I have the following PDE,

$$\tau u_{ttt} + \alpha u_{tt} - c^2 \Delta u - b\Delta u_t = 0$$

and I am trying to prove that $I(u) = E_1(u) + E_2(u)$ where

$$E_1(u) = \frac{b}{2}|\nabla(u_t+c^2b^{-1}u)|^2 + \frac{\tau}{2}|u_{tt}+c^2b^{-1}u_t|^2 + \frac{c^2\gamma}{2b}|u_t|^2$$


$$E_2(u) = \frac{\alpha}{2}|u_t|^2 + \frac{c^2}{2}|\nabla u|^2$$

is the Energy Functional for the above PDE. Here $\gamma = \alpha - \frac{c^2\tau}{2}$.

I tried to verify this by computing the first variation $\delta I$ of this functional but, unless I have made some mistake, it is not getting me what I need. I also tried to verify it by calculating direct the Euler-Lagrange equation but I could not figure out what were the variables I had to work with.

What I need is to show that the given PDE is the Euler-Lagrange equation for the given functional.

I have computated the first variation of $I$ and what I got was

$$\delta I(u) = -b(\Delta u_t,\phi_t) - c^2(\Delta u_t,\phi_t) - c^2(\Delta u,\phi_t) -\frac{c^2}{b}(c^2+b)(\Delta u,\phi) + \tau (u_{tt},\phi_{tt}) + \frac{\tau c^2 }{ b} (u_{tt},\phi_t) + \frac{\tau c^2 }{ b}(\phi_{tt},u_t) + \frac{\alpha }{b}(b+c^2) (u_t,\phi_t). $$

But now, somehow I have to prove that

$$\delta I(u) = (\tau u_{ttt} + \alpha u_{tt} - c^2 \Delta u - b\Delta u_t,\phi)$$

because then I will be able to use the Fundamental Lemma of the Calculus of Variations. I think I can also use it if I write the formula with $\phi_t$ instead of $\phi$, but I could not get what I need with neither of them because if I try with $\phi$ I have a term $\Delta u_{tt}$ which is not a part of my equation and if I try with $\phi_t$ the term $u_{tt}$ is canceled out. I am afraid I am messing up with the integrations by parts...

Does anybody has a hint about how to do that? You guys dont need to give me an exact answer, only give me a hint about how to verify it!

Thank you very much!


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