Guess four functions given their boundary derivatives

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For some reason I am loking for four functions $a(x,y),b(x,y),c(x,y),u(x,y)$

Where $a+b$ is harmonic i.e.

$$\frac{\partial^2 (a+b)}{\partial x^2}+\frac{\partial^2 (a+b)}{\partial y^2}=0$$

and

$$\frac{\partial a}{\partial x} - \frac{\partial c}{\partial y} = A$$

$$\frac{\partial b}{\partial y} - \frac{\partial c}{\partial x} = B$$

where $A$ and $B$ are some given constants.

And most importantly

$$\frac{\partial u}{\partial y}(0,y)+a(0,y)=0$$ $$\frac{\partial u}{\partial y}(1,y)+a(1,y)=0$$ $$\frac{\partial u}{\partial x}(0,y)+c(0,y)=0$$ $$\frac{\partial u}{\partial x}(1,y)+c(1,y)=0$$

These functions can be arbitrary but the simpler the better.

I would be very glad if you have any suggestions for a solution.

Thank you.

Note that I am loking for some particular solution. This problem arised when I computed some partial differential equation.

EDIT:

That means it will be for example sufficient for me to tell me function u(x,y) which satisfies these boundary conditions: (I have chosen $a=Ax$,$b=By$).

$$\frac{\partial u}{\partial y}(0,y)=0$$ $$\frac{\partial u}{\partial y}(1,y)=-A$$ $$\frac{\partial u}{\partial x}(0,y)=0$$ $$\frac{\partial u}{\partial x}(1,y)=0$$

Is there such $u(x,y)$?