→ Laplace Variational Iteration Method Let us consider the following general partial differential equation: Lu(x,t)+ Ru(x,t) + Nu(x,t) = g(x,t), where L (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: Mats (t) = Bu(t) + 2(t) (Lun (t) + R n (t) + Na(t) = g(1) dr. (2) ог - Un+1(t) = u(t) + [d(x − t) (Lun(t) + Rdx(t) + N (r)-g(1)) dr. (3) Taking Laplace transform of (3), we obtain L{un+1(t) = Lua(t)) + L{{ (t−1)(L™³ un(t) + Run(t) + Nün(t) − g(†) de મ L(Uns(t) = L{un(t)) + L{(c)) L{L (t) + Rün(t) + N Bu(t) − g(t)}. Taking the variation of (4), which is given by 8 (L{un+1(e))) = 8 (L{un(t)}) (DO +8 (L{2(t)) L{L¹³un (t) + Rū₁(t) + Nū,(t) − g(t)}). (5) By using computation of (5), we get 5 (L{un+1(t)}) = 5 (L(x(t)}) + L{λ(t)) & ( L{L™)u(t)}) = 0. (6) Hence, from (6) we get where 1+s™ L{(t)) = 0. -1- 8 (L{L (c)}) = 8 (5L (t)) – sm−¹u₂(0) — =s8 (L{u,(t))). Therefore, we get L(2(t)) Therefore, we have the following iteration algorithm: (m-1) Lun(e)) L{un(t))- Sm Lu(t) + Ru(t) + Nu₁(t) − g(t)} L{un(t)) - =L{un(t)) - L{(t)}L(Run(t) + Nun(t) − g(t)) - (sLun())-su(0) - LR u(t) + Nu(t) - g(t)) (-1) (0)) (0)+(0)+ ++ (m-1) -(0)- LR u,(t) + Nu,(t)-g(t)) Finally, the approximate solution is given by u(x,t) = lim u(x,t)
→ Laplace Variational Iteration Method Let us consider the following general partial differential equation: Lu(x,t)+ Ru(x,t) + Nu(x,t) = g(x,t), where L (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: Mats (t) = Bu(t) + 2(t) (Lun (t) + R n (t) + Na(t) = g(1) dr. (2) ог - Un+1(t) = u(t) + [d(x − t) (Lun(t) + Rdx(t) + N (r)-g(1)) dr. (3) Taking Laplace transform of (3), we obtain L{un+1(t) = Lua(t)) + L{{ (t−1)(L™³ un(t) + Run(t) + Nün(t) − g(†) de મ L(Uns(t) = L{un(t)) + L{(c)) L{L (t) + Rün(t) + N Bu(t) − g(t)}. Taking the variation of (4), which is given by 8 (L{un+1(e))) = 8 (L{un(t)}) (DO +8 (L{2(t)) L{L¹³un (t) + Rū₁(t) + Nū,(t) − g(t)}). (5) By using computation of (5), we get 5 (L{un+1(t)}) = 5 (L(x(t)}) + L{λ(t)) & ( L{L™)u(t)}) = 0. (6) Hence, from (6) we get where 1+s™ L{(t)) = 0. -1- 8 (L{L (c)}) = 8 (5L (t)) – sm−¹u₂(0) — =s8 (L{u,(t))). Therefore, we get L(2(t)) Therefore, we have the following iteration algorithm: (m-1) Lun(e)) L{un(t))- Sm Lu(t) + Ru(t) + Nu₁(t) − g(t)} L{un(t)) - =L{un(t)) - L{(t)}L(Run(t) + Nun(t) − g(t)) - (sLun())-su(0) - LR u(t) + Nu(t) - g(t)) (-1) (0)) (0)+(0)+ ++ (m-1) -(0)- LR u,(t) + Nu,(t)-g(t)) Finally, the approximate solution is given by u(x,t) = lim u(x,t)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
This image contains the derivation of the Laplace variational literation method. Help me find the derivation of the Elzaki variational literation method in the same method, please
Transcribed Image Text:→ Laplace Variational Iteration Method
Let us consider the following general partial differential equation:
Lu(x,t)+ Ru(x,t) + Nu(x,t) = g(x,t),
where L
(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:
Mats (t) = Bu(t) + 2(t) (Lun (t) + R n (t) + Na(t) = g(1) dr. (2)
ог
-
Un+1(t) = u(t) + [d(x − t) (Lun(t) + Rdx(t) + N (r)-g(1)) dr. (3)
Taking Laplace transform of (3), we obtain
L{un+1(t) = Lua(t)) + L{{ (t−1)(L™³ un(t) + Run(t) + Nün(t) − g(†) de
મ
L(Uns(t) = L{un(t)) + L{(c)) L{L (t) + Rün(t) + N Bu(t) − g(t)}.
Taking the variation of (4), which is given by
8 (L{un+1(e))) = 8 (L{un(t)})
(DO
+8 (L{2(t)) L{L¹³un (t) + Rū₁(t) + Nū,(t) − g(t)}).
(5)
By using computation of (5), we get
5 (L{un+1(t)}) = 5 (L(x(t)}) + L{λ(t)) & ( L{L™)u(t)}) = 0.
(6)
Hence, from (6) we get
where
1+s™ L{(t)) = 0.
-1-
8 (L{L (c)}) = 8 (5L (t)) – sm−¹u₂(0) —
=s8 (L{u,(t))).
Therefore, we get
L(2(t))
Therefore, we have the following iteration algorithm:
(m-1)
Lun(e)) L{un(t))-
Sm
Lu(t) + Ru(t) + Nu₁(t) − g(t)}
L{un(t))
-
=L{un(t))
-
L{(t)}L(Run(t) + Nun(t) − g(t))
-
(sLun())-su(0) -
LR u(t) + Nu(t) - g(t))
(-1) (0))
(0)+(0)+ ++
(m-1)
-(0)-
LR u,(t) + Nu,(t)-g(t))
Finally, the approximate solution is given by
u(x,t)
= lim u(x,t)
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