Computational Science Asked by Lucas Vieira on October 24, 2020
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?
The identity you're missing from Gauss' divergence theorem is:
$$ int_Omega nabla varphi cdotmathbf{v} = -int_Omega varphinablacdotmathbf{v} +int_{partialOmega}varphimathbf{vcdot n} $$
where I've written $varphi$ as an arbitrary scalar field. So, using the divergence of $mathbf{u}$ as the scalar field you'd get
$$ -int_Omega(lambda+mu) nabla(nablacdotmathbf{u}) cdotmathbf{v} = int_Omega (lambda+mu)(nablacdotmathbf{u})nablacdotmathbf{v} -int_{partialOmega}(lambda+mu)(nablacdotmathbf{u})mathbf{vcdot n} $$
and you can complete your weak formulation.
Note the divergence of the product (scalar*vector) $$nablacdot(varphimathbf{v})=nablavarphicdotmathbf{v}+varphinablacdotmathbf{v}$$ Rearrange to get $$nablavarphicdotmathbf{v}=nablacdot(varphimathbf{v})-varphinablacdotmathbf{v}$$ And plug it into that integral $$int_Omeganablavarphicdotmathbf{v} = int_Omeganablacdot(varphimathbf{v})-int_Omegavarphinablacdotmathbf{v}$$ Apply Gauss' divergence theorem for vector fields in that second integral $$int_Omega nablacdotmathbf{v}=int_{partialOmega}mathbf{vcdot n} quadRightarrowquadint_Omega nablacdotmathbf{varphi v}=int_{partialOmega}varphimathbf{vcdot n} qquadRightarrow$$
$$int_Omeganablavarphicdotmathbf{v} = int_{partialOmega}varphimathbf{vcdot n}-int_Omegavarphinablacdotmathbf{v}$$ Remember that $varphi=nablacdotmathbf{u}$, enter Lamé's parameters, and voila: $$(lambda+mu)int_Omeganabla(nablacdotmathbf{u})cdotmathbf{v} =(lambda+mu)left( int_{partialOmega}(nablacdotmathbf{u})mathbf{vcdot n}-int_Omega(nablacdotmathbf{u})nablacdotmathbf{v}right)$$
Correct answer by lima on October 24, 2020
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 $$ int_Omega V_i frac{partial sigma_{ij}(U)}{partial x_j} dx = int_{partialOmega} V_i sigma_{ij}(U) n_i ds - int_Omega sigma_{ij}(U) frac{partial V_i}{partial x_j} dx $$ 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 $$ int_Omega V_i frac{partial sigma_{ij}(U)}{partial x_j} dx = int_{partialOmega} V_i sigma_{ij}(U) n_i ds - int_Omega sigma_{ij}(U) varepsilon_{ij}(V) dx $$ 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
Answered by cfdlab on October 24, 2020
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