# How to express the derivatives with respect to the initial coordinates {x,y,z} when 3D coordinate system is mapped into 2D manifold {e,n}?

This is often required in the stiffness matrix evaluation. Thanks to this thread I found out how to perform integrals over 2D surface. I've tried to evaluate the integral of the single shape function and got 0.3333.

Now I want to have the expression which is being integrated to be more complicated. Namely I want to perform this integral:

$\displaystyle\int_S {\frac{\partial{H_1(e,n)}}{\partial{x}}}^2 + {\frac{\partial{H_1(e,n)}}{\partial{y}}}^2 + {\frac{\partial{H_1(e,n)}}{\partial{z}}}^2\ ds$

The derivatives of the first shape function with respect to {x,y,z} are not the same as the derivatives with respect to {e,n}. I should use the inversed Jacobian matrix to translate the values of the derivatives with respect to {e,n} at the integration point to the derivatives with respect to {x,y,z}.

But the Jacobian matrix isn't square in this case.

This post uses the same notation and preliminaries as an answer I wrote to an earlier post.

Let's again suppose that you're integrating over a smooth manifold $\mathcal{M}$, and that you have a diffeomorphism $\varphi: [-1,1]^{2} \rightarrow \mathcal{M}$ that maps $[-1,1]^{2}$ to your manifold. Since $\varphi$ is a diffeomorphism, it has the following properties:

• it is continuously differentiable
• it is invertible
• its inverse is continuously differentiable

We will need all of these properties.

I'm going to assume that $H_{1}$ is a scalar-valued function, and for simplicity of notation, I'm going to drop the subscript and refer to it as $H$.

Let's suppose further that the arguments $(e,n)$ refer to coordinates in $[-1,1]^{2}$, and the coordinates $(x,y,z)$ refer to points in $\mathbb{R}^{3}$, where $\mathcal{M} \subset \mathbb{R}^{3}$.

From your question, it looks like you have $H: [-1,1]^{2} \rightarrow \mathbb{R}$, which is to say that the function you've stated is given in terms of the arguments $(e,n)$. We'll assume that $H$ is continuously differentiable.

Let $G: \mathcal{M} \rightarrow \mathbb{R}$ be defined by $G = H \circ \varphi^{-1}$, or $G(x,y,z) = H(\varphi^{-1}(x,y,z))$. We're using $\varphi^{-1}$ as a "change of coordinates" here, mapping $(x,y,z)$ coordinates of points in $\mathcal{M}$ to $(e,n)$ coordinates of points in $[-1,1]^{2}$.

It doesn't make sense to differentiate $H$ with respect to $(x,y,z)$, but it does make sense to differentiate $G$ with respect to $(x,y,z)$. I think the integral you want to express above is:

\begin{align*} \int_{\mathcal{M}} \left(\frac{\partial{G}}{\partial{x}}\right)^{2}(x,y,z) + \left(\frac{\partial{G}}{\partial{y}}\right)^{2}(x,y,z) + \left(\frac{\partial{G}}{\partial{z}}\right)^{2}(x,y,z)\,\mathrm{d}S, \end{align*}

more compactly expressed as:

\begin{align*} \int_{\mathcal{M}} \|\mathrm{D}G^{T}(x,y,z)\|^{2}\,\mathrm{d}S = \int_{\mathcal{M}} \|\nabla G(x,y,z)\|^{2}\,\mathrm{d}S. \end{align*}

By the chain rule,

\begin{align*} \mathrm{D}G(x,y,z) = (\mathrm{D}H \circ \varphi^{-1})(x,y,z) \cdot \mathrm{D}(\varphi^{-1})(x,y,z), \end{align*}

where the $\cdot$ denotes matrix multiplication.

A sanity check is in order here:

• $\mathrm{D}H$ is a 1 by 2 matrix (the Jacobian of a scalar function with respect to multiple variables is a row matrix)

• $\mathrm{D}(\varphi^{-1})$ is a 2 by 3 matrix

• $\mathrm{D}G$ is a 1 by 3 matrix

• the right hand side of the equation is a 1 by 3 matrix (the product of a 1 by 2 matrix and a 2 by 3 matrix

Given all of this information (it's quite a long post already), you should be able to use my answer to your previous post to calculate your integral. I can fill in the additional steps, if need be.

You make an excellent point in your question that:

The derivatives of the first shape function with respect to $(x,y,z)$ are not the same as the derivatives with respect to $(e,n)$. I should use the inverse Jacobian matrix to translate the values of the derivatives with respect to $(e,n)$ at the integration point to the derivatives with respect to $(x,y,z)$.

If $f: X \rightarrow X$, where $X \subset \mathbb{R}^{n}$ for some natural number $n$, and $f$ is a diffeomorphism (again, continuously differentiable function with continuously differentiable inverse), it is true by the chain rule that $\mathrm{D}(f^{-1}) = \mathrm{D}f^{-1}$. The problem with this concept for your problem (as you rightly point out) is that $\varphi$ is a diffeomorphism, but doesn't have the same number of input arguments as output values, so we can't state that $\mathrm{D}(\varphi^{-1}) = \mathrm{D}\varphi^{-1}$ unless we interpret the "inverse" of the nonsquare matrix as some sort of generalized matrix inverse. I hope it is clear from the reasoning above how you get from the "inverse matrix" idea to the more general case.

• Thank you very much again first of all. Do I get it right that I've got to have the $\phi^{-1}$? When building an isotropic finite element I use my form functions $N_i(e,n)$ in order to interpolate $x$,$y$ and $z$ between the nodes. And I have for example $x(e,n)=\sum_{i=1}^{4}N_{i}x_i$ where $x_i$ are the nodal values of $x$ for all four nodes (&i=1,..,4&). I obtain $y(e,n)$ and $z(e,n)$ just the same way. As a result I have a $\phi$. – danny_23 Oct 10 '12 at 14:15
• The form functions I use have the form of $N_i(e,n) = \frac{(1\mp{e})(1\pm{n})}{4}$. – danny_23 Oct 10 '12 at 14:24
• IsoParametric element in the first comment of course. My mistake. – danny_23 Oct 10 '12 at 14:30
• Yes, you need both the mapping from $(e,n)$ to $(x,y,z)$ and its inverse. In the "square" case (equal numbers of inputs and outputs), you can avoid using the inverse function explicitly because all you need is the inverse of the Jacobian matrix. – Geoff Oxberry Oct 10 '12 at 18:47