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This is probably a very trivial question but I have not been able to figure it out myself, so here goes.

Let $\Gamma$ be a smooth boundary in 2-D divided into $N$ quadratic (3-noded), continuous, finite elements. Consider one such 3-noded element on the boundary $\Gamma$. Let $\phi$ be a continuous function on this boundary. Consider a point $\mathbf{x}$ on $\Gamma$ which resides on one of the finite elements. Now we approximate $\phi(\mathbf{x})$ using the conventional FE shape functions, $$ \phi(\mathbf{x}) = \sum_{j=1}^3 N_j \phi_j $$ where $N_j$ are the polynomial shape functions and $\phi_j$ the nodal values of the function $\phi$ on $\Gamma$. Similarly, we can approximate the normal derivative of the continuous function $\phi$ at point $\mathbf{x}$ on $\Gamma$ as, $$ \frac{\partial \phi(\mathbf{x})}{\partial n(\mathbf{x})} = \sum_{j=1}^3 N_j \frac{\partial \phi_j}{\partial n_j} $$ where $\frac{\partial \phi_j}{\partial n_j}$ is simply the nodal value of the normal derivative of $\phi$ at node $j$. The question is can we alternatively write, $$ \frac{\partial \phi(\mathbf{x})}{\partial n(\mathbf{x})} = \sum_{j=1}^3 \frac{\partial N_j}{\partial n_j}\phi_j $$ where the shape functions are differentiated instead of the function $\phi$? Is it necessary that the normal (or other) derivatives of a function be modeled with linear combination of the shape functions multiplied with the nodal values of the derivatives?

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The way you describe the problem makes no sense, and figuring out how to formulate the problem will likely point you in the right direction towards its solution as well. To explain:

  • You describe the function $\phi$ as living only on the boundary. But then, $\partial \phi/\partial n$ doesn't make any sense because defining the derivative would require you to know values away from the boundary. In other words, you can't take the derivative of a function that's only defined on a line in a direction that's perpendicular of the line.

  • The same is true for your shape functions $N_j(x)$ -- the way you describe them, they are only defined on the boundary and consequently you can't take normal derivatives.

  • Your coefficients $\phi_j$ are just numbers, not functions. Consequently, $\partial \phi_j/\partial n = 0$.

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Another take: I would guess from your question that you are interested in BE methods for potential problems. If this is the case, $\phi(x)\rvert_\Gamma$ and $\frac{\partial \phi}{\partial n}\bigr\rvert_\Gamma$ are to be discretized on $\Gamma$ as independent functions, possibly with different shape functions and node positions.

In the BEM case the answer is therefore no: for the method to work you have a set of shape functions and nodal values for $\phi$ and a distinct set of shape functions and nodal values for $\frac{\partial \phi}{\partial n}$ and you cannot link them by differentiation. The reason was clearly stated in the answer by Wolfgang: let me just add that in BE methods this is not a "problem" but the very core of the procedure.

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  • $\begingroup$ Can I caveat this answer slightly? It often seems to me that in BEM's, the whole reason that a given surface $\Gamma$ was chosen as the discretization site in the first place is precisely because one of $\phi$ or $\frac{\partial\phi}{\partial n}$ is zero (e.g. $\Gamma$ is the boundary of a perfectly hard or perfectly soft scatterer). So often you don't have to discretize both. $\endgroup$
    – rchilton1980
    Aug 30 '13 at 15:34
  • $\begingroup$ @rchilton1980 You are right, in the sense that for well posed problems and homogeneous boundary conditions only one of the two fields has to be discretized, since the other one would give raise to a vanishing boundary integral. (Still for non-homogeneous boundary conditions both fields are to be discretized.) My point is that in the BEM both fields (or more precisely their trace on $\Gamma$) enter the problem formulation as independent discretizations, whether or not they are involved in actual computations. $\endgroup$
    – Stefano M
    Aug 30 '13 at 20:29
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  1. Your function $ \phi(x)$ depends on $x$ , the nodal values $\phi_j$ do not. This means only the shape functions depend on the position.
  2. Same thing applies to the normal vector $n(x)$ and the nodal values $n_j$.
  3. I would rather write $(\frac{\partial \phi}{\partial n})(x)= \sum\limits_{j=1}^3 N_j(x) (\frac{\partial \phi}{\partial n})_j $
  4. Consequently, the answer would be "Yes" to your question
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