2.4 Theorems for Linear and Non-Linear ODE PDF

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2.4 Theorems for Linear and Non-Linear ODE...

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Section 2.4: Theorems for Linear and Non-Linear First-Order Equations Today, we focus on theorems that tell us when a first order IVP has a solution (existence), and when there is only one solution to an IVP (uniqueness).

Example (Existence Question for a Linear IVP) In question #3 of the worksheet 2.1, you solved the equation xy ′ − y = x, and have gotten: y (x) = x(ln |x| + C ).

Now, suppose you add an initial condition y (x0 ) = y0 6= 0 to make an initial value problem (IVP). For what value(s) of x0 would the IVP not have a solution?

Solution (Existence for a Linear IVP) The IVP wouldn’t have a solution for x = 0 because ln |0| is not defined and y = 0 is not a solution neither.

Definition (Coefficient Functions) For a linear differential equation in standard from: y ′ (t) + p(t )y (t ) = g (t) the functions p(t) and g (t) are called the coefficient functions.

Theorem (2.4.1) Consider the linear IVP: y ′ (t) + p(t)y (t) = g (t)

y (t0 ) = y0

If p(t) and g (t) are continuous on (α, β) containing t0 , then there exists a unique solution y = f (t) on the entire interval (α, β). Under the proper conditions - continuity of p(t) and g (t ) on (α, β) - a solution exists (existence) and it is the only one (uniqueness). The solution works for α < t < β .

Example (Applying Theorem 2.4.1) For each given initial value problem, determine the relevant quantities and functions from the theorem. Then determine whether the theorem applies, and if it does, what it says. 2. ty ′ + 2y = te 5t , y (0) = −1.

1. ty ′ + 2y = te 5t , y (1) = −1.

Solution (Applying Theorem 2.4.1) 1. First, put the equation in standard form: From here, we can identify:

y ′ + 2t −1 y = e 5t , y (1) = −1.

p(t) = 2t −1 , g (t) = e 5t , t0 = 1, and y0 = −1.

The function p(t) is discontinuous at t = 0, so we can say that p(t) and g (t) are continuous on (0, ∞) or on (−∞, 0). Since t0 = 1, the relevant interval is (0, ∞).

The theorem applies and says there is a unique solution on the interval (0, ∞). 2. The only difference between this example and the previous one is that now t0 = 0.

This means that we can say that p(t) and g (t) are continuous on (0, ∞) or on (−∞, 0). However, neither of these intervals contains t0 = 0. Thus, the theorem does not apply to this initial value problem.

Section 2.4: Theorems for Linear and Non-Linear First-Order Equations Example (Uniqueness Question for a Nonlinear IVP) Check that y (t) =

 1 2 t and y (t) = 0 are both solutions to the IVP for t ≥ 0: 2

y′ =

√

y , y (0) = 0.

Solution (Uniqueness Question for a Nonlinear IVP) The function y (t) = 0 definitely satisfies the initial condition y (0) = 0. It also satisfies the equation √ because y ′ = y = 0. Thus, it is a solution to the IVP.  2 The function y (t) = 21t definitely satisfies   the initial condition y (0) = 0. It also satisfies the equation √ for t ≥ 0 because y = 21 t, and y ′ = 2 21 t 12 = 21 t. Thus, it is a solution to the IVP.

Theorem (2.4.2) Consider the (not necessarily linear) IVP: y ′ (t ) = f (t , y )

y (t0 ) = y0

are continuous in some rectangular region α < t < β , δ < y < γ containing the point If f and ∂f ∂y (t0 , y0 ), then there exists a unique solution, y = F (t) on the interval (t0 − h, t0 + h) for some unknown h > 0, where (t0 − h, t0 + h) is contained in (α, β).

If f (t, y ) is continuous, but but it may not be unique.

∂f ∂y

is not continuous, (in some rectangular region), then a solution exists,

Comments: This theorem also guarantees a unique solution to the IVP, but the result is weaker. The solution depends on both t0 and y0 . This differs from the result in Theorem 2.4.1, which is independent of y0 . The domain of the solution is a sub interval of (α, β) that varies depending on y0 .

Example (Applying Theorem 2.4.2) Determine the relevant quantities and functions from theorem 2.4.2 for the IVP y ′ = Does the theorem apply? If so, what does it tell us?

√

y , y (0) = 0.

Solution (Applying Theorem 2.4.2)

√ For this IVP, we have y0 = t0 = 0, and f (t, y ) = y . The function f (t, y ) is only continuous for y ≥ 0. There is no rectangular region in the ty -plane containing the point (t0 , y0 ) = (0, 0) on which f (t, y ) is continuous. Thus, the theorem does not apply. Note: Although the theorem does not apply, a solution exists to the IVP. It just isn’t a unique solution....