(1.2) Initial-Value Problems

11

INTRODUCTION:

We are often interested in problems in which we seek a solution 𝑦(𝑥) of a differential equation so that 𝑦(𝑥) satisfies prescribed side conditions—that is, conditions imposed on the unknown 𝑦(𝑥) or its derivatives. On some interval 𝐼 containing 𝑥₀ the problem

𝑆𝑜𝑙𝑣𝑒: 𝑑^𝑛𝑦

𝑑𝑥^𝑛 = 𝑓(𝑥, 𝑦, 𝑦^′, … , 𝑦^(𝑛−1))

… … … (1) 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜: 𝑦(𝑥₀) = 𝑦₀, 𝑦^′(𝑥₀) = 𝑦₁, … , 𝑦^(𝑛−1)(𝑥₀) = 𝑦_𝑛−1,

where 𝑦₀, 𝑦₁, … , 𝑦_𝑛−1 are arbitrarily specified real constants, is called an initial-value problem (IVP). The values of 𝑦(𝑥) and its first 𝑛 − 1 derivatives at a single point 𝑥₀, 𝑦(𝑥₀) = 𝑦₀, 𝑦^′(𝑥₀) = 𝑦₁, … , 𝑦^(𝑛−1)(𝑥₀) = 𝑦_𝑛−1, are called initial conditions.

FIRST- AND SECOND-ORDER IVPS:

The problem given in (1) is also called an 𝒏th-order initial-value problem. For example, 𝑆𝑜𝑙𝑣𝑒: 𝑑𝑦

𝑑𝑥 = 𝑓(𝑥, 𝑦)

… … … (2) 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜: 𝑦(𝑥₀) = 𝑦₀

and

𝑆𝑜𝑙𝑣𝑒: 𝑑²𝑦

𝑑𝑥² = 𝑓(𝑥, 𝑦, 𝑦^′)

… … … (3) 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜: 𝑦(𝑥₀) = 𝑦₀, 𝑦^′(𝑥₀) = 𝑦₁

are first- and second-order initial-value problems, respectively.

THEOREM 1.2.1 (Existence of a Unique Solution)

Let 𝑅 be a rectangular region in the 𝑥𝑦-plane defined by 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑 that contains the point (𝑥₀, 𝑦₀) in its interior. If 𝑓(𝑥, 𝑦) and 𝜕𝑓 𝜕𝑦⁄ are continuous on 𝑅, then there exists some interval 𝐼₀: (𝑥₀− ℎ, 𝑥₀ + ℎ), ℎ > 0, contained in [𝑎, 𝑏], and a unique function 𝑦(𝑥), defined on 𝐼₀, that is a solution of the initial-value problem (2).

Roughly speaking, if we have the IVP ^𝑑𝑦

𝑑𝑥 = 𝑓(𝑥, 𝑦), 𝑦(𝑥₀) = 𝑦₀, then we have 3 cases explain the theorem which are follows:

 If 𝑓(𝑥, 𝑦) and 𝜕𝑓/𝜕𝑦 are continuous on a common domain 𝐼

⇒ the solution of the IVP exists and unique.

 If 𝑓(𝑥, 𝑦) is only continuous on a domain 𝐼

⇒ the solution of the IVP exists only. This means that the IVP has infinite solutions

 If 𝑓(𝑥, 𝑦) is discontinuous on a domain 𝐼

⇒ the solution of the IVP does not exist.

12

EXAMPLE 1 (Second-Order IVP) In Example 4 of Section 1.1, we saw that

𝑥 = 𝑐₁cos 4𝑡 + 𝑐₂sin 4𝑡 is a two-parameter family of solutions of

𝑥^′′+ 16𝑥 = 0.

Find a solution of the initial-value problem 𝑥^′′+ 16𝑥 = 0, 𝑥 (𝜋

2) = −2, 𝑥^′(𝜋

2) = 1. … … … (4) SOLUTION:

We first find 𝑥^′ and 𝑥^′′ using the general solution 𝑥 = 𝑐₁cos 4𝑡 + 𝑐₂sin 4𝑡, we obtain 𝑥^′ = −4𝑐₁sin 4𝑡 + 4𝑐₂cos 4𝑡

and

𝑥^′′ = −16𝑐₁cos 4𝑡 − 16𝑐₂sin 4𝑡.

Then, we apply 𝑥 (^𝜋

2) = −2 to the given family of solutions:

𝑥 (𝜋

2) = 𝑐₁cos (4 (𝜋

2)) + 𝑐₂sin (4 (𝜋

2)) = 𝑐₁cos 2𝜋 + 𝑐₂sin 2𝜋 = 𝑐₁(1) + 𝑐₂(0) = 𝑐₁ = −2.

Hence, 𝑐₁ = −2. Similarly, we next apply 𝑥^′(^𝜋₂) = 1 to the two-parameter family 𝑥^′ = −4𝑐₁sin 4𝑡 + 4𝑐₂cos 4𝑡.

So, 𝑥^′(𝜋

2) = −4(−2) sin (4 (𝜋

2)) + 4𝑐₂cos (4 (𝜋

2)) = 8 sin 2𝜋 + 4𝑐₂cos 2𝜋 = 8(0) + 4𝑐₂(1) = 4𝑐₂ = 1.

Thus, 𝑐₂ =¹₄. Hence,

𝑥 = −2 cos 4𝑡 +1 4sin 4𝑡 is a solution of (4).

EXAMPLE 2

According to the Existence and Uniqueness Theorem, determine whether the IVP 𝑑𝑦

𝑑𝑥 = 𝑥𝑦^1/2, 𝑦(0) = 0 has a unique solution.

SOLUTION:

1 − 𝑓(𝑥, 𝑦) = 𝑥𝑦¹² = 𝑥√𝑦 is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ [0, ∞)}.

2 − 𝜕𝑓

𝜕𝑦 = 𝑥

2𝑦¹² = 𝑥

2√𝑦 is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ (0, ∞)}.

Since (0,0) is not in the region defined by 𝑥 ∈ ℝ and 𝑦 > 0, this shows that the IVP has no unique solution but it has infinite solutions.

13

EXAMPLE 3

According to the Existence and Uniqueness Theorem, determine whether the IVP 𝑑𝑦

𝑑𝑥 = 𝑥𝑦^1/2, 𝑦(2) = 1 has a unique solution.

SOLUTION:

1 − 𝑓(𝑥, 𝑦) = 𝑥𝑦¹² = 𝑥√𝑦 is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ [0, ∞)}.

2 − 𝜕𝑓

𝜕𝑦 = 𝑥

2𝑦¹² = 𝑥

2√𝑦 is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ (0, ∞)}.

Since (2,1) is in the region defined by 𝑥 ∈ ℝ and 𝑦 > 0, this shows that the IVP has a unique solution.

14 Exercises 1.2: Page 17

(𝟏𝟖)

Determine a region of the 𝑥𝑦-plane for which the given differential equation would have a unique solution whose graph passes through a point (𝑥₀, 𝑦₀) in the region.

𝑑𝑦

𝑑𝑥 = √𝑥𝑦

SOLUTION:

1 − 𝑓(𝑥, 𝑦) = √𝑥𝑦 is continuous on either {(𝑥, 𝑦): 𝑥 ∈ [0, ∞) and 𝑦 ∈ [0, ∞)} or {(𝑥, 𝑦): 𝑥 ∈ (−∞, 0] and 𝑦 ∈ (−∞, 0]}.

2 − 𝜕𝑓

𝜕𝑦 =1 2 √𝑥

𝑦 is continuous on either {(𝑥, 𝑦): 𝑥 ∈ [0, ∞) and 𝑦 ∈ (0, ∞)} or {(𝑥, 𝑦): 𝑥 ∈ (−∞, 0] and 𝑦 ∈ (−∞, 0)}.

Thus, the differential equation will have a unique solution in any region where 𝑥 > 0 and 𝑦 > 0 or where 𝑥 < 0 and 𝑦 < 0.

--- (𝟐𝟕)

Determine whether Theorem 1.2.1 guarantees that the differential equation

𝑦

^′

= √𝑦

²

− 9

possesses a unique solution through the given point (2, −3). SOLUTION:

1 − 𝑓(𝑥, 𝑦) = √𝑦²− 9 is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ (−∞, −3] ∪ [3, ∞)}.

2 − 𝜕𝑓

𝜕𝑦 = 2𝑦

2√𝑦²− 9= 𝑦

√𝑦²− 9 is continuous on {(𝑥, 𝑦): 𝑥 ∈ ℝ and 𝑦 ∈ (−∞, −3) ∪ (3, ∞)}.

Since (2, −3) is not in either of the regions defined by 𝑦 < −3 or 𝑦 > 3, so there is no guarantee of a unique solution through (2, −3).

15 Exercises 1.2: Page 17 (Homework)

(𝟔)

In the following problem, 𝑦 = 1/(𝑥²+ 𝑐) is a one-parameter family of solutions of the first- order DE 𝑦^′+ 2𝑥𝑦² = 0. Find a solution of the first-order IVP consisting of this differential equation and the given initial condition. Give the largest interval 𝐼 over which the solution is defined.

𝑦 ( 1

2 ) = −4

---

(𝟗)

In the following problem, 𝑥 = 𝑐₁cos 𝑡 + 𝑐₂sin 𝑡 is a two-parameter family of solutions of the second-order DE 𝑥^′′+ 𝑥 = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

𝑥 ( 𝜋 6 ) = 1

2 , 𝑥

^′

( 𝜋

6 ) = 0

---

In the following problems, determine a region of the 𝑥𝑦-plane for which the given differential equation would have a unique solution whose graph passes through a point (𝑥₀, 𝑦₀) in the region.

(𝟏𝟕) 𝑑𝑦

𝑑𝑥 = 𝑦

^2/3

(𝟐𝟐) (1 + 𝑦

³

)𝑦

^′

= 𝑥

²

---

(𝟐𝟓)

In the following problem, determine whether Theorem 1.2.1 guarantees that the differential equation

𝑦

^′

= √𝑦

²

− 9

possesses a unique solution through the given point (1, 4).