Exploring Least Squares Solutions to Impulsive BoundaryValue Problems

Mark LaPointe and Cody Gerres

Concordia University, St. PaulPi Mu Epsilon

lapointm@csp.edu

gerresc@csp.edu

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