1
$\begingroup$

What is the effective 2-form corresponding to the equation

$det Hess v=(v-q_1v_{q_1}-q_2v_{q_2})^4$

you can find the definition of effective forms here

$\endgroup$
  • 1
    $\begingroup$ I removed the newly created tag "nonlinear" as it has no stand-alone meaning. If you think there is a need for a tag for nonlinear differential equations, please create such a tag and not a general add-on "nonlinear". $\endgroup$ – user9072 Mar 10 '13 at 18:00
1
$\begingroup$

Unless I'm mistaken, all the second derivatives of $v$ are isolated in the Hessian on the left-hand side. The notes you linked to already give the effective form that generates the Hessian. To get the effective form on for the right-hand side, simply multiply by $dq_1 \wedge dq_2$. So, I believe, the effective form you are looking for is $$ dp_1 \wedge dp_2 - (v-q_1 p_1-q_2 p_2)^4 dq_1\wedge dq_2 . $$

$\endgroup$
  • $\begingroup$ Dear Igor , why you multiplied the right-hand side to $dq_1∧dq_2$? $\endgroup$ – user21574 Mar 10 '13 at 12:00
  • 1
    $\begingroup$ Actually, both sides of the PDE are multiplied by $dq_1\wedge dq_2$. The left side (containing the determinant of the Hessian) becomes $dp_1\wedge dp_2$ and the right side remains as written. $\endgroup$ – Deane Yang Mar 10 '13 at 16:39
  • $\begingroup$ Why the left side (containing the determinant of the Hessian) becomes $dp_1∧dp_2$. I know the effective form of $detHessv=1 $is $dp_1∧dp_2−dq_1∧dq_2$ so by multiplying dq_1\wedge dq_2$$ on both sides why have we the answer of @Igor? $\endgroup$ – user21574 Mar 10 '13 at 17:33
  • 1
    $\begingroup$ @Hassan, I took the effective form corresponding to the Hessian directly from one of the examples in the notes you linked to: $\det Hess(v) dq_1\wedge dq_2$ corresponds to the form $dp_1\wedge dp_2$. Multiplying both sides of your equation by $dq_1\wedge dq_2$, I think, is the next logical step, and it directly gives the desired effective form. $\endgroup$ – Igor Khavkine Mar 10 '13 at 18:17
  • 1
    $\begingroup$ On a rereading, I noticed that this is exactly what Dean Yang already said in his comment. $\endgroup$ – Igor Khavkine Mar 10 '13 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy