1. NPTEL β Physics β Mathematical Physics - 1
Lecture 18
Eigenvalue Problems
Example
1. Consider a mass-spring system given by,
We shall show that this corresponds to an eigenvalue problem, solution of which
will yield the displacements of the two masses and their frequencies of oscillation. Let π¦
1,2 be the displacement variables for the two masses π1,2.
One can write the differential equations as,
ππ‘ 2
π2π¦1
= β(π + π )π¦ + π π¦
1 2 1 2 2 βΆ for mass π1 (1)
π
π¦
2
2
ππ‘ 2 = π π¦ β π π¦
2 1 2 2 : for mass π (2)
2
Putting values for π1 and π2,
π
π¦
2
1
ππ‘ 2 = β5π¦1 + 2π¦2 (3)
π
π¦
2
2
ππ‘ 2 = 2π¦1 β 2π¦2 (4)
The coupled equations (3) and (4) can be solved using the following eigenvalue equation.
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2. NPTEL β Physics β Mathematical Physics - 1
π2π¦ π¦1
ππ‘2 = π¦Μ = (π¦Μ2
) = π΄π¦ = [
π¦Μ1 β5 2
2 β 2 2
] (π¦ ) (5)
One may try a solution of the form,
π = π΄πππ‘
(6)
Putting this in (5) leads to the eigenvalue equation,
π΄π¦ = πy where π = π2
and π΄ = [ β5 2 ]
2 β 2
A has the
eigenvalues
-
[β5 β π 2
2 β 2 β π
] = 0
β π1 = β1 and π2 = β6
So, π = ββ1 and ββ6
= π and β6π respectively. The
corresponding e-vectors are-
π¦1 = ( ) corresponding to π
1
2
and π¦2 = ( 2 ) corresponding to β6π
β1
Thus we obtain four complex solutions
π¦1πππ‘ = π¦1(πππ π‘ Β± ππ πππ‘)
π¦2ππβ6π‘ = π¦2(πππ β6π‘ Β± ππ ππβ6π‘)
Thus the real solutions are
π¦1πππ π‘, π¦1π πππ‘, π¦2πππ β6π‘, π¦2π ππβ6π‘
A general solution is obtained by taking a linear combination of all the above
solutions.
π¦ = π¦1(π1πππ π‘ + π1π πππ‘) + (π¦2π2πππ β6π‘ + π2π ππβ6π‘)
with arbitrary constants π1, π2, π1, π2 which can be defined from the initial
conditions. From the structure of the vectors π¦1 and π¦2, one can write the final
solutions as-
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3. NPTEL β Physics β Mathematical Physics - 1
π¦1 = π1πππ π‘ + π1π πππ‘ + 2π2πππ β6π‘ + 2π2π ππβ6π‘
π¦2 = 2π1πππ π‘ + 2π1π πππ‘ β π2πππ β6π‘ β π2π ππβ6π‘
These solutions represent the displacement of the coupled
masses.
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