# Taylor Expansion of a Vector Field in Stokes Flow About a Sphere's Surface

So I am going through what could be a pretty simple identity:

$$\frac{1}{4 \pi a^2} \int_S \vec{u}(\vec{x}) dS=\vec{u}(\vec{0})+\frac{a^2}{6} \nabla^2\vec{u}(\vec{x})|_{\vec{x}=\vec{0}}$$

where S is the surface of the sphere, and $$a$$ is the sphere radius. I get the gist that odd terms cancel because of an argument about $$\vec{u}(\vec{x})$$ being an even function, and for a Stokesian flow because of the biharmonic property, even terms with $$\nabla^{2n}$$ for n>1 are zero. But I am still confused about: the expansion of the vector field itself (especially in spherical coordinates), and the details of why we know it is even? I would appreciate any guidance.

• I am assuming $a$ is the radius of the sphere? – JamalS Nov 18 '18 at 21:07
• Right, you are missing a normalization factor. In the left hand side, the integral should be divided by the area of the sphere. – secavara Nov 18 '18 at 21:27
• That being said, your question confuses me, given the fact that you seem to have no trouble accepting the general result. You want this statement to be checked explicitly for spherical coordinates? – secavara Nov 18 '18 at 21:29
• I just don't understand the intuition behind why it is even and how to even attempt expanding the vector field as a Taylor series. Like do you take gradients of vectors to do it? Do you get increasingly higher rank tensors for the nabla terms? – FluidMan Nov 18 '18 at 22:31

Consider a scalar function of several variables $$f(\mathbf{\vec{x}})$$. Let us first use Cartesian coordinates $$\mathbf{\vec{x}} \, = \, x_1 \, \mathbf{\hat{i}} \,+\, x_2 \, \mathbf{\hat{j}} \,+\, x_3 \, \mathbf{\hat{k}} \,$$. A Taylor expansion of $$f(\mathbf{\vec{x}})$$ around $$\mathbf{\vec{y}} \, = \, y_1 \, \mathbf{\hat{i}} \,+\, y_2 \, \mathbf{\hat{j}} \,+\, y_3 \, \mathbf{\hat{k}} \,$$ looks like $$$$f(\mathbf{\vec{x}}) \,= \, f(\mathbf{\vec{y}}) \,+ \, \sum_{i=1}^3 \Delta_i \, \partial_{i} f (\mathbf{\vec{y}}) + \frac{1}{2} \sum_{i=1}^3 \sum_{j=1}^3 \Delta_i \, \Delta_j \, \partial_{i} \partial_{j} f (\mathbf{\vec{y}}) \,+ \, \frac{1}{3!} \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \Delta_i \, \Delta_j \, \Delta_k \, \partial_{i} \partial_{j} \partial_{k} f (\mathbf{\vec{y}}) \,+ \, ...$$$$ where $$$$\Delta_k \,=\, x_k\,-\,y_k \, .$$$$ We should focus on the case $$\mathbf{\vec{y}} \, = \, \mathbf{0} \,$$. Notice that in terms of the $$\mathbf{\vec{x}}$$ dependence, the terms in the expansion are monomials of increasing order. Using spherical coordinates as $$$$\mathbf{\vec{x}} \, = \, r \, \sin \theta \, \cos \phi \, \mathbf{\hat{i}} \,+\, r \, \sin \theta \, \sin \phi \, \mathbf{\hat{j}} \,+\, r \, \cos \theta \, \mathbf{\hat{k}} \,$$$$ we immediately realize that $$\frac{\Delta_i}{r} = \frac{x_i}{r}$$ is a function of only angular variables.
Notice also that the integration of a product of an odd number of these $$x_i$$ over the surface of the sphere vanishes. With this we mean $$$$\left. \int_0^{2\pi}\mathrm{d}\phi \int_0^{\pi} \sin \theta \, \mathrm{d}\theta \, x_{i_1} \, x_{i_2} \cdot \cdot \cdot x_{i_n} \right|_{r\,=\,a} \, = \, 0 \, , \, \mathrm{if} \, n \,\mathrm{is} \, \mathrm{odd} \, .$$$$ Even more, the only terms that lead to non-vanishing integrals are of the form $$\begin{eqnarray} && \left. \int_0^{2\pi}\mathrm{d}\phi \int_0^{\pi} \sin \theta \, \mathrm{d}\theta \, x_1^{2 n_1} x_2^{2 n_2} x_3^{2 n_3} \right|_{r\,=\,a} \\ &&= a^{2n_1+2n_2+2n_3}\int_0^{2\pi}\mathrm{d}\phi \int_0^{\pi} \mathrm{d}\theta \, \sin^{2 n_1 + 2 n_2 +1 } \theta \cos^{2 n_3} \theta \cos^{2 n_1} \phi \sin^{2 n_2} \phi \\ &&= a^{2n_1+2n_2+2n_3} \frac{4\pi}{(2n_1+2n_2+2n_3+1)}\frac{(n_1+n_2+n_3)!}{n_1!n_2!n_3!}\frac{(2n_1)!(2n_2)!(2n_3)!}{(2n_1+2n_2+2n_3)!} \end{eqnarray}$$ with $$n_1,n_2$$ and $$n_3$$ non-negative integers. This is one way of seeing why there is an absence of odd number of derivatives in the expansion. As you see, it does not have anything to do with the parity of the function $$f(\mathbf{\vec{x}})$$. From here, with some combinatorics taking care of the repeated terms being summed, one can deduce that the result for arbitrary $$f(\mathbf{\vec{x}})$$ is $$$$\frac{1}{4 \pi} \left. \int_0^{2\pi}\mathrm{d}\phi \int_0^{\pi} \sin \theta \, \mathrm{d}\theta \, f(\mathbf{\vec{x}}) \, \right|_{r\,=\,a} \, = \sum_{k=0}^{\infty} \frac{a^{2k}}{(2k+1)!} \left( \nabla^2 \right)^k f(\mathbf{0}) \, .$$$$ If $$\left( \nabla^2 \right)^k f = 0$$ for $$k>1$$, the series truncates to your result. We did this for a scalar function but the result naturally extends for vector functions if you apply this process to each Cartesian component.