Demostrar que el jacobiano es constante

Question

Supongamos que $u=u(x,y)$ y $v=v(x,y)$ tienen segundas derivadas parciales continuas. Si para cada $f$ tenemos $$\frac{\partial^2f}{\partial u^2}+\frac{\partial^2f}{\partial v^2}=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2},$$ demostrar que el jacobiano $\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{pmatrix}$ es constante.

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Utilizando la regla de la cadena, $$\frac{\partial^2f}{\partial x^2}=\left(\frac{\partial u}{\partial x}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}+\left(\frac{\partial v}{\partial x}\right)^2\frac{\partial^2f}{\partial v^2}\\ \frac{\partial^2f}{\partial y^2}=\left(\frac{\partial u}{\partial y}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}+\left(\frac{\partial v}{\partial y}\right)^2\frac{\partial^2f}{\partial v^2}$$ y vemos que el jacobiano es ortogonal. Pero, ¿cómo demostrar que es constante?

Answer 1

Un mal cálculo me aleja de resolver esto.

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Como señala @peabody, tenemos $$\frac{\partial^2f}{\partial x^2}=\left(\frac{\partial u}{\partial x}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}+\left(\frac{\partial v}{\partial x}\right)^2\frac{\partial^2f}{\partial v^2}+\frac{\partial^2u}{\partial x^2}\frac{\partial f}{\partial u}+\frac{\partial^2v}{\partial x^2}\frac{\partial f}{\partial v}\\ \frac{\partial^2f}{\partial y^2}=\left(\frac{\partial u}{\partial y}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}+\left(\frac{\partial v}{\partial y}\right)^2\frac{\partial^2f}{\partial v^2}+\frac{\partial^2u}{\partial y^2}\frac{\partial f}{\partial u}+\frac{\partial^2v}{\partial y^2}\frac{\partial f}{\partial v}$$ en lugar de lo que escribí antes.

Utilizando la condición dada (ya que $f$ es arbitraria) obtenemos $$ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0\\ \left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2=1 $$

A partir de la segunda identidad $$ \frac{\partial^2u}{\partial x^2}\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}+\frac{\partial^2u}{\partial x\partial y}\left(\frac{\partial u}{\partial y}\right)^2=0\\ \frac{\partial^2u}{\partial x\partial y}\left(\frac{\partial u}{\partial x}\right)^2+\frac{\partial^2u}{\partial y^2}\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}=0 $$

Sumando obtenemos $$\frac{\partial^2u}{\partial x\partial y}=0.$$

Sea $\frac{\partial u}{\partial x}=A(x)$ , $\frac{\partial u}{\partial y}=B(y)$ . Pon esto en $\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2=1$ obtenemos que ambas son constantes.

Del mismo modo $\frac{\partial v}{\partial x}$ y $\frac{\partial v}{\partial y}$ son constantes. Por lo tanto $\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{pmatrix}$ es constante.