Operador cambio a coordenadas polares

Question

Que $$\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$ be the Laplace operator on the $ (x,y)$-plane. Consider the polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$. Show that $$\Delta=\frac{\partial^2}{\partial r^2}+\frac1r\frac\partial{\partial r}+\frac1{r^2}\frac{\partial^2}{\partial \theta^2}.$$

No sé cómo cambiar las coordenadas para que un operador como éste. ¿Qué es el método que se debe utilizar?

Answer 1

Escribir

$$\frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta}$$

donde $r=\sqrt{x^2+y^2}$ y $\tan{\theta} = y/x$. Entonces $\partial r/\partial x = x/r$ y $\partial \theta/\partial x = -y/r^2$, y

$$\frac{\partial}{\partial x} = \frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta}$$

Del mismo modo, puede mostrar

$$\frac{\partial}{\partial y} = \frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta}$$

El Laplaciano es entonces

$$\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = \left (\frac{x}{r} \frac{\partial}{\partial r} -\frac{y}{r^2} \frac{\partial}{\partial \theta} \right )^2 + \left (\frac{y}{r} \frac{\partial}{\partial r} +\frac{x}{r^2} \frac{\partial}{\partial \theta} \right )^2$$

Esto debería ser suficiente para ir a.