Show that the integral $ f(x,y)=1+\int_0^x\int_0^y f(u,v)dudv$ has at most one continuous solution Show that the integral $$ f(x,y)=1+\int_0^x\int_0^y f(u,v)dudv$$ has at most one continuous solution for $0 \leq x \leq 1$,$ 0\leq y <1$. 
The integral can be written as $$I=1+\int_0^1\int_0^1 f(u,v)dudv.$$
Now if we take $f(u,v)$ to be an integrable function in the given region, certainly, we will get only one continuous solution or outcome. 
How to argue in general case?
Suppose $f(u,v) = g(u,v)$,then  $$\iint f(u,v)dudv = \iint g(u,v)dudv.$$ Therefore,  $$1+\iint f(u,v)dudv = 1+\iint g(u,v)dudv.$$ This implies $$f(x,y)=g(x,y).$$ So only one continuous solution.
Is it suitable argument??
 A: Suppose $f,g$ are continuous solutions of the above. Then, $f-g$ satisfies :
$$
(f-g)(x,y) = \int_{0}^x \int_0^y (f-g)(u,v)dudv
$$
Note that $(f-g)(0,0) = 0$. Note that $f-g$ is continuous on $[0,1]^2$, let its maximum be positive and attained at a point $(t,s)$ where $t \neq 0 , s \neq 0$ (if either is zero, then the function is seen to be zero). Note that $$
(f-g)(t,s) = \int_0^t\int_0^s (f-g)(u,v)dudv \leq_1 \int_0^t \int_0^s (f-g)(t,s) dudv \leq ts (f-g)(t,s) 
$$
Now, suppose $ts < 1$. The above inequality cannot hold looking at the ends, so we have a contradiction. Otherwise, $ts = 1$ at most since $t,s \in [0,1]$, and since equality holds, equality must hold in $\leq_1$. Since the LHS function is dominated pointwise by the RHS function, they must be equal i.e. the function $(f-g)(x,y)$ equals its maximum everywhere. Therefore, it is identically zero, a contradiction to a positive maximum.
In other words, the maximum is zero.

Note that this argument crucially requires the domain to be $[0,1] \times [0,1]$. For more general compact domains in $\mathbb R^n$, this kind of uniqueness should hold, only I think you would require Gronwall's inequality (I am not sure of the details, however).
Existence could be shown in two ways that I know of : you could use the Picard-Lindelof iteration , along with patching, or you could study the Volterra operator associated to this linear equation. I think this should work at least on $[0,1]^2$ with possible extension to compact domains.
