research code2

Exploring Least Squares Solutions to Impulsive BoundaryValue Problems

Mark LaPointe and Cody Gerres

Concordia University, St. PaulPi Mu Epsilon

[email protected]

[email protected]

April 11, 2015

Mark L. and Cody G. (CSP) LSS to Impulsive BVPs April 11, 2015

[email protected]

[email protected]

Overview

Description of the problem

Applications

Examples

Formulation and Theory

Analysis

Results


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.


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


Applications

Several applications to this problem including:

Data Fitting

I PhysicsI StatisticsI EngineeringI Etc.


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.


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.


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.


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.



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



We explored the problem inside a Hilbert Space, which includes:1 Vector Space

2 Inner Product

3 Completeness



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



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.



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)


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


Analysis

< (h, v), (c , c) >=

< h, c > + < v , c >=

∫ 1

0

hc ds + vc

= c

(∫ 1

0

h ds + v

)

= 0


Analysis

< (c , c), (c , c) >=

∫ 1

0

c2 ds + c2 = 2c2 = 1

⇒ c2 = 1/2

⇒ c = ±√

1/2

⇒ c = 1/√

2


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

))


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

))


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)


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


Thank you.


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