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Exploring Least Squares Solutions to Impulsive BoundaryValue Problems
Mark LaPointe and Cody Gerres
Concordia University, St. PaulPi Mu Epsilon
April 11, 2015
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Overview
Description of the problem
Applications
Examples
Formulation and Theory
Analysis
Results
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Description of Problem
We will be analyzing problems with the following criteria:
x ′(t) = h(t), for t ∈ [0, 1] \ {1/2}
x(1/2+)− x(1/2−) = v
bx(0) + dx(1) = 0.
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Description of Problem
If x ′(t) = h with impulse v , then we have solutions of this form:
x(t) =
∫ t
0h(s) ds + x(0), for 0 ≤ t < 1/2
v +∫ t
0h(s) ds + x(0), for 1/2 < t ≤ 1
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Applications
Several applications to this problem including:
Data Fitting
I PhysicsI StatisticsI EngineeringI Etc.
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Example 1
Consider the following impulsive boundary value problem:
x ′(t) = t3, for t ∈ [0, 1] \ {1/2}
x(1/2+)− x(1/2−) = 3/4
x(0)− x(1) = 0.
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Analysis of Example 1
Since x ′(t) = t3 and v = 3/4, by FTC, we get:
x(t) =
∫ t
0s3 ds + x(0), for 0 ≤ t < 1/2
3/4 +∫ t
0s3 ds + x(0), for 1/2 < t ≤ 1
Evaluating at our boundary conditions:
x(0)− x(1) = 1/4 + 3/4 = 1 6= 0
⇒ This certain problem has no solutions.
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Example 2
Now, consider another impulsive boundary value problem:
x ′(t) = t3, for t ∈ [0, 1] \ {1/2}
x(1/2+)− x(1/2−) = −1/4
x(0)− x(1) = 0.
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Analysis of Example 2
Similar to the first example, we get:
x(t) =
∫ t
0s3 ds + x(0), for 0 ≤ t < 1/2
−1/4 +∫ t
0s3 ds + x(0), for 1/2 < t ≤ 1
Evaluating at our boundary conditions:
x(0)− x(1) = 1/4 + (−1/4) = 0
⇒ This problem has solutions.
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Formulation and Theory
Once we looked more closely at our boundary conditions:
bx(0) + dx(1) = 0
bx(0) + d
(v +
∫ 1
0
h(t) dt + x(0)
)= 0
⇒ (b + d)x(0) = −d(v +
∫ 1
0
h(s) ds
)
⇒ x(0) =−d(v +
∫ 1
0h(s) ds
)b + d
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Formulation and Theory
We explored the problem inside a Hilbert Space, which includes:1 Vector Space
2 Inner Product
3 Completeness
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Formulation and Theory
We explored the problem inside a Hilbert Space, which includes:1 Vector Space
2 Inner Product
3 Completeness
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Formulation and Theory
We explored the problem inside a Hilbert Space, which includes:1 Vector Space
2 Inner Product
3 Completeness
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Formulation and Theory
Visualize: R3 with plane M and a point w outside M, where Mrepresents the solutions to our problem.
P() is our projection of functions outside of M into M.I Meaning the closest point in M to w is P(w)
Suppose M = span{e1, e2, ..., en}, where {e1, e2, ..., en} is anorthonormal basis with norm-1.
I < ei , ei >= ‖e1‖2 = ‖ei‖ = 1I < ei , ej >= 0, when i 6= j
In our problem, w represents the ordered pair (h, v)
Claim P(w) =< e1,w > e1+ < e2,w > e2 + ...+ < en,w > en
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Formulation and Theory
Let y ∈ M , then:
< w − y ,w − y >= ‖w − y‖2 = ‖w − P(w) + P(w)− y‖2
= ‖w − P(w)‖2 + ‖P(w)− y‖2
≥ ‖w − P(w)‖2.
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Formulation and Theory
Define: L : PC ′[0, 1] \ {1/2} ⊂ L2[0, 1] 7→ L2[0, 1]× R, then
L(x) = (x ′, x(1/2+)− x(1/2−))
⇒ L(x) = (h, v)
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Analysis
(h,v) ∈ Rng(L) iff, ∫ 1
0
h(t) dt + v = 0
Let Q denote the projection onto Rng(L)⊥
Consider the equation P = I − Q
Kernel (L(x) = 0)
I Rng(L)⊥ is one-dimensional
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Analysis
< (h, v), (c , c) >=
< h, c > + < v , c >=
∫ 1
0
hc ds + vc
= c
(∫ 1
0
h ds + v
)
= 0
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Analysis
< (c , c), (c , c) >=
∫ 1
0
c2 ds + c2 = 2c2 = 1
⇒ c2 = 1/2
⇒ c = ±√
1/2
⇒ c = 1/√
2
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Results
Define: Q : L2[0, 1]× R 7→ Q : L2[0, 1]× R, then
Q(h, v) =< (h, v), (1/√
2, 1/√
2) > (1/√
2, 1/√
2)
= (1/2)
(∫ 1
0
h ds + v
)(1, 1)
=
((1/2)
(∫ 1
0
h ds + v
), (1/2)
(∫ 1
0
h ds + v
))
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Solving for P
Recall the equation P = I − Q.
P(h, v) = (h, v)−(
(1/2)
(∫ 1
0
h ds + v
), (1/2)
(∫ 1
0
h ds + v
))
=
(h − (1/2)
(∫ 1
0
h ds + v
), v − (1/2)
(∫ 1
0
h ds + v
))
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Example 1 Revisited
x ′(t) = t3, for t ∈ [0, 1] \ {1/2}
x(1/2+)− x(1/2−) = 3/4
x(0)− x(1) = 0.
P(t3, 3/4) = (t3− (1/2)((1/4) + (3/4)), (3/4)− (1/2)((1/4) + (3/4))
= (t3 − 1/2, 1/4)
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Example 1 Revisited
Thus, the closest approximate solutions for this problem look like this:
x(t) =
∫ t
0s3 − 1/2 ds, for 0 ≤ t < 1/2
1/4 +∫ t
0s3 − 1/2 ds, for 1/2 < t ≤ 1
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015
Thank you.
Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015