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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?

amWhy
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Peter
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  • Perhaps you are stumbling over the notational ambiguity of whether the “partial with respect to $x$” means the “partial with respect to the first coordinate” or with respect to the variable denoted by the letter $x$. For example, see https://math.stackexchange.com/questions/2200982/notation-for-partial-derivatives. – Steve Kass Nov 09 '17 at 01:11
  • If I understood well, you mean that what the theorem is requesting is that $\tilde{U}$ and $\tilde{V}$ verify the equations in the new variables (x and let's call y'=y+f(x)) ? – Peter Nov 09 '17 at 01:30
  • I didn’t work through it, but quite possibly yes, that the subscript $x$ in lthe equations” means the first parameter, not the variable called $x$. I don’t know the theorem, so am guessing. – Steve Kass Nov 09 '17 at 01:42
  • Sorry, but there are still non-vanishing terms. Does anyone know how to prove it? – Peter Nov 09 '17 at 21:32

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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.

rafa11111
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