# Thread: [SOLVED] Calculus with two variables

1. ## [SOLVED] Calculus with two variables

Hi,

Could someone show me a proof that the two-dimensional del-squared operator is invariant under a rotation of axes in Cartesian co-ordinates?

Thanks.

2. Originally Posted by RET7
Hi,

Could someone show me a proof that the two-dimensional del-squared operator is invariant under a rotation of axes in Cartesian co-ordinates?

Thanks.
Under a counter-clockwise rotation by angle $\theta$:

$x' = x \cos \theta - y \sin \theta$

$y' = x \sin \theta + y \cos \theta$

where x' and y' are the new Cartesian coordinates. Then:

$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x'} \cdot \frac{\partial x'}{\partial x} + \frac{\partial f}{\partial y'} \cdot \frac{\partial y'}{\partial x}$

$= \frac{\partial f}{\partial x'} \cdot \cos \theta + \frac{\partial f}{\partial y'} \cdot \sin \theta$.

Therefore:

$\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x'} \cdot \cos \theta + \frac{\partial f}{\partial y'} \cdot \sin \theta \right)$

$= \frac{\partial}{\partial x'} \left( \frac{\partial f}{\partial x'} \cdot \cos \theta + \frac{\partial f}{\partial y'} \cdot \sin \theta \right) \cdot \frac{\partial x'}{\partial x} + \frac{\partial}{\partial y'} \left( \frac{\partial f}{\partial x'} \cdot \cos \theta + \frac{\partial f}{\partial y'} \cdot \sin \theta \right) \cdot \frac{\partial y'}{\partial x}$

$= \left( \frac{\partial^2 f}{\partial x'^2} \cdot \cos \theta + \frac{\partial^2 f}{\partial y' \partial x'} \cdot \sin \theta \right) \cdot \cos \theta + \left( \frac{\partial^2 f}{\partial x' \partial y'} \cdot \cos \theta + \frac{\partial^2 f}{\partial y'^2} \cdot \sin \theta \right) \cdot \sin \theta$

$= \frac{\partial^2 f}{\partial x'^2} \cdot \cos^2 \theta + 2 \frac{\partial^2 f}{\partial y' \partial x'} \cdot \sin \theta \cos \theta + \frac{\partial^2 f}{\partial y'^2} \cdot \sin^2 \theta$.

In a similar way you get:

$\frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial y'^2} \cdot \cos^2 \theta - 2 \frac{\partial^2 f}{\partial y' \partial x'} \cdot \sin \theta \cos \theta + \frac{\partial^2 f}{\partial x'^2} \cdot \sin^2 \theta$.

Now add the two expressions together and you have $\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial x'^2} +\frac{\partial^2 f}{\partial y'^2} \,$ ,

since $\cos^2 \theta + \sin^2 \theta = 1$.