# Biharmonic equation

I know the gradient $$\Delta f = \sum_{i=1}^n \frac {\partial f}{\partial x_i}$$

Then looking at the Laplace operator $$\Delta^2 f = \sum_{i=1}^n \frac {\partial^2f}{\partial x^2_i}$$

Now my first thought for $$\Delta^4 f$$ would be $$= \sum_{i=1}^n \frac {\partial^4f}{\partial x^4_i}$$

However this isn't the case the Biharmonic equation has a different equation: $$\Delta^4\varphi=\sum_{i=1}^n\sum_{j=1}^n\partial_i\partial_i\partial_j\partial_j \varphi$$

Anyone got a clue why this is the case?

Well, the $$i$$th component of the gradient of the Laplacian is $$\partial_i\left(\sum\limits_{j=1}^n \partial_j^2 f\right)=\sum\limits_{j=1}^n \partial_i\partial_j^2 f.$$ Taking the divergence of the corresponding vector field gives the biharmonic. That is, it would be obtained by taking $$(\partial_1,\partial_2,\cdots,\partial_n)$$ and dotting it with gradient with the above components. Doing so gives $$\sum\limits_{i=1}^n \partial_i^2\left(\sum\limits_{j=1}^n \partial_j^2f\right)=\sum\limits_{i,j=1}^n \partial_i^2\partial_j^2 f.$$