Laplace Variational Iteration Method Let us consider the following general partial differential equation: where L Lu(x,t) + Ru(x, t) + N u(x,t) = g(x,t), (1) R denotes linear operator, N denotes nonlinear operator, and g(x,t) is the source term. The variational iteration method presents a correction functional for (1) in the form: Mas(t) = (t) + (r)(L!?” (t) + (t) + (t)g(t) dr. (2) Maos(t) = µm (t) + ƒ (t − t) (L²™” un (t) + R ® (t) + N Qn(t) − g(t) dr. (3) Taking Laplace transform of (3), we obtain L{una s(t) = L{un(t) + £{}{ (t − 1)(4,7° una(t) + Run(t) + Nä(t) = g(t) dr L{un+1(t)} = L{un(t)} + L{a(t)} L{L{™³ u« (t) + R &„(t) + Nūn(t) − g(t)}. Taking the variation of (4), which is given by 8 (Lun+1(t))) = 8 (L{u,(t))) +8 (La(t)] L{Lu(t) + R₁₂(t) + Nū₂(t) − g(t)}). 00% (5) By using computation of (5), we get 8 (L{un+1(t)}) = 8 (L{u(t)}) + L{a(t)} & ( L{L¹™³u(t)}) = 0. (6) Hence, from (6) we get where 1+s™ L{a(t)) = 0. -1- ·(0)*-³ – ({*}7 ) (177) Therefore, we get =s&(L{un(t))). L{a(t)) =-1) (0)) Therefore, we have the following iteration algorithm: Lunes (t) = L{un(t)){Lun (t) + Run (t) + Nun(t) − g(t)} - (0)# = (x)** * + (x)* *}* * - (07-10- = L{un(t)) — — — (s™ L{un(t)} – s™¯¹un (0) - L(Ru(t) + Nu(t)-g(t)) +--+ (0) %,n+ (0)"n= - ------) (0)) (−1) (0) — — — L{R u₂(t) + N u(t) − g(t)}
Laplace Variational Iteration Method Let us consider the following general partial differential equation: where L Lu(x,t) + Ru(x, t) + N u(x,t) = g(x,t), (1) R denotes linear operator, N denotes nonlinear operator, and g(x,t) is the source term. The variational iteration method presents a correction functional for (1) in the form: Mas(t) = (t) + (r)(L!?” (t) + (t) + (t)g(t) dr. (2) Maos(t) = µm (t) + ƒ (t − t) (L²™” un (t) + R ® (t) + N Qn(t) − g(t) dr. (3) Taking Laplace transform of (3), we obtain L{una s(t) = L{un(t) + £{}{ (t − 1)(4,7° una(t) + Run(t) + Nä(t) = g(t) dr L{un+1(t)} = L{un(t)} + L{a(t)} L{L{™³ u« (t) + R &„(t) + Nūn(t) − g(t)}. Taking the variation of (4), which is given by 8 (Lun+1(t))) = 8 (L{u,(t))) +8 (La(t)] L{Lu(t) + R₁₂(t) + Nū₂(t) − g(t)}). 00% (5) By using computation of (5), we get 8 (L{un+1(t)}) = 8 (L{u(t)}) + L{a(t)} & ( L{L¹™³u(t)}) = 0. (6) Hence, from (6) we get where 1+s™ L{a(t)) = 0. -1- ·(0)*-³ – ({*}7 ) (177) Therefore, we get =s&(L{un(t))). L{a(t)) =-1) (0)) Therefore, we have the following iteration algorithm: Lunes (t) = L{un(t)){Lun (t) + Run (t) + Nun(t) − g(t)} - (0)# = (x)** * + (x)* *}* * - (07-10- = L{un(t)) — — — (s™ L{un(t)} – s™¯¹un (0) - L(Ru(t) + Nu(t)-g(t)) +--+ (0) %,n+ (0)"n= - ------) (0)) (−1) (0) — — — L{R u₂(t) + N u(t) − g(t)}
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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This picture shows the simplifying steps of how to derive the Laplace variational lteration method with the same steps and the same method. Please derive the Elzaki variational lteration method.
![Laplace Variational Iteration Method
Let us consider the following general partial differential equation:
where L
Lu(x,t) + Ru(x, t) + N u(x,t) = g(x,t),
(1)
R denotes linear operator, N denotes nonlinear operator, and
g(x,t) is the source term.
The variational iteration method presents a correction functional for (1) in the form:
Mas(t) = (t) + (r)(L!?” (t) + (t) + (t)g(t) dr. (2)
Maos(t) = µm (t) + ƒ (t − t) (L²™” un (t) + R ® (t) + N Qn(t) − g(t) dr. (3)
Taking Laplace transform of (3), we obtain
L{una s(t) = L{un(t) + £{}{ (t − 1)(4,7° una(t) + Run(t) + Nä(t) = g(t) dr
L{un+1(t)} = L{un(t)} + L{a(t)} L{L{™³ u« (t) + R &„(t) + Nūn(t) − g(t)}.
Taking the variation of (4), which is given by
8 (Lun+1(t))) = 8 (L{u,(t)))
+8 (La(t)] L{Lu(t) + R₁₂(t) + Nū₂(t) − g(t)}).
00%
(5)
By using computation of (5), we get
8 (L{un+1(t)}) = 8 (L{u(t)}) + L{a(t)} & ( L{L¹™³u(t)}) = 0.
(6)
Hence, from (6) we get
where
1+s™ L{a(t)) = 0.
-1-
·(0)*-³ – ({*}7 ) (177)
Therefore, we get
=s&(L{un(t))).
L{a(t))
=-1) (0))
Therefore, we have the following iteration algorithm:
Lunes (t) = L{un(t)){Lun (t) + Run (t) + Nun(t) − g(t)}
-
(0)# = (x)** * + (x)* *}* * - (07-10-
= L{un(t)) — — — (s™ L{un(t)} – s™¯¹un (0) -
L(Ru(t) + Nu(t)-g(t))
+--+ (0) %,n+ (0)"n=
-
------) (0))
(−1) (0) — — — L{R u₂(t) + N u(t) − g(t)}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd43e050c-4e53-4c88-b771-f608ee0cee74%2F8e3cd1ce-ff43-46df-ba79-6ac04aaf43d7%2Fa6vfxw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Laplace Variational Iteration Method
Let us consider the following general partial differential equation:
where L
Lu(x,t) + Ru(x, t) + N u(x,t) = g(x,t),
(1)
R denotes linear operator, N denotes nonlinear operator, and
g(x,t) is the source term.
The variational iteration method presents a correction functional for (1) in the form:
Mas(t) = (t) + (r)(L!?” (t) + (t) + (t)g(t) dr. (2)
Maos(t) = µm (t) + ƒ (t − t) (L²™” un (t) + R ® (t) + N Qn(t) − g(t) dr. (3)
Taking Laplace transform of (3), we obtain
L{una s(t) = L{un(t) + £{}{ (t − 1)(4,7° una(t) + Run(t) + Nä(t) = g(t) dr
L{un+1(t)} = L{un(t)} + L{a(t)} L{L{™³ u« (t) + R &„(t) + Nūn(t) − g(t)}.
Taking the variation of (4), which is given by
8 (Lun+1(t))) = 8 (L{u,(t)))
+8 (La(t)] L{Lu(t) + R₁₂(t) + Nū₂(t) − g(t)}).
00%
(5)
By using computation of (5), we get
8 (L{un+1(t)}) = 8 (L{u(t)}) + L{a(t)} & ( L{L¹™³u(t)}) = 0.
(6)
Hence, from (6) we get
where
1+s™ L{a(t)) = 0.
-1-
·(0)*-³ – ({*}7 ) (177)
Therefore, we get
=s&(L{un(t))).
L{a(t))
=-1) (0))
Therefore, we have the following iteration algorithm:
Lunes (t) = L{un(t)){Lun (t) + Run (t) + Nun(t) − g(t)}
-
(0)# = (x)** * + (x)* *}* * - (07-10-
= L{un(t)) — — — (s™ L{un(t)} – s™¯¹un (0) -
L(Ru(t) + Nu(t)-g(t))
+--+ (0) %,n+ (0)"n=
-
------) (0))
(−1) (0) — — — L{R u₂(t) + N u(t) − g(t)}
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