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), (1) where LR denotes linear operator, N denotes nonlinear operator, and g(x,r) is the source term. The variational iteration method presents a correction functional for (1) in the form: or Mn+1(t) = u(t) + 2(x − 1)(L™³ un(t) + Run(t) + N ûn(t) − g(t) dr. (3) Taking Laplace transform of (3), we obtain L{Un+s(t)) = L(Un(t)) + L{ 2(x − t) (L" Un (t) + R &n(t) + N @n(t) − ི་ L{un+1(t)} = L{un(t)} + L{a(t)} L{L™³u„ (t) + R ûn(t) + N (t) − g(t)} (4) Taking the variation of (4), which is given by 8 (Lun+(t))) = (L{u,(t))) +8 (L(a(t)) L{L)(t) +R(t) + Nün(t) − g(t)}). By using computation of (5), we get (5) 8 (Lu+1(t))) = 8 (Lun(t)]) + L{a(t)) & (L{L) un(t)}) = 0. Hence, from (6) we get (6) 1+ s™ L{a(e)) = 0. where -1- 8 (LL)(t))) = 8 (L{u(t)}-s-(0)- 8 (L{u(t)]). Therefore, we get L(x(t)} (-1)(0)) Therefore, we have the following iteration algorithm: L(+(t)) = Lu(t)) {L³ (t) + Ru(t) + Nu(t) = g(t)} =L{ (t)}-{L()}-L(Run(t) + Nun(t) − g(t)} =L{un (t)) (s" L{u, (t))-s*¯¹u„(0) - LR u(t) + Nu(t)-g(t)) (-1) (0)) (0)+(0) (m-1), (0) L(Ru(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), (1) where LR denotes linear operator, N denotes nonlinear operator, and g(x,r) is the source term. The variational iteration method presents a correction functional for (1) in the form: or Mn+1(t) = u(t) + 2(x − 1)(L™³ un(t) + Run(t) + N ûn(t) − g(t) dr. (3) Taking Laplace transform of (3), we obtain L{Un+s(t)) = L(Un(t)) + L{ 2(x − t) (L" Un (t) + R &n(t) + N @n(t) − ི་ L{un+1(t)} = L{un(t)} + L{a(t)} L{L™³u„ (t) + R ûn(t) + N (t) − g(t)} (4) Taking the variation of (4), which is given by 8 (Lun+(t))) = (L{u,(t))) +8 (L(a(t)) L{L)(t) +R(t) + Nün(t) − g(t)}). By using computation of (5), we get (5) 8 (Lu+1(t))) = 8 (Lun(t)]) + L{a(t)) & (L{L) un(t)}) = 0. Hence, from (6) we get (6) 1+ s™ L{a(e)) = 0. where -1- 8 (LL)(t))) = 8 (L{u(t)}-s-(0)- 8 (L{u(t)]). Therefore, we get L(x(t)} (-1)(0)) Therefore, we have the following iteration algorithm: L(+(t)) = Lu(t)) {L³ (t) + Ru(t) + Nu(t) = g(t)} =L{ (t)}-{L()}-L(Run(t) + Nun(t) − g(t)} =L{un (t)) (s" L{u, (t))-s*¯¹u„(0) - LR u(t) + Nu(t)-g(t)) (-1) (0)) (0)+(0) (m-1), (0) L(Ru(t) + Nu(t)-g(t)} Finally, the approximate solution is given by u(x,t) = lim u(x,t)
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Derivation of the Laplace transform variational leration method in the same and simplified manner Replace the Laplace transform with the Elzaki transform and derive it in order to obtain a new method Elzaki variational leration method
![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),
(1)
where LR denotes linear operator, N denotes nonlinear operator, and
g(x,r) is the source term.
The variational iteration method presents a correction functional for (1) in the form:
or
Mn+1(t) = u(t) + 2(x − 1)(L™³ un(t) + Run(t) + N ûn(t) − g(t) dr. (3)
Taking Laplace transform of (3), we obtain
L{Un+s(t)) = L(Un(t)) + L{ 2(x − t) (L" Un (t) + R &n(t) + N @n(t) −
ི་
L{un+1(t)} = L{un(t)} + L{a(t)} L{L™³u„ (t) + R ûn(t) + N (t) − g(t)}
(4)
Taking the variation of (4), which is given by
8 (Lun+(t))) = (L{u,(t)))
+8 (L(a(t)) L{L)(t) +R(t) + Nün(t) − g(t)}).
By using computation of (5), we get
(5)
8 (Lu+1(t))) = 8 (Lun(t)]) + L{a(t)) & (L{L) un(t)}) = 0.
Hence, from (6) we get
(6)
1+ s™ L{a(e)) = 0.
where
-1-
8 (LL)(t))) = 8 (L{u(t)}-s-(0)-
8 (L{u(t)]).
Therefore, we get
L(x(t)}
(-1)(0))
Therefore, we have the following iteration algorithm:
L(+(t)) = Lu(t)) {L³ (t) + Ru(t) + Nu(t) = g(t)}
=L{ (t)}-{L()}-L(Run(t) + Nun(t) − g(t)}
=L{un (t)) (s" L{u, (t))-s*¯¹u„(0) -
LR u(t) + Nu(t)-g(t))
(-1) (0))
(0)+(0)
(m-1),
(0)
L(Ru(t) + Nu(t)-g(t)}
Finally, the approximate solution is given by
u(x,t) = lim u(x,t)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7748e914-953a-42cc-b447-0de5f52ada47%2F9c300ac4-d509-4271-a324-83b7fe7cd429%2Fug1q6nj_processed.jpeg&w=3840&q=75)
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),
(1)
where LR denotes linear operator, N denotes nonlinear operator, and
g(x,r) is the source term.
The variational iteration method presents a correction functional for (1) in the form:
or
Mn+1(t) = u(t) + 2(x − 1)(L™³ un(t) + Run(t) + N ûn(t) − g(t) dr. (3)
Taking Laplace transform of (3), we obtain
L{Un+s(t)) = L(Un(t)) + L{ 2(x − t) (L" Un (t) + R &n(t) + N @n(t) −
ི་
L{un+1(t)} = L{un(t)} + L{a(t)} L{L™³u„ (t) + R ûn(t) + N (t) − g(t)}
(4)
Taking the variation of (4), which is given by
8 (Lun+(t))) = (L{u,(t)))
+8 (L(a(t)) L{L)(t) +R(t) + Nün(t) − g(t)}).
By using computation of (5), we get
(5)
8 (Lu+1(t))) = 8 (Lun(t)]) + L{a(t)) & (L{L) un(t)}) = 0.
Hence, from (6) we get
(6)
1+ s™ L{a(e)) = 0.
where
-1-
8 (LL)(t))) = 8 (L{u(t)}-s-(0)-
8 (L{u(t)]).
Therefore, we get
L(x(t)}
(-1)(0))
Therefore, we have the following iteration algorithm:
L(+(t)) = Lu(t)) {L³ (t) + Ru(t) + Nu(t) = g(t)}
=L{ (t)}-{L()}-L(Run(t) + Nun(t) − g(t)}
=L{un (t)) (s" L{u, (t))-s*¯¹u„(0) -
LR u(t) + Nu(t)-g(t))
(-1) (0))
(0)+(0)
(m-1),
(0)
L(Ru(t) + Nu(t)-g(t)}
Finally, the approximate solution is given by
u(x,t) = lim u(x,t)
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