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Consider the following system for $u(x,y)$, $v(x,y)$.

$$2x^2yu_x + 5xy^2u_y + 2x^2y^2v_y + 5xyu + x = yu_y - x^2v_x + u - 2xv = 0 $$

Prove that it is equivalent to a second order semilinear PDE.

Any hints appreciated!

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  • $\begingroup$ Derive the first equation wrt $x$ and the second wrt $y$, we get four equations to play with and perhaps to separate $u$ and its derivatives from $v$ and its derivatives. $\endgroup$ – Rafa Budría Oct 21 '18 at 5:52
  • $\begingroup$ It doesn't work for me or at least I do not see how :( $\endgroup$ – DesmondMiles Oct 21 '18 at 16:19
  • $\begingroup$ Use @DesmondMiles or whatever user name preceded by @ if you want he to receive a notification. Not needed if he is asking the question or the comment is in his answer. $\endgroup$ – Rafa Budría Oct 21 '18 at 17:42
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First derive this one wrt $x$

$2x^2yu_x + 5xy^2u_y + 2x^2y^2v_y + 5xyu + x =0$

$$4xyu_x+2x^2yu_{xx}+5y^2u_y+5xy^2u_{yx}+\color{red}{4xy^2v_y+2x^2y^2v_{yx}}+5yu+5xyu_x+1=0\tag 1$$

Now derive the second wrt $y$

$yu_y - x^2v_x + u - 2xv = 0$

$u_y+yu_{yy}-x^2v_{xy}+u_y-2xv_y=0$, rewrite it as

$$2u_y+yu_{yy}=x^2v_{xy}+2xv_y\tag 2$$

Note that in $(1)$ we have almost the same expression with the partial derivatives of $v$ as in $(2)$. We can substitute them by first multiplying $(2)$ by $2y^2$ (we consider that $v_{xy}=v_{yx}$, of course)

$4y^2u_y+2y^3u_{yy}=2y^2x^2v_{xy}+4y^2xv_y$

$4xyu_x+2x^2yu_{xx}+5y^2u_y+5xy^2u_{yx}+\color{red}{4y^2u_y+2y^3u_{yy}}+5yu+5xyu_x+1=0$

Well, it is a second order semilinear pde in $u$ only.

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  • $\begingroup$ Thanks a lot, nice solution! $\endgroup$ – DesmondMiles Oct 21 '18 at 22:27
  • $\begingroup$ Just interested, how did you think of differentiating the equations in this manner (e.g. why not the first one with respect to y and the second one with respect to x) or it was rather lucky? $\endgroup$ – DesmondMiles Oct 22 '18 at 11:46
  • $\begingroup$ For coupled linear ode systems is a good procedure to derive some of the equations and try to substitute its derivatives with the derivatives from the other(s) equation(s). Furthermore, the very problem says what the solution is! and this is a big clue (how is it possible get a second order eq. from a first order one if not taking derivatives?) And the very problem says it has solution!: to know for sure that a problem has solution, ask whoever you want: it is a very big clue. $\endgroup$ – Rafa Budría Oct 22 '18 at 14:38

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