In this subsection we will study the existence of a solution to the problem ( P m , q ) and the continuous dependence of the solution with respect to the initial data in P ′ .

Theorem 3.1 Let μ > 0 , β > 0 , m and q are even number not a multiple of four, and the distributional problem

( P m , q )   |   u ∈ C ( [ 0, + ∞ ) , P ′ )   ∂ t u − i μ ∂ x m u = β ∂ x q u ∈ P ′   u ( 0 ) = f ∈ P ′ .

then ( P m , q ) has a unique solution u ∈ C 1 ( ( 0, + ∞ ) , P ′ ) . Furthermore, the solution depends continuously on the initial data. That is, given f n , f ∈ P ′ such that f n → P ′ f implies u n ( t ) → P ′ u ( t ) , ∀ t ∈ [ 0, + ∞ ) , where u n is solution of ( P m , q ) with initial data f n and u is solution of ( P m , q ) with initial data f.

Proof.- We have organized the proof as follows.

1) Suppose there exists u ∈ C ( [ 0, + ∞ ) , P ′ ) satisfying ( P m , q ); this will allow us to obtain the explicit form of u. Then taking the Fourier transform to the equation

∂ t u − i μ ∂ x m u = β ∂ x q u ,

we get

− β k q u ^ = β ( i k ) q u ^ = ∂ t u ^ − i μ ( i k ) m u ^ = ∂ t u ^ + i μ k m u ^ ,

which for each k ∈ ℤ is an ODE with initial data u ^ ( k ,0 ) = f ^ ( k ) .

Thus, we propose an uncoupled system of homogeneous first-order ordinary differential equations

( Ω k )         |   u ^ ∈ C ( ( 0, + ∞ ) , S ′ ( ℤ ) )   ∂ t u ^ ( k , t ) + i μ k m u ^ ( k , t ) = − β k q u ^ ( k , t )   u ^ ( k ,0 ) = f ^ ( k )     with     f ^ ∈ S ′ ( ℤ ) ,

∀ k ∈ ℤ   and we get

u ^ ( k , t ) = e − i μ k m t e − β k q t f ^ ( k ) ,

from where we obtain the explicit expression of u, candidate for solution:

u ( t ) = ∑ k = − ∞ + ∞   u ^ ( k , t ) ϕ k = ∑ k = − ∞ + ∞   e − i μ k m t e − β k q t f ^ ( k ) ϕ k (1)

= [ ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ] ∨ . (2)

Since f ∈ P ′ then f ^ ∈ S ′ ( Z ) . Thus, we affirm that

( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ∈ S ′ ( Z ) ,   ∀ t ≥ 0. (3)

Indeed, let t ≥ 0 , since f ^ ∈ S ′ ( Z ) then satisfies: ∃ C > 0 , ∃ N ∈ I N such that | f ^ ( k ) | ≤ C | k | N , ∀ k ∈ Z − { 0 } , using this we get

| f ^ ( k ) e − i μ k m t e − β k q t | = | f ^ ( k ) e − i μ k m t | e − β k q t ︸ ≤ 1 ≤ | f ^ ( k ) | | e − i μ k m t | ︸ = 1 = | f ^ ( k ) | ≤ C | k | N .

Then,

( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ∈ S ′ ( Z ) .

If we define

u ( t ) : = [ ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ] ∨ ,       for   all     t ≥ 0, (4)

we have that u ( t ) ∈ P ′ , ∀ t ≥ 0 , since we apply the inverse Fourier transform to ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ∈ S ′ ( Z ) .

2) We will prove that u defined in (4) is solution of ( P m , q ) and u ∈ C 1 ( ( 0, ∞ ) , P ′ ) .

Evaluating (2) at t = 0 , we obtain

u ( 0 ) = [ ( f ^ ( k ) ) k ∈ Z ] ∨ = [ f ^ ] ∨ = f   .

Also, the following statements are verified.

a) ∂ t   u ( t ) = i μ   ∂ x m   u ( t ) + β   ∂ x q   u ( t ) in P ′ , ∀ t > 0 . That is, we will prove that it is satisfied:

lim h → 0 〈 u ( t + h ) − u ( t ) h , φ 〉 ︸ 〈 ∂ t u ( t ) , φ 〉 : = = i μ 〈 ∂ x m u ( t ) , φ 〉 + β 〈 ∂ x q u ( t ) , φ 〉 ,     ∀ φ ∈ P

and for all t > 0 .

Indeed, let t > 0 , φ ∈ P and 0 < | h | < t , we denote

I h , t : = 〈 u ( t + h ) − u ( t ) h , φ 〉 .

Thus, we get

I h , t = 1 h { 〈 u ( t + h ) , φ 〉 − 〈 u ( t ) , φ 〉 } = 1 h { l i m n → + ∞ 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m ( t + h ) e − β k q ( t + h ) ϕ k , φ 〉     − l i m n → + ∞ 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ϕ k , φ 〉 } = 1 h { l i m n → + ∞ 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ( e − i μ k m h e − β k q h − 1 ) ϕ k , φ 〉 }

= l i m n → + ∞ 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ( e − i μ k m h e − β k q h − 1 h ) ϕ k , φ 〉 = l i m n → + ∞ { ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ( e − i μ k m h e − β k q h − 1 h ) 〈 ϕ k , φ 〉 ︸ = 2 π φ ^ ( − k ) } = l i m n → + ∞ 2 π { ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ( e − i μ k m h e − β k q h − 1 h ) φ ^ ( − k ) } = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t ( e − i μ k m h e − β k q h − 1 h ) φ ^ ( − k ) . (5)

Let h > 0 , we have

e − i μ k m h e − β k q h − 1 = ∫ 0 h [ e − i μ k m s e − β k q s ] ′ d s = ∫ 0 h ( − i μ k m − β k q ) e − i μ k m s e − β k q s d s . (6)

Taking norm to equality (6) we obtain

| e − i μ k m h e − β k q h − 1 | ≤ ∫ 0 h { μ | k | m + β | k | q } | e − i μ k m s | ︸ = 1 e − β k q s ︸ ≤ 1 d s = { μ | k | m + β | k | q } ∫ 0 h   d s ︸ = h = { μ | k | m + β | k | q } h . (7)

That is, from (7) we get

| e − i μ k m h e − β k q h − 1 h | ≤ μ | k | m + β | k | q . (8)

Using the inequality (8) and that f ^ ∈ S ′ ( Z ) we obtain

∑ k = − ∞ + ∞ | f ^ ( k ) | | e − i μ k m t | ︸ = 1 e − β k q t ︸ ≤ 1 | φ ^ ( − k ) | | e − i μ k m h e − β k q h − 1 h | ≤ ∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | { μ | k | m + β | k | q } = μ ∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | | k | m + β ∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | | k | q ≤ C { μ ∑ k = − ∞ + ∞ | k | N + m | φ ^ ( − k ︸ = J ) | + β ∑ k = − ∞ + ∞ | k | N + q | φ ^ ( − k ︸ = J ) | } = C { μ ∑ J = − ∞ + ∞ | J | N + m | φ ^ ( J ) | + β ∑ J = − ∞ + ∞ | J | N + q | φ ^ ( J ) | } < ∞

since φ ^ ∈ S ( Z ) .

If h < 0 we have

| e − i μ k m h e − β k q h − 1 h | ≤ { μ | k | m + β | k | q } e − β k q h . (9)

Using the inequality (9), 0 < | h | < t and that f ^ ∈ S ′ ( Z ) we obtain

∑ k = − ∞ + ∞ | f ^ ( k ) | | e − i μ k m t | ︸ = 1 e − β k q t | φ ^ ( − k ) | | e − i μ k m h e − β k q h − 1 h | ≤ ∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | e − β k q ( t + h ) ︸ ≤ 1 { μ | k | m + β | k | q } = μ ∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | | k | m + β ∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | | k | q ≤ C { μ ∑ k = − ∞ + ∞ | k | N + m | φ ^ ( − k ︸ = J ) | + β ∑ k = − ∞ + ∞ | k | N + q | φ ^ ( − k ︸ = J ) | } = C { μ ∑ J = − ∞ + ∞ | J | N + m | φ ^ ( J ) | + β ∑ J = − ∞ + ∞ | J | N + q | φ ^ ( J ) | } < ∞

since φ ^ ∈ S ( Z ) .

Using the Weierstrass M-Test, the series I h , t is absolute and uniformly convergent. Then we can take limit and get

lim h → 0 I h , t = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t φ ^ ( − k ) lim h → 0 { e − i μ k m h e − β k q h − 1 h } ︸ = − i μ k m − β k q = ( − i μ ) 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t φ ^ ( − k ) k m     − β 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t φ ^ ( − k ) k q . (10)

Using (10) and that 〈 T ( r ) , φ 〉 = ( − 1 ) r 〈 T , φ ( r ) 〉 = 〈 T , φ ( r ) 〉 for φ ∈ P , T ∈ P ′ with r a even number, we have

l i m h → 0 I h , t = ( − i μ ) 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t φ ^ ( − k ) ︸ = 1 2 π 〈 φ , ϕ k 〉 k m − β 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t φ ^ ( − k ) ︸ = 1 2 π 〈 φ , ϕ k 〉 k q = i μ ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t 〈 φ , − k m ϕ k ︸ = ( i k ) m ϕ k 〉 + β ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t 〈 φ , − k q ϕ k ︸ = ( i k ) q ϕ k 〉 = i μ ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t 〈 φ , ϕ k ( m ) 〉 ︸ = 〈 φ ( m ) , ϕ k 〉 + β ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t 〈 φ , ϕ k ( q ) 〉 ︸ = 〈 φ ( q ) , ϕ k 〉 = i μ ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t 〈 ϕ k , φ ( m ) 〉 + β ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t 〈 ϕ k , φ ( q ) 〉

= i μ l i m n → + ∞ ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t 〈 ϕ k , φ ( m ) 〉 + β l i m n → + ∞ ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t 〈 ϕ k , φ ( q ) 〉 = i μ l i m n → + ∞ 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ϕ k , φ ( m ) 〉 + β l i m n → + ∞ 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ϕ k , φ ( q ) 〉 = i μ 〈 u ( t ) , φ ( m ) 〉 + β 〈 u ( t ) , φ ( q ) 〉 = i μ 〈 ∂ x m u ( t ) , φ 〉 + β 〈 ∂ x q u ( t ) , φ 〉 . (11)

Therefore,

〈 ∂ t u ( t ) , φ 〉 = i μ 〈 ∂ x m u ( t ) , φ 〉 + β 〈 ∂ x q u ( t ) , φ 〉 ,   ∀ φ ∈ P ,   ∀ t > 0.

That is,

∂ t u ( t ) = i μ   ∂ x m   u ( t ) + β   ∂ x q   u ( t )       in     P ′   ,   ∀ t > 0.

b) u ∈ C ( [ 0, + ∞ ) , P ′ ) . That is, we will prove that

u ( t + h ) → P ′ u ( t )         when       h → 0,     ∀ t ≥ 0.

In effect, let t > 0 , φ ∈ P , we will prove that

H t , h : = 〈 u ( t + h ) − u ( t ) , φ 〉 → 0   ,       when       h → 0.

We know that if φ ∈ P then φ ^ ∈ S ( Z ) . Using (5) we have

H t , h = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t ( e − i μ k m h e − β k q h − 1 ) φ ^ ( − k ) .

Let 0 < h < 1 , from (8) we get

| e − i μ k m h e − β k q h − 1 | ≤ μ | k | m | h | + β k q | h | < μ | k | m + β | k | q . (12)

Using (12) and that f ^ ∈ S ′ ( Z ) we obtain

∑ k = − ∞ + ∞ | f ^ ( k ) | | e − i μ k m t | ︸ = 1 e − β k q t ︸ ≤ 1 | e − i μ k m h e − β k q h − 1 | | φ ^ ( − k ) | ≤ C μ ∑ k = − ∞ + ∞ | k | N + m | φ ^ ( − k ︸ = J ) | + C β ∑ k = − ∞ + ∞ | k | N + q | φ ^ ( − k ︸ = J ) | = C μ ∑ J = − ∞ + ∞ | J | N + m | φ ^ ( J ) | + C β ∑ J = − ∞ + ∞ | J | N + q | φ ^ ( J ) | < ∞

since φ ^ ∈ S ( Z ) .

Let h < 0 , | h | < 1 from (9) we get

| e − i μ k m h e − β k q h − 1 | ≤ e − β k q h { μ | k | m + β | k | q } . (13)

Using (13), 0 < | h | < t and that f ^ ∈ S ′ ( Z ) we obtain

∑ k = − ∞ + ∞ | f ^ ( k ) | | e − i μ k m t | ︸ = 1 e − β k q t | e − i μ k m h e − β k q h − 1 | | φ ^ ( − k ) | ≤ C μ ∑ k = − ∞ + ∞ | k | N + m | φ ^ ( − k ︸ = J ) | + C β ∑ k = − ∞ + ∞ | k | N + q | φ ^ ( − k ︸ = J ) | = C μ ∑ J = − ∞ + ∞ | J | N + m | φ ^ ( J ) | + C β ∑ J = − ∞ + ∞ | J | N + q | φ ^ ( J ) | < ∞

since φ ^ ∈ S ( Z ) .

Using the Weierstrass M-Test we conclude that the series H t , h converges absolute and uniformly. Then it is possible to take limit and obtain

lim h → 0 H t , h = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t φ ^ ( − k ) l i m h → 0 { e − i μ k m h e − β k q h − 1 } ︸ = 0 = 0. (14)

Doing t = 0 in H t , h with h > 0 , we have

H 0 , h = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) φ ^ ( − k ) { e − i μ k m h e − β k q h − 1 } .

Using (12) and that f ^ ∈ S ′ ( Z ) we obtain

∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | | e − i μ k m h e − β k q h − 1 | ≤ C μ ∑ k = − ∞ + ∞ | J | N + m | φ ^ ( J ) | + C β ∑ k = − ∞ + ∞ | J | N + q | φ ^ ( J ) | < ∞

since φ ^ ∈ S ( Z ) .

Using the Weierstrass M-Test we conclude that the series H 0, h converges absolute and uniformly. Then is possible to take limit and obtain

l i m h → 0 + H 0 , h = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) φ ^ ( − k ) l i m h → 0 + { e − i μ k m h e − β k q h − 1 } ︸ = 0 = 0. (15)

From (14) and (15) we can conclude that

u ∈ C ( [ 0, ∞ ) , P ′ ) .

c) ∂ t u ∈ C ( I R + , P ′ ) . That is, we will prove that

∂ t u ( t + h ) → P ′ ∂ t u ( t )         when       h → 0,     ∀ t ∈ I R + .

In effect, let t ∈ I R + and φ ∈ P , using item a) we have

〈 ∂ t u ( t + h ) , φ 〉 − 〈 ∂ t u ( t ) , φ 〉 = i μ { 〈 ∂ x m u ( t + h ) , φ 〉 − 〈 ∂ x m u ( t ) , φ 〉 } + β { 〈 ∂ x q u ( t + h ) , φ 〉 − 〈 ∂ x q u ( t ) , φ 〉 } = i μ { 〈 u ( t + h ) , φ ( m ) 〉 − 〈 u ( t ) , φ ( m ) 〉 } ︸ → 0 + β { 〈 u ( t + h ) , φ ( q ) 〉 − 〈 u ( t ) , φ ( q ) 〉 } ︸ → 0 → 0 (16)

when h → 0 , since item b) is valid with φ ( r ) ∈ P for r ∈ { m , q } .

From b) and c) we have that u ∈ C 1 ( I R + , P ′ ) .

3) Now, we will prove that the solution depends continuously respect to initial data. That is, if f n → P ′ f we will prove that:

u n ( t ) → P ′ u ( t ) ,   ∀ t ∈ I R + .

We know that if f n → P ′ f then f ^ n → S ′ ( Z ) f ^ , that is

〈 f ^ n − f ^ , ξ 〉 → 0         when       n → + ∞   ,   ∀ ξ ∈ S ( Z ) . (17)

For t ∈ I R + fixed and arbitrary, we want to prove that

〈 u n ( t ) , ψ 〉 → 〈 u ( t ) , ψ 〉         when     n → + ∞   ,   ∀ ψ ∈ P   .

Thus, let t ∈ I R + be fixed and ψ ∈ P , using the generalized Parseval identity, we obtain the following equalities:

〈 u n ( t ) , ψ 〉 = 2 π 〈 ( f ^ n ( k ) e − i μ k m t e − β k q t ) k ∈ Z , ψ ^ ˜ 〉 (18)

〈 u ( t ) , ψ 〉 = 2 π 〈 ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z , ψ ^ ˜ 〉 . (19)

From (18) and (19) we obtain:

〈 u n ( t ) , ψ 〉 − 〈 u ( t ) , ψ 〉 = 2 π ∑ k = − ∞ + ∞ { f ^ n ( k ) − f ^ ( k ) } e − i μ k m t e − β k q t ψ ^ ˜ ( k ) ︸ ξ k : = → 0

when n → + ∞ , since ξ : = ( ξ k ) k ∈ Z ∈ S ( Z ) and (17) holds.

□

Corollary 3.1 Let μ > 0 , β > 0 , m and q are even number not a multiple of four, then the unique solution of ( P m , q ) is

u ( t ) = ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t ϕ k = [ ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ] ∨ ,

where ϕ k ( x ) = e i k x , x ∈ I R .

Let’s remember that P ′ is the topological dual of P, where P is a complete metric space.

In this subsection, we will introduce families of operators { T μ , β ( t ) } t ≥ 0 in P ′ , with μ > 0 and β > 0 ; and we will prove that these operators are continuous in the weak sense. That is, T μ , β ( t ) is continuous from P ′ to P ′ with the weak topology of P ′ , which we will call the weakly continuous operator.

Furthermore, we will prove that T μ , β ( t ) satisfies the semigroup properties.

For simplicity, we will denote this family of operators by { T ( t ) } t ≥ 0 .

Theorem 3.2 Let t ≥ 0 , μ > 0 and β > 0 , we define:

T ( t ) : P ′ → P ′       f → T ( t ) f : = [ ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ] ∨ ∈ P ′ ,

then the following statements are satisfied:

1) T ( 0 ) = I .

2) T ( t ) is I C - linear and weakly continuous ∀ t ≥ 0 . That is, for every t ≥ 0 , if f n → P ′ f then T ( t ) f n → P ′ T ( t ) f .

3) T ( t + r ) = T ( t ) ∘ T ( r ) , ∀ t , r ≥ 0 .

4) T ( t ) f → P ′ f when t → 0 + , ∀ f ∈ P ′ .

That is, for each f ∈ P ′ fixed, the following is satisfied

〈 T ( t ) f , ψ 〉 → 〈 f , ψ 〉 ,         when       t → 0 + ,     ∀   ψ ∈ P .

Proof. - Let f ∈ P ′ then f ^ ∈ S ′ ( Z ) . Then, from (3) we have

( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ∈ S ′ ( Z ) ;

taking the inverse Fourier transform, we obtain

[ ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ] ∨ ︸ = T ( t ) f ∈ P ′ ,   ∀ t ≥ 0.

That is, T ( t ) is well defined for all t ≥ 0 .

1) We easily obtain:

T ( 0 ) f = [ ( f ^ ( k ) e − i μ k m 0 e − β k q 0 ) k ∈ Z ] ∨ = [ ( f ^ ( k ) ) k ∈ Z ] ∨ = [ f ^ ] ∨ = f   ,   ∀ f ∈ P ′   .

2) Let t ∈ I R + , we will prove that T ( t ) : P ′ → P ′ is I C -linear. In effect, let a ∈ I C and ( ϕ , ψ ) ∈ P ′ × P ′ , we have

T ( t ) ( a ϕ + ψ ) = [ ( e − i μ k m t e − β k q t [ a ϕ + ψ ] ∧ ( k ) ) k ∈ Z ] ∨ = [ ( e − i μ k m t e − β k q t [ a ϕ ^ ( k ) + ψ ^ ( k ) ] ) k ∈ Z ] ∨ = [ a ( e − i μ k m t e − β k q t ϕ ^ ( k ) ) k ∈ Z + ( e − i μ k m t e − β k q t ψ ^ ( k ) ) k ∈ Z ] ∨ = a [ ( e − i μ k m t e − β k q t ϕ ^ ( k ) ) k ∈ Z ] ∨ + [ ( e − i μ k m t e − β k q t ψ ^ ( k ) ) k ∈ Z ] ∨ = a   T ( t ) ϕ + T ( t ) ψ .

Now, for t ∈ I R + we will prove that T ( t ) : P ′ → P ′ is weakly continuous. That is, if f n → P ′ f we will prove that T ( t ) f n → P ′ T ( t ) f .

We know that if f n → P ′ f then f ^ n → S ′ f ^ , that is,

〈 f ^ n , ξ 〉 → 〈 f ^ , ξ 〉 ,     when     n → + ∞   ,   ∀ ξ ∈ S ( Z ) .

That is,

〈 f ^ n − f ^ , ξ 〉 → 0   ,       when     n → + ∞   ,   ∀ ξ ∈ S ( Z ) . (20)

We want to prove that:

〈 T ( t ) f n , ψ 〉 → 〈 T ( t ) f , ψ 〉         when     n → + ∞   ,   ∀ ψ ∈ P   .

Thus, let t ∈ I R + fixed and ψ ∈ P , using the generalized Parseval identity, we obtain the following equalities

〈 T ( t ) f n , ψ 〉 = 〈 [ ( f ^ n ( k ) e − i μ k m t e − β k q t ) k ∈ Z ] ∨ , ψ 〉 = 2 π 〈 ( f ^ n ( k ) e − i μ k m t e − β k q t ) k ∈ Z , ψ ^ ˜ 〉 , (21)

〈 T ( t ) f , ψ 〉 = 〈 [ ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z ] ∨ , ψ 〉 = 2 π 〈 ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z , ψ ^ ˜ 〉 . (22)

From (21) and (22) we get

〈 T ( t ) f n , ψ 〉 − 〈 T ( t ) f , ψ 〉 = 2 π { 〈 ( f ^ n ( k ) e − i μ k m t e − β k q t ) k ∈ Z , ψ ^ ˜ 〉 − 〈 ( f ^ ( k ) e − i μ k m t e − β k q t ) k ∈ Z , ψ ^ ˜ 〉 } = 2 π { ∑ k = − ∞ + ∞   f ^ n ( k ) e − i μ k m t e − β k q t ψ ^ ˜ ( k ) − ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t ψ ^ ˜ ( k ) } = 2 π ∑ k = − ∞ + ∞ { f ^ n ( k ) − f ^ ( k ) } e − i μ k m t e − β k q t ψ ^ ˜ ( k ) ︸ ξ k : = → 0

when n → + ∞ , since ξ : = ( ξ k ) k ∈ Z ∈ S ( Z ) and (20) holds, that is 〈 f ^ n − f ^ , ξ 〉 → 0 when n → + ∞ .

3) Let t , r ∈ I R + , we will prove that T ( t ) ∘ T ( r ) = T ( t + r ) . In effect, let ϕ ∈ P ′ ,

T ( t + r ) ϕ = [ ( ϕ ^ ( k ) e − i μ k m ( t + r ) e − β k q ( t + r ) ) k ∈ Z ] ∨ = [ ( ϕ ^ ( k ) e − i μ k m r e − β k q r ︸ ⋅ e − i μ k m t e − β k q t ) k ∈ Z ] ∨ . (23)

Since ϕ ∈ P ′ , using (3) we have that

( ϕ ^ ( k ) e − i μ k m r e − β k q r ) k ∈ Z ∈ S ′ ( Z ) ,   ∀ r ∈ [ 0, + ∞ ) . (24)

Then, taking the inverse Fourier transform, we get:

[ ( ϕ ^ ( k ) e − i μ k m r e − β k q r ) k ∈ Z ] ∨ ∈ P ′   ,   ∀ r ∈ [ 0, + ∞ ) .

Thus, we define:

g r : = [ ( ϕ ^ ( k ) e − i μ k m r e − β k q r ) k ∈ Z ] ∨ ∈ P ′ .

That is,

g r : = T ( r ) ϕ   . (25)

Taking the Fourier transform to g r we get:

g ^ r = ( ϕ ^ ( k ) e − i μ k m r e − β k q r ) k ∈ Z ,

that is,

g ^ r ( k ) = ϕ ^ ( k ) e − i μ k m r e − β k q r ,   ∀ k ∈ Z   . (26)

Using (26) in (23) and from (25) we have:

T ( t + r ) ϕ = [ ( g ^ r ( k ) e − i μ k m t e − β k q t ) k ∈ Z ] ∨ ∈ P ′ = T ( t ) g r = T ( t ) ( T ( r ) ϕ ) = [ T ( t ) ∘ T ( r ) ] ( ϕ ) ,   ∀ t , r ∈ I R + .

So we have proven,

T ( t + r ) = T ( t ) ∘ T ( r ) ,         ∀ t , r ∈ I R + . (27)

If t = 0 or r = 0 then equality (27) is also true, with this we conclude the proof of

T ( t + r ) = T ( t ) ∘ T ( r ) ,   ∀ t , r ∈ [ 0, + ∞ ) . (28)

4) Let f ∈ P ′ , we will prove that:

T ( t ) f → P ′ f       when     t → 0 + .

That is, we will prove that

〈 T ( t ) f , φ 〉 → 〈 f , φ 〉       when     t → 0 + ,   ∀ φ ∈ P .

In effect, for t > 0 and φ ∈ P , we have

H t : = 〈 T ( t ) f , φ 〉 − 〈 f , φ 〉 = l i m n → + ∞ { 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ϕ k , φ 〉 − 〈 ∑ k = − n n   f ^ ( k ) ϕ k , φ 〉 } = l i m n → + ∞ 〈 ∑ k = − n n   f ^ ( k ) ( e − i μ k m t e − β k q t − 1 ) ϕ k , φ 〉 = l i m n → + ∞ ∑ k = − n n   f ^ ( k ) ( e − i μ k m t e − β k q t − 1 ) 〈 ϕ k , φ 〉 = l i m n → + ∞ 2 π ∑ k = − n n   f ^ ( k ) ( e − i μ k m t e − β k q t − 1 ) φ ^ ( − k ) = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) ( e − i μ k m t e − β k q t − 1 ) φ ^ ( − k ) . (29)

Since t > 0 , from (7) we get

| e − i μ k m t e − β k q t − 1 | ≤ { μ | k | m + β | k | q } t   . (30)

From (30) we obtain

| e − i μ k m t e − β k q t − 1 | ≤ { μ | k | m + β | k | q } t   ,   ∀ t ∈ [ 0, + ∞ ) . (31)

From (31) with 0 < t < 1 , we have

| e − i μ k m t e − β k q t − 1 | ≤ μ | k | m + β | k | q . (32)

Then using (32) and that f ∈ P ′ , we obtain

∑ k = − ∞ + ∞ | f ^ ( k ) | | e − i μ k m t e − β k q t − 1 | | φ ^ ( − k ) | ≤ C { μ ∑ k = − ∞ + ∞ | k | N + m | φ ^ ( − k ︸ = J ) | + β ∑ k = − ∞ + ∞ | k | N + q | φ ^ ( − k ︸ = J ) | } = C { μ ∑ J = − ∞ + ∞ | J | N + m | φ ^ ( J ) | + β ∑ J = − ∞ + ∞ | J | N + q | φ ^ ( J ) | } < ∞

since φ ^ ∈ S ( Z ) .

Using the Weierstrass M-Test we conclude that the H t series converges absolute and uniformly. So,

lim t → 0 + H t = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) φ ^ ( − k ) lim t → 0 + { e − i μ k m t e − β k q t − 1 } ︸ = 0 = 0.

Thus, we have proved

lim t → 0 + 〈 T ( t ) f , φ 〉 = 〈 f , φ 〉 .

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Theorem 3.3 For each f ∈ P ′ fixed and the family of operators { T ( t ) } t ≥ 0 from Theorem 3.2, then the application

ζ : [ 0, + ∞ ) → P ′     t → T ( t ) f

is continuous in [ 0, + ∞ ) . That is,

T ( t + h ) f → P ′ T ( t ) f       when     h → 0   ,     ∀ t ∈ [ 0, + ∞ ) . (33)

(is the continuity at t).

That is, (33) tell us that for each t ∈ ( 0, + ∞ ) fixed, the following is satisfied

〈 T ( t + h ) f , ψ 〉 → 〈 T ( t ) f , ψ 〉 ,         when       h → 0   ,     ∀   ψ ∈ P   ,

and if t = 0 , we have the continuity of ζ at 0 on the right, which is item 4) of Theorem 3.2.

Proof.- Let t > 0 , arbitrary fixed and f ∈ P ′ , then g : = T ( t ) f ∈ P ′ , using item 4) of Theorem 3.2 we have that T ( h ) g → P ′ g when h → 0 + . That is,

T ( h ) ( T ( t ) f ) ︸ = [ T ( h ) ∘ T ( t ) ] f ︸ = T ( h + t ) f → P ′ T ( t ) f       when     h → 0 +   ,

where we use item 3) of Theorem 3.2. With this we have proved that

T ( t + h ) f → P ′ T ( t ) f       when     h → 0 +   ,     ∀ t ∈ ( 0, + ∞ ) . (34)

Now, we will prove that T ( t + ν ) f → P ′ T ( t ) f       when     ν → 0 − . That is, we will demonstrate

〈 T ( t − h ) f , φ 〉 → 〈 T ( t ) f , φ 〉       when     h → 0 +   ,   ∀ φ ∈ P   . (35)

In effect, for t > h > 0 and φ ∈ P , we have

L t , h : = 〈 T ( t − h ) f , φ 〉 − 〈 T ( t ) f , φ 〉 = l i m n → + ∞ { 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m ( t − h ) e − β k q ( t − h ) ϕ k , φ 〉     − 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ϕ k , φ 〉 } = l i m n → + ∞ 〈 ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ( e i μ k m h e β k q h − 1 ) ϕ k , φ 〉

= l i m n → + ∞ ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ( e i μ k m h e β k q h − 1 ) 〈 ϕ k , φ 〉 = l i m n → + ∞ 2 π ∑ k = − n n   f ^ ( k ) e − i μ k m t e − β k q t ( e i μ k m h e β k q h − 1 ) φ ^ ( − k ) = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t ( e i μ k m h e β k q h − 1 ) φ ^ ( − k ) . (36)

In the series (36), we need to delimit the expression e − i μ k m h e − β k q h − 1 . So, we have

e i μ k m h e β k q h − 1 = ∫ 0 h [ e ( i μ k m + β k q ) s ] ′ d s = ( i μ k m + β k q ) ∫ 0 h   e ( i μ k m + β k q ) s d s   . (37)

Taking norm to equality (37) and using: ∫ 0 h   e ( β k q ) s d s ≤ e ( β k q ) h h for h > 0 , we obtain

| e i μ k m h e β k q h − 1 | ≤ | i μ k m + β k q | ∫ 0 h   e ( β k q ) s d s ≤ { μ | k | m + β | k | q } e ( β k q ) h ⋅ h ≤ { μ | k | m + β | k | q } e ( β k q ) h (38)

whenever 0 < h < 1 .

Using inequality (38) and e − β k q ( t − h ) ≤ 1 for 0 < h < t with h ≪ 1 , we have

∑ k = − ∞ + ∞ | f ^ ( k ) | | e − i μ k m t | e − β k q t | e i μ k m h e β k q h − 1 | | φ ^ ( − k ) | ≤ ∑ k = − ∞ + ∞ | f ^ ( k ) | e − β k q ( t − h ) { μ | k | m + β | k | q } | φ ^ ( − k ) | ≤ ∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | { μ | k | m + β | k | q }

≤ μ ∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | | k | m + β ∑ k = − ∞ + ∞ | f ^ ( k ) | | φ ^ ( − k ) | | k | q ≤ C { μ ∑ k = − ∞ + ∞ | k | N + m | φ ^ ( − k ) | + β ∑ k = − ∞ + ∞ | k | N + q | φ ^ ( − k ) | } ≤ C { μ ∑ J = − ∞ + ∞ | J | N + m | φ ^ ( J ) | + β ∑ J = − ∞ + ∞ | J | N + q | φ ^ ( J ) | } < ∞ (39)

since φ ^ ∈ S ( Z ) .

Using the Weierstrass M-Test we obtain that the series L h , t is absolute and uniformly convergent.

Then, we can take limit and get:

l i m h → 0 + L h , t = 2 π ∑ k = − ∞ + ∞   f ^ ( k ) e − i μ k m t e − β k q t l i m h → 0 + { e i μ k m h e β k q h − 1 } ︸ = 0 φ ^ ( − k ) = 0,

with this (35) is proved.

From (34) and (35) we conclude that

T ( t + h ) f → P ′ T ( t ) f       when     h → 0   ,     ∀ t ∈ ( 0, + ∞ ) . (40)

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Remark 3.1 The results obtain in Theorems 3.2 and 3.3 are also valid for the family of operators { S ( t ) } t ≥ 0 , defined as

S ( t ) : P ′ → P ′       f → S ( t ) f : = [ ( e i μ k m t e − β k q t f ^ ( k ) ) k ∈ Z ] ∨ ,

for t ∈ [ 0, + ∞ ) . Its proof is similar.

We improve the statement of theorem 3.1, using a family of weakly continuous Operators { T ( t ) } t ≥ 0 .

Theorem 3.4 Let f ∈ P ′ and the family of operators { T ( t ) } t ≥ 0 from Theorem 3.2, defining u ( t ) : = T ( t ) f ∈ P ′ , ∀ t ∈ [ 0, + ∞ ) , then u ∈ C ( [ 0, + ∞ ) , P ′ ) is the unique solution of ( P m , q ). Furthermore, u continuously depends on f. That is, given f n , f ∈ P ′ with f n → P ′ f implies u n ( t ) → P ′ u ( t ) , ∀ t ∈ [ 0, + ∞ ) , where u n ( t ) : = T ( t ) f n , ∀ t ∈ [ 0, + ∞ ) (that is, u n is a solution of ( P m , q ) with initial data f n ).

Proof.- It is analogous to the proof of Theorem 3.1.

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Corollary 3.2 Let f ∈ P ′ be fixed and the family of operators { T ( t ) } t ≥ 0 from Theorem 3.4, then ∃ ∂ t T ( t ) f , ∀ t ∈ ( 0, + ∞ ) and the mapping

η : ( 0, + ∞ ) → P ′     t → ∂ t T ( t ) f = i μ ∂ x m T ( t ) f + β   ∂ x q T ( t ) f

is continuous at ( 0, + ∞ ) . That is,

∂ t T ( t + h ) f → P ′ ∂ t T ( t ) f         when     h → 0   ,   ∀ t ∈ ( 0, + ∞ ) . (41)

(41) tells us that for each t ∈ ( 0, + ∞ ) fixed, it holds:

〈 ∂ t T ( t + h ) f , φ 〉 → 〈 ∂ t T ( t ) f , φ 〉         when     h → 0   ,   ∀ φ ∈ P   .

Proof.- Indeed,

〈 ∂ t T ( t + h ) f , φ 〉 − 〈 ∂ t T ( t ) f , φ 〉 = i μ { 〈 ∂ x m T ( t + h ) f , φ 〉 − 〈 ∂ x m T ( t ) f , φ 〉 } + β { 〈 ∂ x q T ( t + h ) f , φ 〉 − 〈 ∂ x q T ( t ) f , φ 〉 } = i μ { 〈 T ( t + h ) f , φ ( m ) 〉 − 〈 T ( t ) f , φ ( m ) 〉 } ︸ → 0 + β { 〈 T ( t + h ) f , φ ( q ) 〉 − 〈 T ( t ) f , φ ( q ) 〉 } ︸ → 0 → 0

when h → 0 , due to Theorem 3.3 with ψ : = φ ( r ) ∈ P for r ∈ { m , q } .

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Corollary 3.3 Let f ∈ P ′ be fixed and the family of operators { T ( t ) } t ≥ 0 from Theorem 3.4, then the solution of ( P m , q ): u ( t ) : = T ( t ) f , ∀ t ∈ [ 0, + ∞ ) , satisfies u ∈ C 1 ( ( 0, + ∞ ) , P ′ ) .

Proof.- It comes out as a consequence of Corollary 3.2.

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