How to go about solving this first-order nonlinear differential equation?

In summary: I would have gotten the same answer. In summary, the functional to minimise is \int^a_{-a}(2\pi y-\lambda)\sqrt{1+y'^2}dx where λ is a Lagrange multiplier. The first integral of the Euler-Lagrange equation gives a different ODE from what I got above, which has the solution y(x)=k(\cosh\frac{x}{k}-\cosh\frac{a}{k}) for some k depending on l, as required.
  • #1
Yondaime5685
27
0
I saw this post at stackexchange:
The fact that curve length is fixed amounts to a functional constraint. So we need to use both the Euler-Lagrange equation and Lagrange multipliers. The functional to minimise is [itex]\int^a_{-a}(2\pi y-\lambda)\sqrt{1+y'^2}dx[/itex], where λ is a Lagrange multiplier. For this functional, the first integral of the Euler-Lagrange equation gives a different ODE from what I got above, which has the solution [itex]y(x)=k(\cosh\frac{x}{k}-\cosh\frac{a}{k})[/itex] for some k depending on l, as required.

I ran across this post when trying to solve a homework problem. But I have no idea how he got that solution for that. When I use the Euler-Lagrange, I get this diff eq below.

Here is the simplest form I have managed to get it in:
[itex]\frac{\alpha}{2\pi}=\frac{\eta \quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}[/itex]

Where [itex]\alpha[/itex] is a constant. For the simplification, [itex]\eta = (y-y_0)[/itex] & [itex]λ \equiv -2\pi y_0[/itex] is used from the quote above.

I don't know where to even begin to solve this. Any help would be good.

Thanks.

-edit
Some details that I thought I should add.

Since [itex]h=f+\lambda g≠h(x)[/itex], I used the special case of the E-L equation:
[itex]\dot{y}\frac{\partial h}{\partial \dot{y}}-h=const.[/itex]

But after the substituions to trim it up the special E-L turns into this:
[itex]\dot{\eta}\frac{\partial h}{\partial \dot{\eta}}-h=const.=\alpha[/itex]
Where [itex]h[/itex] now is: [itex]h=2\pi\eta\sqrt{1+\dot{\eta}^2}[/itex]
 
Last edited:
Physics news on Phys.org
  • #2
Yondaime5685 said:
[itex]\frac{\alpha}{2\pi}=\frac{\eta \quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}[/itex]

That should be [tex]
\frac{\alpha}{2\pi}=\frac{\eta \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\eta\sqrt{1+\dot{\eta}^2} = \eta [\frac{\quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}]
[/tex]

Then you square the equation: [tex]
(\frac{\alpha}{2\pi})^2= \eta^2 [\frac{\quad \dot{\eta}^4}{1+\dot{\eta}^2} -2 \dot{\eta}^2 + 1 + \dot{\eta}^2] = \eta^2 [\frac{\quad \dot{\eta}^4}{1+\dot{\eta}^2} -\dot{\eta}^2 + 1]

\\ (\frac{\alpha}{2\pi})^2= \eta^2 \frac{\quad \dot{\eta}^4 -\dot{\eta}^2(1+\dot{\eta}^2) + (1+\dot{\eta}^2)} {1+\dot{\eta}^2}
= \eta^2 \frac{\quad \dot{\eta}^4 -\dot{\eta}^2 - \dot{\eta}^4 + 1+\dot{\eta}^2} {1+\dot{\eta}^2}
= \eta^2 \frac{1} {1+\dot{\eta}^2}[/tex]

I assume you can continue from here.
 
  • #3
Thanks a lot! That made the problem a lot simpler.

I can't believe I didn't carry that [itex]\eta[/itex] over.
 
Last edited:

What is a first-order nonlinear differential equation?

A first-order nonlinear differential equation is an equation that involves the first derivative of a function and contains nonlinear terms. This means that the equation cannot be solved using standard methods for solving linear differential equations.

How do I know if an equation is a first-order nonlinear differential equation?

To determine if an equation is a first-order nonlinear differential equation, check if it contains the first derivative of a function and if it has any nonlinear terms, such as squared or cubed variables. If both of these conditions are met, then the equation is a first-order nonlinear differential equation.

What are some common techniques for solving first-order nonlinear differential equations?

Some common techniques for solving first-order nonlinear differential equations include separation of variables, substitution, and the use of integrating factors. These techniques may also involve using algebraic manipulation, partial fractions, and integration by parts.

Are there any special cases or exceptions when solving first-order nonlinear differential equations?

Yes, there are some special cases and exceptions when solving first-order nonlinear differential equations. For example, some equations may require the use of a particular method, such as substitution or integrating factors, in order to be solved. Additionally, some equations may not have a closed-form solution and instead require numerical methods to approximate a solution.

How can I check if my solution to a first-order nonlinear differential equation is correct?

To check if your solution to a first-order nonlinear differential equation is correct, you can plug it back into the original equation and see if it satisfies the equation. You can also check if your solution matches any known solutions or if it satisfies any initial conditions given in the problem.

Similar threads

  • Differential Equations
Replies
5
Views
2K
  • Differential Equations
Replies
4
Views
1K
  • Differential Equations
Replies
1
Views
2K
  • Differential Equations
Replies
1
Views
1K
  • Differential Equations
Replies
1
Views
676
Replies
1
Views
1K
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
2
Views
884
  • Differential Equations
Replies
1
Views
2K
Replies
5
Views
1K
Back
Top