k (k,() = (-) g(e + k) f(e+ k). %3D (5.195)

Linear Algebra: A Modern Introduction
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ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 70EQ
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5.5.5 Example E
Assume that c= 0 in equation (5.183) and continue to assume that F(k,l) is
a solution to the homogeneous equation. Under these conditions, we have
(aE1 + bE2)z(k, l) = F(k, l)
(5.189)
and
(aE1 + bE2)F(k, l) = 0.
(5.190)
The solution to the last equation is
F(k, €) = (-b/a)* f(l+ k),
(5.191)
where f is an arbitrary function of l+k.
Examination of the left-hand side of equation (5.189) shows that it is of a
form such that Laplace's method can be used to obtain a solution. If we let
k +l = m = constant,
Vk = 2(k, l) = z(k, m – k),
(5.192)
then vk satisfies the first-order inhomogeneous equation
avk+1 + bvk = (-b/a)* f(m),
(5.193)
where we have used the results of equations (5.191) and (5.192) to replace the
right-hand side of equation (5.189). Note that f(m) is a constant. Solving for
Vk gives
k
k
Vk = A
(5.194)
a
where A is an arbitrary constant. Replacing m by l+k and A by an arbitrary
function of l + k gives the complete solution to equation (5.189), under the
assumption of equation (5.190),
k
k
= (-)" g(e + k) –
f(l+ k).
(5.195)
Transcribed Image Text:5.5.5 Example E Assume that c= 0 in equation (5.183) and continue to assume that F(k,l) is a solution to the homogeneous equation. Under these conditions, we have (aE1 + bE2)z(k, l) = F(k, l) (5.189) and (aE1 + bE2)F(k, l) = 0. (5.190) The solution to the last equation is F(k, €) = (-b/a)* f(l+ k), (5.191) where f is an arbitrary function of l+k. Examination of the left-hand side of equation (5.189) shows that it is of a form such that Laplace's method can be used to obtain a solution. If we let k +l = m = constant, Vk = 2(k, l) = z(k, m – k), (5.192) then vk satisfies the first-order inhomogeneous equation avk+1 + bvk = (-b/a)* f(m), (5.193) where we have used the results of equations (5.191) and (5.192) to replace the right-hand side of equation (5.189). Note that f(m) is a constant. Solving for Vk gives k k Vk = A (5.194) a where A is an arbitrary constant. Replacing m by l+k and A by an arbitrary function of l + k gives the complete solution to equation (5.189), under the assumption of equation (5.190), k k = (-)" g(e + k) – f(l+ k). (5.195)
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