Other classes of self-referred equations

Note di Matematica 29, n. 1, 2009, 61–72.

Other classes of self-referred equations

Michele Miranda Jr.

Dipartimento di Matematica “Ennio De Giorgi”

Universit`a del Salento C.P. 193, 73100 Lecce, Italy michele.miranda@unile.it Eduardo Pascali

Dipartimento di Matematica “Ennio De Giorgi”

Universit`a del Salento C.P. 193, 73100 Lecce, Italy eduardo.pascali@unile.it

Received: 10/09/2007; accepted: 18/04/2008.

Abstract. In a previous note [4] we have established some existence and uniqueness theorem for a new type of differential equation that we have defined hereditary and self-referred type.

In this note we continue the investigation by proving local existence and uniqueness theorems for other classes of self-referred differential equations.

The classes of equations that we are going to consider are:

∂

∂tu(x, t) = u

t

0 u(x, s)ds + ψ(u(x, t)), t

 ,

∂²

∂t²u(x, t) = u

_x+δ(x,t)

x−δ(x,t)

∂

∂tu(ξ, t)dξ, t)

 .

Keywords:Non-linear evolution equations; Self-referred differential equations; Iteration MSC 2000 classification:47J35; 45G10

1 Introduction

In a previous note [4] we have started the study of a new class of functional differential equations that can be adapted phenomena which evolution presents self-reference with respect to his history.

Phenomena of hereditary type have been treated since the beginning of XX- th century ( [10], [3] and [2]), but the attention for phenomena of self-reference type with respect to the state itself of the phenomenon is recent [1], [9] and [8].

In this note, by continuing the inspection started in [4], and by essentially using the same line of proof, we introduce and study some new type of func- tional differential equations establishing for them theorems of local existence and uniqueness. Also for these equations there are several open problems, as already signalized in [4].

Note Mat. 29 (2009), n. 1, 61-72 ISSN 1123-2536, e-ISSN 1590-0932 DOI 10.1285/i15900932v29n1p61

http://siba-ese.unisalento.it, © 2009 Università del Salento

__________________________________________________________________

2 A class of hereditary and self-referred equations

Let X be the space of continuous functions u : R

²

→ R and ψ : R → R.

1 Theorem. If ψ satisfies the following condition:

∃L

ψ

≥ 0 such that:

|ψ(ξ

1

) − ψ(ξ

2

) | ≤ L

ψ

|ξ

1

− ξ

2

| ∀ξ

1

, ξ

₂

∈ R (1) and if u

₀

satisfies the following conditions: ∃L

0

≥ 0 such that

|u

0

(x) − u

0

(y)| ≤ L

0

|x − y|, ∀x, y ∈ R (2) and

u

0



∞

< ∞, (3)

then there exist T

₀

> 0 and a unique u continuous on R × [0, T

0

] such that:

(1) u is bounded in R × [0, T

0

]

(2) u is Lipschitz with respect to x uniformly with respect to t ∈ [0, T

0

] (and Lipschitz in t uniformly with respect to x) and moreover, for x ∈ R and t ∈ [0, T

0

]:

u(x, t) = u

₀

(x) +



t 0

u



_τ

0

u(x, s)ds + ψ(u(x, τ )), τ



dτ ; (4)

whence

⎧ ⎪

⎪ ⎨

⎪ ⎪

⎩

∂

∂t u(x, t) = u



_t

0

u(x, s)ds + ψ(x, t), t



, x ∈ R, t ∈ [0, T

0

]

u(x, 0) = u

₀

(x).

(5)

Moreover the solution u can be prolonged for all t ≥ T

0

.

Proof. Let us consider the sequence (u

n

), defined by recurrence:

u

1

(x, t) = u

0

(x) +



t 0

u

0



_τ

0

u

0

(x)ds + ψ(u

0

(x))

 dτ

= u

₀

(x) +



_t

0

u

₀

(u

₀

(x)τ + ψ(u

₀

(x))) dτ,

(6)

and for n ≥ 1

u

_n+1

(x, t) = u

₀

(x) +



_t

0

u

n



_τ

0

u

n

(x, s)ds + ψ(u

n

(x, τ )), τ



dτ. (7)

Directly from the definition of u

₁

we can deduce that:

|u

1

(x, t)| ≤ u

0



∞

+ tu

0



∞

= u

0



∞

(1 + t), ∀x ∈ R, t ≥ 0. (8)

and then, in general, for every n ∈ N:

|u

n

(x, t)| ≤ u

0



∞



1 + t + · · · + t

ⁿ

n!



≤ u

0



∞

e

^t

, ∀x ∈ R, t ≥ 0. (9)

Moreover

|u

1

(x, t) − u

0

(x)| ≤ tu

0



∞

:= A

₁

(t), and also:

|u

1

(x, t) − u

1

(y, t)| ≤L

0

|x − y| + 

t 0

L

₀

%% %



τ

0

u

₀

(x)ds+

+ ψ(u

₀

(x)) −



_τ

0

u

₀

(y)ds − ψ(u

0

(y)) %% %dτ ≤

≤L

0

|x − y| + 

t 0

&

L

₀

'

_τ

0

L

₀

|x − y|ds + L

ψ

L

₀

|x − y| ( ) dτ =

= &

L

₀

+



_t

0

L

₀

'

_τ

0

L

₀

ds + L

_ψ

L

₀

(

dτ )

|x − y| =

=(L

₀

+ L

²₀

t

²

2 + L

ψ

L

²₀

t)|x − y|.

(10) Now, if we define

L

₁

(t) := L

₀

+ 1 2



_t

0

L

₀

dτ



2

+ L

_ψ



_t

0

L

²₀

dτ, (11)

or, more generally:

L

_n+1

(t) := L

₀

+ 1 2



_t

0

L

n

(τ )dτ



2

+ L

ψ



t

0

L

n

(τ )

²

dτ, (12) for every n ≥ 1, we can prove, by induction:

|u

n

(x, t) − u

n

(y, t)| ≤ L

n

(t)|x − y| ∀x ∈ R ∀t ≥ 0. (13)

Moreover, since:

u

_n+1

(x, t) − u

n

(x, t) =



t 0

&

u

n



_τ

0

u

n

(x, s)ds + ψ(u

n

(x, τ )), τ

 +

− u

n−1



_τ

0

u

_n

(x, s)ds + ψ(u

_n

(x, τ )), τ

 +

+ u

_n−1



_τ

0

u

n

(x, s)ds + ψ(u

n

(x, τ )), τ

 +

− u

n−1



_τ

0

u

_n−1

(x, s)ds + ψ(u

_n−1

(x, τ )), τ  ) dτ we can deduce that:

|u

n+1

(x, t) − u

n

(x, t)| ≤

≤ 

t 0

&

A

n

(τ ) + L

_n−1

(τ )



_τ

0

A

n

(s)ds + L

ψ

A

n

(τ )  )

dτ, (14) if we define, by recurrence, for every n ∈ N:

A

_n+1

(t) :=



t 0

&

A

n

(τ ) + L

_n−1

(τ )



_τ

0

A

n

(s)ds + L

ψ

A

n

(τ )  )

dτ. (15) We remark that:

0 ≤ L

n+1

(t) − L

0

= 1 2



_t

0

L

n

(τ )dτ



2

+ L

ψ



t

0

L

n

(τ )

²

dτ.

We fix k

₀

> 0 and h < 1; we can then choose T

₀

> 0 in such a way that

t ∈ [0, T

0

] = ⇒

⎧ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎪

⎨

⎪ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎩ 1

2 L

²₀

t

²

+ L

_ψ

L

²₀

t ≤ k

0

1

2 k

₀²

t

²

+ L

_ψ

(k

₀

+ L

₀

)

²

t ≤ k

0

0 ≤ t + (k

0

+ L

₀

)

 t

²

2 + L

_ψ

t



≤ h < 1.

(16)

Hence:

0 ≤ L

1

(t) − L

0

= 1

2 L

²₀

t

²

+ L

ψ

L

²₀

t ≤ k

0

,

0 ≤ L

2

(t) − L

0

= 1 2 (



t

0

L

₁

(τ )dτ )

²

+ L

ψ



t

0

L

₁

(τ )

²

dτ

≤ 1

2 k

²₀

t

²

+ L

ψ

(k

₀

+ L

₀

)

²

t ≤ k

0

and then, by induction:

0 ≤ L

n

(t) − L

0

≤ k

0

∀n ∈ N, ∀t ∈ [0, T

0

].

Then

0 ≤ L

n

(t) ≤ L

0

+ k

0

:= k

1

∀n ∈ N. (17) From (15), (16) and (17) there follows:

0 ≤ A

n+1

(t) ≤ 

t

0

(A

n

 + k

1

(A

n

τ + L

ψ

A

n

)) dτ

= A

n



 t + k

1

 t

²

2 + L

ψ

t



≤ hA

n

.

As a consequence,

A

n+1



A

n

 ≤ h, ∀n ∈ N;

then there follows that the series with general term given by A

n

(t) is totally convergent and so there exists u

_∞

: R × [0, T

0

] → R such that:

n→+∞

lim u

_n

(x, t) = u

_∞

(x, t)

uniformly in R ×[0, T

0

]; moreover u

_∞

is bounded and Lipschitz continuous with respect to the variable x with Lipschitz constant L

_∞

(t) ≤ k

1

(and u

_∞

is, of course, Lipschitz in t uniformly with respect to x). Minding that:

%% %u

n



_t

0

u

n

(x, s)ds + ψ(u

n

(x, t)), t



− u

∞



_t

0

u

_∞

(x, s)ds + ψ(u

_∞

(x, t)), t  %

%%

≤u

n

− u

∞



L^∞(R×[0,T0])

+ + L

_∞

(t)k



t

0

|u

n

(x, s) − u

∞

(x, s) |ds + |ψ(u

n

(x, t)) − ψ(u

∞

(x, t)) |

≤u

n

− u

∞



L^∞(R×[0,T0])

+ k

₁



_t

0

u

n

− u

∞



L^∞(R×[0,T0])

ds+

+ L

ψ

u

n

− u

∞



L^∞(R×[0,T0])

, from (16) and (17), we can deduce that:

u

_∞

(x, t) = u

₀

(x) +



t 0

u

_∞



_τ

0

u

_∞

(x, s)ds + ψ(u

_∞

(x, t)), τ



dτ (18)

and then u

_∞

is a solution of the problem:

⎧ ⎪

⎪ ⎨

⎪ ⎪

⎩

∂

∂t u

_∞

(x, t) = u

_∞



_t

0

u

_∞

(x, s)ds + ψ(u

_∞

(x, t)), t



u

_∞

(x, 0) = u

₀

(x), x ∈ R, t ∈ [0, T

0

].

In order to prove the uniqueness, we assume that there exists a real continuous function on R × [0, T

0

], say v, such that, for every x ∈ R and t ∈ [0, T

0

]:

v(x, t) = u

₀

(x) +



_t

0

v



_τ

0

v(x, s)ds + ψ(v(x, s)), τ

 dτ.

Then for every compact K ⊂ R, if x ∈ K:

|v(x, t) − u

∞

(x, t) | ≤



_t

0

%% %v(



_τ

0

v(x, s)ds + ψ(v(x, τ )), τ )+

− u

∞

(



_τ

0

v(x, s)ds + ψ(v(x, τ )), τ )+

+ u

_∞

(



_τ

0

v(x, s)ds + ψ(v(x, τ )), τ )+

− u

∞

(



τ

0

u

∞

(x, s)ds + ψ(u

∞

(x, τ )), τ ) %% %dτ

≤ 

t 0

&

v − u

∞



L^∞(K×[0,T0])

+

+ k

₁

%% %



τ

0

(v − u

∞



L^∞(K×[0,T0])

+ + L

ψ

v − u

∞



L^∞(K×[0,T0])

) %% % )

dτ ≤

≤v − u

∞



L^∞(K×[0,T0])



t

0

(1 + k

₁

τ + L

_ψ

L

_∞

)dτ

≤hv − u

∞



L^∞(K×[0,T0])

< v − u

∞



L^∞(K×[0,T0])

. Therefore u

_∞

= v in K × [0, T

0

], and then the uniqueness follows.

We remark that: v

₀

(x) = u(x, T

₀

) verifies the following conditions:

|v

0

(x) − v

0

(y) | ≥ k

1

|x − y|; ||v

o

||

∞

≤ ||u

0

||∞e

^T0

;

hence the local solution u = u(x, t) can be prolonged.

^QED

3 A class of self-referred equation

In this section we establish a theorem, analogue to the one proved in Section 2, concerning the following class of equations:

∂

²

∂t

²

u(x, t) = u



_x+δ(x,t)

x−δ(x,t)

∂

∂t u(ξ, t)dξ, t

. (19)

The Theorem that we are going to prove is the following:

2 Theorem. Let α

∈ L

^∞

(R, R) ∩ Lip(R, R) and β ∈ L

^∞

(R, R) be given functions and let δ : R

²

→]0, +∞[ be a real and bounded function, Lipschitz continuous in the x variable uniformly with respect to t, that is there exists L

δ

> 0 such that

|δ(x, t) − δ(y, t)| ≤ L

δ

|x − y|, ∀x, y ∈ R, ∀t;

then there exists T

₀

> 0 and there exists a unique u : R × [0, T

0

] → R continu- ous, bounded and Lipschitz in x uniformly with respect to t (and Lipschitz in t uniformly with respect to x) such that:

u(x, t) = α(x) + tβ(x) +



t 0



τ 0

u



_x+δ(x,s)

x−δ(x,s)

∂

∂s u(ξ, s)dξ, s

dsdτ,

that is

⎧ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎨

⎪ ⎪

⎪ ⎪

⎪ ⎪

⎪ ⎪

⎩

∂

²

∂t

²

u(x, t) = u



_x+δ(x,t)

x−δ(x,t)

∂

∂t u(x, t)dξ, t

u(x, 0) = α(x)

∂

∂t u(x, 0) = β(x),

x ∈ R, t ∈ [0, T

0

]

Moreover the solution u can be prolonged for all t ≥ T

0

.

Proof. Let us consider the following sequence, defined by recurrence:

u

_n+1

(x, t) = α(x) + tβ(x)+

+



t 0



τ 0

u

n



_x+δ(x,s)

x−δ(x,s)

∂

∂s u

n

(ξ, s)dξ, s

dsdτ (20)

with:

u

₁

(x, t) = α(x) + tβ(x) +



t 0



τ 0

α



_x+δ(x,s)

x−δ(x,s)

β(ξ)dξ

dsdτ.

Since:

|u

1

(x, t) | ≤ α

∞

+ β

∞

t + α

∞

t

²

2 ,

we get that:

|u

2

(x, t)| ≤α

∞

+ β

∞

t +



t 0



τ 0



α

∞

+ β

∞

s + α

∞

s

²

2

 ds

≤α

∞

 1 + t

²

2! + t

⁴

4!

 + β

∞

 t + t

³

3!



≤(α

∞

+ β

∞

)



1 + t + t

²

2 + t

³

3! + t

⁴

4!

 .

(21)

and then, in general:

|u

n+1

(x, t) | ≤ (α

∞

+ β

∞

)e

^t

, ∀n ∈ N, ∀t > 0. (22) In the same way it is possible to prove that

%% %% ∂

∂t u

n

(x, t) %%

%% ≤ (α

^∞

+ β

∞

)e

^t

, ∀n ∈ N, ∀t > 0. (23) We remark that:

|u

1

(x, t) − α(x)| ≤ tβ

∞

+ α

∞

t

²

2 := A

₁

(t),

%% %% ∂

∂t u

₁

(x, t) − β(x) %%

%% ≤ α

^∞

t := B

₁

(t).

Moreover:

|u

2

(x, t) − u

1

(x, t) | ≤

≤ 

t 0



τ 0

&

A

₁

(s) + L

_α



_x+δ(x,s)

x−δ(x,s)

%% %% ∂

∂s u

₁

(ξ, s) − β(ξ) %%

%%dξ ) dsdτ

≤ 

t 0



τ 0



A

₁

(s) + L

α



_x+δ(x,s)

xδ(x,s)

B

₁

(s)dξ

dsdτ

=



_t

0



_τ

0

(A

₁

(s) + L

α

2δ(x, s)B

₁

(s))dsdτ

≤



_t

0



_τ

0

(A

₁

(s) + L

α

M B

₁

(s))dsdτ := A

₂

(t),

(24) where L

α

denotes the Lipschitz constant Lipschitz of the function α e M = 2δ

∞

. In general we define, for n ≥ 1:

A

n+1

(t) =



t 0



τ

0

(A

n

(s) + L

n−1

(s)M B

n

(s))dsdτ (25)

and

B

n+1

(t) =



t

0

(A

n

(s) + M L

n−1

(s)B

n

(s))ds, (26) and we note that:

%% %% ∂

∂t u

₂

(x, t) − ∂

∂t u

₁

(x, t) %%

%% =

= %% %



_t

0

u

₁



_x+δ(x,s)

x−δ(x,s)

∂

∂s u

₁

(ξ, s)dξ, s

ds −



_t

0

α



_x+δ(x,s)

x−δ(x,s)

β(ξ)dξ

ds %% %

≤ 

t

0

(A

1

(s) + L

α

M B

1

(s))ds. (27) Minding that, if we denote by L

β

the Lipschitz constant of the function β:

|u

1

(x, t) − u

1

(y, t) | ≤(L

α

+ tL

β

) |x − y|

+



_t

0



_τ

0

L

α

%% %



_x+δ(x,s)

x−δ(x,s)

β(ξ)dξ+

− 

_y+δ(y,s)

y−δ(y,s)

β(ξ)dξ %% %dsdτ

≤(L

α

+ tL

β

)|x − y|+

+



t 0



τ

0

L

α

2(1 + L

δ

)β

∞

|x − y|

=



L

_α

+ tL

_β

+ 2L

_α

(1 + L

_δ

) β

∞

t

²

2



|x − y|, we can define the Lipschitz constant of u

₁

by:

L

₁

(t) = L

_α

+ tL

_β

+ 2L

_α

(1 + L

_δ

) β

∞

t

²

2 . We fix k

₀

and we choose T

₀

> 0 in such a way that

tL

_β

+ 4(1 + L

_δ

)e

^T0

( α

∞

+ β

∞

)(L

α

+ k

₀

) t

²

2 ≤ k

0

. (28) Now in general, we can deduce that for n ≥ 1:

|u

n+1

(x, t) − u

n+1

(y, t)| ≤ L

n+1

(t)|x − y|, ∀x, y ∈ R, ∀t > 0, where:

L

n+1

(t) = L

α

+ tL

β

+

+ 2(1 + L

δ

)



t 0



τ 0

L

n

(s) 

 ∂

∂t u

n

 

L^∞(R×[0,T0])

dsdτ. (29)

Then:

L

_n+1

(t) ≤ L

α

+ tL

_β

+

+ 4(1 + L

_δ

)( α

∞

+ β

∞

)



_t

0



_τ

0

e

^T0

L

_n

(s)dsdτ ≤

≤ L

α

+ tL

_β

+ 4(1 + L

_δ

)e

^T0

( α

∞

+ β

∞

)



_t

0



_τ

0

L

_n

(s)dsdτ. (30) From (28), we obtain that:

0 ≤ L

1

(t) − L

α

≤ tL

β

+ 4(1 + L

_δ

)e

^T0

( α

∞

+ β

∞

)L

_α

t

²

2 ≤ k

0

, (31) for t ∈ [0, T

0

]. In the same way:

0 ≤ L

2

(t) − L

α

≤tL

β

+ 4(1 + L

_δ

)( α

∞

+ β

∞

)e

^T0



_t

0



_τ

0

L

₁

(s)

≤tL

β

+ 4e

^T0

(1 + L

δ

)(α

∞

+ β

∞

)(k

₀

+ L

α

) t

²

2

≤k

0

,

t ∈ [0, T

0

]. Hence, by induction, we can prove that:

0 ≤ L

n

(t) ≤ k

0

+ L

_α

:= k

₁

, ∀n ∈ N, ∀t ∈ [0, T

0

]. (32) Using the estimate (32), from (25) we can deduce that, ∀n ∈ N and ∀t ∈ [0, T

0

]:

A

_n+1

(t) ≤



_t

0



_τ

0

(A

_n

(s) + M (k

₀

+ L

_α

)B

_n

(s))ds; (33) and from (26), ∀n ∈ N, ∀t ∈ [0, T

0

]:

B

_n+1

(t) ≤ 

t

0

(A

n

(s) + M (k

₀

+ L

α

)B

n

(s))ds, (34) for every n ∈ N. We have obtained then that:

A

n+1

(t) ≤ (A

n

 + Mk

1

B

n

) t

²

2 , ∀n ∈ N, ∀t ∈ [0, T

0

]. (35) B

_n+1

(t) ≤ (A

n

 + Mk

1

B

n

)t, ∀n ∈ N, ∀t ∈ [0, T

0

]. (36) If at this point we assume that T

₀

has been choosen in such a way that for t ∈ [0, T

0

] we have that

0 ≤ (1 + Mk

1

)

 t

²

2 + t



≤ 2h < 1, (37)

we obtain that A

_n+1

(t) + B

_n+1

(t) ≤

 t

²

2 + t



( A

n

 + Mk

1

B

n

) ≤

≤

 t

²

2 + t



(M k

1

+ 1)(A

n

 + B

n

) ≤ h(A

n

 + B

n

). (38) We have then proved that the series

 (A

_n+1

(t) + B

_n+1

(t)) , t ∈ [0, T

0

]

is totally convergent and, then, the same is true for the series

 A

_n+1

(t),  B

_n+1

(t).

There follows that:

u

n

⇒ u

∞

∂

∂t u

_n

⇒ ∂

∂t u

_∞

,

in R × [0, T

0

].

On the other hand, u

∞

is Lipschitz continuous in x uniformly with respect to t ∈ [0, T

0

] with Lipschitz constant L

∞

(t) ≤ k

1

(and Lipschitz in t uniformly with respect to x); moreover

%% %%

% u

n



_x+δ(x,t)

x−δ(x,t)

∂

∂t u

n

(ξ, t)dξ, t

− u

∞



_x+δ(x,t)

x−δ(x,t)

∂

∂t u

_∞

(ξ, t)dξ, t %% %% % ≤

≤ u

n

− u

∞



L^∞(R×[0,T0])

+ k

₁



_x+δ(x,t)

x−δ(x,t)

%% %% ∂

∂t u

_n

(ξ, t) − ∂

∂t u

_∞

(ξ, t) %%

%%dξ

≤ u

n

− u

∞



L^∞(R×[0,T0])

+ k

₁



 ∂

∂t u

n

− ∂

∂t u

_∞





L^∞(R×[0,T0])

2M −→ 0.

We can then easily deduce that:

u

_∞

(x, t) = α(x) + tβ(x) +



t 0



τ 0

u

_∞



_x+δ(x,s)

x−δ(x,s)

∂

∂t u

_∞

(ξ, s)dξ, s

dsdτ for every x ∈ R, t ∈ [0, T

0

]. Uniqueness of the solution follows easily, arguing as in Theorem 1.

We remark that:

|u(x, T

0

) | ≤ (||α|| + ||β||)e

^T0

; | ∂

∂t u(x, T

₀

) | ≤ (||α|| + ||β||)e

^T0

≤ (||α|| + ||β||)e

^T0

and

|u(x, T

0

) − u(y, T

0

) | ≤ k

1

|x − y|,

hence the solution u = u(x, t) can be prolonged.

^QED

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