Prandtl's transposition theorem Suppose that $U(x,y)$ and $V(x,y)$ satisfy the following equations:
$U_x+V_y=0$
$U U_x+V U_y=g(x)+U_{yy}$
Let
$\tilde{U}=U(x,y+f(x));$
$\tilde{V}=V(x,y+f(x))-f'(x) U(x,y+f(x))$
Prove that $\tilde{U}$ and $\tilde{V}$ also verify the same set of equations.
Note: I try to prove it by computing $\tilde{U}_x+\tilde{V}_y$ but I arrive to $\frac{V_x-f'' U}{f'}$ which is not zero. Do you have any idea?
 A: The continuity equation is
$$
\frac{\partial U}{\partial x} + \frac{\partial V}{\partial y} = 0,
$$
and the momentum conservation equation is
$$
U \frac{\partial U}{\partial x} + V \frac{\partial U}{\partial y} = g(x) + \frac{\partial^2 U}{\partial y^2},
$$
in which $g(x)$ is the pressure gradient. Our transposed variables are
$$
\tilde{U} = U(x,\tilde{y}), \ \ \ \tilde{V} = V(x,\tilde{y}) - f'(x) U(x,\tilde{y}),
$$
in which $\tilde{y} = y + f(x)$. The partial derivatives in terms of the transposed variables can be written as
$$
\frac{\partial \tilde{U}}{\partial x} = \frac{\partial U}{\partial x} + \frac{\partial \tilde{U}}{\partial \tilde{y}} \frac{\partial \tilde{y}}{\partial x} = \frac{\partial U}{\partial x} + f'(x) \frac{\partial \tilde{U}}{\partial \tilde{y}},
$$
$$
\frac{\partial \tilde{V}}{\partial \tilde{y}} = \frac{\partial V}{\partial y} - f'(x) \frac{\partial \tilde{U}}{\partial \tilde{y}}.
$$
It is easy to see that
$$
\frac{\partial \tilde{U}}{\partial x} + \frac{\partial \tilde{V}}{\partial \tilde{y}} = \frac{\partial U}{\partial x} + \frac{\partial V}{\partial y} = 0,
$$
i.e., the transposed variables satisfies the continuity equation. Substituting
$$
V = \tilde{V} + f'(x) U,
$$
$$
\frac{\partial U}{\partial x} = \frac{\partial \tilde{U}}{\partial x} - f'(x) \frac{\partial \tilde{U}}{\partial \tilde{y}}
$$
and
$$
\frac{\partial V}{\partial y} = \frac{\partial \tilde{V}}{\partial \tilde{y}} + f'(x) \frac{\partial \tilde{U}}{\partial \tilde{y}}
$$
in the momentum conservation equation leads to
$$
U \left[ \frac{\partial \tilde{U}}{\partial x} - f'(x) \frac{\partial \tilde{U}}{\partial \tilde{y}} \right] + \left[ \tilde{V} + f'(x) U \right] \frac{\partial \tilde{U}}{\partial \tilde{y}} = g(x) + \frac{\partial^2 \tilde{U}}{\partial \tilde{y}^2},
$$
which, since $U=\tilde{U}$ (i.e., the change of variables do not change the function itself), simplifies to
$$
\tilde{U} \frac{\partial \tilde{U}}{\partial x} + \tilde{V} \frac{\partial \tilde{U}}{\partial \tilde{y}} = g(x) + \frac{\partial^2 \tilde{U}}{\partial \tilde{y}^2},
$$
i.e., the transposed variables satisfies the momentum conservation equation.
