Finding Maximum Solution of Slope Field

Introduction to Differential Equations

Martha L. Abell , James P. Braselton , in Introductory Differential Equations (Fifth Edition), 2018

Exercises 1.2

In Exercises 1–4, use the slope field to determine if the indicated path is that of a solution to the differential equation.

1.

$\frac{dy}{dx}=-y/x$

2.

$\frac{dy}{dx}=x^2+y^2$

3.

$\frac{dy}{dx}=x/y$

4.

$\frac{dy}{dx}=x^2-y$

In Exercises 5–8, use the slope field to sketch the solutions of the differential equation that pass through the given points.

5.

$\frac{dy}{dx}=x/y$

6.

$\frac{dy}{dx}=x^2-y$

7.

$\frac{dy}{dx}=x-y$

8.

$\frac{dy}{dx}=y^2-x^2$

9.

Graph the slope field for $dy/dt=\sin y$ . Determine ${\lim }_{t\to \infty }y(t)$ if $y(0)=y_0$ where (a) $y_0=-3\pi /2$ ; (b) $y_0=-\pi /2$ ; (c) $y_0=\pi /2$ ; (d) $y(0)=3\pi /2$ .

10.

Graph the slope field for $dy/dx=\sin x$ . Does ${\lim }_{x\to \infty }y(x)$ exist for any initial condition $y(0)=y_0$ ? Solve the ODE and find ${\lim }_{x\to \infty }y(x)$ . Does this match the graphical result?

11.

Graph the slope field for $dy/dx=e^{-x}$ . Does ${\lim }_{x\to \infty }y(x)$ exist for any initial condition $y(0)=y_0$ ? Solve the ODE and find ${\lim }_{x\to \infty }y(x)$ . Does this match the graphical result?

12.

Graph the slope field for $dy/dx=1/(x^2+1)$ . Does ${\lim }_{x\to \infty }y(x)$ exist for any initial condition $y(0)=y_0$ ? Solve the ODE and find ${\lim }_{x\to \infty }y(x)$ . Does this match the graphical result?

In Exercises 13–14, use the direction field of the given system to sketch the graph of the solution that satisfies the indicated initial conditions. Determine ${\lim }_{t\to \infty }x(t)$ and ${\lim }_{t\to \infty }y(t)$ in each case (if they exist).

13.

System: $\{\begin{array}{c}dx/dt=y\hfill \\ dy/dt=x\hfill \end{array}$ ; (a) $x(0)=0$ , $y(0)=2$ ; (b) $x(0)=-2$ , $y(0)=0$ ; (c) $x(0)=-2$ , $y(0)=2$ ;

14.

System: $\{\begin{array}{c}dx/dt=3x-2y\hfill \\ dy/dt=4x-y\hfill \end{array}$ ; $x(0)=0$ , $y(0)=5$ ;

In Exercises 15–20, write the second order equation as a system of first order equations.

15.

$d^{2}x/d{t}^2+4x=0$

16.

$d^{2}x/d{t}^2-5dx/dt=0$

17.

$d^{2}x/d{t}^2+4dx/dt+13x=0$

18.

$x^{″}-6x^{\prime }+7x=0$ ( $\prime =d/dt;x=x(t)$ )

19.

$x^{″}+16x=\sin t$

20.

$x^{″}+4x^{\prime }+13x=e^{-t}$

21.

(a) Use a computer algebra system to solve $dy/dx=e^{-x^2}$ . (b) Graph the solution to the IVP $dy/dx=e^{-x^2}$ , $y(0)=a$ for $a=-2,-1,0,12$ . (c) Graph the slope field of the differential equation together with the solutions in (b). (d) Do the solutions appear to match the results described at the beginning of the section?

22.

(a) Use a computer algebra system to graph the slope field for $dy/dx=\sin (2x-y)$ . (b) Use the slope fie to graph the solution $y(x)$ that satisfies the initial condition $y(0)=5$ . Does ${\lim }_{x\to \infty }y(x)$ appear to exist?

23.

Consider the systems

$\{\begin{array}{c}dx/dt=-y\hfill \\ dy/dt=x\hfill \end{array}\text{and}\{\begin{array}{c}dx/dt=-y\hfill \\ dy/dt=-x\hfill \end{array}.$

(a)

For any given initial conditions, in a brief paragraph explain why you think that the solutions of the systems are similar or different.

(b)

Fig. 1.12 shows the direction field associated with each system. Use the direction field to help you graph the solutions that satisfy these initial conditions (i) $x(0)=0.5,y(0)=0$ ; (ii) $x(0)=-0.25,y(0)=0$ ; (iii) $x(0)=0,y(0)=0.75$ ; and (iv) $x(0)=0,y(0)=-0.5$ .

Figure 1.12. Figure for Exercise 23.

(c)

How do your graphs affect your conjecture in (a)?

24.

Consider the systems

$\{\begin{array}{c}dx/dt=x/2\hfill \\ dy/dt=y\hfill \end{array}\text{and}\{\begin{array}{c}dx/dt=-x/2\hfill \\ dy/dt=-y\hfill \end{array}.$

(a)

For any given initial conditions, in a brief paragraph explain why you think that the solutions of the systems are similar or different.

(b)

Fig. 1.13 shows the direction field associated with each system. Use the direction field to help you graph the solutions that satisfy these initial condition (i) $x(0)=0.5,y(0)=0.25$ ; (ii) $x(0)=-0.25$ , $y(0)=-0.5$ ; (iii) $x(0)=-0.5$ , $y(0)=0.75$ ; and (iv) $x(0)=0.75$ , $y(0)=-0.5$ .

Figure 1.13. Figure for Exercise 24.

(c)

How do your graphs affect your conjecture in (a)?

25.

(Competing Species) The system of equations $\{\begin{array}{c}dx/dt=x(a-b_{1}x-b_{2}y)\hfill \\ dy/dt=y(c-d_{1}x-d_{2}y)\hfill \end{array}$ where a, $b_1$ , $b_2$ , c, $d_1$ , and $d_2$ represent positive constants can be used to model the size of the population of two species, represented by $x(t)$ and $y(t)$ , competing for a common food supply.

(a)

Fig. 1.14A shows the direction field for the system if $a=1$ , $b_1=2$ , $b_2=1$ , $c=1$ , $d_1=0.75$ , and $d_2=2$ . (i) Use the direction field to graph various solutions if both $x(0)$ and $y(0)$ are positive. (ii) Use the direction field and your graphs to approximate ${\lim }_{t\to \infty }x(t)$ and ${\lim }_{t\to \infty }y(t)$ .

Figure 1.14. (A) Figure for Exercise 25 (a). (B) Figure for Exercise 25 (b).

(b)

Fig. 1.14B shows the direction field for the system if $a=1$ , $b_1=1$ , $b_2=1$ , $c=0.67$ , $d_1=0.75$ , and $d_2=1$ . (i) Use the direction field to graph various solutions if both $x(0)$ and $y(0)$ are positive. (ii) Use the direction field and your graphs to determine the fate of the species with population $y(t)$ . What happens to the species with population $x(t)$ ?

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Introduction to Differential Equations

Martha L. Abell , James P. Braselton , in Introductory Differential Equations (Fourth Edition), 2014

Systems of ODEs and Direction Fields

We can also consider systems of differential equations. In Chapter 6, we will learn how to solve systems of first-order ODEs of the form

$\left\{\begin{array}{l}\text{d}x∕\text{d}t=\mathrm{ax}+\mathrm{by},\\ \text{d}y∕\text{d}t=\mathrm{cx}+\text{d}y,\end{array}\right.$

where A, B, C, and d are given constants. In the case of this system, we solve for x = x(t) and y = y(t). For example, if we consider the system

$\left\{\begin{array}{l}\text{d}x∕\text{d}t=y,\\ \text{d}y∕\text{d}t=-x,\end{array}\right.$

we can verify that the parametric equations

$\left\{\begin{array}{l}x(t)=sint\\ y(t)=cost\end{array}\right.$

satisfy the system because

$\begin{array}{l}\begin{array}{l}\frac{\text{d}x}{\text{d}t}=\frac{\text{d}}{\text{d}t}(sint)=cost=y\text{and}\\ \frac{\text{d}y}{\text{d}t}=\frac{\text{d}}{\text{d}t}(cost)=-sint=-x.\end{array}\end{array}$

We can graph each function separately as we do in Figure 1.10(a). (The graph of x(t) is the solid curve; that of y(t)is dashed.) Another option is to graph them as a pair of parametric equations as in Figure 1.10(b). As we recall, the graph of this pair of parametric equations is a circle of radius 1 centered at the origin because $x^2={sin}^{2}t$ and $y^2={cos}^{2}t$ , so that $x^2+y^2={sin}^{2}t+{cos}^{2}t=1$ . However, we must indicate the orientation of the curve (the direction of increasing parameter value t). For this pair of parametric equations, we find that at t = 0, $x(0)=sin0=0$ and $y(0)=cos0=1$ . Therefore, the point (0,1) corresponds to t = 0. Similarly, the point (1,0) corresponds to t = π/2 because $x(\pi ∕2)=sin\pi ∕2=1$ and $y(\pi ∕2)=cos\pi ∕2=0$ . This means that the solution moves from (0,1) to (1,0) as t increases. To determine if the orientation is clockwise or counter-clockwise, we test a t-value between 0 and π/2. Choosing t = π/4, we find the $x(\pi ∕4)=sin\pi ∕4=1∕\sqrt{2}$ and $y(\pi ∕4)=cos\pi ∕4=1∕\sqrt{2}$ so the orientation is clockwise. The parametric equations $\{x(t)=sint,y(t)=cost\}$ satisfy theIVP

Figure 1.10. (a) Graphs of $x(t)=sint$ and $y(t)=cost$ for 0 ≤ t ≤ 2π. (b) Graph of the parametric equations $x(t)=sint$ and $y(t)=cost$ for 0 ≤ t ≤ 2π.

$\left\{\begin{array}{ll}\text{d}x∕\text{d}t=y,& x(0)=0\\ \text{d}y∕\text{d}t=-x,& y(0)=1\end{array}\right.$

because they satisfy the system of differential equations as well as the two initial conditions. Notice that in the case of an IVP involving a system of differential equations, an initial condition is given for each of the variables x and y that depend on t. In the parametric plot, the solution to this IVP passes through the point (x(0),y(0)) = (0,1).

Another way to view a system of two ODEs is through the use of a direction field , which is similar to a slope field. For example, for the first-order system

$\left\{\begin{array}{l}\text{d}x∕\text{d}t=\mathrm{ax}+\mathrm{by},\\ \text{d}y∕\text{d}t=\mathrm{cx}+\text{d}y,\end{array}\right.$

we first write it as a first-order equation with

$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y∕\text{d}t}{\text{d}x∕\text{d}t}=\frac{\mathrm{cx}+\text{d}y}{\mathrm{ax}+\mathrm{by}}.$

Observe that the procedure described here can be used for any system of the form $\left\{\begin{array}{l}\text{d}x/\text{d}t=f(x,y),\hfill \\ \text{d}y/\text{d}t=g(x,y).\hfill \end{array}\right.$

Then, we can consider the slope field associated with this differential equation. For example, if we refer back to the system

$\left\{\begin{array}{l}\text{d}x∕\text{d}t=y,\\ \text{d}y∕\text{d}t=-x,\end{array}\right.$

we obtain the first-order equation dy/dx = −x/y, so we can determine the slope of tangent lines to solutions at points in the xy-plane. For example, at the point $(1∕\sqrt{2},1∕\sqrt{2})$ , the solution to this system that passes through this point has slope $(-1∕\sqrt{2})∕(1∕\sqrt{2})=-1$ . In a similar manner, we can find the slope at other points in the plane. However, as we mentioned in our earlier discussion, we must indicate the orientation when we graph parametric equations, so we consider the vector 〈dx/dt,dy/dt〉 = dx/dt i + dy/dt j with components from the system of differential equations. In the case of this system, we consider 〈dx/dt,dy/dt〉 = 〈y,−x〉. At the point $(1∕\sqrt{2},1∕\sqrt{2})$ , we obtain the vector $⟨1∕\sqrt{2},-1∕\sqrt{2}⟩$ . This means that the solution through $(1∕\sqrt{2},1∕\sqrt{2})$ has tangent vector $⟨1∕\sqrt{2},-1∕\sqrt{2}⟩$ . The direction field is made up of tangent vectors such as $⟨1∕\sqrt{2},-1∕\sqrt{2}⟩$ to solutions at points in the plane, so it is similar to the slope field for dy/dy = −x/y shown in Figure 1.11(a), except that vectors are used to indicate the orientation of solutions. In Figure 1.11(b), we show the direction field for this system. The vectors in the direction field indicate that solutions to this system are circles in the xy-plane that are directed clockwise. We graph several solutions along with the direction

Figure 1.11. (a) Slope field for dy/dx = −x/y. (b) Direction field for dx/dt = y, dy/dt = −x. (c) Direction field for dx/dt = y, dy/dt = −x and several solution curves.

Generally, we will use standard mathematical notation throughout Introductory Differential Equations so i = 〈1,0〉 and j = 〈0,1〉.

field in Figure 1.11(c). A collection of solutions in the xy-plane is called the phase portrait of the system. Notice that at points in the first quadrant, where x > 0 and y > 0, dx/dt = y > 0 and dy/dt = −x < 0. This means that x(t) increases and y(t) decreases along solutions in the first quadrant. In the second quadrant, where x < 0 and y > 0, dx/dt = y > 0 and dy/dt = −x > 0. Therefore, x(t) and y(t) increase along solutions in the second quadrant. We can perform a similar analysis for points in the other two quadrants.

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Systems of linear differential equations

Henry J. Ricardo , in A Modern Introduction to Differential Equations (Third Edition), 2021

6.4.1 Phase portraits for systems of equations

Suppose we have an autonomous system of the form

(6.4.1) $\frac{dx}{dt}=f(x,y)\frac{dy}{dt}=g(x,y).$

It is customary to assume that f and g are continuous for all values of x and y and have continuous partial derivatives with respect to x and y. These are realistic assumptions for most applications.

For example, let's take the system

$$\begin{gathered} \frac{dx}{dt}=y \\ \frac{dy}{dt}=-17x-2y \end{gathered}$$

and work with it throughout our initial discussions.

First, we can eliminate the variable t by dividing the second equation by the first equation:

(6.4.2) $\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{g(x,y)}{f(x,y)}.$

(See Section A.2 for a reminder of the Chain Rule used in this process.) For our example $g(x,y)=-17x-2y$ and $f(x,y)=y$ in (6.4.2), and we get

$\frac{dy}{dx}=\frac{-17x-2y}{y}.$

Now we have a single first-order differential equation $\frac{dy}{dx}=\frac{g(x,y)}{f(x,y)}$ in the variables y and x, and we can construct and analyze the slope field corresponding to this first-order equation (6.4.2). (Go back to Fig. 6.2 to see one such slope field.) Fig. 6.4a shows the slope field for our current example.

Figure 6.4a. Slope field for $\frac{dy}{dx}=\frac{-17x-2y}{y}$ , 0 ≤t ≤ 5; −1 ≤x ≤ 1, −4 ≤y ≤ 4

If we could solve Eq. (6.4.2) for y in terms of x, or even implicitly, we would have a solution curve in the x-y plane. The plane of the variables x and y (with x- and y-axes) is called the phase plane of the original system of differential equations with t as a parameter. As we saw in Section 1.3, a solution of the system (6.4.1) consists of a pair of functions $x=x(t)$ , $y=y(t)$ . We can think of a solution curve defined parametrically as a set of points $(x(t),y(t))$ as t varies. Each solution curve in the phase plane is called a trajectory (or orbit) of the system of equations. Although the independent variable t is not present explicitly, the passage of time is represented by the direction that a point $(x(t),y(t))$ takes on a particular trajectory. The way the curve is followed as the values of t increase is called the positive direction on the trajectory. The collection of plots of the trajectories is called the system's phase portrait or phase-plane diagram. The concepts of phase plane and phase portrait are natural extensions of the qualitative concepts discussed in Section 2.5.

Usually much less than the complete phase portrait is required (or of interest) for applications. By choosing good representative initial conditions, points through which we want the trajectories to pass, we can draw a fairly useful picture of the system's solutions.

Even if we can't solve the system, we can still look at the slope field of the single Eq. (6.4.2), the outline of the phase portrait of the system.

Let's do this for the system we've been discussing.

Example 6.4.1 Phase Portrait—One Trajectory

Our system is

$$\begin{gathered} \frac{dx}{dt}=y \\ \frac{dy}{dt}=-17x-2y, \end{gathered}$$

which gives us the first-order equation $\frac{dy}{dx}=\frac{-17x-2y}{y}$ when we eliminate the variable t . Using a calculator or CAS to draw a piece of the slope field for this first-order equation can result in an incorrect plot (see Problem 13 in Exercises 6.4). Your technological device may have a problem at points $(x,y)$ with $y=0$ . If you draw the slope field by hand, be sure to place vertical tangent line segments along the x-axis (where $y=0$ ), corresponding to an undefined (or infinite) slope when $y=0$ . It is better to use technology that takes the pair of equations given previously as input.

Fig. 6.4a shows the slope field, and Fig. 6.4b shows a single trajectory satisfying the initial condition $x(0)=4$ , $y(0)=0$ —that is, a trajectory passing through the point $(4,0)$ in the x-y (phase) plane—superimposed on the slope field.

Figure 6.4b. Trajectory for $\{\frac{dx}{dt}=y,\frac{dy}{dt}=-17x-2y;x(0)=4,y(0)=0\}$ , 0 ≤t ≤ 5; −2 ≤x ≤ 4, −12 ≤y ≤ 7

Because the trajectory starts at $(4,0)$ , we can see that the positive direction on the trajectory is clockwise, and the curve seems to spiral into the origin. (Try using technology to draw the trajectory for $0\le t\le b$ , letting b get larger and larger.) To get an accurate phase portrait, you may want to use the slope field to suggest good initial points to use. Each dynamical system has its own appropriate range for t.

Now let's look at a more elaborate phase portrait, one showing several trajectories.

Example 6.4.2 Phase Portrait—Several Trajectories

The system consists of the two equations (1) $\frac{dx}{dt}=x+y$ and (2) $\frac{dy}{dt}=-x+y$ . Whatever quantities these equations describe, certain details should be obvious from the nature of the equations. First of all, from Eq. (1), the growth of quantity x depends on itself and on the other quantity y in a positive way. On the other hand, Eq. (2) indicates that quantity y depends on itself positively, but its growth is hampered by the presence of quantity x—a larger value of x leads to a slowdown in the growth of y.

Let's look at the phase portrait corresponding to this problem. For our system, Eq. (6.4.2) looks like

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-x+y}{x+y}.$

This first-order equation is neither separable nor linear, but it is homogeneous (see the explanation for Problems 15–18 of Exercises 2.1 ) and can be solved implicitly. Fig. 6.5a shows several trajectories, obtained by specifying nine initial points $(x(0),y(0))$ , superimposed on the slope field for $\frac{dy}{dx}=\frac{-x+y}{x+y}$ . Because points on a trajectory are calculated by numerical methods, your CAS may allow you (or require you) to specify a step size and the actual numerical approximation method to be used. Numerical methods for systems of differential equations were discussed in Section 6.3.

Figure 6.5a. Trajectories for $\{\frac{dx}{dt}=x+y,\frac{dy}{dt}=-x+y\}$ ; (x(0),y(0))=(−1,−1), (−1,0), (−1,1), (0,−1), (0,0), (0,1), (1,−1), (1,0), and (1,1), 0 ≤t ≤ 4

Each point on a particular curve in Fig. 6.5a represents a state of the system. For each value of t, the point $(x(t),y(t))$ on the curve provides a snapshot of this dynamical system. If the variables x and y are supposed to represent animal or human populations, for example, then the proper place to view the trajectories is the first quadrant. Fig. 6.5b describes the first quadrant of the phase plane for our problem, with four trajectories determined by four initial points.

Figure 6.5b. Trajectories for $\{\frac{dx}{dt}=x+y,\frac{dy}{dt}=-x+y\}$ ; (x(0),y(0))=(0,60), (25,100), (50,140), and (0,160), 0 ≤t ≤ 1.2

Using technology again in our preceding example, we can graph $x(t)$ and $y(t)$ in the t-x and t-y planes, respectively. Figs. 6.6a and 6.6b show solution curves with $x(0)=50$ and $y(0)=140$ , respectively.

Figure 6.6a. x(t); x(0)=50, 0 ≤t ≤ 2.355

Figure 6.6b. y(t); y(0)=140, 0 ≤t ≤ 1

These graphs show clearly that the quantity y reaches a maximum of about 164 when $t\approx 0.4$ and that the x quantity hits a peak of about 800 when $t\approx 2$ . Note that the horizontal and vertical scales are different for Figs. 6.6a and 6.6b.

Viewed another way, the system in the preceding example had three variables: the independent variable t and the dependent variables x and y. To be precise about all this, we state that a solution of our system is a pair of functions $x=x(t)$ , $y=y(t)$ , and the graphical representation of such a solution is a curve in three-dimensional t-x-y space—a set of points of the form $(t,x(t),y(t))$ . Fig. 6.7 shows what the solution with initial point $(0,50,140)$ looks like for our problem.

Figure 6.7. Solution of $\{\frac{dx}{dt}=x+y,\frac{dy}{dt}=-x+y;x(0)=50,y(0)=140\}$ , 0 ≤t ≤ 1

Your CAS may allow you to manipulate the axes and get different views of this space curve. Fig. 6.6a represents the projection of several such space curves onto the x-y plane, a much less confusing way of viewing the behavior of the system. These projections can be thought of as the shadows that would be cast by the space curves if a very bright light were shining on them from the front (the x-y face) of Fig. 6.7.

Note that because the system we started with in the preceding example is autonomous, the solution curves are independent of the starting time. This means that if you pick a starting point (x*, y*) at time t*, then the path of a population starting at this point is the same as the path of a population starting at the same point at any other time t**. Geometrically, this says that there is only one path (or trajectory) through each point of the x-y plane. This is a consequence of an Existence and Uniqueness Theorem for systems that we saw in Section 6.2. (Look back at Section 2.8 for the theorem that applies to first-order ODEs.)

These trajectories tell us that for the initial points chosen, the quantity y increases to a maximum value and then decreases to zero, while the quantity x also increases until it reaches its maximum level after quantity y has disappeared.

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DENSITY OF STATES AND Tc OF DISORDERED A15 COMPOUNDS1

A.K. Ghosh , Myron Strongin , in Superconductivity in D- and F-Band Metals, 1980

II EXPERIMENTAL DETAILS

Samples of A15 superconductors were prepared by e-beam co-evaporation onto heated single crystal sapphire substrates, which were masked to give an appropriate geometry for resistivity measurements. These samples were analyzed by x-ray diffraction and microprobed to determine the composition and the phases present. Table I lists the properties of the "as deposited" samples whose data are presented in the subsequent section. Defect disorder was introduced in the samples by means of 2.5 MeV α⁺⁺-particle irradiation at the BNL small Van de Graaff. After each irradiation the T_c, ρ_o, and the critical field slope (dH_C2/dT) T_C≡ ${\text{H}}_{\text{C}}^{\prime }2$ (T_C) was measured.

TABLE I. Some Parameters of Evaporated A15

Material	Tc(K) d	Δ Tc(K)	ρ o(μΩcm)	H′c2 (KOe/K)	η	<hΩ p > (ev)
Nb3Sn	17.88	0.09	14.6	21.6	1.21	–
Nb3Ge	20.9	0.9	50.0	25.2	1.23	3.7 a
Nb3Al	16.3	0.4	53.8	26.8	1.19	3.7
V3Si	16.0	0.3	8.4	21.5	1.14	3.1 b
Mo3Ge	1.45	0.01	0.14	–	1.02	–
Mo3Ge c	1.76	0.2	9.9	0.8	1.02	–

a

P.B. Allen et al., 1978.

b

L. F. Matheiss et al., 1978.

c

Sputtered sample.

d

T_C(K) indicates midpoint of resistive transition.

In earlier publication (Ghosh et al., 1978b) we have shown how one can estimate the electronic specific heat coefficient γ from the critical field slope near T_c. The expression for γ is:

(1) $\text{γ}=\frac{1}{{\text{ρ}}_{\text{o}}\text{η}}\left(2.2\times {10}^4{\text{H}}_{\text{C}2}^{\prime }-\frac{7.17\times {10}^{17}{\text{T}}_{\text{C}}\text{η}}{{\text{v}}_{\text{F}}^{*2}}\right)$

where ${\text{H}}_{\text{C}}^{\prime }2$ is in KOe/K, ρ_o is in μΩ-cm, γ is in ergs/cm³-K², ${\text{v}}_{\text{F}}^{*}$ is the renormalized averaged Fermi velocity in cm/sec, and η is a strong-coupling correction (Rainer and Bergmann, 1974) that depends on T_c/<ω>, and is of the order of unity. It was shown (Ghosh et al.,.1978) that for Nb₃Sn, making the assumption that either <N(o)v_F> = constant or <N(o) ${\text{v}}_{\text{F}}^2$ > = constant leads essentially to similar results. In this paper we have assumed that <N(o) ${\text{v}}_{\text{F}}^2$ > is held constant as the sample is damaged. This implies that the plasma energy ${\text{Ω}}_{\text{p}}^2$ remains constant, which is a fairly good assumption since in the A15's its variation with energy near the Fermi surface is very much weaker than either N(ε) or v_F (Mattheiss et al., 1978; Pickett et al., 1979). Then eq. (1) reduces to

(2) $\text{γ}=\frac{2.2\times {10}^4{\text{H}}_{\text{C}2}^{\prime }}{{\text{ηρ}}_{\text{o}}}{\left(1+\frac{4.844{\text{T}}_{\text{C}}\left(1+\text{λ}\right)}{{\text{ρ}}_{\text{o}}{\text{Ω}}_{\text{p}}^2}\right)}^{-1}$

where Ω_p is in ev. The term in the square brackets represents the correction term which becomes important when ℓ≳ξ_o where ξ_o is the B.C.S. coherence length. The values of Ω_p used in this paper were taken from the literature and is listed in Table I. Once γ is calculated from the data, the value of N(o) = (3/2) (γ/π² ${\text{k}}_{\text{B}}^2$ )/(1+λ) is obtained, λ being calculated from the measured T_c using either the Allen-Dynes equation (Allen and Dynes, 1975) with f₁ f₂ equal to unity or the McMillan equation (McMillan, 1968).

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First-order differential equations

Henry J. Ricardo , in A Modern Introduction to Differential Equations (Third Edition), 2021

Summary

An easy type of first-order ODE to solve is a separable equation, one that can be written in the form $\frac{dy}{dx}=f(x)g(y)$ , where f denotes a function of x alone and g denotes a function of y alone. "Separating the variables" leads to the equation $\int \frac{dy}{g(y)}=\int f(x)dx$ . It is possible that you cannot carry out one of the integrations in terms of elementary functions or you may wind up with an implicit solution. Furthermore, the process of separation of variables may introduce singular solutions.

Another important type of first-order ODE is a linear equation, one that can be written in the form $a_1(x)y^{\prime }+a_0(x)y=f(x)$ , where $a_1(x)$ , $a_0(x)$ , and $f(x)$ are functions of the independent variable x alone. The standard form of such an equation is $\frac{dy}{dx}+P(x)y=Q(x)$ . The equation is called homogeneous if $Q(x)\equiv 0$ and nonhomogeneous otherwise. Any homogeneous linear equation is separable.

After writing a first-order linear equation in the standard form $\frac{dy}{dx}+P(x)y=Q(x)$ , we can solve it by the method of variation of parameters or by introducing an integrating factor, $\mu (x)=e^{\int P(x)dx}$ .

A typical first-order differential equation can be written in the form $\frac{dy}{dx}=f(x,y)$ . Graphically, this tells us that at any point $(x,y)$ on a solution curve of the equation, the slope of the tangent line is given by the value of the function f at that point. We can outline the solution curves by using possible tangent line segments. Such a collection of tangent line segments is called a direction field or slope field of the equation. The set of points $(x,y)$ such that $f(x,y)=C$ , a constant, defines an isocline, a curve along which the slopes of the tangent lines are all the same (namely, C). In particular, the nullcline (or zero isocline) is a curve consisting of points at which the slopes of solution curves are zero. A differential equation in which the independent variable does not appear explicitly is called an autonomous equation. If the independent variable does appear, the equation is called nonautonomous. For an autonomous equation the slopes of the tangent line segments that make up the slope field depend only on the values of the dependent variable. Graphically, if we fix the value of the dependent variable, say x, by drawing a horizontal line $x=C$ for any constant C, we see that all the tangent line segments along this line have the same slope, no matter what the value of the independent variable, say t. Another way to look at this is to realize that we can generate infinitely many solutions by taking any one solution and translating (shifting) its graph left or right. Even when we can't solve an equation, an analysis of its slope field can be very instructive. However, such a graphical analysis may miss certain important features of the integral curves, such as vertical asymptotes.

An autonomous first-order equation can be analyzed qualitatively by using a phase line or phase portrait. For an autonomous equation the points x such that $\frac{dy}{dx}=f(x)=0$ are called critical points. We also use the terms equilibrium points, equilibrium solutions, and stationary points to describe these key values. There are three kinds of equilibrium points for an autonomous first-order equation: sinks, sources, and nodes. An equilibrium solution y is a sink (or asymptotically stable solution) if solutions with initial conditions "sufficiently close" to y approach y as the independent variable tends to infinity. On the other hand, if solutions "sufficiently close" to an equilibrium solution y are asymptotic to y as the independent variable tends to negative infinity, then we call y a source (or unstable equilibrium solution). An equilibrium solution that shows any other kind of behavior is called a node (or semistable equilibrium solution). The First Derivative Test is a simple (but not always conclusive) test to determine the nature of equilibrium points.

Suppose that we have an autonomous differential equation with a parameter α. A bifurcation point ${\alpha }_0$ is a value of the parameter that causes a change in the nature of the equation's equilibrium solutions as α passes through the value ${\alpha }_0$ . There are three main types of bifurcation for a first-order equation: (1) pitchfork bifurcation; (2) saddle-node bifurcation; and (3) transcritical bifurcation.

When we are trying to solve a differential equation, especially an IVP, it is important to understand whether the problem has a solution and whether any solution is unique. The Existence and Uniqueness Theorem provides simple sufficient conditions that guarantee that there is one and only one solution of an IVP. A standard proof of this result involves successive approximations, or Picard iterations.

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d- AND f-BAND SUPERCONDUCTIVITY - SOME EXPERIMENTAL ASPECTS1

T.H. Geballe , in Superconductivity in D- and F-Band Metals, 1980

INTRINSIC PROPERTIES

The intrinsic superconducting properties of type II transition metal superconductors become of intense interest following the discovery of the existence of high-field, high-current superconductivity by Kunzler and coworkers (1961). The possibility of finding improved properties that can directly be put to practical use offers support for this field of research. Again the large homogeneous compositional ranges available make systematic studies and quantitative comparison with the theory of type II superconductivity possible. Recent work of Orlando et al. (to be published) for Nb₃Sn shows that it is necessary to bring strong-coupling and the electron phonon enhancement in on an equal footing with the spin-orbit scattering before the effects of Pauli paramagnetism at high fields can be understood on a quantitative basis.

The maximum known H_c2 as a function of time has increased drastically since 1960, It is perhaps no coincidence that the highest known critical fields and critical field slopes are found for ternary compounds, for it is in these compounds with large unit cells which contain electrically inactive sites that unexpected behavior in high magnetic fields is found. This is true for the Chevrel phases (Fischer, 1978), the RE borides (Fertig et al., 1977) and the intercalated layered metal dichalcogenides (Prober et al., 1975). The field is able to penetrate perpendicular to the c-axis of the dichalcogenides with so little cost in energy that enormous critical field slopes

$\frac{{}^{\text{d}}{\text{H}}_{\text{c2}}}{\text{dT}}>2\times {10}^5$

are found. This enormous slope has motivated theoretical studies of critical fields of the layered compounds which show how novel the superconducting properties of such open structures can be (Klemm et al., 1974). Possibly related, but still not understood are the negative curvatures found in the low field portion of the H_c2-T curves (Woollam, 1974). In the Chevrel and rare earth boride phases it is not so straightforward from the structure alone to see why the critical fields can be so high; however, the intercluster linkage must play an important role.

It is clear that we have a long way to go before magnets and other kinds of devices will be limited by an intrinsic inability to remain superconducting in high magnetic fields. The limitation is more with the mechanical properties of the superconductors, the intrinsic properties of the various kinds of pinning centers and the interaction of the flux-line-lattice with distributed pinning centers. While the fabrication of composites seeking to optimize simultaneously all these properties is more nearly at the center of gravity of the Applied Superconducting Conference, there is plenty of work still to be done of interest to this conference, in developing a quantitative understanding of each of the parameters involved. Quantitative understanding can eventually lead to much higher performance superconductors.

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First-Order Equations

Martha L. Abell , James P. Braselton , in Introductory Differential Equations (Fourth Edition), 2014

Euler's Method

Suppose that we wish to approximate the solution to

$\text{d}y/\text{d}x=f(x,y),y(x_0)=y_0$

over the interval x ₀ ≤ x ≤ X, where x is a specified value of x. For example, we may wish to approximate the solution to an IVP on an interval like 0 ≤ x ≤ 1. This indicates that the initial condition is specified at x ₀ = 0 and that X = 1 is the end of the interval. Now, we partition x ₀ ≤ x ≤ X into n subintervals of equal length h, where h = (X − x ₀)/N. Therefore, the nth value of x in the partition is x _n = x ₀ + nh, n = 1, 2, … , n. This means that x ₁ = x ₀ + h,

In this text, we frequently use t as our independent variable. In this section, we exclusively use x to represent the independent variable and y to represent the dependent variable to make programming these numerical methods into most graphing calculators easier for those students who wish to do so.

x ₂ = x ₀ + 2h, … , x _N = x ₀ + Nh = x ₀ + N × (X − x ₀)/N = X. If we are solving the IVP dy/dx = f(x,y), y(x ₀) = y ₀, then the point (x ₀,y ₀ ) is on the solution curve. Therefore, as we discussed in relation to slope fields, we can follow the line tangent to the solution curve at ( x ₀,y ₀) to approximate the value of the solution at the next value of x, x = x ₁. Because the slope of this tangent line is f(x ₀,y ₀), we find that an equation for the tangent line is y − y ₀ = f(x ₀,y ₀)(x − x ₀) or y = f(x ₀,y ₀)(x − x ₀) + y ₀. Therefore, the approximate value of the solution at x = x ₁ is

$y_1=f(x_0,y_0)(x_1-x_0)+y_0=hf(x_0,y_0)+y_0.$

Notice that we refer to the approximate value of y at x = x _i as y _i while we refer to the exact value of the solution at x = x _i as y(x _i ). This means that y _i approximates y(x _i ) (see Figure 2.25(a)). We remember from calculus that this gives a good approximation if h is a small number. Next, we assume that the point (x ₁, y ₁) is on the solution curve. If we determine the equation of the line with slope f(x ₁, y ₁) that passes through (x ₁, y ₁), we obtain

Figure 2.25. (a) Approximating (x ₁,y ₁). (b) Successive approximations using Euler's method.

$y-y_1=f(x_1,y_1)(x-x_1),$

so we find the approximate value of y at x = x ₂ is

$y_2=f(x_0,y_0)(x_2-x_1)+y_1=hf(x_0,y_0)+y_1.$

We continue this process until we reach y _N , the approximate value of y(x _N ) = y(X). In doing so, we obtain a sequence of points of the form (x _n , y _n ), n = 1, 2, … , n. We show several points of this type along with actual values of y in Figure 2.25(b).

Theorem 7 (Euler's Method). The solution of the IVP

$\text{d}y/\text{d}x=f(x,y),y(x_0)=y_0$

is approximated at the sequence of points (x_n,y_n), n = 1, 2, …, where y_N is the approximate value of y(x_n) by computing

$y_n=hf\left(x_{n-1},y_{n-1}\right)+y_{n-1},n=1,2,\dots ,$

where x_n = x₀ + nh and h is the selected stepsize.

Example 2.6.1

Use Euler's method with h = 0.1 and with h = 0.05 to approximate the solution of y′ = xy, y(0) = 1 on 0 ≤ x ≤ 1. Determine the exact solution and compare the results.

Solution

First, we note that f(x,y) = xy, x ₀ = 0, and y ₀ = 1. With h = 0.1, we have the formula

$y_n=hf(x_{n-1},y_{n-1})+y_{n-1}=0.1x_{n-1}{y}_{n-1}+y_{n-1}.$

Then, for x ₁ = x ₀ + h = 0.1, we have

$y_1=0.1x_0{y}_0+y_0=0.1\times 0\times 1+1=1.$

Similarly, for x ₂ = x ₀ + 2h = 0.2,

$y_2=0.1x_1{y}_1+y_1=0.1\times 0.1\times 1+1=1.01.$

In Table 2.1, we show the results of this sequence of approximations. From this, we see that y(1) is approximately 1.54711.

Table 2.1. Euler's Method with h = 0.1

x n	y n	x n	y n
0.0	1.0	0.6	1.15873
0.1	1.0	0.7	1.22825
0.2	1.01	0.8	1.31423
0.3	1.0302	0.9	1.41937
0.4	1.06111	1.0	1.54711
0.5	1.10355

First, h = 0.05, we use

$y_n=hf(x_{n-1},y_{n-1})+y_{n-1}=0.05x_{n-1}{y}_{n-1}+y_{n-1}$

to obtain the values given in Table 2.2. With this stepsize, the approximate value of y(1) is 1.59594.

Table 2.2. Euler's Method with h = 0.05

x n	y n	x n	y n	x n	y n
0.0	1.0	0.35	1.05361	0.70	1.2523
0.05	1.0	0.40	1.07204	0.75	1.29613
0.10	1.0025	0.45	1.09348	0.80	1.34474
0.15	1.00751	0.50	1.11809	0.85	1.39853
0.20	1.01507	0.55	1.14604	0.90	1.45796
0.25	1.02522	0.60	1.17756	0.95	1.52357
0.30	1.03803	0.65	1.21288	1.00	1.59594

The exact solution to the IVP, which is found with separation of variables, is $y=e^{x^2/2}$ so the exact value of y(1) is e^1/2 ≈ 1.64872. The smaller value of h, therefore, yields a better approximation. We graph the approximations obtained with h = 0.1 and h = 0.05 as well as the graph of $y=e^{x^2/2}$ in Figure 2.26. Notice from these graphs that the approximation is more accurate when h is decreased.

Figure 2.26. (a) h = 0.1 and (b) h = 0.05.

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Systems of nonlinear differential equations

Henry J. Ricardo , in A Modern Introduction to Differential Equations (Third Edition), 2021

7.1 Equilibria of nonlinear systems

Recall that an equilibrium point of a differential equation or a system of differential equations is a constant solution. If we look at the two (somewhat similar) equations

$(1)y^{\prime }=-y(2)y^{\prime }=-y(1-y),$

we see some important differences between linear and nonlinear equations.

Eq. (1) is linear, but more fundamentally it is separable, so it is easy to find the general solution, $y=C{e}^{-t}$ , where C is an arbitrary constant. (We recognize that $C=y(0)$ , the initial state of the system being modeled by the equation.)

Now Eq. (2) is nonlinear and separable, and its general solution is $y=\frac{1}{1+C{e}^t}$ , where $C=1/y(0)-1$ . (Verify the solutions to both equations.)

Let's examine some typical solution curves for Eq. (1). Fig. 7.1 shows that there is only one equilibrium solution, $y\equiv 0$ , and this is a sink. (Review Section 2.6 if necessary.) If an object described by the equation starts off at zero (that is, if $C=0$ ), it remains at zero for all time. If the object's initial state is not zero, then the object will approach the solution $y\equiv 0$ as its asymptotically stable solution (or sink).

Figure 7.1. Solutions of y′ = −y; y(0)=5,3,1,−2,−4

On the other hand, Fig. 7.2 shows the same kind of information for Eq. (2). For such a nonlinear equation there can be more than one equilibrium solution, in this case $y\equiv 0$ and $y\equiv 1$ . Also note that some solutions of a nonlinear equation may become unbounded as t approaches some finite value. (If $y(0)>1$ , for what value of t does the denominator of the general solution to Eq. (2) vanish?)

Figure 7.2. Solutions of y′ = −y(1 −y)

In contrast, all solutions of a linear equation or a system of linear equations are defined for all values of the independent variable. Finally, looking closely at the behavior of solutions of Eq. (2) with different initial values, we see that the solutions starting off above 1 behave differently from those solutions with initial values less than 1. The equilibrium solution $y\equiv 0$ is a sink if $y(0)<1$ and $y\equiv 1$ is a source if $y(0)>1$ . Furthermore, for solutions with initial values C greater than 1, the line $t=\ln (-\frac{1}{C})$ is a vertical asymptote. We say that the solution "blows up in finite time." (Note that when $y(0)>1$ , $C=1/y(0)-1<0$ , so that the blow-up point t is the logarithm of a positive quantity.) The last three types of behavior cannot occur when we are dealing with a linear equation. You should expect that the situation with nonlinear systems is appropriately complicated.

Let's look at an example of a nonlinear system and its behavior near its equilibrium points.

Example 7.1.1 Stability of a Nonlinear System

The nonlinear system

$$\begin{gathered} x^{\prime }=x-x^2-xy \\ {y}^{\prime }=-y-y^2+2xy \end{gathered}$$

represents two populations interacting in a predator-prey relationship. This is essentially a Lotka–Volterra system (see Section 7.4) with "crowding" terms (the squared terms) added for both species.

To calculate the equilibrium points of this system, we solve the system $\{x^{\prime }=0,y^{\prime }=0\}$ , which is the same as the nonlinear algebraic system

$$\begin{gathered} \text{(A)}x(1-x-y)=0 \\ \text{(B)}y(-1-y+2x)=0. \end{gathered}$$

Clearly, the origin, $x=y=0$ , is an equilibrium point. Logically, there are only three other cases to examine: (1) $x=0$ , $y\ne 0$ ; (2) $x\ne 0$ , $y=0$ ; and (3) $x\ne 0$ , $y\ne 0$ . Assuming case 1, we can eliminate Eq. (A) and examine (B), which becomes $y(-1-y)=0$ . Because $y\ne 0$ , we conclude that $-1-y=0$ , or $y=-1$ . Thus, our second equilibrium point is $(0,-1)$ . Moving to case 2, we can ignore Eq. (B) and focus on (A), which now looks like $x(1-x)=0$ . Because we are assuming in case 2 that $x\ne 0$ , we can see that $x=1$ , which gives us the third equilibrium point (1, 0). Finally, if $x\ne 0$ and $y\ne 0$ , our system of algebraic equations becomes

$$\begin{gathered} \text{(A2)}x+y=1 \\ \text{(B2)}y-2x=-1. \end{gathered}$$

(We have divided out x and y in (A) and (B) and then rearranged the terms of each equation.) Subtracting (B2) from (A2) gives us $3x=2$ , or $x=\frac{2}{3}$ . Substituting this value of x in (A2) yields $y=\frac{1}{3}$ . Therefore, the last equilibrium point is $(\frac{2}{3},\frac{1}{3})$ .

In terms of a population problem, the only interesting equilibrium point is the last one we found. (Why is this so? ) If we look at a slope field for the original system of nonlinear differential equations near the point $(\frac{2}{3},\frac{1}{3})$ , we see an interesting pattern (Fig. 7.3a).

Figure 7.3a. Slope field for x′ =x −x ² −xy, y′ = −y −y ² + 2xy near $(\frac{2}{3},\frac{1}{3})$

The apparent spiraling of solutions into the equilibrium point can be seen more clearly if we show some numerically generated solution curves (Fig. 7.3b). Fig. 7.3b represents a predator-prey population that is stabilizing. If the units are thousands of creatures, then the X population is heading for a steady population of about 667, whereas the Y population has 333 as its stable value.

Figure 7.3b. Phase portrait for x′ =x −x ² −xy, y′ = −y −y ² + 2xy near $(\frac{2}{3},\frac{1}{3})$ ; (x(0),y(0))=(0.2,1), (0.8,0.8), (0.8,0.5), (1,0.7), (1,0.2), (0.5,1)

Mathematically, however, we should look at the entire phase portrait to understand the complex behavior of nonlinear systems. We'll return for a detailed analysis of this system in Example 7.3.1.

Exercises 7.1

A

Find all the equilibrium points for each of the systems in Problems 1–14, using technology if necessary.

1.

$x^{\prime }=-x+xy$ , $y^{\prime }=-y+2xy$

2.

$x^{\prime }=x-xy$ , $y^{\prime }=y-xy$

3.

$x^{\prime }=x^2-y^2$ , $y^{\prime }=x-xy$

4.

$x^{\prime }=1-y^2$ , $y^{\prime }=1-x^2$

5.

$x^{\prime }=x+y+2xy$ , $y^{\prime }=-2x+y+y^3$

6.

$x^{\prime }=y(1-x^2)$ , $y^{\prime }=-x(1-y^2)$

7.

$x^{\prime }=x-x^2-xy$ , $y^{\prime }=3y-xy-2y^2$

8.

$x^{\prime }=1-y$ , $y^{\prime }=x^2-y^2$

9.

$x^{\prime }=(1+x)\sin y$ , $y^{\prime }=1-x-\cos y$

10.

$x^{\prime }=3y-e^x$ , $y^{\prime }=2x-y$ [Hint: There are two equilibrium points. Use your CAS to approximate these points.]

11.

$x^{\prime }=y^2-x^2$ , $y^{\prime }=x-1$

12.

$x^{\prime }=x^2-y$ , $y^{\prime }=y^2-x$

13.

$x^{\prime }=xy(1-x)$ , $y^{\prime }=y(1-\frac{y}{x})$

14.

$x^{\prime }=y$ , $y^{\prime }=-\sin x-3y$

B

15.

Use technology to find all the equilibrium points of the system

$x^{\prime }=-y,y^{\prime }=(x^4+4x^3-x^2-4x+y)/8.$

16.

A two-mode laser produces two different kinds of photons, whose numbers are $n_1$ and $n_2$ . The equations governing the rates of photon production are

$$\begin{gathered} {\dot{n}}_1=G_{1}N{n}_1-k_1{n}_1 \\ {\dot{n}}_2=G_{2}N{n}_2-k_2{n}_2, \end{gathered}$$

where $N(t)=N_0-a_1{n}_1-a_2{n}_2$ is the number of excited atoms. The parameters $G_1,G_2,k_1,k_2,a_1,a_2$ , and $N_0$ are all positive. Use a CAS "solve" command to find all the equilibrium points of the system.

17.

A chemostat is a device for growing and studying bacteria by supplying nutrients and maintaining convenient levels of the bacteria in a culture. One model of a chemostat is the nonlinear system

$$\begin{gathered} \frac{dN}{dt}=a_1\left(\frac{C}{1+C}\right)N-N \\ \frac{dC}{dt}=-\left(\frac{C}{1+C}\right)N-C+a_2, \end{gathered}$$

where $N=N(t)$ denotes the bacterial density at time t; $C=C(t)$ denotes the concentration of nutrient; and $a_1,a_2$ are positive parameters. Use technology to find all the equilibrium solutions $(N^{⁎},C^{⁎})$ of the system.

18.

In the absence of damping and any external force, the motion of a pendulum is described by the equation $\frac{d^2\theta }{d{t}^2}+\frac{g}{L}\sin \theta =0$ , where θ is the angle between the pendulum and the downward vertical, g is the acceleration due to gravity, and L is the length of the pendulum.

a.

Write this equation as a system of two first-order equations.

b.

Describe all the equilibrium points of the system.

C

19.

Use technology to find all the equilibrium solutions of the system

$x^{\prime }=x-y^2+a+bxy,y^{\prime }=0.2y-x+x^3,$

where $a=1.28$ and $b=1.4$ . (Round to the nearest thousandth.)

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Finding Maximum Solution of Slope Field

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