Download Direction Fields & Existence-Uniqueness of Solutions for First-Order DEs and more Lab Reports Mathematics in PDF only on Docsity!
2.4 Direction Fields; Existence and Uniqueness
First-order differential equations of the general form
y′^ = f (x, y) (1)
can be solved only in very special cases. We have looked at two such cases, linear equations and separable equations. Two other cases, Bernoulli equations and homogeneous equations, were introduced in the exercise sets. It is important to understand that there are no methods for solving equation (1) in general.
If a given first-order equation does not fall into one of the special cases for which there is a solution method, then some other approach must be used. In such situations numerical methods or approximation methods are typically employed. These methods are studied in more advanced courses. In this section we give a geometric interpretation of equation (1) and then consider the basic questions of existence and uniqueness of solutions.
Direction Fields
Here we introduce a geometric approach to the first order differential equation (1) that enables us to produce sketches of solution curves without actually solving the equation. The approach does not produce equations in x and y; it produces pictures, pictures from which we can gather information on the qualitative behavior of solutions, behavior such as boundedness, concavity, possible maxima and minima, and so forth.
If a solution curve for equation (1) y = y(x) passes through the point (x 0 , y 0 ) then it does so with slope f (x 0 , y 0 ) since y′(x 0 ) = f (x 0 , y 0 ). We can indicate this by drawing a short line segment through (x 0 , y 0 ) with slope f (x 0 , y 0 ). By repeating this process over and over, we construct a direction field for the differential equation (1). That is, we select a grid of points (xi, yi), i = 1, 2 ,... , n, and draw at each of these points a short line segment with slope f (xi, yi). While this is tedious to do by hand, it is a simple task for a computer algebra/graphing utility system (e.g., Mathematica, Maple, MatLab). Such systems typically include a feature for sketching direction fields.
Example 1. The direction field for the differential equation
y′^ = x − y
is
−4 −4 −3 −2 −1 0 1 2 3 4
−
−
−
0
1
2
3
4
x
y
y ’ = x−y
We can use a direction field to sketch the solution of an initial-value problem:
y′^ = f (x, y), y(a) = b.
We start at the point (a, b) and follow the line segments in both directions. A sketch of the solution of y′^ = x − y that satisfies the initial condition y(0) = 1 is shown in the next figure.
−4 −4 −3 −2 −1 0 1 2 3 4
−
−
−
0
1
2
3
4
x
y
y ’ = x − y
The differential equation in this case is linear and so you can find the general solution. As you can check, the general solution is y = x − 1 + Ce−x^ , and the solution satisfying the initial condition is y = x − 1 + 2e−x. The graph of is shown below.
-1 1 2 3 4 x
1
2
3
4 y
Example 2. There is no method for finding the general solution of the differential equation
y′^ = y^2 − xy + 2x
However, we can draw the direction field for the equation and get some idea about the solutions and their behavior. The direction field is
−2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −1.
− −0. 0
1
2
3
x
y
y ’ = y^2 − x y + 2 x
To apply the initial condition (a), we set x = 0, y = 1 in the general solution. This gives
1 =
C · 0 − 1 = 0.
We conclude that there is no value of C such that y(0) = 1; there is no solution of the
initial-value problem y′^ = − y
2 x^2 , y(0) = 1.
Next we apply the initial condition (b) by setting x = 0, y = 0 in the general solution. In this case we obtain the equation
0 =
C · 0 − 1 = 0
which is satisfied by all values of C. The initial-value problem y′^ = − y^2 x^2 , y(0) = 0^ has infinitely many solutions.
Finally, we apply the initial condition (c) by setting x = 1, y = 1 in the general solution: 1 = (^) C · 11 − 1 which implies C = 2.
This initial-value problem y′^ = − y
2 x^2 , y(1) = 1 has a unique solution, namely
y = x/(2x − 1).
Existence and Uniqueness Theorem Given the initial-value problem
y′^ = f (x, y) y(a) = b. (1)
If f and ∂f /∂y are continuous on a rectangle R : a − α ≤ x ≤ a + α, b − β ≤ y ≤ b + β, α, β > 0, then there is an interval a − h ≤ x ≤ a + h, h ≤ α on which the initial-value problem (2) has a unique solution y = y(x).
Going back to our example, note that f (x, y) = −y^2 /x^2 is not continuous on any rectangle that contains (0, b) in its interior. Thus, the existence and uniqueness theorem does not apply in the cases y(0) = 1 and y(0) = 0.
In the case of the linear differential equation
y′^ + p(x)y = q(x)
where p and q are continuous functions on some interval I = [α, β], we have
f (x, y) = q(x) − p(x)y and ∂f ∂y = p(x)
and these functions are continuous on every rectangle R of the form α ≤ x ≤ β, −γ ≤ y ≤ γ where γ is any positive number; that is f and ∂f /∂y are continuous on the “infinite” rectangle α ≤ x ≤ β, −∞ < y < ∞. Thus, every linear initial-value problem has a unique solution.
Exercises 2.
- Given the initial-value problem y′^ = y; y(0) = 1. (a) Draw a direction field in the rectangle R : − 3 ≤ x ≤ 1. 5 , − 1 ≤ y ≤ 3. (b) Use this direction field to sketch the solution curve that satisfies the initial condi- tion. Experiment with other rectangles to obtain additional views of the solution curve. (c) Find the exact solution of the initial-value problem using the methods of this chapter and then compare the graph of your solution with the curve you obtained in part (b).
- Given the initial-value problem y′^ = x + 2y; y(0) = 1. (a) Draw a direction field in the rectangle R : − 1 ≤ x ≤ 2 , − 1 ≤ y ≤ 9. (b) Use this direction field to sketch the solution curve that satisfies the initial condi- tion. Experiment with other rectangles to obtain additional views of the solution curve. (c) Find the exact solution of the initial-value problem using the methods of this chapter and then compare the graph of your solution with the curve you obtained in part (b).
- Given the initial-value problem y′^ = 2xy; y(0) = 1. (a) Draw a direction field in the rectangle R : − 1. 5 ≤ x ≤ 3 , − 1 ≤ y ≤ 8. (b) Use this direction field to sketch the solution curve that satisfies the initial condi- tion. Experiment with other rectangles to obtain additional views of the solution curve. (c) Find the exact solution of the initial-value problem using the methods of this chapter and then compare the graph of your solution with the curve you obtained in part (b).
- Given the initial-value problem y′^ = − 4 x/y; y(1) = 1. (a) Draw a direction field in the rectangle R : − 2 ≤ x ≤ 2 , − 3 ≤ y ≤ 3. (b) Use this direction field to sketch the solution curve that satisfies the initial condi- tion. Experiment with other rectangles to obtain additional views of the solution curve.
