Weak form of the Navier-Cauchy equation

Question

I am trying to obtain the weak form of the Navier-Cauchy equation, which is

$$- \rho \omega ^2 \textbf{U} - \mu \nabla ^2 \textbf{U} - (\mu + \lambda) \nabla (\nabla \cdot \textbf{U}) = \textbf{F}$$

and can be written in the component form

$$-(2 \mu +\lambda) \frac{\partial ^2 U_1}{\partial x_1 ^2} - \mu \frac{\partial ^2 U_1}{\partial x_2 ^2} - (\mu + \lambda) \frac{\partial ^2 U_2}{\partial x_1 \partial x_2} - \rho \omega ^2 U_1 = F_1$$

$$-(2 \mu +\lambda) \frac{\partial ^2 U_2}{\partial x_2 ^2} - \mu \frac{\partial ^2 U_2}{\partial x_1 ^2} - (\mu + \lambda) \frac{\partial ^2 U_1}{\partial x_1 \partial x_2} - \rho \omega ^2 U_2 = F_2$$

The general procedure is to multiply the PDE by a test function $\textbf{v}$ in the space $\textbf{V}$, or $v$ in the space $V$, and integrate it over the domain $\Omega$. I will proceed with the component form, for I believe it is easier for me to understand. Setting $\textbf{F} = 0$ and rearranging the terms

$$-(2 \mu +\lambda) \int_\Omega v \left[ \frac{\partial ^2 U_1}{\partial x_1 ^2} + \frac{\partial ^2 U_2}{\partial x_2 ^2} \right]dxdy - \mu \int_\Omega v \left[ \frac{\partial ^2 U_1}{\partial x_2 ^2} + \frac{\partial ^2 U_2}{\partial x_1 ^2} \right]dxdy -(\mu + \lambda)\int_\Omega v \left[ \frac{\partial ^2 U_2}{\partial x_1 \partial x_2} + \frac{\partial ^2 U_1}{\partial x_1 \partial x_2} \right]dxdy - \rho \omega ^2 \int_\Omega v \left[ U_1+U_2 \right]dxdy = 0$$

From Green's theorem I know that $$ \int_{\Omega} \left(v \frac{\partial ^2 u}{\partial x ^2} \right)dxdy = \int_\Gamma \left(v \frac{\partial u}{\partial x} \hat{n}_x \right)ds - \int_{\Omega} \left( \frac{\partial v}{\partial x} \frac{\partial u}{\partial x} \right)dxdy$$

Which is sufficient to deal with the first and second integrals. However, I do not know how to proceed with the cross derivatives $\partial ^2 / \partial x_1 \partial x_2$ of the third integral. Can someone help me with this?

Answer 1

The identity you're missing from Gauss' divergence theorem is:

$$ \int_\Omega \nabla \varphi \cdot\mathbf{v} = -\int_\Omega \varphi\nabla\cdot\mathbf{v} +\int_{\partial\Omega}\varphi\mathbf{v\cdot n} $$

where I've written $\varphi$ as an arbitrary scalar field. So, using the divergence of $\mathbf{u}$ as the scalar field you'd get

$$ -\int_\Omega(\lambda+\mu) \nabla(\nabla\cdot\mathbf{u}) \cdot\mathbf{v} = \int_\Omega (\lambda+\mu)(\nabla\cdot\mathbf{u})\nabla\cdot\mathbf{v} -\int_{\partial\Omega}(\lambda+\mu)(\nabla\cdot\mathbf{u})\mathbf{v\cdot n} $$

and you can complete your weak formulation.

---

Anyway, here's in a few more steps:

Note the divergence of the product (scalar*vector) $$\nabla\cdot(\varphi\mathbf{v})=\nabla\varphi\cdot\mathbf{v}+\varphi\nabla\cdot\mathbf{v}$$ Rearrange to get $$\nabla\varphi\cdot\mathbf{v}=\nabla\cdot(\varphi\mathbf{v})-\varphi\nabla\cdot\mathbf{v}$$ And plug it into that integral $$\int_\Omega\nabla\varphi\cdot\mathbf{v} = \int_\Omega\nabla\cdot(\varphi\mathbf{v})-\int_\Omega\varphi\nabla\cdot\mathbf{v}$$ Apply Gauss' divergence theorem for vector fields in that second integral $$\int_\Omega \nabla\cdot\mathbf{v}=\int_{\partial\Omega}\mathbf{v\cdot n} \quad\Rightarrow\quad\int_\Omega \nabla\cdot\mathbf{\varphi v}=\int_{\partial\Omega}\varphi\mathbf{v\cdot n} \qquad\Rightarrow$$

$$\int_\Omega\nabla\varphi\cdot\mathbf{v} = \int_{\partial\Omega}\varphi\mathbf{v\cdot n}-\int_\Omega\varphi\nabla\cdot\mathbf{v}$$ Remember that $\varphi=\nabla\cdot\mathbf{u}$, enter Lamé's parameters, and voila: $$(\lambda+\mu)\int_\Omega\nabla(\nabla\cdot\mathbf{u})\cdot\mathbf{v} =(\lambda+\mu)\left( \int_{\partial\Omega}(\nabla\cdot\mathbf{u})\mathbf{v\cdot n}-\int_\Omega(\nabla\cdot\mathbf{u})\nabla\cdot\mathbf{v}\right)$$

Answer 2

The general form of the equation is $$ \frac{\partial \sigma_{ij}}{\partial x_j} + F_i = \rho \frac{\partial^2 U_i}{\partial t^2} $$ where the stress is given by $$ \sigma_{ij} = \sigma_{ij}(U) = 2 \mu \varepsilon_{ij} + \lambda \varepsilon_{kk} \delta_{ij}, \qquad \varepsilon_{ij} = \varepsilon_{ij}(U) = \frac{1}{2}\left( \frac{\partial U_i}{\partial x_j} + \frac{\partial U_j}{\partial x_i}\right) $$ We are using the Einstein summation convention. It is better to derive the weak form here.

If $V_i$ is the test function $$ \int_\Omega V_i \frac{\partial \sigma_{ij}(U)}{\partial x_j} dx = \int_{\partial\Omega} V_i \sigma_{ij}(U) n_i ds - \int_\Omega \sigma_{ij}(U) \frac{\partial V_i}{\partial x_j} dx $$ In this equation, we have summation over both indices i and j. Since $\sigma$ is symmetric tensor, you can show that $$ \sigma_{ij}(U) \frac{\partial V_i}{\partial x_j} = \sigma_{ij}(U) \varepsilon_{ij}(V) $$ Hence you can use this form $$ \int_\Omega V_i \frac{\partial \sigma_{ij}(U)}{\partial x_j} dx = \int_{\partial\Omega} V_i \sigma_{ij}(U) n_i ds - \int_\Omega \sigma_{ij}(U) \varepsilon_{ij}(V) dx $$ The mathematical analysis of the weak formulation should be done in many books, e.g.,

S. Kesavan, Topics in Functional Analysis and Applications, Section 3.2.4