# Integration by parts with cross derivatives to obtain the weak form [duplicate]

I’m trying to write the weak form of the Navier-Cauchy equation in the component form, where $$u_1$$ and $$u_2$$ are the displacement components:

$$-(2 \mu +\lambda) \frac{\partial ^2 u_1}{\partial x_1 ^2} - \mu \frac{\partial ^2 u_1}{\partial x_2 ^2} - (\mu + \lambda) \frac{\partial ^2 u_2}{\partial x_1 \partial x_2} - \rho \omega ^2 u_1 = 0$$

$$-(2 \mu +\lambda) \frac{\partial ^2 u_2}{\partial x_1 ^2} - \mu \frac{\partial ^2 u_2}{\partial x_2 ^2} - (\mu + \lambda) \frac{\partial ^2 u_1}{\partial x_1 \partial x_2} - \rho \omega ^2 u_2 = 0$$

The genral procedure is to multiply these equations by a test function $$q$$ and integrate them over the domain $$\Omega$$. Using Integration by parts and Green's Theorem I believe I can do this. However, I'm getting stuck in the term with the cross derivatives. Can someone please explain how to proceed?

• Your second equation is incorrect, it should read $\iint(v \frac{\partial ^2 u}{\partial x ^2})dxdy = \int (v \frac{\partial u}{\partial x} \hat{n}_x)ds - \iint (\frac{\partial v}{\partial x} \frac{\partial u}{\partial x})dxdy$. Your third equation is also incorrect. – cfdlab Jul 1 at 4:25
• Yes, I've noticed that afterwards, thank you. – Lucas Vieira Jul 1 at 22:53
• I have added an answer in your other question. You can delete this question since it is a duplicate. – cfdlab Jul 4 at 4:03
• I'd like to, but I'm not allowed to do so since there are answers to the question. Could you first delete your answer and then I'll delete my post? – Lucas Vieira Jul 4 at 18:00

The given $$\int_\Omega v \frac{\partial}{\partial x}\left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\right) dx dy = 0$$ becomes $$\int_\Omega \frac{\partial}{\partial x}\left[v\left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\right) \right] dx dy - \int_\Omega \frac{\partial v}{\partial x}\left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\right) dx dy = 0$$ This follows from basic differential calculus $$f = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}, \qquad v \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(vf) - \frac{\partial v}{\partial x} f$$ Converting first integral to a surface integral, we get $$\int_{\partial \Omega} v \left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\right) n_x ds - \int_\Omega \frac{\partial v}{\partial x}\left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y}\right) dx dy = 0$$ But where did you get such an equation $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial x \partial y} = 0$$ which is not elliptic. It can be written as $$\nabla \cdot (A \nabla u) = 0, \qquad A = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$$ but the matrix $$A$$ is not positive definite.

Under some conditions you may be able to integrate it once $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = f(y)$$ which is a hyperbolic equation. You can solve this using method of characteristics.

• Your first equation is correct. In your second equation, v appears on the right side of ∂/∂x, which does not match the question posed by the OP. – Charlie S Jun 30 at 8:44
• It follows from basic differential calculus, see my updated answer. – cfdlab Jul 1 at 4:07
• Thank you @cfdlab. I added more details to my question in the hope it will clarify where I got this equation from. – Lucas Vieira Jul 1 at 23:11
• Have you deleted your answer @CharlieS ? There was something I wanted to check out one more time. – Lucas Vieira Jul 1 at 23:25

I'm trying to obtain the weak form of the Navier-Cauchy equation in the component form

$$-(2 \mu +\lambda) \frac{\partial ^2 u_1}{\partial x_1 ^2} - \mu \frac{\partial ^2 u_1}{\partial x_2 ^2} - (\mu + \lambda) \frac{\partial ^2 u_2}{\partial x_1 \partial x_2} - \rho \omega ^2 u_1 = 0$$

$$-(2 \mu +\lambda) \frac{\partial ^2 u_2}{\partial x_1 ^2} - \mu \frac{\partial ^2 u_2}{\partial x_2 ^2} - (\mu + \lambda) \frac{\partial ^2 u_1}{\partial x_1 \partial x_2} - \rho \omega ^2 u_2 = 0$$

What I did was multiply these equations by a test function $$q$$ and integrated them over the domain, but I got stuck in the $$\frac{\partial ^2 u}{\partial x \partial y}$$ part. Now I believe I can continue with it. I'm not sure where the non elliptic equation issue enters to it though.

• Can I suggest that you delete the contents of your first question and post this instead. – cfdlab Jul 2 at 10:33
• Lucas, I previously assumed $u$ was a vector quantity which created some commotion. Admittedly, if $u$ wasn't a vector, then my answer was totally invalid. Here is the wiki link I used previously: en.wikipedia.org/wiki/… – Charlie S Jul 2 at 14:16
• Yes of course @cfdlab, it's done. I'd be very thankful if you could give an answer to the problem in the referred context. – Lucas Vieira Jul 2 at 20:34