This is from PDE Evans, 2nd edition: Chapter 8, Exercise 3:

The elliptic regularization of the heat equation is the PDE $$ u_t - \Delta u -\epsilon u_{tt}=0 \quad \text{in }U_T, \tag{$*$}$$ where $\epsilon > 0$ and $U_T = U \times (0,t]$. Show that $(*)$ is the Euler-Lagrange equation corresponding to an energy functional $I_\epsilon[w] := \iint_{U_T} L_\epsilon(Dw,w_t,w,x,t) \, dx \, dt$.

(Hint: Look for a Lagrangian with an exponential term involving $t$.)

In my previous question with a different EL equation, I had to find a Lagrangian in terms of only three variables (that is, find $L=L(Du,u,x)$).

Now for this problem, I have to find a Lagrangian that is based with five variables and epsilon (that is, find $L=L_\epsilon(Du,u_t,u,x,t)$).

When the hint says "look for a Lagrangian with an exponential term involving $t$, do they mean something like $$e^{-t}(|Du|^2 +\epsilon u_t)$$ or similar?

Now, I got confused again because, unlike my previous question, for this one I am asked to work with the time variable $t$ sin addition to the spatial variable $x$. If the exponential term involving $t$ is indeed $e^{-t}$ times something, then I'm assuming upon differentiation with respect to $t$, product rule will be used to find our way back to the given EL equation.

If any answerers have any ideas, a good hint works best for me.


2 Answers 2


Edit2:Let's start with the following energy functional $$ I(u) =\int_{U \times (0,T]} \frac{e^{- t/\epsilon}}{2} ( | D u|^2 + \epsilon | u_t | ^2 ) dx dt $$ Recall that the Euler-Lagrange Equation is just an critical solution to the first variation $\delta I$. The first variation is defined as $$ \delta I = \lim_{h \to 0}\frac{ I( u + h \xi ) - I (u) }{h} $$ and a critical solution is $u_*$ s.t. $\delta I(u_*) =0$ for all test functions $\xi$.This amounts to finding the first order term in $h$. The expansion computation goes as follows: $$ |D ( u + h \xi ) |^2 = | Du + h D \xi |^2 = (Du + h D \xi) \cdot (D u + h D \xi ) =|Du|^2 + 2h Du \cdot D \xi + h^2 |D\xi|^2 $$ Then similarly
$$ |\frac{d}{dt} ( u + h \xi ) |^2 = |u_t + h \xi _t |^2 = |u_t|^2 + 2h u_t \cdot \xi_t + h^2 |D \xi|^2 $$ Thus plugging these two expressions into the variation formula we see $$ \delta I = \int_{U \times (0,T]} e^{- t /\epsilon} ( Du \cdot D \xi + \epsilon \cdot u_t \cdot \xi_t) dx dt $$ Recall the integration by parts formula for space: $$ \int_U D u \cdot D \xi dx = \int_{ \partial U} \underbrace{\xi Du \cdot n dS}_{\xi \equiv 0} - \int_U \xi \Delta u d x $$ where $n$ is the normal on the boundary. Now due to the product structure of the measure and Fubini's Theorem we have $$ \int_{U\times (0,T] }e^{- t /\epsilon} ( Du \cdot D \xi) dx dt = \int_{(0,T]} e^{-t / \epsilon } \left ( \int_U Du \cdot D \xi dx \right ) dt =-\int_{(0,T]} e^{-t / \epsilon } \left ( \int_U \xi \Delta u dx \right ) dt = -\int_{U\times (0,T] }e^{-t / \epsilon }\xi \Delta u dx dt$$ Now recall integration by parts in one variable $$ \int _a^b f g'dt = f g \Big |^b_a - \int_a^b f' g dt $$ Here we have $f = \epsilon e^{- t / \epsilon } u_t $ and $g = \xi$. Thus we see a similar thing happens to the 2nd term. $$ \int_{(0,T]} \epsilon e^{- t / \epsilon} u_t \xi _t dt = -\int_{(0,T]} \epsilon \xi \frac{d}{dt} (e^{- t / \epsilon} u_t) dt = \int_{(0,T]} e^{-t / \epsilon} \xi (u_t - \epsilon u_{tt}) dt $$ Thus we see $$ \delta I = \int_{U \times (0,T] } \xi e^{- t / \epsilon } ( - \Delta u +u_t - \epsilon u_{tt} ) dx dt $$ Now since we need $\delta I =0$ for all $\xi$, this implies that $$- \Delta u +u_t - \epsilon u_{tt} = 0$$ since the exponential is never zero.

Original Post: Well, my go to approach is to take the inner product against the solution. i.e. if $\mathcal{L}(u) =u_t - \Delta u - \epsilon u_{tt}$, then we look at $$(\mathcal{L}(u),u) = \int_{U_t} \mathcal{L}(u) u dx dt = \int_{U_t} (u_t u - u\Delta u - \epsilon u_{tt} u)dx dt $$ (space and time!). If we integrate by parts in space on the middle guy and time on the last guy ( noticing the derivative of $t$ on the magnitude on the first guy) we see it should look something like: $$ (u_t u - u\Delta u - \epsilon u_{tt} u) \approx \frac{1}{2}\frac{d}{dt} |u|^2 + \frac{1}{2} |Du|^2 + \frac{1}{2} \epsilon | u_t|^2 $$

See what happens if you try $$ L(u) = e^{-c(\epsilon)t}\left (\frac{1}{2}|Du|^2 + \frac{1}{2} \epsilon | u_t|^2 \right ) $$ The $e^{-c(\epsilon)t}$ will give us a product rule situation without affecting the derivative term.

Hint: When we perform a variation, we're solving $$ \delta I = \lim_{h \to 0} \frac{ I( u + h \xi ) - I ( u) }{h}$$ for any suitable test function $\xi$(notably dies on boundary). This amounts to finding first order terms in $h$ from the perturbation. i.e. $$L( u + h \xi ) = e^{- c( \epsilon) t } \left ( \frac{1}{2} | D ( u + h \xi )|^2 + \frac{1}{2} \epsilon | u_t + h \xi_t |^2 \right) = L(u) + h [e^{ -c(\epsilon)t} \left( D u \cdot D \xi + \epsilon u_t \cdot \xi_t\right) ] + \mathcal{O}(h^2)$$ Thus we see $$ \delta I = \int_{U_t} e^{ -c(\epsilon)t} \left( D u \cdot D \xi + \epsilon u_t \cdot \xi_t \right) dx dt $$ Now perform the integration by parts I mentioned in the motivation and match $c( \epsilon)$ to the equation you want. You should end up with $$\delta I = \int_{U_t} e^{-c(\epsilon)t} (- \Delta u + \epsilon c(\epsilon) u_t - \epsilon u_{tt} )\cdot \xi dx dt $$ Thus we see $u$ is critical if $$- \Delta u + \epsilon c(\epsilon) u_t - \epsilon u_{tt} =0$$ by the fundamental lemma of variational calculus.

  • $\begingroup$ what you've written doesn't make sense.... have a look at the hint I've added $\endgroup$
    – Jeb
    Jun 9, 2015 at 18:35
  • $\begingroup$ I'm guessing we must have $c(\epsilon)=\frac 1{\epsilon}$ to match up with the heat equation given in the problem. $\endgroup$
    – Cookie
    Jun 9, 2015 at 18:45
  • $\begingroup$ Exactly. So the Lagrangian you wanted was very close to Daniela's suggestion. $\endgroup$
    – Jeb
    Jun 9, 2015 at 18:46
  • $\begingroup$ What part would you like me to explain? The integration by parts ? $\endgroup$
    – Jeb
    Jun 10, 2015 at 12:36
  • $\begingroup$ Yes, and this time, if possible, please provide a complete solution. I would award you the second bounty if you do, provided I look over and understand it thoroughly myself. (Also, I am accustomed to Evans' textbook notation, and your notation is substantially different, which is partly why I became confused again.) $\endgroup$
    – Cookie
    Jun 10, 2015 at 18:12

I can't comment, so I am posting a solution.

This is based in 5 variables, but the variable $t$ is, like more one variable in $x$, this is, we can think $x \in \mathbb{R}^{n+1}$.

So the Euler-Lagrange equation has the same form that when we work with only 3 variables.

If you consider $L(p,q,z,x,t) = ( \frac{1}{2} |p|^2 + \frac{1}{2} \epsilon q^2) e^{-\frac{1}{\epsilon} t}$, I think you'll have the answer.

P.S: If I am wrong, say me please. I'm studying this chapter in this moment too.

  • $\begingroup$ Your guess isn't quite right. $\endgroup$
    – Jeb
    Jun 9, 2015 at 19:52
  • $\begingroup$ why? My answer was the same that yours. $\endgroup$
    – Daniela
    Jun 10, 2015 at 7:17
  • $\begingroup$ You have the wrong sign on the second term $\endgroup$
    – Jeb
    Jun 10, 2015 at 12:33
  • $\begingroup$ @Jeb, you are right. Thanks. $\endgroup$
    – Daniela
    Jun 10, 2015 at 19:31
  • $\begingroup$ @lesguimauves I don't know how to justify it well either. Sorry I can't help you. I only thought about the problem, and I tried to obtain a answer. $\endgroup$
    – Daniela
    Jun 10, 2015 at 19:36

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