# Proving that the bi-laplacian of a radial basis function is the dirac delta

According to equation (2.14) of the paper "The Uniform Convergence of Thin Plate Spline Interpolation in Two Dimensions" a radial basis function $\phi(\parallel x \parallel)$ has the property

$$\nabla^4 \phi( \parallel x \parallel ) = 8 \pi \delta(x)$$

I'd like help proving this statement.

PS: I'm sorry but I'm not sure what are the appropriate tags for this question.

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You should probably include the definition of the radial basis function in the statement of your question. –  Willie Wong Apr 16 '11 at 1:47
Definition taken from the Wikipedia entry: a real-valued function whose value depends only on the distance from the origin. (en.wikipedia.org/wiki/Radial_basis_function ) –  Olumide Apr 16 '11 at 1:51
As stated, this is wrong as any constant function is radial and its derivate is not delta –  Vobo Apr 16 '11 at 4:35
You should edit your question: $\phi$ is a very specific function in the mentioned article, basicly the fundamental solution $\phi(r) = r \log r$ of the bi-Laplace operator in two dimensions. –  Vobo Apr 16 '11 at 4:37

See e.g. L. Schwartz, Théorie des Distributions, Example 2.3.2, for a fundamental solution of the $m$-times interated Laplaceoperator in $R^n$, i.e. some $S \in D'$ with $\Delta^m S = \delta$. It is $S(x) = C_{m,n} ||x||^{2m-n}$ if $2m-n$ is odd and $S(x) = C_{m,n} ||x||^{2m-n} \log(||x||)$ in the other case.
Apply this to your specific $\phi$ and check the constants.
Thanks. Unfortunately, I neither have access to "Théorie des Distributions" nor can I read French. Anyway, what I'm trying to show is that $S(x)$ satisfies the PDE $\Delta^m S = \delta$. However it also seems to me that $S(x)$ is obtained as the solution to this PDE. In which case, my question is where does this PDE arise from? On the other hand, if $\Delta^m S = \delta$ is just a condition that $S(x)$ satisfies repeated differentiation of $S(x)$ is sufficient to prove the assertion. What I'm trying to say is that I'm still confused and I'm not sure how to properly think of your answer. –  Olumide Apr 16 '11 at 15:36
Do you know the importance of a "fundamental solution"? If $P$ is a differential operator and $S \in D'$ a solution of $PS=\delta$, then $S \ast f$ is a solution of $PS = f$ (if $f$ is reasonable). –  Vobo Apr 16 '11 at 16:00