5.3.5 Example E Applying the method of Lagrange to the partial difference equation z(k +1, €) – 2z(k, l + 1) – 3z(k, l) = 0 (5.103) LINEAR PARTIAL DIFFERENCE EQUATIONS 179 gives = 2µ + 3, and special solutions having the form Zp(k, l) = (2µ+3)* µº. (5.104) General solutions can be obtained by summing over u; doing this gives z(k, l) = C(H)(2µ + 3)*µ', (5.105) where C(u) is an arbitrary function of u. Now, k (2µ + 3)k = 3k (1+ 24 k(k – 1) ( 2µ 3k + k 3 (5.106) 2! 3 ()] k-1 2µ () +...+ k + 3 and, therefore, equation (5.105) can be written z(k, l) = 3* [H(€) + 2/3kH(l +1) + 2/gk(k – 1)H(l+2) +...+ (2/3)* H(e+ k)], (5.107) where H(l) is an arbitrary function of l and we have used the results H(C) =CH)', (5.108) H(l + m) = C(H)µ+m. Applying the method of separation of variables gives, for Ck and De, the equations Ck+1 Ck 2De+1+3De = a, (5.109) De the solutions of which are Ck = Aa", (De a – 3 B (5.110) 2 where A and B are arbitrary constants. If we let a - 3 (5.111) 2 then z(k, l) = CkDe becomes z(k, l) = AB(3+ 2µ)*µ°, (5.112) %3D which is the same as equation (5.104). Therefore, the separation-of-variables method gives the same solution as presented in equation (5.107).
5.3.5 Example E Applying the method of Lagrange to the partial difference equation z(k +1, €) – 2z(k, l + 1) – 3z(k, l) = 0 (5.103) LINEAR PARTIAL DIFFERENCE EQUATIONS 179 gives = 2µ + 3, and special solutions having the form Zp(k, l) = (2µ+3)* µº. (5.104) General solutions can be obtained by summing over u; doing this gives z(k, l) = C(H)(2µ + 3)*µ', (5.105) where C(u) is an arbitrary function of u. Now, k (2µ + 3)k = 3k (1+ 24 k(k – 1) ( 2µ 3k + k 3 (5.106) 2! 3 ()] k-1 2µ () +...+ k + 3 and, therefore, equation (5.105) can be written z(k, l) = 3* [H(€) + 2/3kH(l +1) + 2/gk(k – 1)H(l+2) +...+ (2/3)* H(e+ k)], (5.107) where H(l) is an arbitrary function of l and we have used the results H(C) =CH)', (5.108) H(l + m) = C(H)µ+m. Applying the method of separation of variables gives, for Ck and De, the equations Ck+1 Ck 2De+1+3De = a, (5.109) De the solutions of which are Ck = Aa", (De a – 3 B (5.110) 2 where A and B are arbitrary constants. If we let a - 3 (5.111) 2 then z(k, l) = CkDe becomes z(k, l) = AB(3+ 2µ)*µ°, (5.112) %3D which is the same as equation (5.104). Therefore, the separation-of-variables method gives the same solution as presented in equation (5.107).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Explain the determine blue
![5.3.5 Example E
Applying the method of Lagrange to the partial difference equation
z(k +1, €) – 2z(k, l + 1) – 3z(k, l) = 0
(5.103)
LINEAR PARTIAL DIFFERENCE EQUATIONS
179
gives A = 2µ + 3, and special solutions having the form
Zp(k, l) = (2µ +3)*µº.
(5.104)
General solutions can be obtained by summing over u; doing this gives
2(k, l) = C(H)(2µ +3)*µ",
(5.105)
where C(u) is an arbitrary function of u. Now,
2µ
(2µ + 3)* = 3* (1+
3
k(k – 1)
2µ
3k
+ k
3
(5.106)
2!
k
k-1
2µ
+...+ k
and, therefore, equation (5.105) can be written
z(k, e) = 3* [H(e) + 2/3kH (l + 1) + 2/9k(k – 1)H(l+2)
+..+ (2/3)* H (l + k)],
(5.107)
where H(l) is an arbitrary function of l and we have used the results
H(C) = C(H)µ°',
(5.108)
H(l + m) = c (µ)µl+m.
Applying the method of separation of variables gives, for Ck and De, the
equations
2De+1+ 3De
= a,
De
Ck+1
(5.109)
Ck
the solutions of which are
Ck = Aa*,
(D,
(5.110)
В
2
where A and B are arbitrary constants. If we let
a – 3
(5.111)
2
then z(k, l) = CkDe becomes
z(k, l) = AB(3+ 2µ)*µ°,
(5.112)
which is the same as equation (5.104). Therefore, the separation-of-variables
method gives the same solution as presented in equation (5.107).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc0e39b00-7c5c-4b9e-ab5d-9170cd8d91e5%2F64371e64-8481-4b8f-8479-fe8c78dff0b4%2Ffan2z6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5.3.5 Example E
Applying the method of Lagrange to the partial difference equation
z(k +1, €) – 2z(k, l + 1) – 3z(k, l) = 0
(5.103)
LINEAR PARTIAL DIFFERENCE EQUATIONS
179
gives A = 2µ + 3, and special solutions having the form
Zp(k, l) = (2µ +3)*µº.
(5.104)
General solutions can be obtained by summing over u; doing this gives
2(k, l) = C(H)(2µ +3)*µ",
(5.105)
where C(u) is an arbitrary function of u. Now,
2µ
(2µ + 3)* = 3* (1+
3
k(k – 1)
2µ
3k
+ k
3
(5.106)
2!
k
k-1
2µ
+...+ k
and, therefore, equation (5.105) can be written
z(k, e) = 3* [H(e) + 2/3kH (l + 1) + 2/9k(k – 1)H(l+2)
+..+ (2/3)* H (l + k)],
(5.107)
where H(l) is an arbitrary function of l and we have used the results
H(C) = C(H)µ°',
(5.108)
H(l + m) = c (µ)µl+m.
Applying the method of separation of variables gives, for Ck and De, the
equations
2De+1+ 3De
= a,
De
Ck+1
(5.109)
Ck
the solutions of which are
Ck = Aa*,
(D,
(5.110)
В
2
where A and B are arbitrary constants. If we let
a – 3
(5.111)
2
then z(k, l) = CkDe becomes
z(k, l) = AB(3+ 2µ)*µ°,
(5.112)
which is the same as equation (5.104). Therefore, the separation-of-variables
method gives the same solution as presented in equation (5.107).
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