Hello,
I was trying to solve the following system of PDEs analytically. Though it looks quite simple, I got stuck. Maybe someone can help me?
dT1/dx = a * (T2-T1)
dT2/dy = b * (T1-T2)
with Tk(x,0) and Tk(0,y) given as boundary conditions (k = 1,2).
Thank you in advance.
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
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 . . .
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