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 + V1+ E2)(E1 – /1+ E2)z(k, e) = 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+ E2 does not have a well-defined meaning in terms of an infinite-series expansion in E2. From equation (5.51), we obtain z(k, l) = (-)'(1 – E)°$(k), (5.53) where ø(k) is an arbitrary function of k. Now, (1 – E†)' = 1 – - ()i + ()E -'Ee-1) + (-)'E?“, (5.54) e-1 +...+ (-) l – 1 and, therefore, =(k. ) = (-1" (614) – (1)«1e +: $(k + 2) + (k + 4) + ... (5 55)

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Example D
The equation
2(k + 2, 2) = 2(k, l+1) + z(k, l)
is second order in k and first order in l. In operator form, it can be written
(5.50)
as either
(E – E2 – 1)z(k, l) = 0
(5.51)
or
(E1 + V1+ 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+ E2 does not have a well-defined meaning in terms of an infinite-series
expansion in E2.
From equation (5.51), we obtain
z(k, l) = (-)'(1 – E)'$(k),
(5.53)
where ø(k) is an arbitrary function of k. Now,
()=
(1 – E)' = 1 –
—1 г2(€—1)
(-)e-1
l – 1
and, therefore,
<(k. €) = (-)* 6(k) – (1) o(k + 2) +
O(k + 4) + ·..
(5.55)
+() (---"o[k + 2(e – 1)] + (-)'ø(k +
Again, we point out that equation (5.44) is first order in l and thus the single
arbitrary function that appears in equation (5.55).
Transcribed Image Text:Example D The equation 2(k + 2, 2) = 2(k, l+1) + z(k, l) is second order in k and first order in l. In operator form, it can be written (5.50) as either (E – E2 – 1)z(k, l) = 0 (5.51) or (E1 + V1+ E2)(E1 – /1+ E2)z (k, l) = 0. (5.52) We will obtain a solution using the first form. Our main reason for not attempt- ing to proceed with the second operator form is that, in general, the operator V1+ E2 does not have a well-defined meaning in terms of an infinite-series expansion in E2. From equation (5.51), we obtain z(k, l) = (-)'(1 – E)'$(k), (5.53) where ø(k) is an arbitrary function of k. Now, ()= (1 – E)' = 1 – —1 г2(€—1) (-)e-1 l – 1 and, therefore, <(k. €) = (-)* 6(k) – (1) o(k + 2) + O(k + 4) + ·.. (5.55) +() (---"o[k + 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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