# Integrate the integral. Where $({x_1},{y_1}),({x_2},{y_2})\in[0,a] \times[0,b],x_1<x_2,y_1<y_2$

What is the integration of the below integral?

$\|N(u)(x_1,y_1)-N(u)(x_2,y_2)\|\le\|\mu(x_1,y_1)-\mu(x_2,y_2)\|+L_1\|u(x_1,y_1)-u(x_2,y_2)\|+\|\dfrac{L_2}{\Gamma (r_1)\Gamma (r_2)}\int_0^{x_1}\int_0^{y_1}[(x_2-s)^{r_1-1}(y_2-t)^{r_2-1}-(x_1-s)^{r_1-1}(y_1-t)^{r_2-1}]\times\ u(s,t)dtds+\dfrac{L_2}{\Gamma (r_1)\Gamma (r_2)}\int_{x_1}^{x_2}\int_{y_1}^{y_2}(x_2-s)^{r_1-1}(y_2-t)^{r_2-1}u(s,t)dtds+\dfrac{L_2}{\Gamma (r_1)\Gamma (r_2)}\int_0^{x_1}\int_{y_1}^{y_2}(x_2-s)^{r_1-1}(y_2-t)^{r_2-1}u(s,t)dtds+\dfrac{L_2}{\Gamma (r_1)\Gamma (r_2)}\int_{x_1}^{x_2}\int_0^{y_1}(x_2-s)^{r_1-1}(y_2-t)^{r_2-1}u(s,t)dtds \|$

• Please check your braces, seems like that at least one "[" is missing. Maybe instead of [ and ], you might use \Big( and \Big). Commented Jul 8, 2014 at 12:40
• sorry for that, Now it is ok. Commented Jul 8, 2014 at 13:00
• what is $u(t,s)$? is it general function? Commented Jul 8, 2014 at 13:05
• Please read my answer, everything is in there so please explain how we get the solution of the last line of my answer. Commented Jul 9, 2014 at 12:42
• Mr. mwomath, Are u understand of my Question? If you are then please do effort on this Qus. Commented Jul 9, 2014 at 18:19

Since $(Nu)(x,y) =\mu(x,y)+f(x,y,I_\theta^r u(x,y),u(x,y)), (Nu)(x,y)=u(x,y)$by fixed point theorem, where $I_\theta^r$ is the left-sided mixed Riemann-Lioville integral of order r, $r=(r_1,r_2)\in(0,\infty)\times(0,\infty), \theta\in(0,0)$ and $u\in L^1(J),L^1(J)$ is the space of Labesgue-integrable functions from $J=[0,a]\times[0,b]$ into $R^n.$ $I_\theta^r$ is defined as follows.

$(I_\theta^ru)(x,y)=\dfrac{1}{\Gamma (r_1)\Gamma (r_2)}\int_0^{x}\int_0^{y}(x-s)^{r_1-1}(y-t)^{r_2-1}u(s,t)dtds$

Now there exist constants $L_1,L_2>0$ such that

$\|f(x,y,u,v)-f(x,y,\bar u,\bar v)\|\le L_2\|u-\bar u\|+L_1\|v-\bar v\|$

and $\displaystyle\|Nu\|_E=\sup_{(x,y)\in J}{\|(Nu)(x,y)\|}e^{-\lambda(x+y)}$, $\|u\|_E\le M$

Set $f^*(x,y) =f(x,y,0,0)$

$\|(Nu)\|_E \le \|\mu\|_E+\|f^*\|_E+ ML_1+\dfrac{ML_2a^{r_1}b^{r_2}} {\Gamma(1+r_1)\Gamma(1+r_2)} =:\eta$

Now we come to the point $({x_1},{y_1}),({x_2},{y_2})\in J=[0,a] \times[0,b],x_1<x_2,y_1<y_2$ and $u\in B_\eta$

where $N$ transform the ball $B_\eta= \{u\in E:\|u\|_E\}$ into itself and $E$ is space of functions $u:J\to R^n.$

Then $\|N(u)(x_1,y_1)-N(u)(x_2,y_2)\|\le\|\mu(x_1,y_1)-\mu(x_2,y_2)\|+L_1\|u(x_1,y_1)-u(x_2,y_2)\|+\|\dfrac{L_2}{\Gamma (r_1)\Gamma (r_2)}\int_0^{x_1}\int_0^{y_1}[(x_2-s)^{r_1-1}(y_2-t)^{r_2-1}-(x_1-s)^{r_1-1}(y_1-t)^{r_2-1}]\times\ u(s,t)dtds+\dfrac{L_2}{\Gamma (r_1)\Gamma (r_2)}\int_{x_1}^{x_2}\int_{y_1}^{y_2}(x_2-s)^{r_1-1}(y_2-t)^{r_2-1}u(s,t)dtds+\dfrac{L_2}{\Gamma (r_1)\Gamma (r_2)}\int_0^{x_1}\int_{y_1}^{y_2}(x_2-s)^{r_1-1}(y_2-t)^{r_2-1}u(s,t)dtds+\dfrac{L_2}{\Gamma (r_1)\Gamma (r_2)}\int_{x_1}^{x_2}\int_0^{y_1}(x_2-s)^{r_1-1}(y_2-t)^{r_2-1}u(s,t)dtds \|$

The answer of the above equation is as follows

$\|N(u)(x_1,y_1)-N(u)(x_2,y_2)\|\le\|\mu(x_1,y_1)-\mu(x_2,y_2)\|+L_1\|u(x_1,y_1)-u(x_2,y_2)\|+\dfrac{L_2\eta}{\Gamma(1+r_1)\Gamma(1+r_2)}[2y_2^{r_2}(x_2-x_1)^{r_1}+2x_2^{r_1}(y_2-y_1)^{r_2}+x_1^{r_1}y_1^{r_2}-x_2^{r_1}y_2^{r_2}-2(x_2-1)^{r_1}(y_2-y_1)^{r_2}]$