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I am reading Evans' book Partial differential equations. but I am really curious about how he define the appropriate energy? Is there any principle or rule to do this things? Because I notice that although the energies he defined in this book are always contain kinetic energy and potential energy, they are always contain other terms. I mean the details are really different. For example, in Chapter 12 of his book on nonlinear wave equations. He defined the energy of equation $$ u_{tt}-\Delta u+f(u)=0 $$ as $$ E(t):=\int_{R^n}\frac{1}{2} (u_t^2+|Du|^2)+F(u) dx $$ where $$ F(z):=\int_{0}^{z}f(w)dw $$

However, He also define the energy of equation $$ u_{tt}-\Delta u+f(Du,u_t,u)=0 $$ as $$ E(t):=\frac{1}{2}\int_{R^n} u_t^2+|Du|^2+u^2 dx $$

I wonder why he didn't define the same energy (might not get the what we want?), or I should say why he defined the second energy just like the first one or vice versa.

Hoping your answers. Many thanks.

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  • $\begingroup$ It looks like in the second case, $f=u$, so $F(u)=\dfrac{1}{2}u^2$, which when substituted into the first equation gives the desired result. I.e., $$F(u)=\int^u_0wdw=\dfrac{1}{2}u^2$$ So, your equation 1 is $$E(t):=\int_\mathbb{R}^n\left(\dfrac{1}{2}\left(u_t^2+|Du|^2\right)+F(u)\right)dx$$ Substituting our earlier obtained result gives $$E(t):=\int_\mathbb{R}^n\left(\dfrac{1}{2}\left(u_t^2+|Du|^2\right)+\dfrac{1}{2}u^2\right)dx=\dfrac{1}{2}\int_\mathbb{R}^n\left(u_t^2+|Du|^2+u^2\right)dx$$ $\endgroup$
    – user122283
    Commented Mar 10, 2014 at 0:24
  • $\begingroup$ @SanathDevalapurkar: But why is that $f=u$? $\endgroup$
    – user99914
    Commented Mar 10, 2014 at 0:31
  • $\begingroup$ @John I don't know - it just seems like that should be the case by looking at the problem. It seems like the second energy of equation is a special case of the first. $\endgroup$
    – user122283
    Commented Mar 10, 2014 at 0:31
  • $\begingroup$ The second equation requires that f is a function of $Du,u_t,u$, so I don't think f could be $u$. Actually just because this $f$ depending on all $Du,u_t,u$ in the second case, he use the according energy form. I want to know how to choose a proper energy facing different kinds of PDEs. $\endgroup$
    – chuck
    Commented Mar 10, 2014 at 0:45

1 Answer 1

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Physical interpretation can be very helpful, but ultimately we need $E$ to be such that we can

  1. infer something about $dE/dt$ from the PDE
  2. infer what we want about the solution from $E$

The consideration usually begins with 1. Note that $u_tu_{tt} = \frac{1}{2}(u_t)^2$. So, if we have a hyperbolic PDE $u_{tt}=F(x,u,Du,D^2u)$, it is reasonable to multiply both sides by $u_t$ and try to see if a part of the right hand-side can also be turned into the $t$-derivative of something, possibly after integration by parts in the $x$ variable. This works for the Laplacian: $$ \int u_t \Delta u = - \int D_x(u_t)\, D_xu = \frac{d}{dt} \int \frac12 |Du|^2 $$ Also works for anything of the form $f(u)$, because $u_t f(u)= \frac{\partial }{\partial t} F(u)$ by the chain rule, if $F$ is an antiderivative of $f$.

However, the term $f(Du,u_t,u)$ is not so simple. A superficial reason is that there is no "antiderivative" $F$, as $f$ is a function of three variables. More substantial reason (again from physics) is that $f$ is a force that depends on derivatives $Du$ and $u_t$, and such a force is not conservative — it does not have an energy potential.

Evans includes $u^2$ in the short-time existence proof for the nonlinear wave equation (Theorem 3 in Chapter 12) not because it makes for a neat identity/inequality for $dE/dt$ (there isn't one) but because $\int u^2$ also needs to be estimated to prove the theorem (see item 2 above).

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  • $\begingroup$ Excellent answer! Thanks a lot! But in the proof of Theorem 3 (Chapter 12), there is still one thing I don't understand that is how to obtain $|F(Du,u_t,u)|\le C(|Du|+|u_t|+|u|)$, I can't see it. $\endgroup$
    – chuck
    Commented Mar 10, 2014 at 2:09

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