z^k}{k!} = 1 + z + \frac{z^2}{2!} + \dots + \frac{z^p}{p!}$$

For classical RK4:

$$R_{\text{RK4}}(z) = 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24}$$

The interval of absolute stability on the real axis is approximately $(-2.78, 0)$, and it contains a segment of the imaginary axis $[-2\sqrt{2}i, 2\sqrt{2}i]$, making RK4 capable of integrating undamped wave oscillators.

§7.2 Stiff Differential Equations, Stiffness Ratio, A-Stability & L-Stability

1. The Phenomenon of Stiffness

Consider a linear system of differential equations:

$$\frac{d\vec{y}}{dt} = A \vec{y}, \quad A \in \mathbb{R}^{d \times d}$$

with eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_d$ such that $\text{Re}(\lambda_i) < 0$.

Definition: Stiffness Ratio

The system is called stiff if:

  1. $\text{Re}(\lambda_i) < 0$ for all $i = 1, \dots, d$.
  2. The stiffness ratio:
$$S = \frac{\max_i |\text{Re}(\lambda_i)|}{\min_i |\text{Re}(\lambda_i)|} \gg 1 \quad (\text{typically } 10^3 \text{ to } 10^9)$$

Physical Reality: Stiff systems describe multi-scale dynamics—such as chemical kinetics, combustion, or structural mechanics—where fast transient modes decay almost instantaneously while slow master modes dominate the physical trajectory.

The Failure of Explicit Methods:

To keep explicit methods from experiencing exponential divergence, the step size $h$ is severely constrained by the fastest decaying eigenvalue:

$$h < \frac{C}{\max_i |\lambda_i|}$$

Even after the fast transient has decayed to zero ($10^{-10}$), an explicit integrator is forced to take millions of tiny steps purely to satisfy numerical stability, even though accuracy would permit a step size $10^6$ times larger!


2. A-Stability and L-Stability

To solve stiff problems efficiently, we require methods whose stability region encompasses the entire physics of decaying systems.

Definition 1: A-Stability (Dahlquist, 1963)

A numerical method is A-stable if its stability region contains the entire open left half-plane:

$$\mathbb{C}^- = \{ z \in \mathbb{C} : \text{Re}(z) < 0 \} \subset \mathcal{S}$$

For an A-stable method, any step size $h > 0$ produces bounded solutions, eliminating the stability bottleneck completely!

Definition 2: L-Stability (Stiff Decay)

While the Trapezoidal rule is A-stable, its stability function satisfies:

$$\lim_{\text{Re}(z) \to -\infty} R_{\text{Trap}}(z) = \lim_{z \to -\infty} \frac{1 + z/2}{1 - z/2} = -1$$

As a consequence, very stiff modes do not decay to zero; they oscillate wildly between $\pm y_0$!

A method is L-stable if:

  1. It is A-stable.
  2. Its stability function vanishes at infinity:
$$\lim_{z \to -\infty} |R(z)| = 0$$

Backward Euler is L-stable since $\lim_{z \to -\infty} \frac{1}{1 - z} = 0$. Stiff solvers in production (e.g., Radau IIA, BDF) are L-stable, ensuring that stiff transients are damped out instantly.

§7.3 Linear Multi-Step Methods (LMMs): Adams Families & Predictor-Corrector Schemes

1. General Linear Multi-Step Framework

A linear $k$-step method uses numerical values and function derivatives from the $k$ preceding points $(y_{n+j}, f_{n+j})$ to compute $y_{n+k}$:

$$\sum_{j=0}^k \alpha_j y_{n+j} = h \sum_{j=0}^k \beta_j f_{n+j}$$

where $\alpha_k \ne 0$ and $|\alpha_0| + |\beta_0| > 0$. By standard convention, normalize $\alpha_k = 1$.

  • If $\beta_k = 0$, the method is explicit ($y_{n+k}$ appears only on the left).
  • If $\beta_k \ne 0$, the method is implicit ($y_{n+k}$ appears on both sides inside $f(t_{n+k}, y_{n+k})$).

2. The Adams Family of Integrators

Integrating $y' = f(t, y)$ from $t_{n+k-1}$ to $t_{n+k}$:

$$y(t_{n+k}) = y(t_{n+k-1}) + \int_{t_{n+k-1}}^{t_{n+k}} f(t, y(t)) \, dt$$

Replace $f(t, y(t))$ by an interpolating polynomial $P(t)$:

1. Adams-Bashforth Methods (Explicit):

$P(t)$ interpolates past values $\{f_{n}, f_{n+1}, \dots, f_{n+k-1}\}$ at $k$ points:

  • AB1 (Euler): $y_{n+1} = y_n + h f_n$
  • AB2: $y_{n+2} = y_{n+1} + \frac{h}{2} (3f_{n+1} - f_n)$
  • AB3: $y_{n+3} = y_{n+2} + \frac{h}{12} (23f_{n+2} - 16f_{n+1} + 5f_n)$
  • AB4: $y_{n+4} = y_{n+3} + \frac{h}{24} (55f_{n+3} - 59f_{n+2} + 37f_{n+1} - 9f_n)$
2. Adams-Moulton Methods (Implicit):

$P(t)$ interpolates $\{f_{n}, \dots, f_{n+k-1}, f_{n+k}\}$ at $k+1$ points:

  • AM1 (Backward Euler): $y_{n+1} = y_n + h f_{n+1}$
  • AM2 (Trapezoidal): $y_{n+1} = y_n + \frac{h}{2} (f_{n+1} + f_n)$
  • AM3: $y_{n+2} = y_{n+1} + \frac{h}{12} (5f_{n+2} + 8f_{n+1} - f_n)$
  • AM4: $y_{n+3} = y_{n+2} + \frac{h}{24} (9f_{n+3} + 19f_{n+2} - 5f_{n+1} + f_n)$

3. Predictor-Corrector Architecture (PECE)

To avoid solving nonlinear systems for implicit Adams-Moulton methods, modern solvers use a Predict-Evaluate-Correct-Evaluate (PECE) sequence:

1. P (Predict): Compute initial guess $y_{n+1}^{(0)}$ using explicit Adams-Bashforth:

$$y_{n+1}^{(0)} = y_n + \frac{h}{24}(55f_n - 59f_{n-1} + 37f_{n-2} - 9f_{n-3})$$

2. E (Evaluate): Evaluate $f_{n+1}^{(0)} = f(t_{n+1}, y_{n+1}^{(0)})$.

3. C (Correct): Refine using implicit Adams-Moulton:

$$y_{n+1} = y_n + \frac{h}{24}(9f_{n+1}^{(0)} + 19f_n - 5f_{n-1} + f_{n-2})$$

4. E (Evaluate): Final update $f_{n+1} = f(t_{n+1}, y_{n+1})$ for the next time step.

This achieves order 4 accuracy with only two function evaluations per step, compared to RK4's four evaluations!

§7.4 Characteristic Polynomials, Dahlquist Root Condition & Stability Barriers

1. Characteristic Polynomials

Associated with the linear multi-step method $\sum_{j=0}^k \alpha_j y_{n+j} = h \sum_{j=0}^k \beta_j f_{n+j}$ are the two characteristic polynomials:

$$\rho(r) = \sum_{j=0}^k \alpha_j r^j, \qquad \sigma(r) = \sum_{j=0}^k \beta_j r^j$$
Consistency Conditions:

A linear multi-step method is consistent if and only if:

$$\rho(1) = 0 \quad \text{and} \quad \rho'(1) = \sigma(1)$$

2. Zero-Stability and the Dahlquist Root Condition

Applying the method to the trivial differential equation $y' = 0$ yields the difference equation:

$$\sum_{j=0}^k \alpha_j y_{n+j} = 0$$

whose general solution is a linear combination of $r_m^n$, where $r_m$ are roots of $\rho(r) = 0$.

Definition (Dahlquist Root Condition / Zero-Stability): A linear multi-step method is zero-stable if all roots $r$ of the first characteristic polynomial $\rho(r) = 0$ satisfy:

  1. $|r| \le 1$ (all roots lie within or on the closed unit disk).
  2. Any root lying on the unit circle ($|r| = 1$) is simple (multiplicity 1).

Theorem (Dahlquist Equivalence Theorem): For any linear multi-step method:

$$\text{Convergence of order } p \iff \text{Consistency of order } p \;+\; \text{Zero-Stability}$$

3. The Dahlquist Stability Barriers

Germund Dahlquist established two profound mathematical bounds that dictate what numerical integrators can and cannot do:

Dahlquist's First Barrier (1956):

The order of convergence $p$ of a zero-stable $k$-step linear multi-step method cannot exceed:

$$p \le \begin{cases} k + 1 & \text{if } k \text{ is odd} \\ k + 2 & \text{if } k \text{ is even} \\ k & \text{if the method is explicit } (\beta_k = 0) \end{cases}$$
Dahlquist's Second Barrier (1963):
  1. No explicit linear multi-step method can be A-stable.
  2. An A-stable linear multi-step method cannot have order of accuracy exceeding $p = 2$.
  3. Among all second-order A-stable linear multi-step methods, the Implicit Trapezoidal Rule has the smallest asymptotic error constant:
$$C_3 = -\frac{1}{12}$$

This theorem proves that high-order A-stable methods cannot be achieved with linear multi-step formulas, motivating the development of Backward Differentiation Formulas (BDF) (which sacrifice A-stability for $A(\alpha)$-stability at orders $k \le 6$) and Implicit Runge-Kutta methods (which overcome the barrier by utilizing multiple internal stages).

Tiered Solved Practice Problems & Examination Proofs

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Algorithmic, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Concept Example 7.1: Absolute Stability Boundary Analysis for Euler vs. Trapezoidal Methods

Consider the test equation $y' = \lambda y$ where $\lambda < 0$ is a negative real eigenvalue. 1. Determine the maximum allowable step size $h_{\max}$ for Forward Euler, Backward Euler, and the Trapezoidal Rule when $\lambda = -50$. 2. If $h = 0.05$, compute the numerical factor $R(h\lambda)$ for each method and determine whether the solution decays, oscillates stably, or blows up to infinity.

Step 1: Maximum Allowable Step Size for $\lambda = -50$

1. Forward Euler:

Stability requires $|1 + h\lambda| < 1 \iff -2 < h\lambda < 0$. With $\lambda = -50$:

$$-2 < -50 h < 0 \implies h < \frac{2}{50} = 0.04$$

Thus, $h_{\max} = 0.04$.

2. Backward Euler:

Stability requires $|R(z)| = \left|\frac{1}{1 - h\lambda}\right| < 1 \iff |1 - h\lambda| > 1$. Since $\lambda = -50$, $1 - h(-50) = 1 + 50h > 1$ for all $h > 0$. Thus, Backward Euler is unconditionally stable ($h_{\max} = \infty$).

3. Trapezoidal Rule:

Stability requires $|R(z)| = \left|\frac{1 + h\lambda/2}{1 - h\lambda/2}\right| < 1 \iff |1 - 25h| < |1 + 25h|$. This holds strictly for all $h > 0$. Thus, the Trapezoidal Rule is unconditionally stable ($h_{\max} = \infty$).


Step 2: Analysis at $h = 0.05$ Here $z = h\lambda = 0.05 \times (-50) = -2.5$.

1. Forward Euler:

$$R(z) = 1 + z = 1 + (-2.5) = -1.5$$

Since $|R(z)| = 1.5 > 1$, the numerical solution is:

$$y_n = (-1.5)^n y_0$$

The solution diverges with alternating sign and blows up exponentially to $\pm \infty$!

2. Backward Euler:

$$R(z) = \frac{1}{1 - z} = \frac{1}{1 - (-2.5)} = \frac{1}{3.5} \approx 0.2857$$

Since $|R(z)| = 0.2857 < 1$, the solution decays monotonically to zero without oscillations.

3. Trapezoidal Rule:

$$R(z) = \frac{1 + z/2}{1 - z/2} = \frac{1 - 1.25}{1 + 1.25} = \frac{-0.25}{2.25} = -\frac{1}{9} \approx -0.1111$$

Since $|R(z)| = 0.1111 < 1$, the solution decays stably to zero, with small damped oscillations due to the negative sign.

Tier 2: Advanced Algorithmic Example 7.2: Derivation and Local Error of the 3-Step Adams-Bashforth Method

Derive the 3-step explicit Adams-Bashforth formula:

$$y_{n+3} = y_{n+2} + h \left( \beta_2 f_{n+2} + \beta_1 f_{n+1} + \beta_0 f_n \right)$$

by integrating the Newton backward difference interpolating polynomial for $f(t, y(t))$ over $[t_{n+2}, t_{n+3}]$. Determine the exact fractional coefficients $(\beta_0, \beta_1, \beta_2)$ and compute the leading local truncation error constant $C_4$.

Step 1: Integral Formulation Integrating $y' = f(t, y)$ from $t_{n+2}$ to $t_{n+3}$:

$$y(t_{n+3}) - y(t_{n+2}) = \int_{t_{n+2}}^{t_{n+3}} f(t, y(t)) \, dt$$

Introduce the change of variable $t = t_{n+2} + s h \implies dt = h \, ds$, where $s \in [0, 1]$.

Step 2: Interpolating Polynomial in Backward Differences The quadratic polynomial interpolating $f_{n+2}, f_{n+1}, f_n$ is:

$$P(t_{n+2} + sh) = f_{n+2} + s \nabla f_{n+2} + \frac{s(s+1)}{2} \nabla^2 f_{n+2}$$

where:

$$\nabla f_{n+2} = f_{n+2} - f_{n+1}$$
$$\nabla^2 f_{n+2} = f_{n+2} - 2f_{n+1} + f_n$$

Step 3: Integration Over $s \in [0, 1]$ Evaluating the integrals:

  1. $\int_0^1 1 \, ds = 1$
  2. $\int_0^1 s \, ds = \frac{1}{2}$
  3. $\int_0^1 \frac{s(s+1)}{2} \, ds = \frac{1}{2} \left[ \frac{s^3}{3} + \frac{s^2}{2} \right]_0^1 = \frac{1}{2} \left( \frac{1}{3} + \frac{1}{2} \right) = \frac{1}{2} \cdot \frac{5}{6} = \frac{5}{12}$

Thus:

$$\int_{t_{n+2}}^{t_{n+3}} P(t) \, dt = h \left[ f_{n+2} + \frac{1}{2} \nabla f_{n+2} + \frac{5}{12} \nabla^2 f_{n+2} \right]$$

Step 4: Expressing in Terms of Function Values Substitute backward differences:

$$\begin{aligned} \int_{t_{n+2}}^{t_{n+3}} P(t) \, dt &= h \left[ f_{n+2} + \frac{1}{2}(f_{n+2} - f_{n+1}) + \frac{5}{12}(f_{n+2} - 2f_{n+1} + f_n) \right] \\ &= h \left[ \left(1 + \frac{1}{2} + \frac{5}{12}\right) f_{n+2} + \left(-\frac{1}{2} - \frac{10}{12}\right) f_{n+1} + \frac{5}{12} f_n \right] \\ &= h \left[ \left(\frac{12 + 6 + 5}{12}\right) f_{n+2} - \left(\frac{6 + 10}{12}\right) f_{n+1} + \frac{5}{12} f_n \right] \\ &= \frac{h}{12} \left[ 23 f_{n+2} - 16 f_{n+1} + 5 f_n \right] \end{aligned}$$

Thus, $\beta_2 = \frac{23}{12}$, $\beta_1 = -\frac{16}{12} = -\frac{4}{3}$, $\beta_0 = \frac{5}{12}$.

Step 5: Local Truncation Error Constant The next term in the backward difference expansion is:

$$\frac{s(s+1)(s+2)}{6} \nabla^3 f_{n+2} \approx \frac{s^3 + 3s^2 + 2s}{6} h^3 y^{(4)}(\xi)$$

Integrating:

$$\int_0^1 \frac{s^3 + 3s^2 + 2s}{6} \, ds = \frac{1}{6} \left[ \frac{1}{4} + 1 + 1 \right] = \frac{1}{6} \cdot \frac{9}{4} = \frac{9}{24} = \frac{3}{8}$$

Multiplying by $h$ and adding to the order $h^3$ gives:

$$d_{n+3} = \frac{3}{8} h \cdot h^3 y^{(4)}(\xi) = \frac{3}{8} h^4 y^{(4)}(\xi) = \frac{9}{24} h^4 y^{(4)}$$

Local truncation error per step is $\tau = \frac{3}{8} h^3 y^{(4)}$, verifying third-order accuracy ($p = 3$).

Tier 3: Honors / Proof Challenge Example 7.3: Proof of Dahlquist's Root Condition and Second Stability Barrier
  1. For the general linear multi-step method $\sum_{j=0}^k \alpha_j y_{n+j} = h \sum_{j=0}^k \beta_j f_{n+j}$, prove that zero-stability requires that all roots of the first characteristic polynomial $\rho(r) = \sum_{j=0}^k \alpha_j r^j$ satisfy $|r| \le 1$, with any root on $|r| = 1$ being simple. 2. Using the conformal mapping $z = \frac{r+1}{r-1}$, prove Dahlquist's Second Barrier: an A-stable linear multi-step method cannot have order of accuracy $p > 2$.

Part 1: Necessity of the Root Condition for Zero-Stability

Consider the ODE $y' = 0$ with true solution $y(t) \equiv y_0$. The linear multi-step method reduces to the homogeneous linear recurrence:

$$\sum_{j=0}^k \alpha_j y_{n+j} = 0 \iff \rho(E) y_n = 0$$

where $E$ is the forward shift operator $E y_n = y_{n+1}$. The general solution to this difference equation is:

$$y_n = \sum_{m=1}^s P_m(n) r_m^n$$

where $r_1, \dots, r_s$ are the distinct roots of $\rho(r) = 0$, and $P_m(n)$ is a polynomial in $n$ of degree $(\mu_m - 1)$, with $\mu_m$ being the algebraic multiplicity of root $r_m$.

For the method to be stable, the numerical solution must remain bounded as $n \to \infty$ for any bounded initial perturbations:

  • Case 1: $|r_m| > 1$.

Then $|r_m^n| = |r_m|^n \to \infty$ exponentially as $n \to \infty$. A perturbation in the initial condition is amplified without bound. Hence, we must have $|r_m| \le 1$.

  • Case 2: $|r_m| = 1$ with multiplicity $\mu_m \ge 2$.

Then $P_m(n)$ contains terms proportional to $n, n^2, \dots, n^{\mu_m - 1}$. Consequently, $|y_n| \sim C n |r_m|^n = C n \to \infty$ as $n \to \infty$. Even though $|r_m| = 1$, the secular polynomial factor causes algebraic unbounded growth!

Therefore, boundedness requires:

  1. Every root satisfies $|r| \le 1$.
  2. Any root with $|r| = 1$ must have $\mu = 1$ (simple root).

This establishes Dahlquist's Root Condition. $\blacksquare$


Part 2: Proof of Dahlquist's Second Barrier ($p \le 2$ for A-stability)

Let the method have order $p \ge 1$. Then consistency requires $\rho(1) = 0$ and $\rho'(1) = \sigma(1) \ne 0$. The Dahlquist test equation $y' = \lambda y$ gives the recurrence:

$$(\rho - z \sigma)(E) y_n = 0, \quad z = h\lambda$$

The method is A-stable if and only if for all $z \in \mathbb{C}$ with $\text{Re}(z) < 0$, all roots of the characteristic equation:

$$\rho(r) - z \sigma(r) = 0$$

lie strictly inside the unit disk: $|r| < 1$.

Equivalently, this means the rational function:

$$z(r) = \frac{\rho(r)}{\sigma(r)}$$

maps the exterior of the unit disk $\{r \in \mathbb{C} : |r| > 1\}$ into the right half-plane $\{z \in \mathbb{C} : \text{Re}(z) \ge 0\}$.

Now apply the standard bilinear Möbius transformation from the unit disk to the left half-plane:

$$r = \frac{w + 1}{w - 1} \iff w = \frac{r + 1}{r - 1}$$

which maps the unit disk $|r| < 1$ to the left half-plane $\text{Re}(w) < 0$, and the unit circle $|r| = 1$ to the imaginary axis $\text{Re}(w) = 0$.

Under this transformation, the order conditions require matching the Taylor expansion of the logarithm:

$$z\left( \frac{w+1}{w-1} \right) = \ln\left( \frac{w+1}{w-1} \right) = 2 \left( \frac{1}{w} + \frac{1}{3 w^3} + \frac{1}{5 w^5} + \dots \right)$$

For A-stability, $z(w)$ must be a positive real function (Herglotz function), meaning $\text{Re}(z(w)) \ge 0$ whenever $\text{Re}(w) \ge 0$.

However, the continued fraction expansion of a positive real rational function can only match the series expansion of $\ln\left(\frac{w+1}{w-1}\right)$ up to the first term $\frac{2}{w}$. Matching the third-order term $\frac{2}{3 w^3}$ forces the rational function to have poles in the right half-plane, which strictly violates the positive real condition (A-stability)!

Therefore, no linear multi-step method can match the logarithmic expansion beyond degree 2 while remaining a positive real mapping. Consequently, the maximum order of any A-stable linear multi-step method is $p = 2$. $\blacksquare$