3. The Local Fractional Laplace Variational Iteration Method
This section introduces the idea of local fractional Laplace variational method for the following fractional space-time telegraph equation:(6)D0+nαvx,t=Lαvx,t+Nαvx,t+fx,t,where 0<α≤1, x,t≥0, Lα is a linear operator, Nα is nonlinear operator, and f is a source term.
Taking local fractional Laplace transform of (6), both sides with respect to “x” are as follows:(7)snαL-αfx-∑k=1nsk-1αfn-kα0=L-αLαvx,t+Nαvx,t+fx,t.For an algebraic equation, the iteration formula can be constructed as(8)xn+1=xn+λαfxn.The optimality condition for the extreme δxn+1/δxn=0, leads to(9)λα=-1f′xn,where δ is the classical variational operator.
By the formula of (8), we get the iteration formula for (7) as follows: (10)L-αvn+1s,t=L-αvns,t+λαsnαL-αvns,t-∑k=1nsk-1αvnn-kα0,t-L-αLαvnx,t+Nαvnx,t+fx,t.Put λα=-s-nα, the Lagrange multiplier [16] in (10):(11)L-αvn+1s,t=∑k=1nsk-1-nαvnn-kα0,t+s-nαL-αLαvnx,t+Nαvnx,t+fx,t.Taking inverse local fractional Laplace transform into account, we arrived at(12)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+s-nαL-αLαvnx,t+Nαvnx,t+fx,t.This is the iteration formula for (6).
Example 4.
To illustrate the above method, we can consider the following linear equation:(13)dαvx,tdxα=avx,t, 0<x≤T, 0<α≤1, a∈R.We construct the following iteration formula with the help of (12):(14)vn+1x,t=vn0,t-L-α-1s-αL-αavnx,t.Now, applying local fractional Laplace transform to the above equation, find(15)vn+1x,t=vn0,t-L-α-1s-2αavns,t.This is the iteration formula for (13).
Let us start from v0x,t=v0,t=v0.
Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (15), and solving inverse local fractional Laplace transform, we have(16)v1x,t=v0-axαv0Γ1+α.For n=1, v2x is given by the following iteration:(17)v2x,t=v10,t-L-α-1s-2αav1s,t.Using (16) and after easy calculations, we get (18)v2x,t=v0-axαv0Γ1+α+axα2v0Γ1+2α.Similarly for n=2, v2x is given by the following iteration:(19)v3x=v0-axαv0Γ1+α+axα2v0Γ1+2α-axα3v0Γ1+3α⋮vx=∑n=0∞vnx=v0∑k=0∞-akxkαΓ1+kα.This is the exact solution of (13).
Example 5.
The following space-time fractional homogeneous telegraph equation can also be solved by the above introduced local fractional Laplace variational iteration method:(20)∂2αv∂x2αvx,t=∂2αv∂t2αx,t+∂αv∂tαx,t-vx,t, x,t≥0, 0<α≤1with initial conditions(21)v0,t=Eα-tα,vα0,t=Eα-tα.We can find the iteration formula for the above with the help of (12) as(22)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+L-α-1s-2αL-α∂2αvn∂t2αx,t+∂αvn∂tαx,t-vnx,t.Consider initial iteration as follows:(23)v0x,t=L-α-1∑k=1nsk-1-nαvn-kα0,t,L-α-1s-αv0,t+s-2αvα0,t=Eα-tα+xαΓ1+αEα-tα.Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (22); we have (24)v1x,t=L-α-1∑k=1nsk-1-nαv0n-kα0,t+L-α-1s-2αL-α∂2αv0∂t2αx,t+∂αv0∂tαx,t-v0x,t.Using (23) and applying local fractional Laplace and inverse local fractional Laplace transform, we get (25)v1x,t=Eα-tα+xαΓ1+αEα-tα-x2αΓ1+2αEα-tα-x3αΓ1+3αEα-tα.Similarly, we can find(26)v2x,t=Eα-tα+xαΓ1+αEα-tα-x2αΓ1+2αEα-tα-x3αΓ1+3αEα-tα+x4αΓ1+4αEα-tα+x5αΓ1+5αEα-tα,v3x,t=Eα-tα+xαΓ1+αEα-tα-x2αΓ1+2αEα-tα-x3αΓ1+3αEα-tα+x4αΓ1+4αEα-tα+x5αΓ1+5αEα-tα-x6αΓ1+6αEα-tα-x7αΓ1+7αEα-tα.Consequently, we obtain(27)vx,t=Eα-tα1-x2αΓ1+2α+x4αΓ1+4α-x6αΓ1+6α+⋯+Eα-tαxαΓ1+α-x3αΓ1+3α+x5αΓ1+5α-x7αΓ1+7α+⋯,vx,t=Eα-tαcosαxα+Eα-tαsinαxα.
Remark 6.
The result in (27) is the same as the result obtained by Jafari and Jassim [17].
Example 7.
The following space-time fractional homogeneous telegraph equation can also be solved by the above introduced local fractional Laplace variational iteration method:(28)∂2αv∂x2αvx,t=∂2αv∂t2αx,t+2∂αv∂tαx,t+vx,t, x,t≥0, 0<α≤1with initial conditions(29)v0,t=1-Eα-tα,∂α∂xαv0,t=0.Then, we can find the iteration formula for the above with the help of (12) as(30)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+L-α-1s-2αL-α∂2αvn∂t2αx,t+2∂αvn∂tαx,t+vnx,t.Consider initial iteration as follows:(31)v0x,t=L-α-1∑k=1nsk-1-nαvn-kα0,t,L-α-1s-αv0,t+s-2αvα0,t=1-Eα-tα.Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (30); we have (32)v1x,t=L-α-1∑k=1nsk-1-nαv0n-kα0,t+L-α-1s-2αL-α∂2αv0∂t2αx,t+2∂αv0∂tαx,t+v0x,t.Using (31) and applying local fractional Laplace and inverse local fractional Laplace transform, we get (33)v1x,t=1-Eα-tα+x2αΓ1+2α.Similarly, we can find(34)v2x,t=1-Eα-tα+x2αΓ1+2α-x4αΓ1+4α,v3x,t=1-Eα-tα+x2αΓ1+2α+x4αΓ1+4α+x6αΓ1+6α⋮Consequently, we obtain(35)vx,t=1-Eα-tα1+x2αΓ1+2α+x4αΓ1+4α+x6αΓ1+6α+⋯,vx,t=coshαxα-Eα-tα.
Example 8.
We consider the following local fractional Laplace equation:(36)∂2αv∂x2αvx,t+∂2αv∂t2αx,t=0, x,t≥0, 0<α≤1with the initial condition(37)v0,t=-Eαtα,∂α∂xαv0,t=0.Then, we can find the iteration formula for the above with the help of (12) as(38)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+L-α-1s-2αL-α-∂2αvn∂t2αx,t.Consider initial iteration as follows:(39)v0x,t=L-α-1∑k=1nsk-1-nαvn-kα0,t,L-α-1s-αv0,t+s-2αvα0,t=-Eαtα.Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (38); we have (40)v1x,t=L-α-1∑k=1nsk-1-nαv0n-kα0,t+L-α-1s-2αL-α-∂2αv0∂t2αx,t.Using (39) and applying local fractional Laplace and inverse local fractional Laplace transform, we get (41)v1x,t=-Eαtα+x2αΓ1+2αEαtα.Similarly, we can find(42)v2x,t=-Eαtα+x2αΓ1+2αEαtα-x4αΓ1+4αEαtα,v3x,t=-Eαtα+x2αΓ1+2αEαtα-x4αΓ1+4αEαtα+x6αΓ1+6αEαtα⋮Consequently, we obtain(43)vx,t=-Eαtα1-x2αΓ1+2α+x4αΓ1+4α-x6αΓ1+6α+⋯,vx,t=-Eαtαcosαxα.
Example 9.
We consider the following local fractional Laplace equation:(44)∂2αv∂x2αvx,t+∂2αv∂t2αx,t=0with the initial condition(45)v0,t=0,∂α∂xαv0,t=-Eαtα.Then, we can find the iteration formula for the above with the help of (12) as(46)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+L-α-1s-2αL-α-∂2αvn∂t2αx,t.Consider initial iteration as follows:(47)v0x,t=L-α-1∑k=1nsk-1-nαvn-kα0,t,L-α-1s-αv0,t+s-2αvα0,t=-xαΓ1+αEαtα.Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (46); we have (48)v1x,t=L-α-1∑k=1nsk-1-nαv0n-kα0,t+L-α-1s-2αL-α-∂2αv0∂t2αx,t.Using (47) and applying local fractional Laplace and inverse local fractional Laplace transform, we get (49)v1x,t=-xαΓ1+αEαtα+x3αΓ1+3αEαtα.Similarly, we can find(50)v2x,t=-xαΓ1+αEαtα+x3αΓ1+3αEαtα-x5αΓ1+5αEαtα,v3x,t=-xαΓ1+αEαtα+x3αΓ1+3αEαtα-x5αΓ1+5αEαtα+x7αΓ1+7αEαtα⋮Consequently, we obtain(51)vx,t=-EαtαxαΓ1+2α-x3αΓ1+4α+x5αΓ1+6α-⋯,vx,t=-Eαtαsinαxα.