Solving a 2nd-Order ODE for Conservation of Energy

[Jazradel]

Homework Statement

Consider a mechanical system describe by the conservative 2nd-order ODE
$\frac{\partial^{2}x}{\partial t^{2}}=f(x)$
(which could be non linear). If the potential energy is $V(x)=-\int^{x}_{0} f(\xi) d \xi$, show that the system satisfies conservation of energy $\frac{1}{2}x^{2}+V(x)=E$ (E is a constant).

Homework Equations

As above.

The Attempt at a Solution

I've missed almost everything we've done on ODEs, so I don't really have any idea how to being. Even knowing what to call the problem, or a link to some notes/worked examples/relevant textbook would be great. I think the start is to define:
$x(t)=\begin{array}{c} <br /> x_{1}(t) \\<br /> x_{2}(t) \\ <br /> \end{array}$
Then sub into the first equation:
$\frac{\partial^{2} x_{1}}{\partial t^{2}}=f_{1}(x_{1},x_{2})$
$\frac{\partial^{2} x_{2}}{\partial t^{2}}=f_{2}(x_{1},x_{2})$
Now I think I should use the chain rule, and integrate equation 2, but I can't see how.

 

[vela]

Homework Statement

Consider a mechanical system describe by the conservative 2nd-order ODE
$\frac{\partial^{2}x}{\partial t^{2}}=f(x)$
(which could be non linear). If the potential energy is $V(x)=-\int^{x_0} f(\xi) \xi$, show that the system satisfies conservation of energy $\frac{1}{2}x^{2}+V(X)=E$ (E is a constant).

I think you mean $\frac{1}{2}\dot{x}^2 + V(x) = E$. Just integrate the original equation with respect to x. You'll want to use the chain rule to evaluate the integral on the LHS.

 

[Jazradel]

I just checked, it's definitely $\frac{1}{2}x^{2}+V(x)=E$. I did have to fix the V(X) and the domain of the integration.

You're saying do this?
$\int f(x) dx = \int \frac{\partial^{2}x}{\partial t^{2}} dx$
$\int \frac{\partial^{2}x}{\partial t^{2}} dx = \int \frac{\partial}{\partial t} ( \frac{\partial x}{\partial t} ) dx$
The problem is I have no idea how to apply the chain rule to this case.

 

[vela]

I just checked, it's definitely $\frac{1}{2}x^{2}+V(x)=E$.

That can't be right. It would only hold if V(x)=-1/2 x²+V₀. The first term is supposed to be the kinetic energy, so it needs to depend on v², not x².

You're saying do this?
$\int f(x) dx = \int \frac{\partial^{2}x}{\partial t^{2}} dx$
$\int \frac{\partial^{2}x}{\partial t^{2}} dx = \int \frac{\partial}{\partial t} ( \frac{\partial x}{\partial t} ) dx$
The problem is I have no idea how to apply the chain rule to this case.

You want to use
$$\frac{\partial}{\partial t} = \frac{\partial x}{\partial t} \frac{\partial}{\partial x}$$

 

[Jazradel]

Ah thanks, that should be great help.

You can view the assignment here http://www.maths.utas.edu.au/People/Forbes/KYA314Ass2in2011.pdf . Question 2. (a) is the one in question. I think I've typed it correctly though.

Edit: I have confirmed it is not a typo.

 

[vela]

That's a typo for sure.

 

[HallsofIvy]

For problems like these, "conservation of energy" arises from using "quadrature" on the differential equation.

That is, if x''= f(x), a function of x, only, we can let v= x' and then use the chain rule:
$$\frac{d^2x}{dt^2}= \frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}$$
so that our equation x''= f(x) becomes v v'= f(x) where the differentiation is now with respect to x:
$$v\frac{dv}{dx}= f(x)$$
$$v dv= f(x)dx$$
$$\frac{1}{2}v^2= \int_0^x f(x)dx+ C$$
$$\frac{1}{2}v^2- \int_0^x f(x)dx= C$$