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.

  • $\begingroup$ 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 Mar 10 '14 at 0:24
  • $\begingroup$ @SanathDevalapurkar: But why is that $f=u$? $\endgroup$ – user99914 Mar 10 '14 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 Mar 10 '14 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 Mar 10 '14 at 0:45

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).

  • $\begingroup$ 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 Mar 10 '14 at 2:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.