Euler-Lagrange equation for continuous systems I'm having a little trouble with wrapping my head around a part of a method which is fairly 'new' in some fashions to me.  I imagine it should be fairly obvious, but I am not seeing something at the moment.
Let's take the Lagrangian density to be 
$$\mathcal{ L}={1\over{2}}\left(\mu\dot{\phi}^2-Y(\partial_x\phi)^2\right),$$
where  $\mu$ is mass density of rod and $Y$ is Young's modulus.
One "should" be able to put this into the equation
$${\partial\over \partial t}\left({\partial \mathcal{L}\over{\partial(\partial_t \phi)}}\right)+{\partial\over \partial x}\left({\partial \mathcal{L}\over{\partial(\partial_x \phi)}}\right)={\partial \mathcal{L}\over{\partial \phi}}$$
to get the equation of motion for the rod 
$$\mu\ddot\phi=Y\partial_{xx}\phi,$$
 right?
The trouble is I'm not exactly seeing where to go with the LHS of the E-L equation applied to $\mathcal L$ at the moment.  I'm hoping someone would be so kind as to demonstrate that in a few lines.
 A: Comment to the question (v6): For field theory in 1+1 dimensions, the Lagrangian density 
$$\tag{1} {\cal L}\left(\phi(x,t), \partial_t\phi(x,t),\partial_x\phi(x,t), x,t\right)$$ depends on 5 arguments
$$\tag{2} \phi(x,t),~ \partial_t\phi(x,t),~\partial_x\phi(x,t),~ x,~t,$$
if we exclude theories with higher derivatives.
It should be stressed that the outer differentiations in the corresponding Euler-Lagrange equations 
$$\tag{3} \frac{d}{d t}\frac{\partial {\cal L}}{\partial(\partial_t \phi)}
+\frac{d}{d x}\frac{\partial {\cal L}}{\partial(\partial_x \phi)}
~=~\frac{\partial {\cal L}}{\partial \phi}$$
involve the total derivatives 
$$\tag{4}\frac{d}{d t}~=~\frac{\partial}{\partial t}+ \partial_t\phi~\frac{\partial}{\partial \phi}+\partial^2_t \phi~\frac{\partial}{\partial(\partial_t \phi)}+\partial_t\partial_x \phi~\frac{\partial}{\partial(\partial_x \phi)}$$
and 
$$\tag{5}\frac{d}{d x}~=~\frac{\partial}{\partial x}+ \partial_x\phi~\frac{\partial}{\partial \phi}+\partial^2_x \phi~\frac{\partial}{\partial(\partial_x \phi)}+\partial_x\partial_t \phi~\frac{\partial}{\partial(\partial_t \phi)}$$
rather than the partial/explicit derivatives
$$\tag{6} \frac{\partial}{\partial t} \quad\text{and}\quad \frac{\partial}{\partial x} .$$
This follows from the derivation of the Euler-Lagrange equations via the stationary action principle, cf. e.g. Ref. 1.
Finally, it should be mentioned that many authors confusingly typeset the outer differentiations in Euler-Lagrange equations (3) as partial derivatives (6). Such notation does not change the fact that they should be understood as total derivatives.
References:


*

*Herbert Goldstein, Classical Mechanics; last chapter.

A: When doing field theory, we work with a Lagrangian density instead of a Lagrangian:
$$S=\int dt L,\hspace{.5cm} L=\int d^3x \mathcal{L}\longrightarrow S=\int d^4x \mathcal{L}$$
The corresponding Euler-Lagrange equations are easily derived in a fashion exactly analogous to the derivation of the Euler-Lagrange equations in e.g. classical mechanics. The result is:
$$\partial_\mu \frac{\partial \mathcal{L} }{\partial (\partial_\mu \phi)}=\frac{\partial \mathcal{L}}{\partial \phi}$$
where I have defined $$\partial_\mu=\frac{\partial}{\partial x^\mu},\hspace{1cm}x^\mu=(t,x,y,z)$$
and used the Einstein summation convention. The point is that the $L$'s in your second equation should be $\mathcal{L}$'s.
A: I assume that what you are asking here is more an operational question rather than a conceptual question. Bear in mind that $\phi$ depends on $x$ and $t$, i.e. $\phi = \phi(x,t)$, and that with two variables $t$ and $x = x(t)$:
$D_t = \partial_t + \partial_x \dot{x}$
You can then expand the $\dot{\phi}^2(x,t)$ term into
$\dot{\phi}^2(x,t) = (\partial_t \phi)^2 + 2 (\partial_t \phi) (\partial_x \phi) \dot{x} + (\partial_x \phi)^2 \dot{x}^2$
Now going back to the Lagrangian density $\mathcal{L} = \frac{1}{2} [\mu \dot{\phi}^2(x,t) - Y\partial_x \phi(x,t)]$. Bear in mind that the partial derivatives operates on explicit dependence of variables in question. As an example, let me do the first term of the EL equation here:
$\begin{align*}
\frac{\partial \mathcal{L}}{\partial (\partial_t \phi)} &= \frac{1}{2} \mu \frac{\partial}{\partial (\partial_t \phi)} [(\partial_t \phi)^2 + 2 (\partial_t \phi) (\partial_x \phi) \dot{x} + (\partial_x \phi)^2\dot{x}^2] + \frac{1}{2} Y \frac{\partial}{\partial (\partial_t \phi)} (\partial_x \phi)^2\\
&=\frac{1}{2} \mu [\frac{\partial}{\partial (\partial_t \phi)} (\partial_t \phi)^2 + 2\frac{\partial}{\partial (\partial_t \phi)} (\partial_t \phi) (\partial_x \phi) \dot{x} + \frac{\partial}{\partial (\partial_t \phi)} (\partial_x \phi)^2 \dot{x}^2] + 0 \\
&=\frac{1}{2} \mu [2 (\partial_t \phi) + 2(\partial_x \phi) \dot{x} + 0]\\
&=\mu [(\partial_t \phi) + (\partial_x \phi) \dot{x}]
\end{align*}
$
So far, the point here is that you should treat $\partial_t \phi$ to be an explicit variable, and apply partial derivatives as you normally would just like $x$ or $t$. The term $\frac{1}{2} Y \frac{\partial}{\partial (\partial_t \phi)} (\partial_x \phi)^2$ in the first line, for instance, evaluates to be zero, for there is no dependence on $\partial_t \phi$.
You can halt at this stage and carry on to evaluate the derivative $\frac{\partial \mathcal{L}}{\partial (\partial_x \phi)}$ first, and you should yield something like $\mu [(\partial_t \phi) \dot{x} + (\partial_x \phi) \dot{x}^2] - Y \partial_x \phi$. And since the Lagrangian density does not depend on $\phi$, $\frac{\partial \mathcal{L}}{\partial \phi} = 0$. (Do try to calculate them yourself! I may be wrong here :p)
Combining what you have, you should be able to get the EL equation to be:
$\partial_t \mu [\partial_t \phi + (\partial_x \phi) \dot{x}] + \partial_x \mu [\partial_t \phi + (\partial_x \phi) \dot{x}] \dot{x} - Y \partial_{xx} \phi = 0 $
The third term is obviously the RHS of the equation that you need to prove. What is perhaps messy is the first two terms, but you should refer back to the first equation that we had discussed previously, i.e. $D_t = \partial_t + \partial_x \dot{x}$; with this, it is pretty straightforward to show that
$\mu[\partial_t \dot{\phi} + (\partial_x \dot{\phi}) \dot{x}] = Y \partial_{xx} \phi$
Do the simplification once more and you will obtain the equation that you need to prove :)
A: So you have given the Langrangian Density to be:
$$\mathcal{ L}={1\over{2}}\left(\mu\dot{\phi}^2-Y(\partial_x\phi)^2\right),$$
The Euler-Lagrange equation is :
$${\partial\over \partial t}\left({\partial \mathcal{L}\over{\partial\dot{\phi}}}\right)+{\partial\over \partial x}\left({\partial \mathcal{L}\over{\partial(\partial_x \phi)}}\right)={\partial \mathcal{L}\over{\partial \phi}}$$
So for the your first equation $\partial \mathcal{L}\over{\partial\dot{\phi}}$ = $2$ x $1\over2$x $ \mu\dot\phi$
Then you proceed to take the time derivative of this which is $\mu\ddot\phi$.
You follow the same steps to get ${\partial\over \partial x}\left({\partial \mathcal{L}\over{\partial(\partial_x \phi)}}\right)$ 
noting that ${\partial \mathcal{L}\over{\partial \phi}} = 0$ as there is no $\phi$ dependence in your Lagrangian Density.
Also keep in mind that $\partial_x\phi$ means $\partial\phi\over \partial x$. Hopefully its more clear to you how the equations of motion are obtained now.
A: Figured it out...
It should be apparent that the function $\phi$ is analagous to the coordiate x for the continuous distribution.  And $\dot{\phi}$  Also it should be apparent that this is a non-relativistic system.  But... one thing to keep in mind is that if time were treated as a coordinate there would be an additional equation to account for, and an additionian function $\psi$ which would be analogous to t.  Its time derivative in the non-relativistic limit is analagous to 1.  I will show why that is helpful in solution.
Start by evaluating the result of plugging $\dot\phi^2$ into LHS of E-L in given form.
1)${\partial\over\partial x}{\partial (\dot\phi\dot\phi)\over{\partial(\partial_x\phi)}}\rightarrow2{\partial\over\partial x}\dot\phi{\partial (\dot\phi\partial_x\phi+"1"\partial_t\phi)\over{\partial(\partial_x\phi)}}$
where $\dot\phi="v"\cdot\nabla\phi$ has been used and $"v"\leftrightarrow (\dot{\psi},\dot{phi})$ as $\dot{\psi}\rightarrow 1$ in non-relativistic limit.
so the first part gives $2{\partial_x\dot{\phi}}\dot{\phi}$
2)We do analgous analysis for the $"\partial_t"$ part which gives
$2\partial_t\dot{\psi}\dot{phi}$
3) sum parts 1 an 2 to get $2(\partial_t\dot{\psi}\dot{\phi}+{\partial_x\dot{\phi}}\dot{\phi})$ which taking the $\partial_t$ and $\partial_x$ operators to commute with $\dot\psi$ and $\dot{\phi}$ respectively is the same as $2"v"\cdot\nabla \phi\rightarrow 2\ddot{\phi}$
4) From these considerations it is apparent that plugging the ${1\over{2}}\mu\dot\phi^2$ portion of density $\mathcal{L}$ into the L-H-S of the E-L equation provides the term $\mu\ddot{\phi}$ of the final equation of motion.
