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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?

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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 $$\sum\limits_{i=1}^n \partial_i^2\left(\sum\limits_{j=1}^n \partial_j^2f\right)=\sum\limits_{i,j=1}^n \partial_i^2\partial_j^2 f.$$

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