0
$\begingroup$

I am trying to solve a system of differential equations with partial derivatives. The initial equation is just one

$$ \frac{(v-1)^3}{(a(u-1) u v +b (v-1))^2} \Big(\frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial v}{\partial x} \frac{\partial u}{\partial y} \Big) = 2 \frac{x^3 y^3 (x-1)}{(a(x-1) (y-1) + b x (1-xy))^2} $$

But since a and b are independent, this can be rearranged into 3 different equations: passing the denominators multiplying to each side, and finding the coefficient of $a^2$, $b^2$, and $a b$, we find the three different equations. However, when I feed it to DSolve, the input stays unevaluated... Perhaps Mathematica cannot solve this system?

eq1 = 2*(-1 + x)*x^3*y^3*(-1 + v[x, y])^2 == 
  x^2*(-1 + x*y)^2*(-1 + v[x, y])^3*(Derivative[0, 1][v][x, y]*
      Derivative[1, 0][u][x, y] - 
     Derivative[0, 1][u][x, y]*Derivative[1, 0][v][x, y])

eq2 = 2*(-1 + x)*x^3*y^3*(-1 + u[x, y])^2*u[x, y]^2*
   v[x, y]^2 == (-1 + x)^2*(-1 + y)^2*(-1 + 
      v[x, y])^3*(Derivative[0, 1][v][x, y]*
      Derivative[1, 0][u][x, y] - 
     Derivative[0, 1][u][x, y]*Derivative[1, 0][v][x, y])

eq3 = 4*(-1 + x)*x^3*y^3*(-1 + u[x, y])*u[x, y]*(-1 + v[x, y])*
   v[x, y] == -2*(-1 + x)*
   x*(-1 + y)*(-1 + 
     x*y)*(-1 + v[x, y])^3*(Derivative[0, 1][v][x, y]*
      Derivative[1, 0][u][x, y] - 
     Derivative[0, 1][u][x, y]*Derivative[1, 0][v][x, y])


DSolve[{eq1, eq2, eq3}, {u[x, y], v[x, y]}, {x, y}]

Any help or suggestion would be much appreciated

$\endgroup$
2
  • 2
    $\begingroup$ Didt you not get an error message: "DSolve::overdet: There are fewer dependent variables than equations, so the system is overdetermined." In fact, you can not solve 3 equations with only 2 variables: u and v $\endgroup$ May 3, 2021 at 11:59
  • $\begingroup$ Thank you for the comment. I do not get an error. The output is the same as the input. Could you check if my syntax is ok? I also tried to feed it only eq1 and eq2, getting the same (output = input). $\endgroup$
    – Albercoc
    May 3, 2021 at 12:04

1 Answer 1

3
$\begingroup$

Rewrite your equations, so that all the derivatives are on the left side and then display them in a column:

Column[{eq1, eq2, eq3}]

enter image description here

You see that the left sides are all identical, but the right sides are different. That means, the equations contradict each other.

I think you must go back to where the equations come from and search for an error.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.