Let $u_1,u_2 : (0,1)\to \mathbb{R}$, and given that $$u_1(0) = u_1(1),u_1'(0)= u_1'(1)$$ and $$u_2(0) = u_2(1),u_2'(0)=u_2'(1)$$ and also $$(|u_1'|u_1')' = \lambda_1|u_1|u_1$$ and $$(|u_2'|u_2')' = \lambda_2|u_2|u_2$$ $\lambda_1\ne\lambda_2,\lambda_1,\lambda_2 \in (0,\infty)$ . I need to prove/disprove that $\int\limits_0^1|u_1|u_1u_2dx + \int\limits_0^1|u_2|u_2u_1dx =0 $
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$\begingroup$ Noticing that $|u|u=\pm u^2$, you can solve by quadrature your equations by incorporating the $\pm$ sign in the $\lambda$ on intervals over which the respective function $u$ and its derivative don't vanish. You obtain something defined implicitly as $F(u(x))=d+x$ where $\frac{dF}{du}=(c+0.5\lambda u^3)^{-1/3}$ for two constants $c, d$ whose value is given by the initial values you provide. Don't know if that helps, though. $\endgroup$– Loïc TeyssierCommented Sep 19, 2017 at 20:09
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$\begingroup$ @LoïcTeyssier : Agree. Can you please look at my attempt below. $\endgroup$– Rajesh DCommented Sep 21, 2017 at 2:56
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Let $$I = \int\limits_0^1|u_1|u_1u_2dx + \int\limits_0^1|u_2|u_2u_1dx$$
Let $$I_1 = \int_0^1|u_1|u_1u_2dx$$ Integrating by parts, $$I_1 = u_2\int|u_1|u_1 \Big|_0^1 - \int_0^1u_2'\int|u_1|u_1dx$$ using the ODE relation on $u_1$ and boundary conditions,$$I_1 = \frac{1}{\lambda_1}u_2|u_1'|u_1'\Big|_0^1 - \frac{1}{\lambda_1}\int_0^1 u_2'|u_1'|u_1' = -\frac{1}{\lambda_1}\int u_2'|u_1'|u_1'dx$$
Using Similarity $$I = I_1+I_2 = -\int_0^1u_1'u_2'(\frac{|u_1'|}{\lambda_1}+\frac{|u_2'|}{\lambda_2})dx$$
Need some help, how to proceed from here...