Dear Giuseppe Papari,

I would suggest to start with the following NECESSARY  equation:

                        div(T_1 T_2 )   = (b-a)  (1,-1) \cdot (T_1, T_2)

which in the particular case of 

                                                 a = b 

implies that  the vector field (T_1, T_2)  HAS TO BE a gradient of some potential. Thus your boundary conditions define the field uniquely (possibly, I am wrong, plese consult works/textbooks on electrostatics). This would imply necessity of the  consistency  with your equations -  which is rather impossible unless sever restrictions on the boundary conditions are imposed.  

EXAMPLE:  the constant wector field:   T_1(x,y) = 1  T_2(x,y) = 0  is the (probably)  unique source-free field with  T_1(0,y) = T_1(x,0) = 1  and  T_2(0,y) = T_2(x,0) = 0   and it does not satisfy the given equations unless  the common value  A:=a=b equals 0. Indeed, we have then  d T_1/dx = d T_2/dy = 0, while on the RHSs we have -A and A, respectively. 

Best regards

You can refer to the book by Prof. Polyanin.

Dear Sir,

The solution is of the form:

T1 = c * exp(-k*x -l *y)

I have no time to do the math now. You need to fill this in into the 2 equations, and calculate k and l from them, and k and l and c from the border conditions.

Cheers,

Henri.

given, dT1/dx = a(T2-T1)   and  dT2/dy = b (T1-T2)

Hence,    (1/a)*(dT1/dx) + (1/b)*(dT2/dy) =0   or, dT2/dy = (-b/a)(dT1/dx)

Now, differentiating first equation w.r.t. y, we get:

    d^2 T1/dxdy = a(dT2/dy - dT1/dy)

Putting the value of dT2/dy, we get:

 d^2 T1/dxdy = a [ (-b/a)(dT1/dx)-dT1/dy]

or, d^2T1/dxdy = -b*(dT1/dx) -a*(dT1/dy)

or, d^2T1/dxdy + b*(dT1/dx)+a*(dT1/dy)=0

Clearly, this is a hyperbolic pde. I hope you can solve it from here.

The complexity of the problem comes from the difficulty in taking into account the boundary conditions.  According to Rajkamal suggestion (which imples Henri's examples) the a one-parameter family of (seemingly the simplest)  solutions is given by (excuse, please some non-substantial for the explanation errors, which certainly  are possible),

T_1(A; x,y) =  \exp( (A - 1) a x + (B - 1) b y ),

T_2(A; x,y) =  A \exp( (A - 1) a x + (B - 1) b y )

where A is  arbitrary,  A \ne 0  A \cdot B = 1, provided  a \cdot b \ne 0 (let us omit the other cases). By linearity every suitably generalized linear combination of these is a solution to the original problem. In particular, one can use it in the form of integral

T_j(\mu; x,y)  :=  \int_{R} T_j (A; x,y)  d\mu(A)  for  j=1,2.

which leads then to a kind of Laplace transform for the boudary functions. When admitting complex values for  A  (in particular purely imaginary) then the solution obtains a form of Fourier-Laplace transform. Further details are not clear for me - should we restrict to purely imaginary or purely real parameters, how about the possibility of using arbitrary  boudary functions  . . .

You can use runge kutta method