The subject of linis paper is the Darboux problem for functions 1 equations of hyperbolic type. Such problems were first studied by D.V.Ionescu [4] and many contributions are to be found in some recent papers $[1],[2],[5]-[8]$$[1],[2],[5]-[8]$[1],[2],[5]-[8][1],[2],[5]-[8]$[1],[2],[5]-[8]$.

We consider equations of the following type
(1) $u_{xy}(x,y)=f(x,y,u(x,y),u(g(x,y)))$$u_{xy}(x,y)=f(x,y,u(x,y),u(g(x,y)))$u_(xy)(x,y)=f(x,y,u(x,y),u(g(x,y)))u_{x y}(x, y)=f(x, y, u(x, y), u(g(x, y)))$u_{xy}(x,y)=f(x,y,u(x,y),u(g(x,y)))$
(2) $u_{xy}(x,y)=f(x,y,u(x,y),u_x(x,y),u_y(x,y),u(E_1(x,y)),u_x(E_2(x,y))$$u_{xy}(x,y)=f\left(x,y,u(x,y),u_x(x,y),u_y(x,y),u\left(E_1(x,y)\right),u_x\left(E_2(x,y)\right)\right.$quadu_(xy)(x,y)=f(x,y,u(x,y),u_(x)(x,y),u_(y)(x,y),u(E_(1)(x,y)),u_(x)(E_(2)(x,y)):}\quad u_{x y}(x, y)=f\left(x, y, u(x, y), u_{x}(x, y), u_{y}(x, y), u\left(E_{1}(x, y)\right), u_{x}\left(E_{2}(x, y)\right)\right.$u_{xy}(x,y)=f(x,y,u(x,y),u_x(x,y),u_y(x,y),u(E_1(x,y)),u_x(E_2(x,y))$, $u_y({\tilde{s}}_z(x,y)))$$\left.u_y\left({\tilde{s}}_z(x,y)\right)\right)${:u_(y)( tilde(s)_(z)(x,y)))\left.u_{y}\left(\tilde{s}_{z}(x, y)\right)\right)$u_y({\tilde{s}}_z(x,y)))$.

For $T_1,T_2⩾0$$T_1,T_2⩾0$T_(1),T_(2) >= 0T_{1}, T_{2} \geqslant 0$T_1,T_2⩾0$ and $\alpha ,\beta >0$$\alpha ,\beta >0$alpha,beta > 0\alpha, \beta>0$\alpha ,\beta >0$ we denote
(3) $D=[0,\alpha ]\times [0,\beta ]$$D=[0,\alpha ]\times [0,\beta ]$D=[0,alpha]xx[0,beta]D=[0, \alpha] \times[0, \beta]$D=[0,\alpha ]\times [0,\beta ]$ and $D_0=([-T_1,0]\times [-T_2,\beta ])\cup ([0,\alpha ]x\times [-I_2,0])$$D_0=\left(\left[-T_1,0\right]\times \left[-T_2,\beta \right]\right)\cup ([0,\alpha ]x\left.\times \left[-I_2,0\right]\right)$D_(0)=([-T_(1),0]xx[-T_(2),beta])uu([0,alpha]x{: xx[-I_(2),0])D_{0}=\left(\left[-T_{1}, 0\right] \times\left[-T_{2}, \beta\right]\right) \cup([0, \alpha] x \left.\times\left[-I_{2}, 0\right]\right)$D_0=([-T_1,0]\times [-T_2,\beta ])\cup ([0,\alpha ]x\times [-I_2,0])$
(4) $E=[0,\mathrm{\infty }{)}^2$$E=[0,\mathrm{\infty }{)}^2$E=[0,oo)^(2)E=[0, \infty)^{2}$E=[0,\mathrm{\infty }{)}^2$ and $E_0=([-T_1,0]\times [-T_2,\mathrm{\infty }))\cup ([0,\mathrm{\infty })\times [-T_2,0])$$E_0=\left(\left[-T_1,0\right]\times \left[-T_2,\mathrm{\infty }\right)\right)\cup \left([0,\mathrm{\infty })\times \left[-T_2,0\right]\right)$E_(0)=([-T_(1),0]xx[-T_(2),oo))uu([0,oo)xx[-T_(2),0])E_{0}=\left(\left[-T_{1}, 0\right] \times\left[-T_{2}, \infty\right)\right) \cup\left([0, \infty) \times\left[-T_{2}, 0\right]\right)$E_0=([-T_1,0]\times [-T_2,\mathrm{\infty }))\cup ([0,\mathrm{\infty })\times [-T_2,0])$.
A. We study at first the case when the equation is considered on the compact $D$$D$DD$D$ with boundary conditions on $D_0$$D_0$D_(0)D_{0}$D_0$ ( not only on ( $[0,\mathbb{C}]\times \times \{0\})\cup (\{0\}\times [0,\beta ])$$[0,\mathbb{C}]\times \times \{0\})\cup (\{0\}\times [0,\beta ])$[0,C]xx xx{0})uu({0}xx[0,beta])[0, \mathbb{C}] \times \times\{0\}) \cup(\{0\} \times[0, \beta])$[0,\mathbb{C}]\times \times \{0\})\cup (\{0\}\times [0,\beta ])$, which corresponds to $T_1=T_2\mathrm{\&}0$$T_1=T_2\mathrm{\&}0$T_(1)=T_(2)&0T_{1}=T_{2} \& 0$T_1=T_2\mathrm{\&}0$ ) in order to admit also retarded arguments. The boundary condition is
(5) $u(x,y)=\phi (x,y),(x,y)\in D_0$$u(x,y)=\phi (x,y),(x,y)\in D_0$u(x,y)=varphi(x,y),(x,y)inD_(0)u(x, y)=\varphi(x, y),(x, y) \in D_{0}$u(x,y)=\phi (x,y),(x,y)\in D_0$.

A solution of the problem (1) (or (2)) with the condition (5) is a function $U\in C(D\cup D_0)($$U\in C\left(D\cup D_0\right)\left(\right.$U in C(D uuD_(0))(:}U \in C\left(D \cup D_{0}\right)\left(\right.$U\in C(D\cup D_0)($ or $U\in C^l(D\cup D_0))$$\left.U\in C^l\left(D\cup D_0\right)\right)${:U inC^(l)(D uuD_(0)))\left.U \in C^{l}\left(D \cup D_{0}\right)\right)$U\in C^l(D\cup D_0))$ such that $U_{xy}$$U_{xy}$U_(xy)U_{x y}$U_{xy}$ exists on $D$$D$DD$D$ and $U$$U$UU$U$ satisfies the equation (1) (or (2)) for any $(x,y)\in D$$(x,y)\in D$(x,y)in D(x, y) \in D$(x,y)\in D$ and the

Cencition (5) on Do.
All the functions whose range is not mentioned hav real values.
THEOREI 1. II
(a) $r\in C(D\times R^2)$$r\in C\left(D\times R^2\right)$quad r in C(D xxR^(2))\quad r \in C\left(D \times R^{2}\right)$r\in C(D\times R^2)$ satisfies a Lipcchite condition with respect to
the last two variables

(b) $\phi \in C^1\left(D_0\right)$$\phi \in C^1\left(D_0\right)$varphi inC^(1)(D_(0))\varphi \in C^{1}\left(D_{0}\right)$\phi \in C^1\left(D_0\right)$
(c) $\delta \in C(D),\xi :D\to D\cup D_0,g=(h,k)$$\delta \in C(D),\xi :D\to D\cup D_0,g=(h,k)$delta in C(D),xi:D rarr D uuD_(0),g=(h,k)\delta \in C(D), \xi: D \rightarrow D \cup D_{0}, g=(h, k)$\delta \in C(D),\xi :D\to D\cup D_0,g=(h,k)$ with $h(x,y)+k(x,y)⩽x+y+a$$h(x,y)+k(x,y)⩽x+y+a$h(x,y)+k(x,y) <= x+y+ah(x, y)+k(x, y) \leqslant x+y+a$h(x,y)+k(x,y)⩽x+y+a$
sind $a<sup\{\frac{1}{t}\frac{t^2-I_1}{I_2}:t>\sqrt{I_3}\}$$a<sup\left\{\frac{1}{t}\frac{t^2-I_1}{I_2}:t>\sqrt{I_3}\right\}$a < s u p{(1)/(t)(t^(2)-I_(1))/(I_(2)):t > sqrt(I_(3))}a<\sup \left\{\frac{1}{t} \frac{t^{2}-I_{1}}{I_{2}}: t>\sqrt{I_{3}}\right\}$a<sup\{\frac{1}{t}\frac{t^2-I_1}{I_2}:t>\sqrt{I_3}\}$,
then the problem (1) with the condition (5) has in CSUD 2 d unique Solution.

Proof. We consider the Banach space $C(D\cup D_0)$$C\left(D\cup D_0\right)$C(D uuD_(0))C\left(D \cup D_{0}\right)$C(D\cup D_0)$ with the Bielecki type norm
(6) $\Vert u\Vert =-max\{|u(x,y)|\exp (-t(x+y)):(x,y)\in D\cup D_0\}$$\Vert u\Vert =-max\left\{|u(x,y)|\exp (-t(x+y)):(x,y)\in D\cup D_0\right\}$||u||=-max{|u(x,y)|exp(-t(x+y)):(x,y)in D uuD_(0)}\|u\|=-\max \left\{|u(x, y)| \exp (-t(x+y)):(x, y) \in D \cup D_{0}\right\}$\Vert u\Vert =-max\{|u(x,y)|\exp (-t(x+y)):(x,y)\in D\cup D_0\}$
and the operator $T:C(D\cup D_0)\to C(D\cup D_0)$$T:C\left(D\cup D_0\right)\to C\left(D\cup D_0\right)$T:C(D uuD_(0))rarr C(D uuD_(0))T: C\left(D \cup D_{0}\right) \rightarrow C\left(D \cup D_{0}\right)$T:C(D\cup D_0)\to C(D\cup D_0)$ given by
(7) $\{\begin{array}{l}\mathrm{Tu}(x,y)={\int }_0^2{\int }_0^{y}f(x,s,u(x,s),u(g(x,s)))drds+\psi (x,y;,\text{if}(x,y)\in D\\ \mathrm{Tu}(x,y)=\phi (x,y),\text{for}(x,y)\in D_0,\end{array}$$\left\{\begin{array}{l}\mathrm{Tu}(x,y)={\int }_0^2 {\int }_0^y f(x,s,u(x,s),u(g(x,s)))drds+\psi (x,y;,\text{ if }(x,y)\in D\\ \mathrm{Tu}(x,y)=\phi (x,y),\text{ for }(x,y)\in D_0,\end{array}\right.${[Tu(x","y)=int_(0)^(2)int_(0)^(y)f(x","s","u(x","s)","u(g(x","s)))drds+psi(x","y;","" if "(x","y)in D],[Tu(x","y)=varphi(x","y)","" for "(x","y)inD_(0)","]:}\left\{\begin{array}{l}\operatorname{Tu}(x, y)=\int_{0}^{2} \int_{0}^{y} f(x, s, u(x, s), u(g(x, s))) d r d s+\psi(x, y ;, \text { if }(x, y) \in D \\ \operatorname{Tu}(x, y)=\varphi(x, y), \text { for }(x, y) \in D_{0},\end{array}\right.$\{\begin{array}{l}\mathrm{Tu}(x,y)={\int }_0^2{\int }_0^{y}f(x,s,u(x,s),u(g(x,s)))drds+\psi (x,y;,\text{ if }(x,y)\in D\\ \mathrm{Tu}(x,y)=\phi (x,y),\text{ for }(x,y)\in D_0,\end{array}$
where
(3) $\psi (x,y)=\phi (x,0)+\phi (0,y)-\phi (0,0),(x,y)\in D_0$$\psi (x,y)=\phi (x,0)+\phi (0,y)-\phi (0,0),(x,y)\in D_0$psi(x,y)=varphi(x,0)+varphi(0,y)-varphi(0,0),(x,y)inD_(0)\psi(x, y)=\varphi(x, 0)+\varphi(0, y)-\varphi(0,0),(x, y) \in D_{0}$\psi (x,y)=\phi (x,0)+\phi (0,y)-\phi (0,0),(x,y)\in D_0$

The problem is equivalent to the equation $Tu=u$$Tu=u$Tu=uT u=u$Tu=u$. We prove that If is a contraction and then the Banach fixed point theorem guarantees the existence and uniqueness of a fixed point for $T$$T$TT$T$.
. Let $u,v\in C(D\cup D_0)$$u,v\in C\left(D\cup D_0\right)$u,v in C(D uuD_(0))u, v \in C\left(D \cup D_{0}\right)$u,v\in C(D\cup D_0)$. If $(x,y)\in D_0$$(x,y)\in D_0$(x,y)inD_(0)(x, y) \in D_{0}$(x,y)\in D_0$, we have $\mathrm{Tu}(x,y)-\mathrm{Tv}(x,y)==0$$\mathrm{Tu}(x,y)-\mathrm{Tv}(x,y)==0$Tu(x,y)-Tv(x,y)==0\operatorname{Tu}(x, y)-\operatorname{Tv}(x, y)= =0$\mathrm{Tu}(x,y)-\mathrm{Tv}(x,y)==0$. For $(x,y)\in D$$(x,y)\in D$(x,y)in D(x, y) \in D$(x,y)\in D$,

$+I_2|u(g(r,g))-v(g(r,s))|)drds\le {\int }_0^y{\int }_0^y(I_1\Vert u-v\Vert \exp (t(r+s))+$$\left.+I_2|u(g(r,g))-v(g(r,s))|\right)drds\le {\int }_0^y {\int }_0^y \left(I_1\Vert u-v\Vert \exp (t(r+s))+\right.${:+I_(2)|u(g(r,g))-v(g(r,s))|)drds <= int_(0)^(y)int_(0)^(y)(I_(1)||u-v||exp(t(r+s))+:}\left.+I_{2}|u(g(r, g))-v(g(r, s))|\right) d r d s \leq \int_{0}^{y} \int_{0}^{y}\left(I_{1}\|u-v\| \exp (t(r+s))+\right.$+I_2|u(g(r,g))-v(g(r,s))|)drds\le {\int }_0^y{\int }_0^y(I_1\Vert u-v\Vert \exp (t(r+s))+$
$+I_2\Vert u-v\Vert \exp (t(h(x,s)+k(x,s)))dxds\le {\int }_0^x{\int }_0^y(I_2\mid lu-v\Vert \exp (t(x+3))++L_2\Vert u-v\Vert \exp (t(r+s+e))drds\le \frac{1}{t^2}(I_1+I_2\exp \left(t_a\right))\Vert u-v\Vert \exp (t(x+y))$$+I_2\Vert u-v\Vert \exp (t(h(x,s)+k(x,s)))dxds\le {\int }_0^x {\int }_0^y \left(I_2\mid lu-v\Vert \exp (t(x+3))+\right.+L_2\Vert u-v\Vert \exp (t(r+s+e))drds\le \frac{1}{t^2}\left(I_1+I_2\exp \left(t_a\right)\right)\Vert u-v\Vert \exp (t(x+y))$+I_(2)||u-v||exp(t(h(x,s)+k(x,s)))dxds <= int_(0)^(x)int_(0)^(y)(I_(2)∣lu-v||exp(t(x+3))+:}+L_(2)||u-v||exp(t(r+s+e))drds <= (1)/(t^(2))(I_(1)+I_(2)exp(t_(a)))||u-v||exp(t(x+y))+I_{2}\|u-v\| \exp (t(h(x, s)+k(x, s))) d x d s \leq \int_{0}^{x} \int_{0}^{y}\left(I_{2} \mid l u-v \| \exp (t(x+3))+\right. +L_{2}\|u-v\| \exp (t(r+s+e)) d r d s \leq \frac{1}{t^{2}}\left(I_{1}+I_{2} \exp \left(t_{a}\right)\right)\|u-v\| \exp (t(x+y))$+I_2\Vert u-v\Vert \exp (t(h(x,s)+k(x,s)))dxds\le {\int }_0^x{\int }_0^y(I_2\mid lu-v\Vert \exp (t(x+3))++L_2\Vert u-v\Vert \exp (t(r+s+e))drds\le \frac{1}{t^2}(I_1+I_2\exp \left(t_a\right))\Vert u-v\Vert \exp (t(x+y))$ and therefore

The bypothesis on a makes $\frac{1}{t^2}(L_1+L_2\exp \left(t_a\right))<1$$\frac{1}{t^2}\left(L_1+L_2\exp \left(t_a\right)\right)<1$(1)/(t^(2))(L_(1)+L_(2)exp(t_(a))) < 1\frac{1}{t^{2}}\left(L_{1}+L_{2} \exp \left(t_{a}\right)\right)<1$\frac{1}{t^2}(L_1+L_2\exp \left(t_a\right))<1$ ．
R⿱䒑⿰⺝刂心裣 i．The maximum of the expression that denotes a majorant for a is attained in the case when t．is the unique colution greater than $\sqrt{L_1}$$\sqrt{L_1}$sqrt(L_(1))\sqrt{L_{1}}$\sqrt{L_1}$ of the equation

If the right member of the equation（1）contains only the func tion having modified argument，i．e．the equation becomes
（9）$u_{xy}(x,y)=f(x,y,u(g(x,y)))$$u_{xy}(x,y)=f(x,y,u(g(x,y)))$u_(xy)(x,y)=f(x,y,u(g(x,y)))u_{x y}(x, y)=f(x, y, u(g(x, y)))$u_{xy}(x,y)=f(x,y,u(g(x,y)))$ ，
them we have
COROLLARY 1．If
（a）$f\in C(D\times R)$$f\in C(D\times R)$f in C(D xx R)f \in C(D \times R)$f\in C(D\times R)$ is Lipschitz with respect to the last variable

（b）$\phi \in G^1\left(D_0\right)$$\phi \in G^1\left(D_0\right)$varphi inG^(1)(D_(0))\varphi \in G^{1}\left(D_{0}\right)$\phi \in G^1\left(D_0\right)$
（c）$g\in C(D),g:D\to D\cup D_D,g=(h,k)$$g\in C(D),g:D\to D\cup D_D,g=(h,k)$g in C(D),g:D rarr D uuD_(D),g=(h,k)g \in C(D), g: D \rightarrow D \cup D_{D}, g=(h, k)$g\in C(D),g:D\to D\cup D_D,g=(h,k)$ with $h(x,y)+k(x,y)\le \le x+y+a,a<\frac{2}{\rho /I^{\prime }}$$h(x,y)+k(x,y)\le \le x+y+a,a<\frac{2}{\rho /I^{\prime }}$h(x,y)+k(x,y)≤≤x+y+a,a < (2)/(rho//I^('))h(x, y)+k(x, y) \leq \leq x+y+a, a<\frac{2}{\rho / I^{\prime}}$h(x,y)+k(x,y)\le \le x+y+a,a<\frac{2}{\rho /I^{\prime }}$, then the problem（9）with the condition（5）bas in $C(D\cup D_0)$$C\left(D\cup D_0\right)$C(D uuD_(0))C\left(D \cup D_{0}\right)$C(D\cup D_0)$ a unique solution．

Proof，We may consider that $f$$f$ff$f$ satisfies（a）from Theorem 1 with $I_{1_1}=0$$I_{1_1}=0$I_(1_(1))=0I_{1_{1}}=0$I_{1_1}=0$ and $I_2=I$$I_2=I$I_(2)=II_{2}=I$I_2=I$ ．The equation in Remaxk 1 becomes

2－ $\ln \frac{x^2}{I^2}=0$$\ln \frac{x^2}{I^2}=0$ln((x^(2))/(I^(2)))=0\ln \frac{x^{2}}{I^{2}}=0$\ln \frac{x^2}{I^2}=0$ with the positive solution $t=e[I_y$$t=e\left[I_y\right.$t=e[I_(y):}t=e\left[I_{y}\right.$t=e[I_y$ for which
$\frac{1}{t}2n\frac{t^2}{t}=\frac{2}{e\sqrt{L}}$$\frac{1}{t}2n\frac{t^2}{t}=\frac{2}{e\sqrt{L}}$(1)/(t)2n(t^(2))/(t)=(2)/(esqrtL)\frac{1}{t} 2 n \frac{t^{2}}{t}=\frac{2}{e \sqrt{L}}$\frac{1}{t}2n\frac{t^2}{t}=\frac{2}{e\sqrt{L}}$ and all the conditions in Theorem 1 are satisfied．It follows that the problem（9）with the condition（5）has a unique
solution in S(DUDo).

(a) $i\in C(D\times R^6)$$i\in C\left(D\times R^6\right)$i in C(D xxR^(6))i \in C\left(D \times R^{6}\right)$i\in C(D\times R^6)$ satistion a Lipschitz condition with respect to the last six variables

(b) $\phi \in C^l\left(D_0\right)$$\phi \in C^l\left(D_0\right)$varphi inC^(l)(D_(0))\varphi \in C^{l}\left(D_{0}\right)$\phi \in C^l\left(D_0\right)$ and ${\phi }_{xy}$${\phi }_{xy}$varphi_(xy)\varphi_{x y}${\phi }_{xy}$ exists, $f(0,y,z_1,\dots ,z_6)={\phi }_{xy}(0,y)$$f\left(0,y,z_1,\dots ,z_6\right)={\phi }_{xy}(0,y)$f(0,y,z_(1),dots,z_(6))=varphi_(xy)(0,y)f\left(0, y, z_{1}, \ldots, z_{6}\right)=\varphi_{x y}(0, y)$f(0,y,z_1,\dots ,z_6)={\phi }_{xy}(0,y)$, $f(x,0,z_1,\dots ,z_{\sigma })={\phi }_{xy}(x,0),(x,y)\in D,(z_1,\dots ,z_{\sigma })\in R^6$$f\left(x,0,z_1,\dots ,z_{\sigma }\right)={\phi }_{xy}(x,0),(x,y)\in D,\left(z_1,\dots ,z_{\sigma }\right)\in R^6$f(x,0,z_(1),dots,z_(sigma))=varphi_(xy)(x,0),(x,y)in D,(z_(1),dots,z_(sigma))inR^(6)f\left(x, 0, z_{1}, \ldots, z_{\sigma}\right)=\varphi_{x y}(x, 0),(x, y) \in D,\left(z_{1}, \ldots, z_{\sigma}\right) \in R^{6}$f(x,0,z_1,\dots ,z_{\sigma })={\phi }_{xy}(x,0),(x,y)\in D,(z_1,\dots ,z_{\sigma })\in R^6$
(C) $g_i\in C\left(D_0\right),E_i:D\to D\cup D_0,g_i=(h_i,k_i)$$g_i\in C\left(D_0\right),E_i:D\to D\cup D_0,g_i=\left(h_i,k_i\right)$quadg_(i)in C(D_(0)),E_(i):D rarr D uuD_(0),g_(i)=(h_(i),k_(i))\quad g_{i} \in C\left(D_{0}\right), E_{i}: D \rightarrow D \cup D_{0}, g_{i}=\left(h_{i}, k_{i}\right)$g_i\in C\left(D_0\right),E_i:D\to D\cup D_0,g_i=(h_i,k_i)$ with $h_i(x,y)+k_i(x,y)⩽⩽x+y+a,i=\sqrt{1,3}$$h_i(x,y)+k_i(x,y)⩽⩽x+y+a,i=\sqrt{1,3}$h_(i)(x,y)+k_(i)(x,y)⩽⩽x+y+a,i=sqrt(1,3)h_{i}(x, y)+k_{i}(x, y) \leqslant \leqslant x+y+a, i=\sqrt{1,3}$h_i(x,y)+k_i(x,y)⩽⩽x+y+a,i=\sqrt{1,3}$ and

then the problem (2) with the condition (5) has in $C^1\left(DU{D}_0\right)$$C^1\left(DU{D}_0\right)$C^(1)(DUD_(0))C^{1}\left(D U D_{0}\right)$C^1\left(DU{D}_0\right)$ a unique solution.

Proof. In the Banach space $C^l(D\cup D_0)$$C^l\left(D\cup D_0\right)$C^(l)(D uuD_(0))C^{l}\left(D \cup D_{0}\right)$C^l(D\cup D_0)$ endowed with the norm $\Vert u{\Vert }_1=\Vert u\Vert +\Vert u_x\Vert +\Vert u_y\Vert$$\Vert u{\Vert }_1=\Vert u\Vert +‖u_x‖+‖u_y‖$||u||_(1)=||u||+||u_(x)||+||u_(y)||\|u\|_{1}=\|u\|+\left\|u_{x}\right\|+\left\|u_{y}\right\|$\Vert u{\Vert }_1=\Vert u\Vert +\Vert u_x\Vert +\Vert u_y\Vert$, with $\Vert \cdot \Vert$$\Vert \cdot \Vert$||*||\|\cdot\|$\Vert \cdot \Vert$ given by (6) we consider the operator i : $C^1(D\cup D_0)\to C(D\cup D_0)$$C^1\left(D\cup D_0\right)\to C\left(D\cup D_0\right)$C^(1)(D uuD_(0))rarr C(D uuD_(0))C^{1}\left(D \cup D_{0}\right) \rightarrow C\left(D \cup D_{0}\right)$C^1(D\cup D_0)\to C(D\cup D_0)$ given by
$(1.0)\{\begin{array}{rl}Tu(x,y)=& {\int }_0^x{\int }_0^{y}f(x,s,u(x,s),u_x(r,s),u_y(x,s),u(g_1(r,s)),\\ & u_x(g_2(r,s)),u_y({\sigma }_3(r,s)))drds+\psi (x,y),for(x,y)\in D\\ Iu(x,y)=& \phi (x,y),for(x,y)\in D_0\end{array}$$(1.0)\left\{\begin{array}{r}Tu(x,y)={\int }_0^x {\int }_0^y f\left(x,s,u(x,s),u_x(r,s),u_y(x,s),u\left(g_1(r,s)\right),\right.\\ \left.u_x\left(g_2(r,s)\right),u_y\left({\sigma }_3(r,s)\right)\right)drds+\psi (x,y),for(x,y)\in D\\ Iu(x,y)=\phi (x,y),for(x,y)\in D_0\end{array}\right.$(1.0){[Tu(x","y)=int_(0)^(x)int_(0)^(y)f(x,s,u(x,s),u_(x)(r,s),u_(y)(x,s),u(g_(1)(r,s)),:}],[{:u_(x)(g_(2)(r,s)),u_(y)(sigma_(3)(r,s)))drds+psi(x","y)","for(x","y)in D],[Iu(x","y)=varphi(x","y)","for(x","y)inD_(0)]:}(1.0)\left\{\begin{aligned} T u(x, y)= & \int_{0}^{x} \int_{0}^{y} f\left(x, s, u(x, s), u_{x}(r, s), u_{y}(x, s), u\left(g_{1}(r, s)\right),\right. \\ & \left.u_{x}\left(g_{2}(r, s)\right), u_{y}\left(\sigma_{3}(r, s)\right)\right) d r d s+\psi(x, y), f o r(x, y) \in D \\ I u(x, y)= & \varphi(x, y), f o r(x, y) \in D_{0}\end{aligned}\right.$(1.0)\{\begin{array}{rl}Tu(x,y)=& {\int }_0^x{\int }_0^{y}f(x,s,u(x,s),u_x(r,s),u_y(x,s),u(g_1(r,s)),\\ & u_x(g_2(r,s)),u_y({\sigma }_3(r,s)))drds+\psi (x,y),for(x,y)\in D\\ Iu(x,y)=& \phi (x,y),for(x,y)\in D_0\end{array}$
where $\psi$$\psi$psi\psi$\psi$ is defined by (8).
We have to show that the range of $T$$T$TT$T$ is contained in $C^l\left(DU{D}_0\right)$$C^l\left(DU{D}_0\right)$C^(l)(DUD_(0))C^{l}\left(D U D_{0}\right)$C^l\left(DU{D}_0\right)$. We obtain
$(Tu{)}_x(x,y)={\int }_0^{y}f(x,s,u(x,s),\dots ,u_y(g_z(x,s)))ds+{\phi }_x(x,0)$$(Tu{)}_x(x,y)={\int }_0^y f\left(x,s,u(x,s),\dots ,u_y\left(g_z(x,s)\right)\right)ds+{\phi }_x(x,0)$(Tu)_(x)(x,y)=int_(0)^(y)f(x,s,u(x,s),dots,u_(y)(g_(z)(x,s)))ds+varphi_(x)(x,0)(T u)_{x}(x, y)=\int_{0}^{y} f\left(x, s, u(x, s), \ldots, u_{y}\left(g_{z}(x, s)\right)\right) d s+\varphi_{x}(x, 0)$(Tu{)}_x(x,y)={\int }_0^{y}f(x,s,u(x,s),\dots ,u_y(g_z(x,s)))ds+{\phi }_x(x,0)$, for
$(x,y)\in (0,\alpha ]\times (0,\beta ]$$(x,y)\in (0,\alpha ]\times (0,\beta ]$(x,y)in(0,alpha]xx(0,beta](x, y) \in(0, \alpha] \times(0, \beta]$(x,y)\in (0,\alpha ]\times (0,\beta ]$ and
$(Xu{)}_x(x,y)={\rho }_x(x,y)$$(Xu{)}_x(x,y)={\rho }_x(x,y)$(Xu)_(x)(x,y)=rho_(x)(x,y)(X u)_{x}(x, y)=\rho_{x}(x, y)$(Xu{)}_x(x,y)={\rho }_x(x,y)$ for $(x,y)\in D_0\mathrm{\setminus }\{(x,y)\in D:x=0$$(x,y)\in D_0\mathrm{\setminus }\{(x,y)\in D:x=0$(x,y)inD_(0)\\{(x,y)in D:x=0(x, y) \in D_{0} \backslash\{(x, y) \in D: x=0$(x,y)\in D_0\mathrm{\setminus }\{(x,y)\in D:x=0$ or $y=0\}$$y=0\}$y=0}y=0\}$y=0\}$.
The limits in the points of the form $(x_0,0)$$\left(x_0,0\right)$(x_(0),0)\left(x_{0}, 0\right)$(x_0,0)$ an $(0,y_0)$$\left(0,y_0\right)$(0,y_(0))\left(0, y_{0}\right)$(0,y_0)$ are the same
for both expressions, in view of the condition (b). We have proved that $(Tu{)}_x\in C(D\cup D_0)$$(Tu{)}_x\in C\left(D\cup D_0\right)$(Tu)_(x)in C(D uuD_(0))(T u)_{x} \in C\left(D \cup D_{0}\right)$(Tu{)}_x\in C(D\cup D_0)$ and a similar argument shows that $(Tu{)}_y\in \in C(D\cup D_0)$$(Tu{)}_y\in \in C\left(D\cup D_0\right)$(Tu)_(y)∈∈C(D uuD_(0))(T u)_{y} \in \in C\left(D \cup D_{0}\right)$(Tu{)}_y\in \in C(D\cup D_0)$, so $Tu\in C^1(D\cup D_0)$$Tu\in C^1\left(D\cup D_0\right)$Tu inC^(1)(D uuD_(0))T u \in C^{1}\left(D \cup D_{0}\right)$Tu\in C^1(D\cup D_0)$.

Let $u,v\in C^1(D\cup D_0)$$u,v\in C^1\left(D\cup D_0\right)$u,v inC^(1)(D uuD_(0))u, v \in C^{1}\left(D \cup D_{0}\right)$u,v\in C^1(D\cup D_0)$. For $(x,y)\in D_0$$(x,y)\in D_0$(x,y)inD_(0)(x, y) \in D_{0}$(x,y)\in D_0$ we have $\mathrm{Tu}(x,y)-\mathrm{Tv}(x,y)=C_y$$\mathrm{Tu}(x,y)-\mathrm{Tv}(x,y)=C_y$Tu(x,y)-Tv(x,y)=C_(y)\operatorname{Tu}(x, y)-\operatorname{Tv}(x, y)=C_{y}$\mathrm{Tu}(x,y)-\mathrm{Tv}(x,y)=C_y$ For $(x,y)\in D$$(x,y)\in D$(x,y)in D(x, y) \in D$(x,y)\in D$ we obtain $|\mathrm{Iu}(x,y)-\mathrm{Iv}(x,y)|⩽$$|\mathrm{Iu}(x,y)-\mathrm{Iv}(x,y)|⩽$|Iu(x,y)-Iv(x,y)| <=|\operatorname{Iu}(x, y)-\operatorname{Iv}(x, y)| \leqslant$|\mathrm{Iu}(x,y)-\mathrm{Iv}(x,y)|⩽$