Euler Lagrange equation assuming once differentiability

Question

Let $$S=\int_0^T L(\varphi(t),\varphi'(t))dt$$

Assume that $\varphi(0)=0$ is the only constraint. Assume that $\varphi$ is once differentiable but not twice. Is there an equivalent of Euler Lagrange equation for minimizing $S$? I know if we have both endpoints and twice differentiability, we get EL equation. But what about this more general situation?

Answer 1

In case where $\varphi$ is not twice differentiable, what we have is that for all compactly supported smooth function $\eta$ on $(0,T)$, we have

$$ \frac{d}{d\epsilon } \bigg|_{\epsilon=0} \int _0^T L(\varphi(t)+ \epsilon \eta(t),\varphi'(t) + \epsilon \eta'(t))\mathrm dt =0.$$

Using Chain rule we obtain

$$ \int_0^T \left( \frac{\partial L}{\partial \varphi} ( \varphi, \varphi') \eta + \frac{\partial L}{\partial \varphi'} ( \varphi, \varphi') \eta'\right) \mathrm dt = 0, $$

or $$\tag{1} -\int_0^T \frac{\partial L}{\partial \varphi} ( \varphi, \varphi') \eta \mathrm d t = \int_0^T \frac{\partial L}{\partial \varphi'} ( \varphi, \varphi') \eta'\mathrm dt. $$

For a nice enough $L$ (Say, when $L$ is $C^1$), we can assume that both $$ \frac{\partial L}{\partial \varphi} ( \varphi, \varphi'),\ \ \ \frac{\partial L}{\partial \varphi'} ( \varphi, \varphi') $$ are continuous. In this case, (1) is another way to say that $ \frac{\partial L}{\partial \varphi'} ( \varphi, \varphi') $ is weakly differentiable with derivative $\frac{\partial L}{\partial \varphi} ( \varphi, \varphi')$.

Thus $\frac{\partial L}{\partial \varphi'} ( \varphi, \varphi')$ has a continuous weak derivative. In particular, it belongs to $W^{1, \infty}(0,T)$ and is a Lipschitz function. In particular, it is differentiable almost everywhere.

Then the conclusion is that one still have the Euler-Language equation

$$ \frac{d}{dt} \frac{\partial L}{\partial \varphi'} ( \varphi, \varphi') = \frac{\partial L}{\partial \varphi} ( \varphi, \varphi')$$

which is satisfied almost everywhere.

Answer 2

You can do the derivation of the Euler Lagrange equations to develop a similar condition. We want to minimize

$$S(\varphi)=\int_0^TL(\varphi(t),\varphi'(t))dt$$

subject to $\varphi(0)=0$. We add a deviation term $\eta$ which satisfies $\eta(0)=0$ and consider

$$f(\epsilon)=S(\varphi+\epsilon\eta)=\int_0^TL(\varphi+\epsilon\eta,\varphi'+\epsilon\eta')dt$$

We treat $L$ as a function of $u$ and $v$ to avoid notational confusion. If $\varphi$ is the minimum, then we have

$$f'(\epsilon)=\int_0^T\eta L_u(\varphi+\epsilon\eta,\varphi'+\epsilon\eta')dt+\int_0^T\eta'L_v(\varphi+\epsilon\eta,\varphi'+\epsilon\eta')dt$$

The minimum occurs then $\epsilon=0$, so we have

$$0=f'(0)=\int_0^T\left(\eta\frac{\partial L}{\partial\varphi}+\eta'\frac{\partial L}{\partial\varphi'}\right)dt$$

From here there is essentially nothing we can do to simplify this without employing more differentiability. A minimal $\varphi$ solves the equation

$$\int_0^T\left(\eta\frac{\partial L}{\partial\varphi}+\eta'\frac{\partial L}{\partial\varphi'}\right)dt=0$$

for any $\eta$ with $\eta(0)=0$.

Answer 3

FWIW, the main issue is that there is no final condition, only an initial condition. This means that OP's variational problem generically is not bounded from below nor above, even if we are only allowed to use $C^{\infty}$-configurations $\varphi$, i.e. the problem is generically under-constrained.

Answer 4

If $L$ is $C^1$ then as Josh. B. already answered that we have that the weak Euler Lagrange equation is satisfied namely: $$ \int_0^T L_\varphi(\varphi, \varphi') \eta' dt = - \int_0^T L_{\varphi'}(\varphi, \varphi') \eta dt$$ holds for all $\eta \in C^1_c$. As he also noted the strong equation is satisfied almost everywhere. Oddly enough this means that the strong Euler Lagrange equation is satisfied everywhere. Why?

Let us take two continuous functions $f$, $g$ satisfying $$ \int_0^T f \eta' dt = - \int_0^T g \eta dt, $$ for all $\eta \in C^1_c$. Define $$G(t) := \int_0^tg(x) dx $$ which is in $C^1$. By partial differentiation you get $$ \int_0^T f \eta' dt = \int_0^T G \eta' dt.$$ But this means by the Du Bois-Reymond lemma that $G - f = c$ for a constant $c$ which in turn implies $f \in C^1$. Applying this to our case we get exactly the strong Euler Lagrange equation.

Alternatively one can argue over the property of weakly differentiable functions in 1D which are exactly the functions where the fundamental theorem of calculus holds. This is the same line of thought, but if you do not know Sobolev spaces or absolutly continuous functions the explanation above is probably more satisfying.

This conclusion is very interesting since it tells us that although $L_{\varphi'}$ does not need to be $C^1$ the composition $L_{\varphi'} \circ (\varphi, \varphi')$ actually is at least once continuous differentiable.