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Chapter 18: Problem 6

Exercise 18.3 .6 Interpret the output of the MATLAB program. $$ \begin{array}{l} \mathrm{A}=[0.770 .1 ; 0.0680 .9] \\ \mathrm{T} 1=[\exp (\mathrm{A}) \operatorname{expm}(\mathrm{A})] \\ \mathrm{T} 2=[\log (\mathrm{A}) \operatorname{logm}(\mathrm{A})] \\ \mathrm{T} 3=[\exp (\log (\mathrm{A})) \operatorname{expm}(\operatorname{logm}(\mathrm{A}))] \\ \mathrm{T} 4=\left[\log \left(\mathrm{A}^{-2}\right) 2 * \log (\mathrm{a})\right] \end{array} $$ \(\mathrm{T} 5=\left[\operatorname{logm}\left(\mathrm{A}^{-2}\right) 2 * \operatorname{logm}\left(\mathrm{A}^{-2}\right)\right]\) \(\mathrm{T} 6=[\exp (2 * \mathrm{~A}) \exp (\mathrm{A}) * \exp (\mathrm{A})]\) \(\mathrm{T} 7=[\operatorname{expm}(2 * \mathrm{~A}) \operatorname{expm}(\mathrm{A}) * \operatorname{expm}(\mathrm{A})]\) $$ \text { Output } $$ $$ \begin{array}{rrrrr} \mathrm{T} 1= & 2.1598 & 1.1052 & 2.1674 & 0.2309 \\ & 1.0704 & 2.4596 & 0.1570 & 2.4676 \\ \mathrm{~T} 2= & -0.2614 & -2.3026 & -0.2666 & 0.1204 \\ & -2.6882 & -0.1054 & 0.0819 & -0.1100 \\ \mathrm{~T} 3= & 0.7700 & 0.1000 & 0.7700 & 0.1000 \\ & 0.0680 & 0.9000 & 0.0680 & 0.9000 \\ \mathrm{~T} 4= & -0.5113 & -1.7898 & -0.5227 & -4.6052 \\ & -2.1754 & -0.2024 & -5.3765 & -0.2107 \\ \mathrm{~T} 5= & -0.5331 & 0.2408 & -0.5331 & 0.2408 \\ & 0.1637 & -0.2201 & 0.1637 & -0.2201 \\ \mathrm{~T} 6= & 4.6646 & 1.2214 & 5.8475 & 5.1052 \\ & 1.1457 & 6.0496 & 4.9444 & 7.2326 \\ \mathrm{~T} 7= & 4.7341 & 1.0703 & 4.7341 & 1.0703 \\ & 0.7278 & 6.1254 & 0.7278 & 6.1254 \end{array} $$

Short Answer

Expert verified
The MATLAB outputs show discrepancies mainly due to numerical precision and operational definitions of matrix functions.

Step by step solution

01

Analyze T1

T1 involves evaluating both \(\exp(A)\) and \(\operatorname{expm}(A)\). In MATLAB, \(\operatorname{expm}(A)\) computes the matrix exponential of \(A\). The output shows that both \(\exp(A)\) and \(\operatorname{expm}(A)\) yield similar structures, though they differ slightly, indicating numerical precision differences.
02

Analyze T2

T2 involves evaluating both \(\log(A)\) and \(\operatorname{logm}(A)\). The matrix \( A \) should have only positive eigenvalues for these operations to return real results. The discrepancy in results might be due to \(\log(A)\) not being well-defined for matrices, whereas \(\operatorname{logm}(A)\) handles matrix logarithms specifically.
03

Analyze T3

T3 involves computing \(\exp(\log(A))\) and \(\operatorname{expm}(\operatorname{logm}(A))\). Ideally, since \(\log(A)\) followed by \(\exp \) should return \( A \), both methods yield \( A \), confirming the consistency of the logarithm and exponential functions.
04

Analyze T4

T4 calculates \(\log(A^{-2})\) and multiplies twice the result of \(\log(a)\). Here \(\logm(A^{-2})\), a matrix operation, and \(2*\log(A)\) produce similar yet not identical results due to their definitions and potential mathematical nuances.
05

Analyze T5

T5 evaluates \(\operatorname{logm}(A^{-2})\) and compares it with \(2*\operatorname{logm}(A^{-2})\). The use of \(\operatorname{logm}(A^{-2})\) ensures consistency in method and results differ from simple multiplication due to complexity in matrix logs.
06

Analyze T6

T6 involves calculating \(\exp(2*A)\) and \(\exp(A)\cdot\exp(A)\). Although ideally these should be equal, numerical precision and computational methods can cause small deviations, as observed.
07

Analyze T7

T7 computes \(\operatorname{expm}(2*A)\) and checks it against \(\operatorname{expm}(A)\cdot\operatorname{expm}(A)\). The use of \(\operatorname{expm}\) handles matrix exponentiation more robustly, resulting in slightly different numerical discrepancies compared to \(T6\), aligning more closely than in \(T6\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Exponential
The matrix exponential is an extension of the scalar exponential function to matrices. In MATLAB, the function `expm()` is specifically designed to compute the matrix exponential. This operation is crucial when dealing with systems of linear differential equations.
The matrix exponential of an input matrix \( A \) is denoted by \( e^A \) and can be represented in MATLAB using `expm(A)`. This measure computes a new matrix that is not simply the element-wise exponential of \( A \); rather, it uses a series expansion similar to the exponential function for scalars:
  • Computing \( ext{expm}(A) \) involves power series expansions.
  • It can be calculated using techniques like Pade approximation.
  • In systems where dynamics are modeled, such as linear transformations, matrix exponentials help describe the continuous transition states.
The output of `expm(A)` in MATLAB is critically dependent on the underlying numerical algorithms, which might result in small differences due to rounding errors and approximations.
Matrix Logarithm
The matrix logarithm is the reverse of the matrix exponential and is used to find the logarithmic form of a matrix that could, when exponentiated, return to the original matrix, more formally expressed as \( A = ext{logm}(B) \) given \( B = ext{expm}(A) \). This operation requires that a matrix have positive eigenvalues to ensure real results, as negative eigenvalues can produce complex results.
The `logm()` function in MATLAB computes this logarithm and specializes in handling matrices, unlike the regular logarithm which doesn't extend well to matrices. Some key points include:
  • Computing matrix logarithms is more complex due to matrix properties like non-commutativity.
  • \( ext{logm}(A) \) can produce numerical differences due to the branch cuts in the complex logarithm.
  • It is primarily used in stability analysis, signal processing, and solving particular matrix equations.
Adjustments to numerical methods and careful handling of eigendata are needed to prevent errors when computing matrix logarithms.
Numerical Precision
Numerical precision refers to the degree of accuracy with which numerical data is expressed in computational systems like MATLAB. This is a key feature of many operations such as matrix exponentiation and logarithms. Even tiny discrepancies can arise from how MATLAB uses numbers stored in binary and floating-point representations.
The discrepancies observed in calculations like `expm(A) * expm(A)` versus `expm(2*A)` illustrate how numerical precision affects calculations:
  • The internal algorithms used by MATLAB, such as Pade approximations, impact results.
  • Factors such as floating-point arithmetic lead to variations when numbers are very large or small.
  • Ensuring precision involves controlling these errors, but minute inconsistencies can still occur.
Understanding limits of precision helps one design better algorithms and interpret results with an appreciation of underlying numerical constraints.
Eigenvalues
Eigenvalues hold a pivotal role when dealing with matrices in both theory and practical computations like those in MATLAB. An eigenvalue is a scalar indicating the factor by which associated eigenvectors are scaled during a linear transformation represented by a matrix.
During operations such as computing the matrix logarithm or inversions, knowing the eigenvalues of a matrix can dictate whether the operations yield real or complex results. Important points include:
  • Matrix operations like logarithms need positive eigenvalues for real outcomes.
  • Eigenvalues are solved from the characteristic equation \( ext{det}(A - ext{I} imes ext{eigenvalue}) = 0 \).
  • Dependence on eigenvalues can influence stability and convergence of iterative methods in computing matrix functions.
Spotting these values helps in understanding matrix behavior during transformations, making their calculation and analysis a common practice in advanced matrix analysis.

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Most popular questions from this chapter

Draw the nullclines and some direction arrows and analyze the equilibria of the following symbiosis models. $$ \begin{aligned} \text { a. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { b. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.8 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.5 y(t)) \\ \text { c. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-0.5 x(t)+0.4 y(t)) \\\ & y^{\prime}(t)=0.1 \times y(t) \times(1+0.4 x(t)-0.2 y(t)) \\ \text { d. } x^{\prime}(t) &=0.2 \times x(t) \times(1-5 x(t)+10 y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+2 x(t)-5 y(t)) \\ \text { e. } \quad & x^{\prime}(t)=0.2 \times x(t) \times(1-1.1 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \\ \text { f. } & x^{\prime}(t)=0.2 \times x(t) \times(1-0.9 x(t)+y(t)) \\ & y^{\prime}(t)=0.1 \times y(t) \times(1+x(t)-y(t)) \end{aligned} $$

Consider two interacting populations, \(x\) and \(y,\) that are mutually symbiotic: the presence of \(x\) enhances the growth of \(y\) and the presence of \(y\) enhances the growth of \(x\). A dynamic relation between \(x\) and \(y\) may take the form $$ \begin{aligned} x^{\prime}(t) &=r_{x} \times x(t) \times(1-a x(t)+b y(t)) \\ y^{\prime}(t) &=r_{y} \times y(t) \times(1+c x(t)-d y(t)) \end{aligned} $$ Describe the roles of the parameters \(a, b, c,\) and \(d\) in Equations \(18.63 .\)

Anderson and May \(^{4}\) give the following model of immune effector cells (helper and cytotoxic T-cells), \(E,\) that limit viral population, \(V\), growth in a human body. $$ \begin{array}{l} d E / d t=\Lambda-\mu E+\epsilon V E \\ d V / d t=r V-\sigma V E \end{array} $$ a. \(\Lambda\) is intrinsic production rate of effector cells from bone marrow. Give similar meaning to each of the other four terms on the RHS of Equations 18.64 . b. Find the equilibrium effector cell population, \(\hat{E},\) in the absence of virus. c. Suppose an inoculum \(V_{0}\) of virus is introduced into the body with \(E=\hat{E}\). Find conditions on \(r\), \(\sigma,\) and \(\hat{E}\) in order that the viral population will increase. d. The Jacobian matrix at any \((E, V)\) is $$ J(E, V)=\left[\begin{array}{rc} -\mu+\epsilon V & \epsilon E \\ -\sigma V & r-\sigma V \end{array}\right] $$ Show that $$ J(\hat{E}, 0)=\left[\begin{array}{rr} -\mu & \sigma \frac{\Lambda}{\mu} \\ 0 & r-\sigma \frac{\Lambda}{\mu} \end{array}\right] $$ e. The characteristic values of the upper diagonal matrix \(J(\hat{E}, 0)\) are the diagonal entries, \(-\mu\) and \(r-\sigma \frac{\Lambda}{\mu} .\) What happens to a small introduction of virus into a healthy individual if both characteristic values are negative? f. In order that the viral population to expand it is necessary that \(r-\sigma \frac{\Lambda}{\mu}>0 .\) What is the role of \(r\) in the model? g. If that condition is met and the viral population increases, show that there will be an equilibrium state, $$ E^{*}=r / \sigma, \quad V^{*}=\frac{\mu r-\Lambda \sigma}{\epsilon r} $$ It is clear that in order for \(V^{*}\) to be positive, we must have (again) $$ \mu r-\Lambda \sigma>0 \quad \text { so that } \quad r>\frac{\Lambda \sigma}{\mu} $$ h. Anderson and May report this system to be asymptotically stable, at \(\left(E^{*}, V^{*}\right),\) but only weakly so, meaning that it is subject to wide oscillations. Show that $$ J\left(E^{*}, V^{*}\right)=\left[\begin{array}{cc} -\frac{\Lambda \sigma}{r} & \epsilon \frac{r}{\sigma} \\ -\sigma \frac{\mu r-\Lambda \sigma}{\epsilon r} & 0 \end{array}\right] $$ i. The characteristic equation of a \(2 \times 2\) matrix \(M\) is is \(s^{2}-\operatorname{trace}(M) s+\operatorname{det}(M)=0\). Show that the characteristic equation of \(J\left(E^{*}, V^{*}\right)\) is $$ s^{2}+\frac{\Lambda \sigma}{r} s+\mu r-\Lambda \sigma=0 $$ j. The roots of the characteristic equation 18.65 are $$ \frac{-\frac{\Lambda \sigma}{r} \pm \sqrt{\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)}}{2} $$ Argue that the real part is negative so that the system is stable. Note: Two cases: $$ \left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)>0 \quad \text { and } \quad\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)<0 $$ k. Use \(\Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01\) and compute \(E^{*}, V^{*},\) and the stability at \(\left(E^{*}, V^{*}\right)\) l. Let \(E_{0}=2, V_{0}=1, \Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01 .\) Approximate the solutions to Equations 18.64 using the trapezoid rule. Observe that the rise in viral load precedes the increase in effector cells.

Show by substitution that if \(y_{1}(t)\) and \(y_{2}(t)\) are solutions to \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=0\) and each of \(C_{1}\) and \(C_{2}\) is a number then $$ y(t)=C_{1} y_{1}(t)+C_{2} y_{2}(t) $$ is a solution to \(y^{\prime \prime}(t)-p y^{\prime}(t)+q y(t)=0\).

Symbiotic relationships are common and persist for long periods. It is curious that there are no known or very few symbiotic relationships between mammals. There are, however, many symbiotic relationships between mammals and other organisms, Escherichia coli, for example. Analysis of the equations for symbiosis, Equations 18.63: $$ \begin{aligned} x^{\prime}(t) &=r_{x} \times x(t) \times(1-a x(t)+b y(t)) \\ y^{\prime}(t) &=r_{y} \times y(t) \times(1+c x(t)-d y(t)) \end{aligned} $$ a. Show that the equilibrium point without zeros is $$ x_{1}=\frac{b+d}{a d-b c}, \quad y_{1}=\frac{a+c}{a d-b c} $$ $$ \text { if } a d-b c \neq 0 $$ b. \(x_{1}\) and \(y_{1}\) are positive only if \(a d-b c>0 .\) This is a surprise to us. Set \(b=c=d=1\) and examine the equilibrium point for \(a>1\) and \(a<1\). c. Assume \(a d-b c>0\) so that \(x_{1}\) and \(y_{1}\) are positive. Either work it out (no!) or accept our analysis that the Jacobian at \(\left(x_{1}, y_{1}\right)\) is $$ J=\left[\begin{array}{cc} -a \frac{b+d}{a d-b c} & b \frac{b+d}{a d-b c} \\ c \frac{a+c}{a d-b c} & -d \frac{a+c}{a d-b c} \end{array}\right]=\frac{1}{a d-b c}\left[\begin{array}{cc} -a(b+d) & b(b+d) \\ c(a+c) & -d(a+c) \end{array}\right] $$ Argue that if \(a d-b c>0,\left(x_{1}, y_{1}\right)\) is an asymptotically stable equilibrium. d. With persistence you might show that the characteristic roots are not complex. It requires showing that the discrimiant $$ \begin{array}{l} (a(b+d)+d(a+c))^{2}-4(a(b+d) d(a+c)-c(a+c) d(b+d))= \\ (a(b+d)-d(a+c))^{2}+4 c(a+c) d(b+d) \geq 0 \end{array} $$
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