EIGENVALUE PROBLEM

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EIGENVALUE PROBLEM
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In this lecture we study Boundary value problems
for Ordinary Differential equations. These
problems arise in Physics when we attempt to
solve the problem of heat conduction, vibrating
strings etc. A Boundary Value problem involving a
parameter ? is known also as an eigenvalue
problem.
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Boundary Value Problems for Ordinary Differential
Equations
Consider a linear differential equation
Here
We want to find solutions of the above equation
valid over the interval a, b satisfying the
boundary conditions
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where ?i, ?i, ?i are constants.
Such a problem is known as a Boundary Value
problem
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Example Consider the boundary value problem
satisfying the boundary conditions
Obviously the trivial solution y 0 is a
solution to the above problem. A nontrivial
solution is clearly
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Another nontrivial solution is
In fact we have an infinite number of non-trivial
solutions
Thus unlike initial-value problems, we find a
Boundary Value problem may have an infinite
number of nontrivial solutions.
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Eigenvalues and Eigenfunctions
Consider the boundary value problem
satisfying the boundary conditions
The above differential equation may be rewritten
as
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Letting U as the space of all twice continuously
differentiable functions y on the interval 0, L
such that y (0) 0, y (L) 0, V as the space of
all continuous functions on 0, L, we thus have
a linear transformation L U ? V defined by L(y)
- y?.
L
V
U
y
Ly -y?
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We want to find all ?s such that there exists a
nonzero function y such that L (y) ? y. In
Linear Algebra language, such a ? is called an
eigenvalue of the linear transformation L and the
corresponding nonzero y is called an eigenvector
corresponding to the eigenvalue ?.
In the theory of Differential equations y is
called an eigenfunction corresponding to the
eigenvalue ?.
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And we refer to the above problem as an
Eigenvalue problem.
Solution to the above problem
Case (i) ? lt 0, say ? - a2, where a gt 0.
The given differential equation becomes
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The general solution to the above d.e. is
where c1, c2 are arbitrary constants.
Hence
No nontrivial solution.
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Case (ii) ? 0.
The given differential equation becomes
The general solution to the above d.e. is
where c1, c2 are arbitrary constants.
No nontrivial solutions.
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Case (iii) ? gt 0, say ? a2, where a gt 0.
The given differential equation becomes
The general solution to the above d.e. is
where c1, c2 are arbitrary constants.
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(c2 ? 0 for nontrivial solutions)
Hence
or
Hence for each n 1, 2, we get an eigenvalue
And a corresponding
eigenfunction is
Note that any
non-zero multiple of yn is also an eigenfunction.
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Problem 2 Solve the eigenvalue problem
Solution
Case (i) ? lt 0, say ? - a2, where a gt 0.
The given differential equation becomes
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The general solution to the above d.e. is
c1, c2 are arb. constants.
Hence
No nontrivial solution.
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Case (ii) ? 0.
The given differential equation becomes
The general solution to the above d.e. is
where c1, c2 are arbitrary constants.
Thus y c2 is a solution for all c2.
Hence ? 0 is an eigenvalue and y 1 is a
corresponding eigenfunction.
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Case (iii) ? gt 0, say ? a2, where a gt 0.
The given differential equation becomes
The general solution to the above d.e. is
where c1, c2 are arbitrary constants.
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(c1 ? 0 for nontrivial solutions)
Hence
or
Hence for each n 0,1, 2, we get an eigenvalue
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And a corresponding eigenfunction is
Note that any non-zero multiple of yn is also an
eigenfunction.
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Problem 3 Solve the eigenvalue problem
Solution
Case (i) ? lt 0, say ? - a2, where a gt 0.
The given differential equation becomes
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The general solution to the above d.e. is
c1, c2 are arb. constants.
i.e.
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Hence c1, c2 satisfy the pair of homogeneous
linear equations
Adding, we get
or
as a gt 0.
Hence
No nontrivial solutions
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Case (ii) ? 0.
The given differential equation becomes
The general solution to the above d.e. is
(c1, c2 are arbitrary constants)
holds for all c2.
Thus y c2 is a solution for all c2.
Hence ? 0 is an eigenvalue and y 1 is a
corresponding eigenfunction.
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Case (iii) ? gt 0, say ? a2, where a gt 0.
The given differential equation becomes
The general solution to the above d.e. is
where c1, c2 are arbitrary constants.
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i.e.
Hence c1, c2 satisfy the pair of homogeneous
linear equations
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For nontrivial solutions for c1, c2 determinant
of the coefficients must be zero
i.e.
or
Hence
Thus
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When a 2n,
are satisfied by all c1, c2
Hence for all c1, c2 (not both zero),
is an eigenfunction corresponding to ? a24n2.
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Hence for each n 1, 2, .
?n 4n2 is an eigenvalue with corresponding LI
eigenfunctions cos 2nx, and sin 2nx.
Note that unlike the previous problems, we see
that to each eigenvalue ?n 4n2 there
corresponds two LI eigenfunctions, namely cos
2nx, and sin 2nx.
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Problem 4 Solve the eigenvalue problem
Hint
Let
Hence the eigenvalues are
with corresponding
eigenfunctions
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Problem 5 Solve the eigenvalue problem
Solution
Auxiliary Equation
i.e.
Roots
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Case (i) ? lt 0, say ? - a2, where a gt 0.
Hence the roots are
The general solution to the given d.e. is
c1, c2 are arb. constants.
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i.e.
Determinant of the coefficients
Hence c1 c2 0.
So no nontrivial solutions.
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Case (ii) ? 0.
The roots are
The general solution to the above d.e. is
(c1, c2 are arb. constants)
Solving we get c1 c2 0
No non-trivial solutions.
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Case (iii) ? gt 0, say ? a2, where a gt 0.
The roots are
The general solution to the above d.e. is
where c1, c2 are arbitrary constants.
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For nontrivial solutions for c1, c2 determinant
of the coefficients must be zero
i.e.
or
Hence
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Now
If
Hence c2 0 and c1 is an arbitrary nonzero
number.
Thus corresponding to the eigenvalue ?n a2
n2?2, n 1, 3, 5, we get an eigenfunction
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If
Hence c1 0 and c2 is an arbitrary nonzero
number.
Thus corresponding to the eigenvalue ?n a2
n2?2, n 2, 4, 6, we get an eigenfunction
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Hence the eigenvalues are
For n 1, 3, 5, the eigenfunction
correspondimg to the eigenvalue n2?2 is
For n 2, 4, 6, the eigenfunction
correspondimg to the eigenvalue n2?2 is
End of lecture