Answers or Hints

Example 12.1. Consider the initial value problem

25. 
    ^′ = sin(xy), y(0) = 1 (12.4)
    

in the rectangle S : |x| ≤ 1/2, |y − 1| ≤ 1/2. To apply Theorem 8.1 we note that a = 1/2, b = 1/2 and max_S |sin(xy)| ≤ 1 ≤ M, and from
Theorem 7.2 the function sin(xy) satisfies the Lipschitz condition (7.3) in
S, and max_S |xcos(xy)| = 1/2 = L. Thus, the problem (12.4) has a unique solution in the interval |x| ≤ h ≤ 1/2.
As an approximation of the initial value problem (12.4), we consider

25. 
    ^′ = xz, z(0) = 1.1, (12.5)
    

which also has a unique solution z(x) = 1.1exp(x²/2) in the interval |x| ≤
1/2. Now by Taylor’s formula, we find
.
Using Theorem 12.1 for the above initial value problems, we obtain an upper error bound for the difference between the solutions y(x) and z(x)
for all .
To emphasize the dependence of the initial point (x₀,y₀), we shall denote the solution of the initial value problem (7.1) as y(x,x₀,y₀). In our next result we shall show that y(x,x₀,y₀) is differentiable with respect to y₀.
Theorem 12.2. Let the following conditions be satisfied:

1. 
   f(x,y) is continuous and bounded by M in a domain D containing the point (x₀,y₀).
   
2. 
   ∂f(x,y)/∂y exists, continuous and bounded by L in D.
   
3. 
   The solution y(x,x₀,y₀) of the initial value problem (7.1) exists in an interval J containing x₀.
   

Then we have that y(x,x₀,y₀) is differentiable with respect to y₀ and z(x) = ∂y(x,x₀,y₀)/∂y₀ is the solution of the initial value problem
(12.6)
z(x₀) = 1. (12.7)
The DE (12.6) is called the variational equation corresponding to the solution y(x,x₀,y₀).
Proof. Let (x₀,y₁) ∈ D be such that the solution y(x,x₀,y₁) of the initial value problem y ^′ = f(x,y), y(x₀) = y₁ exists in an interval J₁. Then for all x ∈ J₂ = J ∩ J₁, Theorem 12.1 implies that
|y(x,x₀,y₀) − y(x,x₀,y₁)| ≤ |y₀ − y₁|e^{L |x − x0 |} ,
i.e., |y(x,x₀,y₀) − y(x,x₀,y₁)| → 0 as |y₀ − y₁| → 0.
Now for all x ∈ J₂ it is easy to verify that

where δ{y(x,x₀,y₀),y(x,x₀,y₁)} → 0 as |y(x,x₀,y₀) − y(x,x₀,y₁)| → 0, i.e., as |y₀ − y₁| → 0.
Hence, we find that
.
Thus, |y(x,x₀,y₀) − y(x,x₀,y₁) − z(x)(y₀ − y₁)| → 0 as |y₀ − y₁| → 0. This completes the proof.
In our next result we shall show that the conditions of Theorem 12.2 are also sufficient for the solution y(x,x₀,y₀) to be differentiable with respect to x₀.
Theorem 12.3. Let the conditions of Theorem 12.2 be satisfied. Then the solution y(x,x₀,y₀) is differentiable with respect to x₀ and z(x) = ∂y(x,x₀,y₀)/∂x₀ is the solution of the variational equation (12.6), satisfying the initial condition
z(x₀) = − f(x₀,y₀). (12.8)
Proof. The proof is similar to that of Theorem 12.2.
We note that the variational equation (12.6) can be obtained directly by differentiating the relation y ^′ (x,x₀,y₀) = f(x,y(x,x₀,y₀)) with respect to y₀ (or x₀). Further, since y(x₀,x₀,y₀) = y₀, differentiation with respect to y₀ gives the initial condition (12.7). To obtain (12.8), we begin with the integral equation

and differentiate it with respect to x₀, to obtain
.
Finally, in this lecture we shall consider the initial value problem
y ^′ = f(x,y,λ), y(x₀) = y₀, (12.9)
where λ ∈ IR is a parameter. The proof of the following theorem is very similar to earlier results.
Theorem 12.4. Let the following conditions be satisfied:

1. 
   f(x,y,λ) is continuous and bounded by M in a domain D ⊂ IR³ containing the point (x₀,y₀,λ₀).
   
2. 
   ∂f(x,y,λ)/∂y, ∂f(x,y,λ)/∂λ exist, continuous and bounded, respectively, by L and L₁ in D.
   

Then the following hold:

1. 
   There exist positive numbers h and ǫ such that given any λ in the interval |λ − λ₀| ≤ ǫ, there exits a unique solution y(x,λ) of the initial value problem (12.9) in the interval |x − x₀| ≤ h.
   
2. 
   For all λ₁, λ₂ in the interval |λ − λ₀| ≤ ǫ, and x in |x − x₀| ≤ h the following inequality holds:
   

. (12.10)

3. 
   The solution y(x,λ) is differentiable with respect to λ and z(x,λ) = ∂y(x,λ)/∂λ is the solution of the initial value problem
   

, (12.11)
z(x₀,λ) = 0. (12.12)
If λ is such that |λ − λ₀| is sufficiently small, then we have a first-order approximation of the solution y(x,λ) given by
(12.13)

We illustrate this important idea in the following example. Example 12.2. Consider the initial value problem
y ^′= λy^2+ 1, y(0) = 0 (λ ≥ 0)	(12.14)

for which the solution y(x,λ) = (1/√λ)tan(√λx) exists in the inter-
val (−π/(2√λ),π/(2√λ)). Let in (12.14) the parameter λ = 0, so that y(x,0) = x. Since ∂f/∂y = 2λy and ∂f/∂λ = y², the initial value problem corresponding to (12.11), (12.12) is
z ^′ (x,0) = x², z(0,0) = 0,
whose solution is z(x,0) = x³/3. Thus, for λ near zero, (12.13) gives the approximation
.
Problems
12.1. Let y(x,λ) denote the solution of the initial value problem
y ^′ + p(x)y = q(x), y(x₀) = λ
in the interval x₀ ≤ x < ∞. Show that for each fixed x > x₀ and for each positive number ǫ there exists a positive number δ such that |y(x,λ+∆λ)− y(x,λ)| ≤ ǫ, whenever |∆λ| ≤ δ, i.e., the solution y(x,λ) is continuous with respect to the parameter λ.

2. 
   Prove Theorem 12.3.
   
3. 
   Prove Theorem 12.4.
   
4. 
   For the initial value problem
   

y ^′ = x + e^x sin(xy), y(0) = 0 = y₀
estimate the variation of the solution in the interval [0,1] if y₀ is perturbed by 0.01.

5. 
   For the initial value problem y ′ = λ + cosy, y(0) = 0 obtain an upper estimate for |y(x,λ₁) − y(x,λ₂)| in the interval [0,1].
   
6. 
   For sufficiently small λ find a first-order approximation of the solution y(x,λ) of the initial value problem y ^′ = y + λ(x + y²), y(0) = 1.
   
7. 
   State and prove an analog of Theorem 12.1 for the initial value problem (7.9).
   
8. 
   Find the error in using the approximate solution
   

y(x) = exp(−x³/6)
for the initial value problem y ′′ +xy = 0, y(0) = 1, y ′ (0) = 0 in the interval |x| ≤ 1/2.
Answers or Hints

1. 
   Let z(x) = y(x,λ+∆λ)−y(x,λ). Then z ′ (x)+p(x)z(x) = 0, z(x₀) =
   

∆λ, whose solution is .

2. 
   The proof is similar to that of Theorem 12.2.
   
3. 
   For the existence and uniqueness of solutions of (12.9) see the remark following Theorem 16.7. To prove inequality (12.10) use Corollary 7.6.
   
4. 
   0.01e^e.
   
5. 
   |λ₁ − λ₂|(e − 1).
   
6. 
   e^x + λ(e^2x − x − 1).
   
7. 
   The result corresponding to Theorem 12.1 for (7.9) can be stated as follows: Let the conditions (i)–(iii) of Theorem 12.1 be satisfied and (iv)′ y(x) and z(x), the solutions of (7.9) and z ′′ = f(x,z)+g(x,z), z(x₁) = z₀, z ^′ (x₁) = z₁, respectively, exist in J = (α,β) containing x₀ and x₁. Then for all x ∈ J, the following inequality holds:
   

where γ = [|y₁ − z₁| + M₁(β − α)]/L(β − α).

8. 
   (1/512)e^1/12.Lecture 13
   
9. 
   Preliminary Results from Algebra and Analysis
   
10. 
    For future reference we collect here several fundamental concepts and results from algebra and analysis.
    
11. 
    A function P_n(x) defined by
    
12. 
    
    
13. 
    where a_i ∈ IR, 0 ≤ i ≤ n, is called a polynomial of degree n in x. If P_n(x₁) = 0, then the number x = x₁ is called a zero of P_n(x). The following fundamental theorem of algebra of complex numbers is valid.
    
14. 
    Theorem 13.1. Every polynomial of degree n ≥ 1 has at least one zero.
    
15. 
    Thus, P_n(x) has exactly n zeros; however, some of these may be the same, i.e., P_n(x) can be written as
    
16. 
    P_n(x) = a_n(x − x₁)^r1(x − x₂)^r2 ···(x − x_k)^rk, r_i ≥ 1, 1 ≤ i ≤ k,
    
17. 
    where , then x_i is called a simple zero, and if r_i >
    
18. 
    1, then multiple zero of multiplicity r_i. Thus, if x_i is a multiple zero of multiplicity r_i, then .
    
19. 
    A rectangular table of m×n elements arranged in m rows and n columns
    
20. 
    
    
21. 
    is called an m × n matrix and in short represented as A = (a_ij). We shall mainly deal with square matrices (m = n), row matrices or row vectors (m = 1), and column matrices or column vectors (n = 1).
    
22. 
    A matrix with a_ij = 0, 1 ≤ i,j ≤ n, is called null or zero matrix, which we shall denote by 0.
    
23. 
    R.P. Agarwal and D. O’Regan, An Introduction to Ordinary Differential Equations, 91 doi: 10.1007/978-0-387-71276-5_13, © Springer Science + Business Media, LLC 2008
    
24. 
    
25. 
    A matrix with is called a diagonal matrix; if, in addition, a_ii = 1, 1 ≤ i ≤ n, then it is an identity matrix which we shall denote by I.
    
26. 
    The transpose of a matrix denoted by A^T is a matrix with elements a_ji. A matrix is called symmetric if A = A^T.
    
27. 
    The sum of two matrices A = (a_ij) and B = (b_ij) is a matrix C = (c_ij) with elements c_ij = a_ij + b_ij.
    
28. 
    Let α be a constant and A be a matrix; then αA is a matrix C = (c_ij) with elements c_ij = αa_ij. Let A and B be two matrices; then the product AB is a matrix C = (c_ij)
    
29. 
    n
    
30. 
    . Note that in general
    
31. 
    ( ) = .
    
32. 
    The trace of a matrix A is denoted by TrA and it is the sum of the diagonal elements, i.e.,
    
33. 
    Tr .
    
34. 
    Associated with an n × n matrix A = (a_ij) there is a scalar called the determinant of A
    
35. 
    .
    
36. 
    An (n − 1) × (n − 1) determinant obtained by deleting ith row and jth column of the matrix A is called the minor a˜_ij of the element a_ij. We define the cofactor of a_ij as α_ij = (−1)^i+ja˜_ij. In terms of cofactors the determinant of A is defined as
    
37. 
    . (13.1)
    
38. 
    Further,
    
39. 
    = 0 if (13.2)
    
40. 
    The following properties of determinants are fundamental:
    

1. 
   If two rows (or columns) of A are equal or have a constant ratio, then detA = 0.
   
2. 
   If any two consecutive rows (or columns) of A are interchanged, then the determinant of the new matrix A₁ is −detA.
   
3. 
   If a row (or column) of A is multiplied by a constant α, then the determinant of the new matrix A₁ is αdet A.
   

1. 
   Preliminary Results from Algebra and Analysis 93
   

4. 
   If a constant multiple of one row (or column) of A is added to another, then the determinant of the new matrix A₁ is unchanged.
   
5. 
   detA^T = detA.
   
6. 
   detAB = (detA)(detB).
   

1. 
   A linear system of n equations in n unknowns is a set of equations of the following form:
   
2. 
   ( 13.3)
   
3. 
   ,
   
4. 
   w hereare given real numbers and n unknowns are u_i, 1 ≤ ≤ n.
   
5. 
   The system (13.3) can be written in a compact form as
   
6. 
   Au = b, (13.4)
   
7. 
   where A is an n×n matrix (a_ij), b is an n×1 vector (b_i), and u is an n×1 unknown vector (u_i). If b = 0, the system (13.4) is called homogeneous, otherwise it is called nonhomogeneous.
   
8. 
   The following result provides a necessary and sufficient condition for the system (13.4) to have a unique solution.
   
9. 
   Theorem 13.2. The system (13.4) has a unique solution if and only if det . Alternatively, if the homogeneous system has only the trivial solution (u = 0), then det .
   
10. 
    If detA = 0 then the matrix A is said to be singular; otherwise, nonsingular. Thus, the homogeneous system has nontrivial solutions if and only if the matrix A is singular. The importance of this concept lies in the fact that a nonsingular matrix A possesses a unique inverse denoted by A − ¹. This matrix has the property that AA− ¹ = A − ¹A = I. Moreover, A − ¹ = (AdjA)/(detA), where AdjA is an n×n matrix with elements α_ji.
    
11. 
    A real vector space (linear space) V is a collection of objects called vectors, together with an addition and a multiplication by real numbers which satisfy the following axioms:
    
12. 
    1. Given any pair of vectors u and v in V there exists a unique vector u+v in V called the sum of u and v. It is required that
    

1. 
   addition be commutative, i.e., u + v = v + u;
   
2. 
   addition be associative, i.e., u + (v + w) = (u + v) + w;
   
3. 
   there exists a vector 0 in V (called the zero vector) such that u+0 = 0 + u = u for all u in V ;