# Recasting integrals from Lagrangian to Eulerian frame

Working on a research problem in the continuum mechanics of fluids. For clarity, uppercase will be used for tensors in the reference configuration, and corresponding spatial items will be in lowercase.

For a moving body, define a map $$\Phi:(X,t)\to{(x,t)}\;$$ from a reference configuration to spatial configuration, whose deformation gradient is $$F=\nabla\Phi$$ with the determinant $$J=\textrm{Det }F$$.

What is the correct approach to convert a surface integral in reference configuration to one in spatial configuration? I have come across conflicting information in multiple sources: e.g. for a reference surface $$A$$ with area element $$dS$$,

(a) Using a change of variables (integration by substitution) i.e. $$\int_A T\;dS = \int_a J t\;ds$$ where $$a=\Phi(A)$$, $$ds=\Phi(dS)$$ and $$t=\Phi(T)$$ for an arbitrary tensor $$T$$ in the reference configuration.

(b) Applying Piola transforms.

Certainly these approaches give very different answers unless I am missing something?

The actual integrals I am trying to convert to spatial integrals have the general forms:

$$I_1 = \int_A (N\cdot{J^k}t^k(F^{-\top})^kN)\;(U^{k+1}{\cdot}N)\;dS$$

$$I_2 = \int_A (U^{k+1}{\cdot}N)(V^{k+1}{\cdot}N)\;dS$$

$$I_3 = \int_A (U^{k+1}{\cdot}V^{k+1})\;dS$$

where $$N$$ is the unit normal to surface area element $$dS$$, $$t$$ is a symmetric second order tensor (in the spatial frame), and $$U$$ and $$V$$ are vectors in the reference frame. The superscripts $$k$$ and $$k+1$$ refer to two separate times.

Please any experienced insights would be very welcome.

I suspect that you hope to find a formula involving $$\mathbf{F}$$, but the deformation gradient is used primarily to re-write integrals over deformed configurations as integrals over the reference configuration. Expressing everything in terms of the reference configuration is what makes the Lagrangian approach to solid mechanics so convenient.

Again, let $$\mathcal{B}$$ be the portion of $$\mathbb{R}^3$$ occupied by the reference configuration, so that $$\Phi_t(\mathcal{B})$$ is the set that those (deformed) material points occupy at time $$t$$.

We use the symbol $$\partial\mathcal{B}$$ for the boundary of $$\mathcal{B}$$ and $$\partial\Phi_t(\mathcal{B})$$ for the boundary of $$\Phi_t(\mathcal{B})$$.

I think that your $$I_3$$ is $$$$\int_{\partial\mathcal{B}}\mathbf{U}(\Phi_{t_{k+1}}(\mathbf{X}))\cdot\mathbf{V}(\Phi_{t_{k+1}}(\mathbf{X}))dS(\mathbf{X}).$$$$ It would be more helpful to know what the vector fields $$\mathbf{U}$$ and $$\mathbf{V}$$ are, as we usually integrate fields whose immediate argument is $$\mathbf{X}$$ (as opposed to $$\mathbf{x} = \Phi_t(\mathbf{X})$$) and use transformations such as Nanson's formula to allow us to integrate functions of $$\mathbf{X}$$ over $$\mathcal{B}$$.

The most general I can get without venturing into differential forms is to note that we assume there are parametrizations of the material points in the reference configuration. Since the surface $$\partial\mathcal{B}$$ is a 2-dimensional surface, there should be (at least locally) some real variables $$s_1$$ and $$s_2$$ (for example, the angles $$\theta$$ and $$\phi$$ on a spherical surface) such that a point on that surface can be described as $$\mathbf{X}(s_1,s_2)$$. Let $$F$$ be a smooth-enough function on $$\partial\mathcal{B}$$. Then in a small area around $$\mathbf{X}(s_1,s_2)$$ the differential area is $$$$dA(\mathbf{X}(s_1,s_2)) = \left\Arrowvert\frac{\partial\mathbf{X}}{\partial s_1}(s_1,s_2)\times\frac{\partial\mathbf{X}}{\partial s_2}(s_1,s_2)\right\Arrowvert ds_1 ds_2,$$$$ where we take the norm of the cross-product of the partial derivates of $$\mathbf{X}(s_1,s_2)$$. The integral of a function $$F$$ over a (perhaps small) patch would be the iterated integral $$$$\int_{a}^{b}\int_{c}^{d}F(\mathbf{X}(s_1,s_2))\left\Arrowvert\frac{\partial\mathbf{X}}{\partial s_1}(s_1,s_2)\times\frac{\partial\mathbf{X}}{\partial s_2}(s_1,s_2)\right\Arrowvert ds_1 ds_2.$$$$
But the co-ordinates $$s_1$$ and $$s_2$$ also parametrize the material points on the surface of the deformed body: $$$$\mathbf{x}(s_1,s_2) = \Phi_t(\mathbf{X}(s_1,s_2)).$$$$ Integrating a function $$f$$ on part of the surface of the deformed body has the same form: $$$$\int_{a}^{b}\int_{c}^{d}f(\mathbf{x}(s_1,s_2))\underbrace{\left\Arrowvert\frac{\partial\mathbf{x}}{\partial s_1}(s_1,s_2)\times\frac{\partial\mathbf{x}}{\partial s_2}(s_1,s_2)\right\Arrowvert ds_1 ds_2}_{da(\mathbf{x}(s_1,s_2))}.$$$$

I could re-write this using facts such as $$$$\frac{\partial\mathbf{x}}{\partial s_i}(s_1,s_2) = \mathbf{F}(\mathbf{X}(s_1,s_2))\cdot\frac{\partial\mathbf{X}}{\partial s_i}(s_1,s_2),$$$$ but that would change this to an integral over the reference configuration, which is not what you seek.

Do you have a parametrization of the reference configuration and some specific form for the kinds of deformations under consideration? Those details could make this discussion much more concrete.

• Thank you so much for all the details! The reference is only parametrized in terms of Euclidean space coordinates and time, that's all. The terms in the integral are: U(X) is a finite element test function associated with the fluid velocity V(X), and t is the fluid stress tensor. F have J are their usual meanings: Jacobian of the deformation, and its determinant respectively. Would I need to apply Piola transforms to each term within the integral? Commented May 26, 2020 at 17:54
• From your comment about $I_3$ I see that the integral is with respect to the reference. I need all integrals with respect to the current config. Would that mean I would simply integrate $(\frac{1}{J(X)})(u(x).v(x))$ with respect to x? I am still unsure how to approach the others, especially $I_1$ which has so many terms in various configurations for space and time. Commented May 26, 2020 at 17:57
• I am confused. t is the Cauchy stress tensor, which is symmetric and maps surface normal vectors in the def config to load/traction vectors in the def config. P = JtF^{-T} is the first Piola-Kirchhoff stress tensor, which maps unit normal vectors in the reference config to load/traction vectors in the deformed config. But in I_1, P is applied to N^k, and k indicates that N^k is a unit vector in the deformed config at time t_k. That's not how P works. Commented May 26, 2020 at 19:51
• I see. In that case, N has no time-dependence, so the superscript should not appear. Commented May 26, 2020 at 20:50
• There is still a problem: P.N is a traction/load vector in the deformed configuration, while N (on the left) is still a unit vector in the reference configuration. We can't have a dot product of N with P.N. Could the N on the left really be n (a t time t_k), so that this is the flux of traction through the deformed boundary? Commented May 26, 2020 at 21:32

BOTTOM LINE UP FRONT: Accounting can make a huge difference in the speed of understanding this material. For example, always note whether a function is a function on the reference configuration or on the deformed configuration. If you work with indices, raised and lowered indices and Einstein's summation convention are very helpful.

My references are Chapter 2 of Non-Linear Elastic Deformations by R. W. Ogden and the Continuum mechanics/Volume change and area change section of Wikiversity.

I will re-write some of the notation to make some expressions simpler. For example, I will use the sumbol $$\Phi_t$$ for the map that transforms a body (or volume of fluid) at time 0 to its configuration at time $$t$$. Then for each $$t$$, $$\Phi_t$$ is a (smooth?) map from one 3-dimensional manifold to another, and for fixed $$t$$ we write $$\mathbf{x} = \Phi_t(\mathbf{X})$$.

Let the reference configuration of the body/volume of fluid be $$\mathcal{B}$$, a subset of $$\mathbb{R}^3$$. This body consists of material points. At time $$t$$, those same material points occupy $$\Phi_t(\mathcal{B})$$ (also a subset of $$\mathbb{R}^3$$). $$\Phi_t$$ is a map from $$\mathcal{B}$$ onto $$\Phi_t(\mathcal{B})$$, but if $$\mathbf{T}$$ is a tensor field on $$\mathcal{B}$$, it is incorrect to claim that the corresponding field $$\mathbf{t}$$ on $$\Phi_t(\mathcal{B})$$ is $$\Phi_t(\mathbf{T})$$. It's not quite correct to claim that $$\mathbf{t}(\mathbf{x}) = \mathbf{T}(\Phi_t^{-1}(\mathbf{x}))$$, as one is constructed from vectors attached to the point $$\mathbf{x}\in\Phi_t(\mathcal{B})$$ while the other is constructed from vectors attached to the point $$\Phi^{-1}(\mathbf{x})\in\mathcal{B}$$.

There are different transformation rules for different kinds of tensor fields as well as for differential forms such as the volume form $$dV$$, area form $$d\mathbf{S}$$, and line form $$d\mathbf{\ell}$$.

Let $$F$$ be a smooth-enough real-valued function on $$\Phi_t(\mathcal{B})$$. How to express the integral of $$F$$ over material points $$\mathbf{x}\in\Phi_t(\mathcal{B})$$ as an integral over material points $$\mathbf{X}\in\mathcal{B}$$ (the reference configuration)? $$$$\int_{\Phi_t(\mathcal{B})}F(\mathbf{x})dv(\mathbf{x}) = \int_{\mathcal{B}}F(\Phi_t(\mathbf{X}))J(\mathbf{X})dV(\mathbf{X}),$$$$ where $$J$$ is the absolute value of the determinant of the deformation gradient $$\mathbf{F}$$: $$$$J ~=~ \left|\det\mathbf{F}\right| ~=~ \left|\det\left(\frac{\partial\mathbf{x}}{\partial\mathbf{X}}\right)\right|.$$$$

Note that we must integrate $$F\circ\Phi_t$$ (not $$F$$) over $$\mathcal{B}$$ because $$F$$ is not a function of the points $$\mathbf{X}\in\mathcal{B}$$. $$(F\circ\Phi_t)(\mathbf{X})$$ is equal to the value of $$F$$ at the "deformed" material point $$\mathbf{x} = \Phi_t(\mathbf{X})$$.

This shows the transformation of the volume form: $$dv \longleftrightarrow JdV$$. I don't write that they are equal because they are forms on different spaces. Many authors are not bothered by this technicality, but I think it is worth noting.

I am torn over whether to claim that $$J$$ is a function on $$\mathcal{B}$$ or on $$\Phi_t(\mathcal{B})$$, since $$\Phi_t$$ is a diffeomorphism between these two sets, allowing us to change our minds if we need to do so.

Note, however, I do like to write $$dV(\mathbf{X})$$ (as opposed to just $$dV$$) and $$dv(\mathbf{x})$$ (as opposed to just $$dv$$) for accounting. Another accounting principle will help remember the side of equation that has $$J$$. Consider dimensions in $$\mathcal{B}$$ and in $$\Phi_t(\mathcal{B})$$ as distinct. Then the dimensions of $$J$$ are $$$$\left[J\right] ~=~ \left[\det\left(\frac{\partial\mathbf{x}}{\partial\mathbf{X}}\right)\right] ~=~ \frac{\textrm{length^3 in \Phi_t(\mathcal{B})}}{\textrm{length^3 in \mathcal{B}}}.$$$$ Hence, $$J$$ has the same dimensions as the non-rigorous expression $$dv/dV$$, and we conclude $$dv \longleftrightarrow JdV$$.

The analogous transformation for the area form is called Nanson's formula: $$$$\mathbf{n}da \longleftrightarrow (\mathbf{F}^{-\mathsf{T}}\cdot\mathbf{N})dA,$$$$ where

• $$da$$ is the differential area in $$\Phi_t(\mathcal{B})$$,
• $$\mathbf{n}$$ is the unit normal vector attached to $$da$$,
• $$dA$$ is the differential area in $$\mathcal{B}$$,
• $$\mathbf{N}$$ is the unit normal vector attached to $$dA$$,
• $$\mathbf{F}^{-\mathsf{T}}$$ is the inverse of the transpose of $$\mathbf{F}$$.

It might be a good time to note that $$\mathbf{F}$$ is a two-point tensor field, and that $$\mathbf{F}(\mathbf{X})$$ maps from the tangent space at $$\mathbf{X}\in\mathcal{B}$$ to the tangent space at $$\Phi_t(\mathbf{X})\in\Phi_t(\mathcal{B})$$. In Cartesian co-ordinates, $$$$\mathbf{F} = \frac{\partial x^i}{\partial X^j}\mathbf{e}_i\otimes\mathbf{E}^j.$$$$ $$\mathbf{F}^{-1}$$ maps in the opposite direction, but $$\mathbf{F}^{-\mathsf{T}} = (\mathbf{F}^{-1})^{\mathsf{T}}$$ maps in the opposite direction from that$$^{\dagger}$$, so it maps from the tangent space at $$\mathbf{X}\in\mathcal{B}$$ to the tangent space at $$\Phi_t(\mathbf{X})\in\Phi_t(\mathcal{B})$$.

• $$\mathbf{F}(\mathbf{X}):(\textrm{tangent space at}~\mathbf{X}\in\mathcal{B})\to(\textrm{tangent space at}~\Phi_t(\mathbf{X})\in\Phi_t(\mathcal{B}))$$
• $$\mathbf{F}^{-1}(\mathbf{x}):(\textrm{tangent space at}~\mathbf{x}\in\Phi_t(\mathcal{B}))\to(\textrm{tangent space at}~\Phi_t^{-1}(\mathbf{x})\in\mathcal{B})$$
• $$\mathbf{F}^{-\mathsf{T}}(\mathbf{X}):(\textrm{tangent space at}~\mathbf{X}\in\mathcal{B})\to(\textrm{tangent space at}~\Phi_t(\mathbf{X})\in\Phi_t(\mathcal{B}))$$

For the sake of accounting, note that the units of a particular entry of $$\mathbf{F}$$ are $$$$\left[F^i_j\right] ~=~ \left[\frac{\partial x^i}{\partial X^j}\right] ~=~ \frac{\textrm{length in \Phi_t(\mathcal{B})}}{\textrm{length in \mathcal{B}}}.$$$$

$$\dagger$$ One might have guessed this by observing that the formula includes $$\mathbf{F}^{-\mathsf{T}}$$ acting a vector attached to a point of $$\mathcal{B}$$.

• Thank you for the detailed explanation! My apologies for my non-rigorous description of the problem. My issue is that in those 3 integrals, there is more than one map here, since we have functions defined on reference at previous time, defined on reference at current time, and defined on spatial at previous time. i.e. maps in displacement as well as in time. The change of variables formula for integrals, I haven't seen anywhere that it applies to multiple maps within the same integral. Commented May 26, 2020 at 1:01
• When integrating the functions over the surface with differential area dS, what is the set over which this integral is computed? Is it integrated over the reference configuration, over the deformed configuration at at time t_k or t_{k+1}, or over space? This is why I liked to add an argument to volume forms (e.g. dV(X) and dv(x)) and differential areas. Commented May 26, 2020 at 1:16
• Yes, the integrals $I_1$, $I_2$ and $I_3$ with differential area $dS$ are over the reference configuration i.e. $dS(X)$ Commented May 26, 2020 at 1:36
• Any idea, please? Commented May 26, 2020 at 16:12
• I found the solution to everything except I3 on page 143-144 of Gonzalez' "A first course in Continuum Mechanics". I just posted a screenshot in the original post. I will be marking your answer as accepted because of all your hard work and explanation! But if you can give a hint on how to approach I3 that would be great. (there is no normal in THAT surface integral) Commented May 27, 2020 at 9:48