1. The Elementary Theory of Initial-Value Problems
Sec:5.1
The Elementary Theory of Initial-Value Problems

2. Example of a Differential Equation
Sec:5.1 The Elementary Theory of Initial-Value Problems
Differential equations are used to model problems in science and engineering that involve the change of some variable with respect to another.
Example of a Differential Equation
𝑑𝑦 𝑑𝑡 =−2𝑦(𝑡)
𝑦 𝑡 = 𝑒 −2𝑡
Initial Value Problem
Find a function 𝑦(𝑡) that satisfies the above DE
Find a function 𝑦(𝑡) that satisfies both the DE and the initial condition
𝑑𝑦 𝑑𝑡 =−2𝑦(𝑡)
𝑦 0 =5
𝑦 𝑡 = 5𝑒 −2𝑡
Sec(5.1) Theory IVP
Sec(5.2) numerical solution for IVP Euler Method
(𝟎, 𝟓)
Sec(5.4) numerical solution for IVP Runge Kutta Method

3. Sec:5.1 The Elementary Theory of Initial-Value Problems
IVP
IVP
IVP
𝑦 ′ = 𝑦 3 +𝑦 3 𝑦 2 +1 𝑡
𝑑𝑦 𝑑𝑡 =−2𝑦(𝑡)
𝑑𝑣 𝑑𝑡 =𝑔− 𝑐 𝑚 𝑣
1≤𝑡≤2
𝑦 1 =1
𝑦 0 =5
𝑣 0 =0
ODE
IVP
𝑑[𝐴] 𝑑𝑡 =− 𝑘 1 𝐴 𝐵 − 2𝑘 2 𝐴 𝐴 𝐶
𝑑[𝐵] 𝑑𝑡 =− 𝑘 1 𝐴 𝐵 + 𝑘 2 𝐴 𝐴 [𝐶]
𝑑[𝐶] 𝑑𝑡 = 𝑘 1 𝐴 [𝐵]− 𝑘 2 𝐴 𝐴 [𝐶]
𝑑[𝐷] 𝑑𝑡 = 𝑘 1 𝐴 [𝐵]
𝑑 𝑑𝑥 𝐶 𝐵 𝜃 =− 𝑃 𝐻 2 𝜅 0 𝐾 0 𝑇 𝑒𝑥𝑝 (−𝑄− 𝐸 𝑎 ) 𝑅 𝑔 𝑇 𝐶 𝐵
𝐶 𝐵 = 𝐶 𝐵 0 at 𝑥=0
IVP
𝑦 ′ =𝑓(𝑡,𝑦)
𝑎≤𝑡≤𝑏
All these IVP’s can be written as a one general form
𝑦 𝑎 =𝛼

4. Sec:5.1 The Elementary Theory of Initial-Value Problems
We need some definitions and results from the theory of ordinary differential equations before considering methods for approximating the solutions to initial-value problems.
An IVP may have no solution, unique solution or infinitely many solutions.
IVP
𝑦 ′ =𝑓(𝑡,𝑦)
𝑎≤𝑡≤𝑏
𝑦′ 𝑦 = 0,
𝑦(0) = 1
𝑦 𝑎 =𝛼
𝑥𝑦’ = 𝑦−1,
𝑦(0) = 1
𝑦 = 1 + 𝛼𝑥
1) Condition on 𝑓(𝑡,𝑦)
IVP have a unique solution
2) Condition on the domain 𝐷= 𝑡,𝑦

5. Sec:5.1 The Elementary Theory of Initial-Value Problems
clear; clc; syms y(t)
ode1 = diff(y,t) == sqrt(1-y^2)
cond1 = y(0) == 0
y1Sol(t) = dsolve(ode1,cond1)
f1 y1Sol(t)
x=0:0.01:2*pi; y1=f1(x);
plot(x,y1,'-r'); grid on
IVP
𝒚 ′ = 𝟏− 𝒚 𝟐
0≤𝑡≤2𝜋
𝒚 𝟎 =𝟎
ode1(t) =
diff(y(t), t) == (1 - y(t)^2)^(1/2)
cond1 =
y(0) == 0
y1Sol(t) =
sin(t)

6. Sec:5.1 The Elementary Theory of Initial-Value Problems

7. Sec:5.1 The Elementary Theory of Initial-Value Problems
We need some definitions and results from the theory of ordinary differential equations before considering methods for approximating the solutions to initial-value problems.
An IVP may have no solution, unique solution or infinitely many solutions.
IVP
𝑦 ′ =𝑓(𝑡,𝑦)
𝑎≤𝑡≤𝑏
𝑦′ 𝑦 = 0,
𝑦(0) = 1
𝑦 𝑎 =𝛼
𝑥𝑦’ = 𝑦−1,
𝑦(0) = 1
𝑦 = 1 + 𝛼𝑥
1) Condition on 𝑓(𝑡,𝑦)
satisfies a Lipschitz condition
2) Condition on the domain 𝐷= 𝑡,𝑦
convex

8. Sec:5.1 The Elementary Theory of Initial-Value Problems
Definition
Special Case
a set is convex provided that whenever two points belong to the set, the entire straight-line segment between the points also belongs to the set.
D = {(t, y) | a ≤ t ≤ b and −∞ < y < ∞} 𝒂 𝒃 𝑡 𝑦
Definition
A set D ⊂ R2 is said to be convex if whenever (t1, y1) and (t2, y2) belong to D, then
( (1 − λ)t1 + λt2 , (1 − λ)y1 + λy2 )
also belongs to D for every λ in [0, 1].
convex

9. |𝒇 (𝒕, 𝒚 𝟏 ) − 𝒇 (𝒕, 𝒚 𝟐 )| ≤ 𝑳| 𝒚 𝟏 − 𝒚 𝟐 |,
Sec:5.1 The Elementary Theory of Initial-Value Problems
Definition
Theorem 5.3
A function f (t, y) is said to satisfy a Lipschitz condition in the variable y on a set 𝐷 ⊂ 𝑅 2 if a constant L > 0 exists with
|𝒇 (𝒕, 𝒚 𝟏 ) − 𝒇 (𝒕, 𝒚 𝟐 )| ≤ 𝑳| 𝒚 𝟏 − 𝒚 𝟐 |,
whenever (t, y1) and (t, y2) are in D. The constant L is called a Lipschitz constant for f .
𝐷 ⊂ 𝑅 2
convex 1
𝜕𝑓 𝜕𝑦 (𝑡, 𝑦) ≤ 𝐿
𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑡, 𝑦) ∈ 𝐷 2
Example
Show that 𝑓 (𝑡, 𝑦) = 𝑡| 𝑦| satisfies a Lipschitz condition on D = {(t, y) | 1 ≤ t ≤ 2 and − 3 ≤ y ≤ 4}.
𝑓 𝑡, 𝑦1 − 𝑓 𝑡, 𝑦2 =
𝑡 𝑦 − 𝑡|𝑦2|
f satisfies a Lipschitz condition on D
= 𝑡 𝑦1 − 𝑦2
= 𝑡 ∙ 𝑦1 − 𝑦2
≤ 2| 𝑦1 − 𝑦2|.

10. Theorem 5.4 (Main Theorem)
Sec:5.1 The Elementary Theory of Initial-Value Problems
Theorem 5.4 (Main Theorem)
IVP 1
D = {(t, y) | a ≤ t ≤ b and −∞ < y < ∞}
𝑦 ′ =𝑓(𝑡,𝑦)
𝑎≤𝑡≤𝑏 2
𝑓 (𝑡, 𝑦) is continuous on D
𝑦 𝑎 =𝛼 3
𝑓 satisfies a Lipschitz condition on D in y
has a unique solution 𝑦(𝑡)
Example 1
D = {(t, y) | 0 ≤ t ≤ 2 and −∞ < y < ∞}
Use Theorem 5.4 to show that there is a unique solution to the initial-value problem
f (t, y) is continuous when 0 ≤ t ≤ 2 and −∞ < y < ∞, 2
𝑦 = 1 + 𝑡 sin 𝑡𝑦
0≤ 𝑡 ≤ 2,
𝑦(0) = 0.
Mean Value Theorem
𝑓 (𝑡, 𝒚 𝟏 ) − 𝑓 (𝑡, 𝒚 𝟐 ) 𝒚 𝟏 − 𝒚 𝟐 ≤ 𝜕𝑓 𝜕𝑦 (𝑡, 𝝃)
= 𝑡 2 𝑐𝑜𝑠(𝑡𝜉)
𝑓 (𝑡, 𝒚 𝟏 ) − 𝑓 (𝑡, 𝒚 𝟐 ) ≤ 𝑡 2 𝑐𝑜𝑠(𝑡𝜉) 𝒚 𝟏 − 𝒚 𝟐
Sol: 3
𝑓 (𝑡, 𝒚 𝟏 ) − 𝑓 (𝑡, 𝒚 𝟐 ) ≤𝟒 𝒚 𝟏 − 𝒚 𝟐
𝑓(𝑡,𝑦) = 1 + 𝑡 sin 𝑡𝑦
𝑎=0, 𝑏=2
Theorem 5.4 implies that a unique solution exists to this initial-value problem.

11. Sec:5.1 The Elementary Theory of Initial-Value Problems
Definition
The initial-value problem
𝑦 ′ =𝑓(𝑡,𝑦)
𝑎≤𝑡≤𝑏
𝑦 𝑎 =𝛼
is said to be a well-posed problem if:
• A unique solution, 𝑦(𝑡), to the problem exists
• small changes in the statement of the problem introduce correspondingly small changes in the solution
IVP
IVP1
𝑦 ′ =𝑡𝑦
𝑦 1 ′ =𝑡 𝑦 1
0≤𝑡≤1
0≤𝑡≤1
𝒚(𝒕)
𝑦 0 =1/3
𝑦 1 0 =0.33
𝑦 𝑡 = 1 3 𝑒 𝑡 2 /2
𝑦 1 𝑡 = 0.33𝑒 𝑡 2 /2
𝟏 𝟑
𝑦 1 𝑡
𝟎.𝟑𝟑

12. Sec:5.1 The Elementary Theory of Initial-Value Problems
Definition
IVP
IVP1
The initial-value problem
𝑦 ′ = 𝟏− 𝑦 2
𝑦 ′ = 𝟏.𝟏− 𝑦 2
𝑦 ′ =𝑓(𝑡,𝑦)
𝑎≤𝑡≤𝑏
𝑦 𝑎 =𝛼
0≤𝑡≤2𝜋
0≤𝑡≤2𝜋
𝑦 0 =𝟎
𝑦 0 =𝟎.𝟏
is said to be a well-posed problem if:
• A unique solution, 𝑦(𝑡), to the problem exists
• small changes in the statement of the problem introduce correspondingly small changes in the solution
𝑦 𝑡 =sin⁡(𝑡)
𝑦 1 𝑡 = sin⁡ 𝑡+ 𝑠𝑖𝑛 −
𝟎.𝟏
𝑦 1 𝑡 𝟎
𝒚(𝒕)

13. Sec:5.1 The Elementary Theory of Initial-Value Problems
clear; clc; syms y(t)
ode1 = diff(y,t) == sqrt(1-y^2)
cond1 = y(0) == 0;
y1Sol(t) = dsolve(ode1,cond1)
f1 y1Sol(t);
ode2 = diff(y,t) == sqrt( y^2) ;
cond2 = y(0) == 0.1;
y2Sol(t) = dsolve(ode2,cond2)
f2 y2Sol(t);
x=0:0.01:2*pi; y1=f1(x); y2=f2(x);
plot(x,y1,'-r',x,y2,'-k'); grid on
IVP
IVP1
𝑦 ′ = 𝟏− 𝑦 2
𝑦 ′ = 𝟏.𝟏− 𝑦 2
0≤𝑡≤2𝜋
0≤𝑡≤2𝜋
𝑦 0 =𝟎
𝑦 0 =𝟎.𝟏

14. Sec:5.1 The Elementary Theory of Initial-Value Problems
Definition
The initial-value problem
𝑦 ′ =𝑓(𝑡,𝑦)
𝑎≤𝑡≤𝑏
𝑦 𝑎 =𝛼
is said to be a well-posed problem if:
• A unique solution, 𝑦(𝑡), to the problem exists
• small changes in the statement of the problem introduce correspondingly small changes in the solution
𝑧 ′ =𝑓 𝑡,𝑧 +𝛿(𝑡)
𝑧 𝑎 =𝛼+ 𝛿 0
𝑎≤𝑡≤𝑏
perturbed problem
𝑦 ′ =𝑓(𝑡,𝑦)
𝑦 𝑎 =𝛼
𝑎≤𝑡≤𝑏
There exist constants ε0 > 0 and k > 0 such that for any ε, with ε0 > ε > 0,
|δ(t)| < ε and |δ0| < ε,
has a unique solution z(t) that satisfies
|z(t) − y(t)| < kε for all t in [a, b].

15. Sec:5.1 The Elementary Theory of Initial-Value Problems
Theorem 5.6 (Well-posed)
IVP 1
D = {(t, y) | a ≤ t ≤ b and −∞ < y < ∞}
𝑦 ′ =𝑓(𝑡,𝑦)
𝑎≤𝑡≤𝑏 2
𝑓 (𝑡, 𝑦) is continuous on D
𝑦 𝑎 =𝛼 3
𝑓 satisfies a Lipschitz condition on D in y
is well-posed
Example
Sol:
𝑓 𝑡,𝑦 =𝑦− 𝑡 2 +1
Show that the initial-value problem
𝑎=0, 𝑏=2 1
D = {(t, y) | 0 ≤ t ≤ 2 and −∞ < y < ∞}
𝑦 ′ =𝑦− 𝑡 2 +1
0≤ 𝑡 ≤ 2,
𝑦(0) = 0.5
f (t, y) is continuous when 0 ≤ t ≤ 2 and −∞ < y < ∞, 2
𝜕𝑓 𝜕𝑦 𝑡, 𝝃 = 𝟏 =𝟏 3
Theorem 5.6 implies that the problem I well-posed.