Application of the Haar wavelet transform to solving integral and ...

Satisfying the boundary conditions, we get u(1) = a 1 + u(0) or
a 1 = B − A. (29)
Since a 1 is now prescribed, we have ∆a 1 = 0; it means that the matrix of
system (22) must be replaced with a reduced matrix for which the first row and
column are deleted. Besides, (23) obtains the form
For further details consult again [ 33 ].
Example 3. Consider the boundary value problem
u ′′ − uu ′ =
∂K
= ∂K
∂a i ∂u p i,1(t) + ∂K
∂u ′ h i(t). (30)
1
4 √ (1 + x) 3 − 1 2 , u(0) = 1, u(1) = √ 2, (31)
which has the exact solution
u = √ 1 + x.
By integrating (31) we obtain
u ′ (x) =
∫ x
0
uu ′ dt + u ′ (0) − 1 2 (1 + x) + 1
2 √ 1 + x
(32)
with the initial condition u(0) = 1.
a 1 = u(1) − u(0) = √ 2 − 1.
According to (28),
where
Now
u ′ (0) =
It follows from boundary conditions that
2M∑
i=1
a i h i (0), (33)
⎧
⎨ 1 if i = 1,
h i (0) = 1 if i = 2 j + 1, j = 0, 1, 2,
⎩
0 elsewhere.
∂K
∂a i
= u ′ (t)p i,1 (t) + u(t)h i (t), i = 2, 3, ..., 2M.
Some numerical calculations were carried out. The errors of the obtained results
are presented in Table 3.
38