Research Article

Abstract: In this paper, we establish some new retarded integral inequalities in two variables. These on the one hand generalize and on the other hand furnish a handy tool for the study of qualitative as well as quantitative properties of solutions of differential equation. Some applications are also given.

Keywords: Explicit bound, two independent variable, integral inequality, partial derivatives, boundedness.

 

Introduction
            In the development of the theory of differential and integral equations integral inequalities which provide explicit bounds on unknown functions take very important place. For instant the explicit bounds given by the well-known Gronwall-Bellman [1,3] inequality and its nonlinear generalization due to Bihari [2] are used to considerable extent in the literature [1-12]. The main purpose of this paper is to establish explicit bounds on retarded Gronwall-Bellman and Bihari-like inequalities in two variables which can be used to study the qualitative behavior of the solutions of certain classes of retarded partial differential equations. Some applications of one of our result are also given.

 

Main Results

 In what follows, R denotes the set of real numbers; R₊ = [0, ) ,  = (0 _, ), R₁ = [1_, ), J₁ = [x₀₁ X), and J₂ = (y₀, Y) are the given subsets of R,  D= J₁ × J₂ .The first order partial derivatives of a function z(x,y)  with respect to x and y are denoted by D₁ z (x,y) and D₂ z(x, y), respectively.

Theorem 2.1:  Let u, g, h Î C (D, R₊) and a Î C¹ (J₁, J₁), b Î C¹ (J₂, J₂) be non decreasing with C¹ (J₂, J₂) be nondecreasing with a (x) < x on J₁,  b (y) <  y on J₂. Let f (x,y) be nondecreasing in (x,y ) Î D If the inequality.

u (x,y) <  f (x,y) +  (s,t) u(s,t) dtds +  (s,t) u(s,t) dtds                                                                                       (2.1)

holds, then

u(x,y) < f (x,y) exp [G(x,y) + H(x,y)]                                                                                                          (2.2)

for (x,y) Î D , where

G (x,y) =  (s_,t) dtds                                                                                                                                           (2.3)

H(x,y) =  (s_,t) dtds                                                                                                                                     (2.4)

Proof : Since f(x,y) is positive and nondecreasing, we can restate (2.1) as

 <  1 +  (s,t) dtds +  (s,t)                                                                                   (2.5)

Let r(x,y) =    then

r(x,y) < 1 + (s,t) r(s,t) dtds +  (s,t) r (s,t) dtds                                                                         (2.6)

Define a function z(x,y) by the right - hand side of (2.6), then we have

z(x,y) = 1 +  (s,t) r (s,t) dtds +  (s,t) r (s,t) dtds                                                                                  (2.7)

Then it is clear that

r(x,y) < z (x,y), z(x₀,y) = z(x,y₀) = 1                                                                                                            (2.8)

Differentiate (2.7) with respect to x, we get

D₁ z (x,y) =  (x,t) r(x,t) dt + ( a (x), t) r ( a (x)₁ t) dt  a^¢ (x)

Using (2.8), we have

≤  (x,t) z(x,t) dt +  ( a(x),t)  z (a (x), t) dt
 ≤ z(x,y)                                                                                         (2.9)

 <    (x,t) dt +  ( ,t) dt                                                                                                  (2.10)

Keeping y fixed in (2.10) , setting x = s and integrating it with respect to s from x₀  to x, x Î J₁ and making change of variable, we get

z(x,y) <  exp [G (x,y) + H (x,y)]

for (x,y) Î D

Using (2.8), we have

r(x,y) <  exp [G (x,y) +H (x,y)]

Hence

u (x, y) <  f(x,y) exp [G (x,y) + H (x,y)]

Theorem 2.2 : Let u,g,h Î  C( D, R₊) and a Î C¹ (J₁, J₁), B Î  C¹ (J₂,J₂) be nondecreasing with a(x) <  x on J₁, b(y)< y on J_2. Let f(x,y) be nondecreasing in (x,y) Î  D and  p > 1 is a constant. If the inequality

u^P (x,y) < f^P (x,y) +  (s,t) u(s,t) dtds +  (s,t)u (s,t) dtds                                                                              (2.11)

holds, then

u(x, y)  < f (x, y)                                                                                        (2.12)

where

Q (x, y) =   (s, t) g(s, t) dtds                                                                                                           (2.13)

and

W (x,y) =  (s,t) h (s,t) dtds                                                                                                                               (2.14)

Proof : Since f(x, y) is positive and non decreasing in both variables equation (2.11) rewrite as

 < 1 +  (s,t) f^1-P (s,t) dtds +  (s,t) f^1-P (s, t) dtds                                                  (2.15)

Let r(x,y) =  then

r^P (x, y) < 1 +  (s,t) f^1-P (s,t) r(s,t) dtds +  (s,t) f^1-P  (s,t) r(s,t) dtds                                                       (2.16)

Define a function z (x,y) by the right hand side of (2.16) then we have

z(x,y) = 1 +  (s,t) f ^1-P(s,t) r (s,t) dtds +  (s,t) f^1-P (s,t) r (s,t) dtds                                                   (2.17)

then it is clear that

r^p (x,y) < z (x,y) ,  z (x₀,y) = z(x,y₀) = 1                                                                                                                 (2.18)

Differentiation (2.17) with respect to x, we have

D₁ z (x,y) = (x,t) f^1-P (x,t) r (x,t) dt + (x),t) f^1-P (x,t) r (x,t) dt a¢(x)

Using (2.18) we get

D₁ z (x,y) <   (x,t) f^1-P (x,t)  (x,t) dt +  (x,t) f^1-P (x,t) (x,t) dt . a¢(x)

<  (x,t) f^1-P (x,t) dt +  (x),t) f^1-P (a(x), dt . a¢(x)                                                                       (2.19)

Keeping y fixed in (2.19), setting x = s and integrating it with respect to s from x₀ to x, x e J₁ and making change of variable, we get

   (x,y) <  +  (s,t) f^1-P (s,t) dtds + (s,t)f^1-P (s,t) dtds

 (x,y) < 1 + [  (s,t) f^1-P (s,t) dtds +  (s,t) f^1-P (s,t) dtds]                                                         (2.20)

Using (2.18) in (2.20) we get

r^P-1 (x,y) < 1 +  [  (s,t) f^1-P (s,t) dtds +  (s,t) f^1-P (s,t) dtds]

r^P-1 (x,y) < 1 +  [Q(x,y) + W (x,y)]

where Q(x,y) and W(x,y) are mentioned in (2.13) & (2.14) .

 Hence  u(x,y) <  f(x,y)                                                                                                (2.21)

Theorem 2×3: Let u_, g_, h Î C (D,R₊),  f Î C(D, ) and a  Î C¹ (J₁,J₁), b Î C¹ (J₂,J₂) be non-decreasing with a (x) <  x on J_1, b (y) < y on J₂. For i = 1,2, let g_i Î C (R₊, R₊) be non-decreasing function with g_i (u) > 0 for u > 0 and

 < g_i   .

If the inequality
u (x,y)  ≤  f(x,y) +  (s_,t) g₁(u (s,t)) dtds +  (s,t) g₂ (u(s,t)) dtds                                              (2.22)

for (x,y) Î D,

(i)         In case g₁ (u)   < g₂ (u)

u (x,y) ≤ f(x,y)  [G₂(1) + G (x,y) + H(x,y)]                                                                                                                   (2.23)

(ii)        In case g₂ (u) ≤ g₁(u)

u(x,y) < f (x,y)  [G₁(1) + G (x,y) + H (x,y)]                                                                                                                  (2.24)

Where G(x,y) and H(x,y)  are defined by (2.3) and (2.4) and for i = 1,2  are the inverse functions of

G_i (r) = , r > 0, r₀ > 0                                                                                                                                   (2.25)

and x Î J₁, y Î J₂ are so chosen that for i = 1,2

G_i (1) + G(x,y) + H (x,y) Î Dom ( )

respectively, for all x,y lying in [x₀,x₁] and [y₀,y₁].

Proof : Since f (x,y) is positive and non-decreasing in both variables (2.22) Can be rewrite as

 < 1+ (s,t)g₁ dtds + (s,t)g₂  dtds                                                                    (2.26)

 

 <1+  (s,t)g₁ dtds +  (s,t) g₂  dtds                                                  (2.27)

Let r (x,y) =  

 Hence we obtain

r(s,t) <1 +  (s₁t) g₁(r(s,t) dtds +  (s,t) g₂ (r (s,t) dtds                                                                           (2.28)

Define z (x, y) by the right hand side of (2.28), we get

z(x,y) = 1 +  (s,t)g₁ (r(s,t) dtds +  (s,t)g₂ (r(s,t) dtds                                                                           (2.29)

 From (2.28), we have

r (s₁t) < z (x,t), z(x,y₀)= z (x₀,y) = 1                                                                                                           (2.30)

Now
D₁ z(x,y)=  (x,t) g₁(u(x,t,))dt + (a (x),t)g₂ (u(a(x),t) dt) a¢(x)                                                                               (2.31)

from (2.31) in (2.30), we get

D₁z (x,y) < (x,y) g₁ (z (x,t))dt + a¢(x)                                                                    (2.32)

D₁ (z (x, y)) < g₁ (z (x,y)) (x,t) dt + g₂ (z(x,y)) ((a (x),t) g₂ (z (a(x), t) dt a¢(x)                                                   (2.33)

 

When g₂ (u) < g₁ (u), then from (2.33), we observed that

<  (x,t)dt +  a¢(x)                                                                             (2.34)

Using (2.25) in (2.34) ,  we get

D₁ G₁ (z(x,y)) <  (x,t)dt + a¢ (x)                                                                                              (2.35)

Keeping y fixed in (2.35) Setting x = s, then integrating with respect to s from x₀ to x, xÎJ₁

and making change of variable, we get

G₁ (z(x,y)) < G₁(1) +  (s,t)dtds +  (s,t) dtds                                                                                      (2.36)

Using the bound on z(x,y) from (2.36) in (2.30), we get required inequality in (2.23).

Applications:

In this section we present applications of the inequality in Theorem 1 to study the boundedness and uniqueness of the solutions of the initial boundary value problem for hyperbolic partial differential equations of the from

D₂ D₁ u (x,y) = F( x,y, u (x,y), u (x-h₁(x), y-h₂(y)),                                                                                                 (3.1)

u (x,y₀) = e₁(x), u (x₀,y) = e₂ (y) e₁ (x₀) = e₂ (y₀) = 0                                                                                    (3.2)

Where FÎ C(D × R², R), e₁ Î C¹ (J,R) e₂ Î C¹ (J₂,R), h₁ Î C¹ (J₁, R+), h₂ Î C¹ (J₂, R₊) such that x-h₁ (x) > 0,

 y₁ - h₂ (y) > 0, h₁¢ (x) < 1, h₂¢ (x) < 1 and h₁(x₀) = h₂ (y₀) = 0.

Our first result gives the bound on the solution of the problem (3.1) - (3.2)

Theorem 3.1: suppose that

|F(x,y,u,v) | <  g(x,y) |u| + k (x,y) |v|                                                                                                                        (3.3)

and |e₁ (x) + e₂(y) | <  p (x,y)                                                                                                                                               (3.4)

Where g, k Î C (D, R+) and k > 0 and let

M₁ =   , M₂ =                                                                                                    (3.5)

If u(x,y) is any solution of (3.1) - (3.2) then

|u(x,y) | ≤ p (x,y) exp (G(x,y)+  (x,y))                                                                                                                  (3.6)

where G(x,y) is defined by (2.3) and

 (x,y) = M₁ M₂                                                                                                    (3.7)

and

for .

Proof : It is easy to see that the solution u(x,y) of the problem (3.1) - (3.2) satisfies the equivalent integral equation

u(x,y) = e₁ (x) + e₂(x) + (s,t,u (x,y), u(x-h₁ (x), y-h₂ (y)))dtds                                                                          (3.8)

Using (3.3), (3.4),(3.5) in (3.8) and making change of variables , we get

|u(x,y)|  <p(x,y) +  (x,y) |u(x,t)| dtds + M₁ M₂                                                        (3.9)

Now suitable application of the inequality in theorem 2.1 to 3.9 yields  (3.6). The right-hand side of (3.6) gives us the bound on solution u(x,y) of (3.1) - (3.2) in terms of known function. Thus if the right-hand side of (3.6) is bounded, then we assent that the solution of (3.1)-(3.2) is bounded.

The next result deals with the uniqueness of the solution the problem (3.1) - (3.2)

 

Theorem 3.2: Suppose that the function F in (3.1) satisfies the condition

| F (x,y,u,v) – F(x,y, , ₎|  <  g (x,y) | u-  | + k(x,y) | v-  |                                                                                   (3.10)

g_,k Î C (D,R) ,  and  let M_1, M_2, α, b ,  be as in theorem 3.1 then for problem (3.1) – (3.2) has at most one solution.
Proof : Let u (x,y) and  (x,y) be two solution of (3.1) – (3.2) on D. Then we have

u(x,y) – (x,y) =                                              (3.11)
Using (3.10) in (3.11) and making change of variables we get

| u  (x,y) - (x,y) | <  (s,t) | z (s,t) -  (s,t))+ M₁ M₂  

Now suitable application of inequality in theorem (2.1) yields

| u  (x,y) -  (x,y) | <  0

Therefore u (x, y) =  (x,y)  i.e there is almost one solution of problem (3.1) - (3.2).

 

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