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In this work, we present four results for the Laplace inverse transform of functions that involve the nth root of a product of linear factors. In order to find the Laplace inverse transform, we considered a branch cut for the nth root and a region of suitable integration, to avoid the branching points. Due to that, the solution is in terms of integrals, we easily approach this solution for some specific parameters.

Modeling phenomena using partial differential equations are attractive to physicists, mathematicians, engineers, etc., this is due to many phenomena such as diffusion of particles, heat diffusion, fluid flow behavior on porous media, among others, are modeled with this type of differential equations (see [

f ( t ) = 1 2 π i ∫ δ − i ∞ δ + i ∞ e s t F ( s ) d s , (1)

to find the function f ( t ) . Here δ > 0 and F ( s ) involve to

( s + a 0 ) ( s + a 1 ) ( s + a 2 ) ⋯ ( s + a k − 1 ) ( s + a k ) n ,

where n is positive integer and a 0 < a 1 < a 2 < ⋯ < a k are real positive. The function f ( t ) is represented in terms of integrals that are easily approximated numerically. To find this function, we consider the branch cut and the integration contour of

In section two, We give analytical examples and also numerically solve the function f ( t ) for some particular cases and compare versus the Stehfest

numerical method. We also show that the Stehfest numerical method does not approximate well to the exact solution near the discontinuities.

In this section, we propose and prove four theorems associated with the Laplace inverse transform for multivalued functions that involve . n . It should be mentioned that in particular Theorem 4 can be used to find the solution of some differential equations that model the fluid flow on porous media, for example [

Theorem 1. Suppose n ∈ ℤ + , t > 0 , 0 = a 0 < a 1 < a 2 < ⋯ < a k < a k + 1 = ∞ and

F ( s ) = 1 s ( s + a 0 ) ( s + a 1 ) ( s + a 2 ) ⋯ ( s + a k − 1 ) ( s + a k ) n ,

then the Laplace inverse transform of the function F ( s ) is given by

f ( t ) = 1 π ∑ j = 0 k sin ( ( j + 1 ) π n ) ∫ a j a j + 1 1 x e − x t u j ( x ) d x , (2)

where

u j ( x ) = ( x − a 0 ) ( x − a 1 ) ⋯ ( x − a j ) ( − x + a j + 1 ) ⋯ ( − x + a k − 1 ) ( − x + a k ) n .

Proof. Using the Laplace inverse transform, we have

f ( t ) = 1 2 π i ∫ δ − i ∞ δ + i ∞ e s x F ( s ) d s .

As a 0 , − a 1 , − a 2 , ⋯ , − a k are branch points of the function F ( s ) , then we consider the region of the

f ( t ) = 1 2 π i ∫ δ − i ∞ δ + i ∞ e s x F ( s ) d s = − 1 2 π i ( ∫ Γ 1 + ∫ Γ ′ 1 + ⋯ + ∫ Γ k + 1 + ∫ Γ ′ k + 1 + ∫ γ 0 ) − 1 2 π i ( ∫ γ 1 + ∫ γ ′ 1 + ⋯ + ∫ γ k + ∫ γ ′ k + ∫ γ k + 1 + ∫ γ k + 2 ) . (3)

It is easy to prove that ∫ γ 0 = 0 when ϵ → 0 . In addition also ∫ γ j → 0 to

j = 1 , 2 , ⋯ , k when ϵ → 0 , this is because if s = − a j + ϵ e i θ with 0 < θ < π then

| ∫ γ j | < ∫ 0 π e ( − a j + ϵ ) t | − a j + ϵ | ( − a j + ϵ + a 0 ) ⋯ ( − a j + ϵ + a j ) ⋯ ( − a j + ϵ + a k ) n ( ϵ ) d θ ,

thus ∫ γ j → 0 when ϵ → 0 . Analogously if s = − a j + ϵ e i θ with − π < θ < 0 we prove that ∫ γ ′ j → 0 when ϵ → 0 .

For ∫ γ k + 1 , we obtain

| ∫ γ k + 1 | < ∫ α π e R cos ( θ ) t ( R + a 0 ) ( R + a 1 ) ⋯ ( R + a k ) n d θ ,

as t > 0 and c o s θ is negative in ( π / 2 , π ) , then e R t c o s ( θ ) → 0 when R → ∞ , thus ∫ γ k + 1 → 0 it is for δ small enough. Similarly for ∫ γ k + 2 .

For integrals over Γ j + 1 and over Γ ′ j + 1 with j = 0 , 1 , 2 , ⋯ , k we analyze as follows: If s = x e i π = − x with x ∈ ( a j , a j + 1 ) we obtain

( s + a k ) 1 / n = { e i π n ( x − a k ) 1 / n for k = 0,1, ⋯ , j , e 0 i n ( − x + a k ) 1 / n for k = j + 1, j + 2, ⋯ , k . (4)

Using Equation (4) we find

u 1 ( x ) = ( s + a 0 ) ( s + a 1 ) ⋯ ( s + a j ) ( s + a j + 1 ) ⋯ ( s + a k ) n = ( s ) 1 / n ( s + a 1 ) 1 / n ⋯ ( s + a j ) 1 / n ( s + a j + 1 ) 1 / n ⋯ ( s + a k ) 1 / n = e i π n x 1 / n e i π n ( x − a 1 ) 1 / n ⋯ e i π n ( x − a j ) 1 / n e i 0 n ( − x + a j + 1 ) 1 / n ⋯ e i 0 n ( − x + a k ) 1 / n = e ( j + 1 ) i π n u j ( x ) ,

then

∫ Γ j + 1 = ∫ a j + 1 a j 1 x e − x t u 1 ( x ) d x = − ∫ a j a j + 1 1 x e − x t u 1 ( x ) d x . (5)

On the other hand, if s = x e − i π = − x con x ∈ ( a j , a j + 1 ) we obtain

u 2 ( x ) = ( s + a 0 ) ( s + a 1 ) ⋯ ( s + a j ) ( s + a j + 1 ) ⋯ ( s + a k ) n = e − ( j + 1 ) i π n u j ( x ) ,

then

∫ Γ ′ j + 1 = ∫ a j a j + 1 1 x e − x t u 2 ( x ) d x . (6)

Thus, for Equation (5) and Equation (6) we find

∫ Γ j + 1 + ∫ Γ ′ j + 1 = − 2 i sin ( ( j + 1 ) π n ) ∫ a j a j + 1 1 x e − x t u j ( x ) d x .

Finally, adding all integrals and replacing in Equation (3), we obtain

f ( t ) = 1 π ∑ j = 0 k sin ( ( j + 1 ) π n ) ∫ a j a j + 1 1 x e − x t u j ( x ) d x .

,

Theorem 2. Suppose n ∈ ℤ + , t > 0 , 0 = a 0 < a 1 < a 2 < ⋯ < a k < a k + 1 = ∞ and

F ( s ) = 1 ( s + a 0 ) ( s + a 1 ) ( s + a 2 ) ⋯ ( s + a k − 1 ) ( s + a k ) n ,

then the Laplace inverse transform of the function F ( s ) is given by

f ( t ) = 1 π ∑ j = 0 k sin ( ( j + 1 ) π n ) ∫ a j a j + 1 e − x t u j ( x ) d x , (7)

where

u j ( x ) = 1 ( x − a 0 ) ( x − a 1 ) ⋯ ( x − a j ) ( − x + a j + 1 ) ⋯ ( − x + a k − 1 ) ( − x + a k ) n .

Proof. Again we use the region

previous Theorem we have ∫ γ 0 → 0 , ∫ γ j → 0 and ∫ γ ′ j → 0 when ϵ → 0 for j = 1 , 2 , ⋯ , k and also ∫ γ k + 1 → 0 and ∫ γ k + 2 → 0 when R → ∞ .

On the other hand we have

∫ Γ j + 1 = ∫ a j + 1 a j e − x t u 1 ( x ) ( − d x ) = ∫ a j a j + 1 e − x t u 1 ( x ) d x ,

∫ Γ ′ j + 1 = − ∫ a j a j + 1 e − x t u 2 ( x ) d x ,

where

u 1 ( x ) = e − ( j + 1 ) i π n u j ( x ) and u 2 ( x ) = e ( j + 1 ) i π n u j ( x ) ,

then

∫ Γ j + 1 + ∫ Γ ′ j + 1 = − 2 i sin ( ( j + 1 ) π n ) ∫ a j a j + 1 e − x t u j ( x ) d x .

Therefore, we add all the integrals so we find the result.,

Theorem 3. Suppose n , m ∈ ℤ + , t > 0 , 0 = a 0 < a 1 < a 2 < ⋯ < a k < a k + 1 = ∞ and

F ( s ) = 1 s ( s + a 0 ) ( s + a 2 ) ( s + a 4 ) ⋯ ( s + a k − 3 ) ( s + a k − 1 ) n ( s + a 1 ) ( s + a 3 ) ( s + a 5 ) ⋯ ( s + a k − 2 ) ( s + a k ) m ,

then the Laplace inverse transform of the function F ( s ) is given by

f ( t ) = 1 π ∑ j = 0 [ k 2 ] sin ( ( j + 1 ) π n − j π m ) ∫ a 2 j a 2 j + 1 1 x e − x t u 2 j ( x ) d x + 1 π ∑ j = 0 [ k 2 ] sin ( ( j + 1 ) π n − ( j + 1 ) π m ) ∫ a 2 j + 1 a 2 j + 2 1 x e − x t u 2 j + 1 ( x ) d x , (8)

where

u j ( x ) = ( x − a 0 ) 1 n ( x − a 1 ) − 1 m ⋯ ( x − a j ) 1 n ( − x + a j + 1 ) − 1 m ⋯ ( − x + a k − 1 ) 1 n ( − x + a k ) − 1 m .

Note that ( x − a j ) 1 / n ( − x + a j + 1 ) − 1 / m can change by ( x − a j ) − 1 / m ( − x + a j + 1 ) 1 / n when k is even or odd.

Proof. The proof is analogous to the previous theorems. ,

Theorem 4. Suppose t > 0 , r > 0 , 0 = a 0 < a 1 < a 2 < ⋯ < a k < a k + 1 = ∞ and

F ( s ) = 1 s K 0 ( r ( s + a 0 ) ( s + a 1 ) ( s + a 2 ) ⋯ ( s + a k − 1 ) ( s + a k ) ) K 0 ( ( s + a 0 ) ( s + a 1 ) ( s + a 2 ) ⋯ ( s + a k − 1 ) ( s + a k ) ) ,

where K 0 is the modified Bessel function of order 0 then the Laplace inverse transform of the function F ( s ) is given by

f ( t ) = 1 + 1 π ∑ j = 1 [ k 2 ] + 1 ∫ a 2 j − 2 a 2 j − 1 1 x e − x t Ψ 2 j − 2 ( r , x ) d x + ∑ j = 1 L ∫ a 2 j − 1 a 2 j 1 x e − x t Ψ 2 j − 1 ( r , x ) d x (9)

where

L = { [ k 2 ] for k even , 1 + [ k 2 ] for k odd ,

Ψ l ( r , x ) = { I 0 ( u l ( x ) ) K 0 ( u l ( x ) r ) − K 0 ( u l ( x ) ) I 0 ( u l ( x ) r ) K 0 ( u l ( x ) ) 2 + π 2 I 0 ( u l ( x ) ) 2 for l = 2 j − 1, Y 0 ( u l ( x ) ) J 0 ( u l ( x ) r ) − J 0 ( u l ( x ) ) Y 0 ( u l ( x ) r ) J 0 ( u l ( x ) ) 2 + Y 0 ( u l ( x ) ) 2 for l = 2 j − 2,

I 0 , Y 0 , J 0 are the corresponding Bessel functions of order 0 and

u j ( x ) = ( x − a 0 ) ( x − a 1 ) ⋯ ( x − a j ) ( − x + a j + 1 ) ⋯ ( − x + a k − 1 ) ( − x + a k ) .

Proof. For the proof, we used the following properties of the Bessel functions (see [

K ν ( z ) = i π 2 e π 2 ν i H ν ( 1 ) ( z e 1 2 π i ) , ( − π < arg z ≤ 1 2 π ) , (10)

K ν ( z ) = − i π 2 e − π 2 ν i H − ν ( 2 ) ( z e − 1 2 π i ) , ( − 1 2 π < arg z ≤ π ) , (11)

K ν ( z e i m π ) = e − i m ν π K ν ( z ) − i π sin ( m ν π ) csc ( ν π ) I ν ( z ) , ( m ∈ ℤ ) , (12)

where H ν ( 1 ) ( z ) and H − ν ( 2 ) ( z ) are the Hankel’s functions of the order ν .

Consider the case when k is odd (for k even the proof is analogous). Also

consider the region of

the integrals on Γ j and on Γ ′ j with j = 1 , ⋯ , k + 1 we obtain the following: if j is of the form 2 j − 1 , we used Equation (10) and Equation (11), then

∫ Γ 2 j − 1 + ∫ Γ ′ 2 j − 1 = − 2 i ∫ a 2 j − 2 a 2 j − 1 1 x e − x t Ψ 2 j − 2 ( r , x ) d x , (13)

where

Ψ 2 j − 2 ( r , x ) = Y 0 ( u 2 j − 2 ( x ) ) J 0 ( u 2 j − 2 ( x ) r ) − J 0 ( u 2 j − 2 ( x ) ) Y 0 ( u 2 j − 2 ( x ) r ) J 0 ( u 2 j − 2 ( x ) ) 2 + Y 0 ( u 2 j − 2 ( x ) ) 2 .

On the other hand, if j is of the form 2 j , we used Equation (12) then

∫ Γ 2 j + ∫ Γ ′ 2 j = − 2 π i ∫ a 2 j − 1 a 2 j 1 x e − x t Ψ 2 j − 1 ( r , x ) d x , (14)

where

Ψ 2 j − 1 ( r , x ) = I 0 ( u 2 j − 1 ( x ) ) K 0 ( u 2 j − 1 ( x ) r ) − K 0 ( u 2 j − 1 ( x ) ) I 0 ( u 2 j − 1 ( x ) r ) K 0 ( u 2 j − 1 ( x ) ) 2 + π 2 I 0 ( u 2 j − 1 ( x ) ) 2 .

Note that u j ( x ) is found using Equation (4) with n = 2 . As k is odd then there are [ k 2 ] + 1 integrals of the form Equation (13) and there are [ k 2 ] + 1 of the form Equation (14). Thus

∫ γ 0 + ∫ Γ 1 + ∫ Γ ′ 1 + ⋯ + ∫ Γ k + 1 + ∫ Γ ′ k + 1 = − 2 π i − 2 i ∑ j = 1 [ k 2 ] + 1 ∫ a 2 j − 2 a 2 j − 1 1 x e − x t Ψ 2 j − 2 ( r , x ) d x − 2 π i ∑ j = 1 [ k 2 ] + 1 ∫ a 2 j − 1 a 2 j 1 x e − x t Ψ 2 j − 1 ( r , x ) d x . (15)

On the other hand, for ∫ γ k + 1 , we have that

| ∫ γ k + 1 | < ∫ α π e R t cos ( θ ) K 0 ( r ( R k + 1 ( a 1 R + 1 ) ( a 2 R + 1 ) ⋯ ) ( a k − 1 R + 1 ) ( a k R + 1 ) ) K 0 ( ( R k + 1 ( a 1 R + 1 ) ( a 2 R + 1 ) ⋯ ) ( a k − 1 R + 1 ) ( a k R + 1 ) ) d θ , (16)

so, ∫ γ k + 1 is 0 when R → ∞ for δ sufficiently small. Similarly to ∫ γ k + 2 . Thus, using ∫ γ 0 = − 2 π i , Equation (15) and Equation (16), we arrive to

f ( t ) = 1 + 1 π ∑ j = 1 [ k 2 ] + 1 ∫ a 2 j − 2 a 2 j − 1 1 x e − x t Ψ 2 j − 2 ( r , x ) d x + ∑ j = 1 [ k 2 ] + 1 ∫ a 2 j − 1 a 2 j 1 x e − x t Ψ 2 j − 1 ( r , x ) d x

,

In this section, we give some analytical examples corresponding to the exact solutions of the previous theorems and solve the integrals numerically for some particular cases.

Example 1. We consider that k = 1 , n = 2 , 0 = a 0 < a 1 < a 2 = ∞ and

F ( s ) = 1 s s ( s + a 1 ) , then using (2)

f ( t ) = 1 π sin ( ( 0 + 1 ) π 2 ) ∫ 0 a 1 1 x e − x t x ( − x + a 1 ) d x + 1 π sin ( ( 1 + 1 ) π 2 ) ∫ a 1 ∞ 1 x e − x t x ( x − a 1 ) d x = 1 π ∫ 0 a 1 1 x e − x t x ( − x + a 1 ) d x .

Furthermore, since

∫ 0 a 1 1 x e − x t x ( − x + a 1 ) d x = a 1 2 e − a 1 t 2 π ( I 0 ( a 1 t 2 ) + I 1 ( a 1 t 2 ) ) ,

with I α is the modified Bessel function of order α , then

f ( t ) = a 1 2 e − a 1 t 2 ( I 0 ( a 1 t 2 ) + I 1 ( a 1 t 2 ) ) .

Example 2. We consider that k = 2 , n = 2 , 0 = a 0 < a 1 < a 2 < a 3 = ∞ and F ( s ) = 1 s ( s + a 1 ) ( s + a 2 ) , then using (7)

f ( t ) = 1 π sin ( ( 0 + 1 ) π 2 ) ∫ 0 a 1 e − x t 1 x ( − x + a 1 ) ( − x + a 2 ) d x + 1 π sin ( ( 1 + 1 ) π 2 ) ∫ a 1 a 2 e − x t 1 x ( x − a 1 ) ( − x + a 2 ) d x + 1 π sin ( ( 2 + 1 ) π 2 ) ∫ a 2 ∞ e − x t 1 x ( x − a 1 ) ( x − a 2 ) d x ,

so

f ( t ) = 1 π ∫ 0 a 1 e − x t 1 x ( − x + a 1 ) ( − x + a 2 ) d x − 1 π ∫ a 2 ∞ e − x t 1 x ( x − a 1 ) ( x − a 2 ) d x

Example 3. We consider that k = 1 , n = 3 , m = 2 , 0 = a 0 < a 1 = π < a 2 = ∞ and F ( s ) = 1 s s 3 ( s + a 1 ) 2 , then using (8)

f ( t ) = 1 π ∑ j = 0 [ 1 2 ] sin ( ( j + 1 ) π 3 − j π 2 ) ∫ a 2 j a 2 j + 1 1 x e − x t u 2 j ( x ) d x + 1 π ∑ j = 0 [ 1 2 ] sin ( ( j + 1 ) π 3 − ( j + 1 ) π 2 ) ∫ a 2 j + 1 a 2 j + 2 1 x e − x t u 2 j + 1 ( x ) d x ,

so

f ( t ) = 1 π sin ( π 3 ) ∫ 0 π 1 x e − x t s 3 − x + π d x + 1 π sin ( − π 6 ) ∫ π ∞ 1 x e − x t s 3 x − π d x = 3 Γ ( 5 / 6 ) t 1 / 6 π F 1 1 ( 1 / 2 , 7 / 6 , − π t ) ,

where 1 F 1 is the hypergeometric function.

Example 4. We consider that r > 0 , k = 1 , a 1 = 1 ϵ and F ( s ) = 1 s k 0 ( r s ( s + 1 / ϵ ) ) k 0 ( s ( s + 1 / ϵ ) ) then using (9)

f ( t ) = 1 + 1 π ∫ 0 1 ϵ 1 x e − x t Ψ 0 ( r , x ) d x + ∫ 1 ϵ ∞ 1 x e − x t Ψ 1 ( r , x ) d x

where

Ψ 0 ( r , x ) = Y 0 ( u 0 ( x ) ) J 0 ( u 0 ( x ) r ) − J 0 ( u 0 ( x ) ) Y 0 ( u 0 ( x ) r ) J 0 ( u 0 ( x ) ) 2 + Y 0 ( u 0 ( x ) ) 2 ,

Ψ 1 ( r , x ) = I 0 ( u 1 ( x ) ) K 0 ( u 1 ( x ) r ) − K 0 ( u 1 ( x ) ) I 0 ( u 1 ( x ) r ) K 0 ( u 1 ( x ) ) 2 + π 2 I 0 ( u 1 ( x ) ) 2 ,

u 0 ( x ) = x ( − x + 1 / ϵ ) and u 1 ( x ) = x ( x − 1 / ϵ ) .

On the other hand, we solve the integrals numerically corresponding to the exact solutions of Theorems 1, 2, 3 and 4 then we compare versus Stehfest numerical method. This method consist in numerically finding the Laplace inverse transform of the F ( s ) using:

f ( t ) = log ( 2 ) t ∑ i = 1 n 1 g ( i ) F ( log ( 2 ) t i )

where

g ( i ) = ( − 1 ) n 1 2 + i ∑ k = [ i + 1 2 ] min { i , n 1 2 } k n 1 2 ( 2 k ) ! ( n 1 2 − k ) ! k ! ( k − 1 ) ! ( i − k ) ! ( 2 k − i ) ! .

In

(7)) respectively with parameters n = 2 , k = 5 , a 1 = 1 , a 2 = π , a 3 = 5 , a 4 = e 2 and a 5 = 10 . In

In this work, we solved the inverse Laplace transform for multivalued functions that involving the nth root of a product of linear factors, we show that results are correct and also give analytical and numerical examples. The numerical examples were compared with Stehfest numerical method, concluding that the curves coincide for values far from the discontinuities of the solution, while for values close to the discontinuity the Stehfest numerical method does not have good approximation.

Salmeron-Rodri- guez, U. (2017) Laplace Inverse Transform for Functions of Type nth Root of a Product of Linear Factors. Open Access Library Jour- nal, 4: e3741. https://doi.org/10.4236/oalib.1103741