(E – E2 – 1)z(k, l) = 0 (E1 + V1+ E2)(E1 – /1+ E2)z(k, l) = 0. %3D

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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Example D
The equation
z(k + 2, l) = z(k, l+1) + z(k, l)
(5.50)
is second order in k and first order in l. In operator form, it can be written
as either
(E – E2 – 1)z(k, l) = 0
(5.51)
or
(E1 + /1+ E2)(E1 – /1+ E2)z (k, l) = 0.
(5.52)
We will obtain a solution using the first form. Our main reason for not attempt-
ing to proceed with the second operator form is that, in general, the operator
V1+ E, does not have a well-defined meaning in terms of an infinite-series
expansion in E2.
From equation (5.51), we obtain
2(k, l) = (-)'(1 – E})'$(k),
(5.53)
where ø(k) is an arbitrary function of k. Now,
(1- E)' -1-(1) 태 + (6)대
2.
—1 2(е-1)
'Ee-1) + (-)'E?“, (5.54)
- 1
and, therefore,
z(k, €) = (-)' #(k) –
().
)o(k + 2) +
()
(k + 4) + · ..
(5.55)
+(1)(--olk + 2(e – 1)] + (-)'ø(k +
Again, we point out that equation (5.44) is first order in l and thus the single
arbitrary function that appears in equation (5.55).
Transcribed Image Text:Example D The equation z(k + 2, l) = z(k, l+1) + z(k, l) (5.50) is second order in k and first order in l. In operator form, it can be written as either (E – E2 – 1)z(k, l) = 0 (5.51) or (E1 + /1+ E2)(E1 – /1+ E2)z (k, l) = 0. (5.52) We will obtain a solution using the first form. Our main reason for not attempt- ing to proceed with the second operator form is that, in general, the operator V1+ E, does not have a well-defined meaning in terms of an infinite-series expansion in E2. From equation (5.51), we obtain 2(k, l) = (-)'(1 – E})'$(k), (5.53) where ø(k) is an arbitrary function of k. Now, (1- E)' -1-(1) 태 + (6)대 2. —1 2(е-1) 'Ee-1) + (-)'E?“, (5.54) - 1 and, therefore, z(k, €) = (-)' #(k) – (). )o(k + 2) + () (k + 4) + · .. (5.55) +(1)(--olk + 2(e – 1)] + (-)'ø(k + Again, we point out that equation (5.44) is first order in l and thus the single arbitrary function that appears in equation (5.55).
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