Beaded Strings - Rice Universitycox/vigre/stringtalk_6_11_08.pdf · 2008. 6. 13. · Beaded Strings...
Transcript of Beaded Strings - Rice Universitycox/vigre/stringtalk_6_11_08.pdf · 2008. 6. 13. · Beaded Strings...
Beaded StringsForward and Inverse Problems
Hunter Gilbert, Walter Kelm, and Brian Leake
Physics of Strings PFUG
11 June 2008
Frequencies → Masses + LengthsMasses + Lengths → Frequencies
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 1 / 36
IntroductionWhy Study Strings?
Consists of a simple physical system
Governed by a system of differential equations
Interesting math for special conditions
Experimental results easy to acquire
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 2 / 36
IntroductionBrief History of the String Lab
Classes of problems we have worked on:
Dirac Damping of String (Sean Hardesty’s Thesis)
Viscous constant damping
Magnetic damping (Last summer)
Network of strings (Jesse Chan)
’http://cnx.org/content/m16177/latest/’
Physical lab for CAAM 335
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 3 / 36
IntroductionThis Summer: Beaded Strings
Point masses, taut masslessstring, fixed at both ends
Forward Problem
Given: masses, lengths, andinitial input
Find: the string’s motion
Inverse Problem
Given: the string’s motion forany input
Find: masses and lengths thatuniquely describe the system
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 4 / 36
Variable Meanings
A Beaded String is uniquely defined by masses and lengths:
mk = Mass of a particular bead.
`k = Distance between the k and k+1 beads.
We will use these expressions to get the equations of motion:
T (y ′) = Total Kinetic Energy as a function of time.
V (y) = Total Potential Energy as a function of time.
yk(t) = Vertical Displacment of a particular bead.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 5 / 36
Dynamics of a Beaded String
To find frequencies of vibration given masses and positions ofthe beads, we will use a system of differential equations.
Kinetic energy is the focus of one of the equations.
T (y ′) =1
2
n∑k=1
mk(y ′k(t))2
Symbol meanings:
T (y ′) = Kinetic energy as a function of time.mk = Mass of the given bead.y ′k = Velocity of given bead.
Gives only part of the required system.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 6 / 36
Dynamics of a Beaded String
To find frequencies of vibration given masses and positions ofthe beads, we will use a system of differential equations.
Kinetic energy is the focus of one of the equations.
T (y ′) =1
2
n∑k=1
mk(y ′k(t))2
Symbol meanings:
T (y ′) = Kinetic energy as a function of time.mk = Mass of the given bead.y ′k = Velocity of given bead.
Gives only part of the required system.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 6 / 36
Dynamics of a Beaded String
To find frequencies of vibration given masses and positions ofthe beads, we will use a system of differential equations.
Kinetic energy is the focus of one of the equations.
T (y ′) =1
2
n∑k=1
mk(y ′k(t))2
Symbol meanings:
T (y ′) = Kinetic energy as a function of time.mk = Mass of the given bead.y ′k = Velocity of given bead.
Gives only part of the required system.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 6 / 36
Dynamics of a Beaded String
Potential Energy is the work done by stretching the string
Work = Force X Distance = Tension X String Elongation
Assume constant tension σ, and measure elongation by:
√`2k + (yk+1 − yk)2 − `k = `k
√1 +
(yk+1 − yk)2
`2k− `k
≈ `k
(1 +
1
2
(yk+1 − yk)2
`2k
)− `k
≈ (yk+1 − yk)2
2`2k.
V (y) =σ
2
n∑k=0
(yk+1 − yk)2
`k
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 7 / 36
Dynamics of a Beaded String
Potential Energy is the work done by stretching the string
Work = Force X Distance = Tension X String Elongation
Assume constant tension σ, and measure elongation by:
√`2k + (yk+1 − yk)2 − `k = `k
√1 +
(yk+1 − yk)2
`2k− `k
≈ `k
(1 +
1
2
(yk+1 − yk)2
`2k
)− `k
≈ (yk+1 − yk)2
2`2k.
V (y) =σ
2
n∑k=0
(yk+1 − yk)2
`k
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 7 / 36
Dyanamics of a Beaded StringConservation of Energy
We can use the Euler-Lagrange Equation to generate a unifiedsystem of differential equations.
d
dt
∂T
∂y ′j+∂V
∂yj= 0, j = 1, ..., n
d
dt
∂T
∂y ′j(t) = mjy
′′j (t)
∂V
∂yj=
(− σ
`j−1
)yj−1 +
(σ
`j−1+σ
`j
)yj +
(−σ`j
)yj+1
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 8 / 36
Dynamics of a Beaded StringMatrix Form of Equations
Doing the necessary algebra provides n equations:
uk − uk+1
`k+
uk − uk−1
`k−1−mkλ
2uk = 0, k = 1, 2, ..., n
We can write this system in matrix form
My ′′(t) = −Ky(t)
Here, M is a diagonal matrix holding the masses of the beads.
K is a tridiagonal matrix that gives the elongation of the string
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 9 / 36
Dynamics of a Beaded String
Essential Linear AlgebraInverse: M−1M = I
Eigenvalues (λ) and eigenvectors (u)
Defined for Matrix A by:
Au = λu
They are fundamental properties that describe the matrixbehaviorFor a beaded string, they correspond to vibration frequency(eigenvalue) and vibration mode shape (eigenvector)
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 10 / 36
The Solution
If we can solve that particular differential equation, we will beable to know the frequencies of vibration.
With the inverse, we can say:
y ′′(t) = −M−1Ky(t)
With the eigenvalues of M−1K , we can rewrite the equation as:
y ′′(t) = −V ΛV−1y(t)
V is the matrix where each column is an eigenvector of M−1K ,and Λ is a diagonal matrix of its eigenvalues.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 11 / 36
The Solution
If we can solve that particular differential equation, we will beable to know the frequencies of vibration.
With the inverse, we can say:
y ′′(t) = −M−1Ky(t)
With the eigenvalues of M−1K , we can rewrite the equation as:
y ′′(t) = −V ΛV−1y(t)
V is the matrix where each column is an eigenvector of M−1K ,and Λ is a diagonal matrix of its eigenvalues.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 11 / 36
The Solution, pt2
Since Λ is a diagonal matrix, the n × n system reduces to nindependent scalar equations.
γ′′j (t) = −ω2j γj(t)
If we presume that the string begins with no initial velocity, wecan state
γj(t) = γj(0)cos(ωjt)
We then have:
y(t) =∑n
j=1 γj(0)cos(ωjt)vj
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 12 / 36
The Solution, pt2
Since Λ is a diagonal matrix, the n × n system reduces to nindependent scalar equations.
γ′′j (t) = −ω2j γj(t)
If we presume that the string begins with no initial velocity, wecan state
γj(t) = γj(0)cos(ωjt)
We then have:
y(t) =∑n
j=1 γj(0)cos(ωjt)vj
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 12 / 36
Summary
The masses are displaced as the superposition of nindependent vectors vibrating at distinct frequencies.These frequencies are in turn given by the eigenvaluesof the matrix M−1K
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 13 / 36
Hearing the Composition of a String
We are able to go from masses and positions to frequencies ofvibration.
The natural question is now to ask if we can hear the positionsand masses of beads on a string from the vibrations it undergoes.
Using the techniques put forward by Gantmacher and Krein, wecan solve this inverse problem.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 14 / 36
Hearing the Composition of a String
We are able to go from masses and positions to frequencies ofvibration.
The natural question is now to ask if we can hear the positionsand masses of beads on a string from the vibrations it undergoes.
Using the techniques put forward by Gantmacher and Krein, wecan solve this inverse problem.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 14 / 36
Recurrence
We wish to find values of M and K to solve Ku = λMu.
By simple matrix multiplication, we can show:(− σ
`j−1
)uj−1 +
(σ
`j−1+σ
`j
)uj +
(−σ`j
)uj+1 = λmjuj
Having a fixed left & right end implies u0 = 0 and un+1 = 0
Rearranging the above, we get :
uj+1 =
(− `j`j−1
)uj−1 +
(1 +
`j`j−1− λ`jmj
σ
)uj
Since we know un+1 = 0 when the conditions for a fixed-fixedstring are met...
We can use this condition to make sure λ is an eigenvalue.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 15 / 36
Recurrence
We wish to find values of M and K to solve Ku = λMu.
By simple matrix multiplication, we can show:(− σ
`j−1
)uj−1 +
(σ
`j−1+σ
`j
)uj +
(−σ`j
)uj+1 = λmjuj
Having a fixed left & right end implies u0 = 0 and un+1 = 0
Rearranging the above, we get :
uj+1 =
(− `j`j−1
)uj−1 +
(1 +
`j`j−1− λ`jmj
σ
)uj
Since we know un+1 = 0 when the conditions for a fixed-fixedstring are met...
We can use this condition to make sure λ is an eigenvalue.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 15 / 36
Recurrence
We wish to find values of M and K to solve Ku = λMu.
By simple matrix multiplication, we can show:(− σ
`j−1
)uj−1 +
(σ
`j−1+σ
`j
)uj +
(−σ`j
)uj+1 = λmjuj
Having a fixed left & right end implies u0 = 0 and un+1 = 0
Rearranging the above, we get :
uj+1 =
(− `j`j−1
)uj−1 +
(1 +
`j`j−1− λ`jmj
σ
)uj
Since we know un+1 = 0 when the conditions for a fixed-fixedstring are met...
We can use this condition to make sure λ is an eigenvalue.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 15 / 36
Polynomial Construction
When j = 1, we can say:
u2 =
(−`1`0
)u0 +
(1 +
`1`0− λ`1m1
σ
)u1
We now create a polynomial p of linear degree, and define itsuch that:u2 ≡ p1(λ)u1
Recalling the formula for a general element of the eigenvector,
uj+1 =
(− `j`j−1
)uj−1 +
(1 +
`j`j−1− λ`jmj
σ
)uj
We can reuse the same trick from above to create a polynomialof degree j.uj+1 ≡ pj(λ)u1
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 16 / 36
Polynomial Construction
When j = 1, we can say:
u2 =
(−`1`0
)u0 +
(1 +
`1`0− λ`1m1
σ
)u1
We now create a polynomial p of linear degree, and define itsuch that:u2 ≡ p1(λ)u1
Recalling the formula for a general element of the eigenvector,
uj+1 =
(− `j`j−1
)uj−1 +
(1 +
`j`j−1− λ`jmj
σ
)uj
We can reuse the same trick from above to create a polynomialof degree j.uj+1 ≡ pj(λ)u1
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 16 / 36
Insight
We can build pn by following the recurrence, which requiresknowledge of lengths & masses.
Or, if we have knowledge of the roots of the polynomialbeforehand, we can construct a polynomial of degree n multipliedby a real coefficient that will provide the same behavior.
We already have this knowledge.
The string is fixed at both ends, requiring u0 = 0 and un+1 = 0
Therefore, we can say
pn(λ) = γ∏n
j=1 (λ− λj)
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 17 / 36
Insight
We can build pn by following the recurrence, which requiresknowledge of lengths & masses.
Or, if we have knowledge of the roots of the polynomialbeforehand, we can construct a polynomial of degree n multipliedby a real coefficient that will provide the same behavior.
We already have this knowledge.
The string is fixed at both ends, requiring u0 = 0 and un+1 = 0
Therefore, we can say
pn(λ) = γ∏n
j=1 (λ− λj)
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 17 / 36
Fixed-Fixed & Fixed-flat
In order to make the solution unique, we need to find moreinformation.We will break our assumption that the string is fixed at bothends.Allow the string to move at one end, but require it to have 0slope at its last node.
There is now another set of eigenvalues, λ′, that represent the
system.The system is no longer underdetermined.
Figure: Fixed-Flat & Fixed-Fixed Strings
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 18 / 36
Fixed-Fixed & Fixed-flat
In order to make the solution unique, we need to find moreinformation.We will break our assumption that the string is fixed at bothends.Allow the string to move at one end, but require it to have 0slope at its last node.There is now another set of eigenvalues, λ
′, that represent the
system.The system is no longer underdetermined.
Figure: Fixed-Flat & Fixed-Fixed StringsHunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 18 / 36
Even More Polynomials
We can develop recurrence relations for the slope of the string,like was done for the positions of the beads.
Rearranging our equation developed from matrix multiplication:
uj+1 − uj
`j=
uj − uj−1
`j−1−(λmj
σ
)uj
Using the polynomials already developed, we can rewrite theequation as:
uj+1 − uj
`j=
pj(λ)u1 − pj−1(λ)u1
`j
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 19 / 36
Even More Polynomials
We can develop recurrence relations for the slope of the string,like was done for the positions of the beads.
Rearranging our equation developed from matrix multiplication:
uj+1 − uj
`j=
uj − uj−1
`j−1−(λmj
σ
)uj
Using the polynomials already developed, we can rewrite theequation as:
uj+1 − uj
`j=
pj(λ)u1 − pj−1(λ)u1
`j
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 19 / 36
Algebra in Anticipation
Using the same semantic trick as earlier, we define a newpolynomial qn that characterizes the system in fixed-flatoperation.
qj(λ) =1
`j(pj(λ)− pj−1(λ))
Equating expressions for the left and right sides of the equationused to define qj gives us:
qj(λ)u1 = qj−1(λ)−(λmj
σ
)pj−1(λ)u1
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 20 / 36
Algebra in Anticipation
Using the same semantic trick as earlier, we define a newpolynomial qn that characterizes the system in fixed-flatoperation.
qj(λ) =1
`j(pj(λ)− pj−1(λ))
Equating expressions for the left and right sides of the equationused to define qj gives us:
qj(λ)u1 = qj−1(λ)−(λmj
σ
)pj−1(λ)u1
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 20 / 36
Continued Fractions
With these characteristic polynomials, we can find masses &displacements.
Using recurrence relationships of the polynomials:
pn(λ)
qn(λ)=`nqn(λ) + pn−1(λ)
qn(λ)
pn(λ)
qn(λ)= `n +
1−mn
σλpn−1(λ) + qn−1(λ)
pn−1(λ)
pn(λ)
qn(λ)= `n +
1
−mn
σλ+
1
`n−1 +1
−mn−1
σλ+ ...+
1
`1 +1
−m1
σλ+
1
`0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 21 / 36
Continued Fractions
With these characteristic polynomials, we can find masses &displacements.
Using recurrence relationships of the polynomials:
pn(λ)
qn(λ)=`nqn(λ) + pn−1(λ)
qn(λ)
pn(λ)
qn(λ)= `n +
1−mn
σλpn−1(λ) + qn−1(λ)
pn−1(λ)
pn(λ)
qn(λ)= `n +
1
−mn
σλ+
1
`n−1 +1
−mn−1
σλ+ ...+
1
`1 +1
−m1
σλ+
1
`0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 21 / 36
Continued Fractions
With these characteristic polynomials, we can find masses &displacements.
Using recurrence relationships of the polynomials:
pn(λ)
qn(λ)=`nqn(λ) + pn−1(λ)
qn(λ)
pn(λ)
qn(λ)= `n +
1−mn
σλpn−1(λ) + qn−1(λ)
pn−1(λ)
pn(λ)
qn(λ)= `n +
1
−mn
σλ+
1
`n−1 +1
−mn−1
σλ+ ...+
1
`1 +1
−m1
σλ+
1
`0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 21 / 36
Continued Fractions
With these characteristic polynomials, we can find masses &displacements.
Using recurrence relationships of the polynomials:
pn(λ)
qn(λ)=`nqn(λ) + pn−1(λ)
qn(λ)
pn(λ)
qn(λ)= `n +
1−mn
σλpn−1(λ) + qn−1(λ)
pn−1(λ)
pn(λ)
qn(λ)= `n +
1
−mn
σλ+
1
`n−1 +1
−mn−1
σλ+ ...+
1
`1 +1
−m1
σλ+
1
`0Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 21 / 36
Practical Problems
With the continued fraction, we have values for masses &lengths.
The process works from a theoretical standpoint...
Fixed-flat is hard to implement.
Fortunately, there is a workaround.
It can be shown that if the beads are symmetric about themidpoint of the string, we can find the fixed-flat eigenvalueswithout any extra work.
.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 22 / 36
Practical Problems
With the continued fraction, we have values for masses &lengths.
The process works from a theoretical standpoint...
Fixed-flat is hard to implement.
Fortunately, there is a workaround.
It can be shown that if the beads are symmetric about themidpoint of the string, we can find the fixed-flat eigenvalueswithout any extra work.
.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 22 / 36
Symmetry
Experimentally measuring eigenvalues on a symmetric beadedstring gives us a new set of eigenvalues, termed Λ.
Λ has some beautiful properties.
It can be shown that:
The odd indexed eigenvalues of Λ are all symmetric about themiddle of the string.The even indexed eigenvalues of Λ are antisymmetric about themidpoint.
This relationship holds for all symmetric strings.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 23 / 36
Symmetry
Experimentally measuring eigenvalues on a symmetric beadedstring gives us a new set of eigenvalues, termed Λ.
Λ has some beautiful properties.
It can be shown that:
The odd indexed eigenvalues of Λ are all symmetric about themiddle of the string.The even indexed eigenvalues of Λ are antisymmetric about themidpoint.
This relationship holds for all symmetric strings.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 23 / 36
The Point
The N eigenvalues of a symmetric beaded stringfixed at both ends exactly match the N/2 fixed-fixedand N/2 fixed-flat eigenvalues associated with half of
the string.
Figure: Eigenvalues of Symmetric Beaded Strings
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 24 / 36
The Inverse Algorithm
Record the eigenvalues for the whole string.
Use the eigenvalues to generate characteristic polynomials pn
and qn.
Use the characteristic polynomials to find the set of masses andlengths.
Figure: The Symmetric Beaded String
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 25 / 36
Non-symmetric ProblemIntroduction
Masses and lengths may vary arbitrarily
Spectra of entrire string no longer sufficient
Clamp string at some interior point between two masses
Leads to three problems with fixed-fixed boundary conditions
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 26 / 36
Non-symmetric ProblemSetup
For n1 masses on the left and n2 masses on the right:uk − uk+1
`k+
uk − uk−1
`k−1−mkλ
2uk = 0, k = 1, 2, ..., n1
uk − uk+1
˜k
+uk − uk−1
˜k−1
− mkλ2uk = 0, k = 1, 2, ..., n2
Whole String Boundary Conditions
un1+1 = un2+1
u0 = 0, u0 = 0
un1+1 − un1
`n1
+un2+1 − un2
˜n2
= 0
Clamped String Boundary Conditions
un1+1 = 0, un2+1 = 0
u0 = 0, u0 = 0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 27 / 36
Non-symmetric ProblemSetup
For n1 masses on the left and n2 masses on the right:uk − uk+1
`k+
uk − uk−1
`k−1−mkλ
2uk = 0, k = 1, 2, ..., n1
uk − uk+1
˜k
+uk − uk−1
˜k−1
− mkλ2uk = 0, k = 1, 2, ..., n2
Whole String Boundary Conditions
un1+1 = un2+1
u0 = 0, u0 = 0
un1+1 − un1
`n1
+un2+1 − un2
˜n2
= 0
Clamped String Boundary Conditions
un1+1 = 0, un2+1 = 0
u0 = 0, u0 = 0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 27 / 36
Non-symmetric ProblemSetup
For n1 masses on the left and n2 masses on the right:uk − uk+1
`k+
uk − uk−1
`k−1−mkλ
2uk = 0, k = 1, 2, ..., n1
uk − uk+1
˜k
+uk − uk−1
˜k−1
− mkλ2uk = 0, k = 1, 2, ..., n2
Whole String Boundary Conditions
un1+1 = un2+1
u0 = 0, u0 = 0
un1+1 − un1
`n1
+un2+1 − un2
˜n2
= 0
Clamped String Boundary Conditions
un1+1 = 0, un2+1 = 0
u0 = 0, u0 = 0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 27 / 36
Non-symmetric ProblemSetup
For n1 masses on the left and n2 masses on the right:uk − uk+1
`k+
uk − uk−1
`k−1−mkλ
2uk = 0, k = 1, 2, ..., n1
uk − uk+1
˜k
+uk − uk−1
˜k−1
− mkλ2uk = 0, k = 1, 2, ..., n2
Whole String Boundary Conditions
un1+1 = un2+1
u0 = 0, u0 = 0
un1+1 − un1
`n1
+un2+1 − un2
˜n2
= 0
Clamped String Boundary Conditions
un1+1 = 0, un2+1 = 0
u0 = 0, u0 = 0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 27 / 36
Non-symmetric ProblemSetup
For n1 masses on the left and n2 masses on the right:uk − uk+1
`k+
uk − uk−1
`k−1−mkλ
2uk = 0, k = 1, 2, ..., n1
uk − uk+1
˜k
+uk − uk−1
˜k−1
− mkλ2uk = 0, k = 1, 2, ..., n2
Whole String Boundary Conditions
un1+1 = un2+1
u0 = 0, u0 = 0
un1+1 − un1
`n1
+un2+1 − un2
˜n2
= 0
Clamped String Boundary Conditions
un1+1 = 0, un2+1 = 0
u0 = 0, u0 = 0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 27 / 36
Non-symmetric ProblemSetup
For n1 masses on the left and n2 masses on the right:uk − uk+1
`k+
uk − uk−1
`k−1−mkλ
2uk = 0, k = 1, 2, ..., n1
uk − uk+1
˜k
+uk − uk−1
˜k−1
− mkλ2uk = 0, k = 1, 2, ..., n2
Whole String Boundary Conditions
un1+1 = un2+1
u0 = 0, u0 = 0
un1+1 − un1
`n1
+un2+1 − un2
˜n2
= 0
Clamped String Boundary Conditions
un1+1 = 0, un2+1 = 0
u0 = 0, u0 = 0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 27 / 36
Three spectra Inverse ProblemThe Problem
Relation between characteristic equations
pwhole(λ) = pleft(λ)qright(λ) + pright(λ)qleft(λ)
The roots of pleft and pright are the eigenvalues of the fixed-fixedproblems for the left and right segments
The roots of qleft and qright are the eigenvalues of the fixed-flatproblems
Problem: we can construct pleft and pright and pwhole up to ascaling factor from measured roots, but we cannot measure theroots of qleft and qright , preventing us from forming thecontinued fractions for the left and right segments.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 28 / 36
Three spectra Inverse ProblemThe Problem
Relation between characteristic equations
pwhole(λ) = pleft(λ)qright(λ) + pright(λ)qleft(λ)
The roots of pleft and pright are the eigenvalues of the fixed-fixedproblems for the left and right segments
The roots of qleft and qright are the eigenvalues of the fixed-flatproblems
Problem: we can construct pleft and pright and pwhole up to ascaling factor from measured roots, but we cannot measure theroots of qleft and qright , preventing us from forming thecontinued fractions for the left and right segments.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 28 / 36
Three spectra Inverse ProblemDefinitions
Through the clever use of polynomials, Boyko and Pivovarchikshow that we do not need to know the roots of qleft and qright toconstruct them
Let λk , k = 1, 2, ..., (n1 + n2), be the spectra of the whole stringand let νk,`, k = 1, 2, ..., n1, and νk,r , k = 1, 2, ..., n2, be thespectra of the left and right parts, respectively
Let L be the length of the whole string and let L` and Lr be thelengths of the left and right segments of the clamped string
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 29 / 36
Three spectra Inverse ProblemDefinitions
Through the clever use of polynomials, Boyko and Pivovarchikshow that we do not need to know the roots of qleft and qright toconstruct them
Let λk , k = 1, 2, ..., (n1 + n2), be the spectra of the whole stringand let νk,`, k = 1, 2, ..., n1, and νk,r , k = 1, 2, ..., n2, be thespectra of the left and right parts, respectively
Let L be the length of the whole string and let L` and Lr be thelengths of the left and right segments of the clamped string
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 29 / 36
Three spectra Inverse ProblemDefinitions
Through the clever use of polynomials, Boyko and Pivovarchikshow that we do not need to know the roots of qleft and qright toconstruct them
Let λk , k = 1, 2, ..., (n1 + n2), be the spectra of the whole stringand let νk,`, k = 1, 2, ..., n1, and νk,r , k = 1, 2, ..., n2, be thespectra of the left and right parts, respectively
Let L be the length of the whole string and let L` and Lr be thelengths of the left and right segments of the clamped string
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 29 / 36
Three spectra Inverse ProblemConstructing Known Polynomials
Construct the polynomials we know, up to a scaling factor:
pwhole(λ) = Ln1+n2∏k=1
(1− λ
λk
)
pleft(λ) = L`
n1∏k=1
(1− λ
νk,`
)
pright(λ) = Lr
n2∏k=1
(1− λ
νk,r
)
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 30 / 36
Three spectra Inverse ProblemFinding qleft and qright
Recall the relationship of the characteristic equations:
pwhole(λ) = pleft(λ)qright(λ) + pright(λ)qleft(λ)
Notice that pleft(νk,`) = 0 and pright(νk,r ) = 0, as these aresimply the roots we used to construct those polynomials.
Plug in νk,` and νk,r into the relation between the characteristicequations to reveal:
qleft(νk,`) =pwhole(νk,`)
pright(νk,`), k = 1, 2, ..., n1
qright(νk,r ) =pwhole(νk,r )
pleft(νk,r ), k = 1, 2, ..., n2
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 31 / 36
Three spectra Inverse ProblemFinding qleft and qright
Recall the relationship of the characteristic equations:
pwhole(λ) = pleft(λ)qright(λ) + pright(λ)qleft(λ)
Notice that pleft(νk,`) = 0 and pright(νk,r ) = 0, as these aresimply the roots we used to construct those polynomials.
Plug in νk,` and νk,r into the relation between the characteristicequations to reveal:
qleft(νk,`) =pwhole(νk,`)
pright(νk,`), k = 1, 2, ..., n1
qright(νk,r ) =pwhole(νk,r )
pleft(νk,r ), k = 1, 2, ..., n2
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 31 / 36
Three spectra Inverse ProblemFinding qleft and qright
Recall the relationship of the characteristic equations:
pwhole(λ) = pleft(λ)qright(λ) + pright(λ)qleft(λ)
Notice that pleft(νk,`) = 0 and pright(νk,r ) = 0, as these aresimply the roots we used to construct those polynomials.
Plug in νk,` and νk,r into the relation between the characteristicequations to reveal:
qleft(νk,`) =pwhole(νk,`)
pright(νk,`), k = 1, 2, ..., n1
qright(νk,r ) =pwhole(νk,r )
pleft(νk,r ), k = 1, 2, ..., n2
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 31 / 36
Three spectra Inverse ProblemFinding qleft and qright (continued)
Recall the continued fraction equation:
pn(λ)
qn(λ)= `n +
1
−mn
σλ+
1
`n−1 +1
−mn−1
σλ+ ...+
1
`1 +1
−m1
σλ+
1
`0
pn(0)
qn(0)= `n + `n−1 + ... + `1 + `0 = L
These equations must hold for the two clamped string segmentsas well, giving us the last points, qleft(0) = 1 and qright(0) = 1,needed to completely define the polynomials
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 32 / 36
Three spectra Inverse ProblemConstructing qleft and qright
Construct qleft and qright :
qleft(λ) =n1∑
k=1
λpwhole(νk,`)
νk,`pright(νk,`)
n1∏j=1,j 6=k
(λ− νj ,`)
(νk,` − νj ,`)+
n1∏k=1
νk,` − λνk,`
qright(λ) =n2∑
k=1
λpwhole(νk,r )
νk,rpleft(νk,r )
n2∏j=1,j 6=k
(λ− νj ,r )
(νk,r − νj ,r )+
n2∏k=1
νk,r − λνk,r
Notice:
qleft(νk,`) =pwhole(νk,`)
pright(νk,`), qright(νk,r ) =
pwhole(νk,r )
pleft(νk,r )
qleft(0) = 1, qright(0) = 1
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 33 / 36
Three spectra Inverse ProblemConstructing qleft and qright
Construct qleft and qright :
qleft(λ) =n1∑
k=1
λpwhole(νk,`)
νk,`pright(νk,`)
n1∏j=1,j 6=k
(λ− νj ,`)
(νk,` − νj ,`)+
n1∏k=1
νk,` − λνk,`
qright(λ) =n2∑
k=1
λpwhole(νk,r )
νk,rpleft(νk,r )
n2∏j=1,j 6=k
(λ− νj ,r )
(νk,r − νj ,r )+
n2∏k=1
νk,r − λνk,r
Notice:
qleft(νk,`) =pwhole(νk,`)
pright(νk,`), qright(νk,r ) =
pwhole(νk,r )
pleft(νk,r )
qleft(0) = 1, qright(0) = 1
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 33 / 36
Three spectra Inverse ProblemFinding {mk}, {`k}, {mk}, and {˜
k}
Unique lengths and masses are then determined by continuedfraction expansion of ratio of p’s and q’s, i.e.
pleft(λ)
qleft(λ)= `n1+
1
−mn1
σλ+
1
`n1−1 +1
−mn1−1
σλ+ ...+
1
`1 +1
−m1
σλ+
1
`0
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 34 / 36
Future Work
Three spectra problem with damping at an interior point
No unique solution
Damping at a mass
Damping at the ends
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 35 / 36
Mentors
Jeffrey HokansonDr. Steve CoxDr. Mark Embree
References
CAAM 335 Lab 9.http://www.caam.rice.edu/ caam335lab/lab9.pdfCAAM 335 Lab 10.http://www.caam.rice.edu/ caam335lab/lab10.pdfBoyko O and Pivovarchik V. ”The inverse three-spectralproblem for a Stieltjes string and the inverse problem withone-dimensional damping.” Inverse Problems 24 (2008) 1-13.Gantmacher F P and Krein M G. Oscillation Matrices andKernels and Small Vibrations of Mechanical Systems. RevisedEdition. Providence, Rhode Island: AMS Chelsea Pub, 2002.
Hunter Gilbert, Walter Kelm, and Brian Leake (Physics of Strings PFUG)Beaded Strings 11 June 2008 36 / 36