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