Why does the coordinate transformation from Cartesian coordinates leads to an additional term in the biharmonic operator in spherical coordinates I am trying to solve a problem in physics where the biharmonic operator is involved. I think that the bihahmonic operator can be obtained by taking twice the Laplace operator, such that $\nabla^4 f = \nabla^2 (\nabla^2 f)$. 
Here $f$ is a function of the polar angle $\phi$ only (i.e. in the axisymmetric case for the sake of simplicity.)
In this way, the Laplace operator reads
$$
\nabla^2 f(\phi) = \frac{1}{a^2 \sin\phi} \frac{\partial}{\partial \phi} \left( \sin\phi \frac{\partial f}{\partial \phi} \right) \, ,
$$
where $a$ being the sphere radius.
In principle, the biharmonic operator is obtained after applying twice the above operation, to obtain
$$
\nabla^4 f(\phi) = \frac{1}{a^4} 
\left( 
\frac{\partial^4 f}{\partial \phi^4} 
+ 2\cot\phi \frac{\partial^3 f}{\partial \phi^3}
-(2+\cot^2 \phi) \frac{\partial^2 f}{\partial \phi^2}
+\cot\phi (1+\cot^2\phi) \frac{\partial f}{\partial \phi}
\right) \, .
$$
However, if we start from the biharmonic equation in Cartesian coordinates, namely
$$
\nabla^4 f(x,y,z) = \frac{\partial^4 f}{\partial x^4} 
+ \frac{\partial^4 f}{\partial y^4} 
+ \frac{\partial^4 f}{\partial z^4}
+ 2 \left( 
\frac{\partial^4 f}{\partial x^2 \partial y^2}
+ \frac{\partial^4 f}{\partial y^2 \partial z^2}
+ \frac{\partial^4 f}{\partial x^2 \partial z^2}
\right) \, ,
$$
and apply the coordinate transformation to spherical, e.g. using Maple PDEchangecoords and of course drop the dependence on $r$ and $\theta$, I get the following biharmonic
$$
\nabla^4 f(\phi) = \frac{1}{a^4} 
\left( 
\frac{\partial^4 f}{\partial \phi^4} 
+ 2\cot\phi \frac{\partial^3 f}{\partial \phi^3}
-\cot^2 \phi\frac{\partial^2 f}{\partial \phi^2}
+\cot\phi (3+\cot^2\phi) \frac{\partial f}{\partial \phi}
\right) \, ,
$$
i.e. the following additional terms appear
$$
\frac{2}{a^4} \left( \frac{\partial^2 f}{\partial \phi^2} + \frac{\partial f}{\partial \phi} \cot \phi \right) \, , 
$$
My question is which resulting biharmonic operator is correct? What is the reason behind the discrepancy between the two results.
Thank you,
a 
 A: Since you did not show your work, I can only conjecture that you forgot to differentiate $a$ in 
$$\Delta f = \frac{1}{a^2 \sin\phi} \frac{\partial}{\partial \phi} \left( \sin\phi \frac{\partial f}{\partial \phi} \right) + \frac{1}{a^2\sin^2\phi} \frac{\partial^2 f}{\partial \theta^2} \, ,
$$
In the next step, if you want to take $\Delta$ again, note then $\Delta f$ is NOT independent of $a$. So you have to use the general formula, which has an extra term:
$$\frac{1}{a^2} \frac{\partial}{\partial a}\left( a^2 \frac{\partial}{\partial a}\right)$$
Since 
$$\left[\frac{1}{a^2} \frac{\partial}{\partial a}\left( a^2 \frac{\partial}{\partial a}\right)\right] \frac{1}{a^2} = \frac{2}{a^4},$$
so if you really missed that, you will be off with the term as stated. 
A: I think the issue is that you are conflating the Biharmonic operator $\Delta^2$ of the sphere (the first one in your question) with the one $\bar \Delta^2$ of  $\mathbb R^3;$ i.e. you are assuming that when $f : \mathbb R^3 \to \mathbb R$ depends only on $\phi, \theta$ that $\Delta^2 f = \bar \Delta^2 f$. The equivalent statement for the Laplacian is true since $\bar \Delta f = \Delta f + \partial_r ^2 f$; but for $\Delta^2$ it simply isn't. The correct expression is
$$ \bar \Delta^2 f = \bar \Delta (\Delta f + \partial_r ^2 f) = \bar \Delta \Delta f = \Delta^2 f + \partial_r^2 \Delta f.$$
The terms $\Delta \partial_r^2 f$ and $\partial_r^4 f$ disappear because $f$ does not depend on $r$, but the $\partial_r^2 \Delta f$ does not. This is what John's answer and comments are getting at - since the coordinate expression for $\Delta f$ depends on the sphere radius, $\bar \Delta^2$ and $\Delta^2$ do not agree even in the symmetric case.
