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I have a nonlinear Diffeq:

$$\frac{d^2x}{dt^2}+\beta \frac{dx}{dt}+\epsilon \times e^{- \lambda x} = f(t) $$

where $f(t)$ is a function that is known, and $\beta$ and $\lambda$ are constants that are known. Also, we know that $\epsilon$ is a constant parameter that is small.

I first need to obtain the zero order solution $x_0$, before finding the first order solution $x_1$

The first thing that I need to do is to use asymptotic expansions to obtain solutions of order $\epsilon=0$ and (TYPO)

Note that general solution for f(t) that will have two unknown constants.

UPDATE: After the first order term is solve, it needs to be plugged back in. The exponential needs to be linearized and things should start cancelling out. I am not sure how to do this, I just know this is what needs to be done.

UPDATE2: Correction, $\epsilon = 1$ was a typo. It should be $\epsilon^1$ I need to find a solution in the form:

$$x(t)=x_0(t)+\epsilon^1x_1(t)+\epsilon^2x_2 (t) + ... $$

So initially, $\epsilon$ needs to be set to 0 in order to obtain $x_0$. To find $x_1$, I need $\epsilon^1$

UPDATE3: I know now that I need to plug:

$$x=x_0+\epsilon_1x_1 $$ back into the original equation

Thus:

$$\frac{d^2}{dt^2}(x_0+\epsilon_1x_1) + \beta\frac{d}{dt}(x_0+\epsilon_1x_1)+\epsilon \times exp(-\lambda(x_0+\epsilon_1x_1)) $$

Then

$$\frac{d^2}{dt^2}x_0+\frac{d^2}{dt^2}\epsilon_1x_1+\beta \frac{d}{dt}x_0 +\beta \frac{d}{dt}\epsilon_1 x_1+\epsilon \times exp(-\lambda x_0))+\epsilon \times exp(-\lambda \epsilon_1 x_1)$$

I think then the $x_0$ terms may cancel with f(t) or something like that? It may be some sort of approximation.

I still need to linearize the exponential. Any help is appreciated.

Update4: Taking the solution a little but further...

We know that:

$$\frac{d^2x_0}{dt^2}+\beta \frac{dx_0}{dt} = f(t) $$

So, those terms all cancel. And now we have: $$\frac{d^2}{dt^2}\epsilon_1x_1 +\beta \frac{d}{dt}\epsilon_1 x_1+\epsilon \times exp(-\lambda(x_0+\epsilon_1x_1))=0$$

But we dont want $\epsilon^2$ terms, to part of the exponential goes away as well.

We are left with: $$\frac{d^2}{dt^2}\epsilon_1x_1 +\beta \frac{d}{dt}\epsilon_1x_1+\epsilon \times exp(-\lambda x_0)=0$$

Where we know $x_0$. This now means that the exponential is no longer a function of arbitrary x. I feel like the solution should be trivial now, but I am having a hard time finding it. Any ideas?

Can this be solved with the method of undetermined coefficients?

UPDATE5: Well I have updated this problem several times with very little response. As a latch ditch effort, is there anyone who can offer any advice on how to solve:

$$\frac{d^2x_1}{dt^2} +\beta \frac{dx_1}{dt} + exp{-\lambda x_0}=0$$

where $x_0$ is known

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  • $\begingroup$ Have you done a similar problem before? A first order equation? Something with a simpler nonlinear term, e.g. $\epsilon \cdot x^2$ instead of $\epsilon \cdot e^{-\lambda x}$? $\endgroup$ – Hans Engler Dec 7 '14 at 1:15
  • $\begingroup$ Look up your notes or a textbook and (re)do a simpler problem. Btw the solution to this problem will still involve an integral involving $f(t)$. $\endgroup$ – Hans Engler Dec 7 '14 at 1:18
  • $\begingroup$ "If you have a problem that you cannot solve, then there is a simpler problem that you cans solve. Find such a problem and solve it." That's advice from George Polya ("How to solve it"). en.wikipedia.org/wiki/How_to_Solve_It $\endgroup$ – Hans Engler Dec 7 '14 at 1:26
  • $\begingroup$ Google3 for "regular perturbation ode". There are several good intro articles that will come up. The ones from ucl.ac.uk, cims.nyu.edu and utwente.nl are all good places to start. Good luck. $\endgroup$ – Hans Engler Dec 7 '14 at 1:39
  • $\begingroup$ You don't have any specifications on $ f(t) $ ? $\endgroup$ – David Cardozo Dec 8 '14 at 19:03
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A possible solution is as follows:

Write the equation as

${\frac {d^{2}}{d{t}^{2}}}x \left( t \right) +\beta\,{\frac {d}{dt}}x \left( t \right) +\epsilon\,{{\rm e}^{-\lambda\,x \left( t \right) }} =f_{{0}} \left( t \right) +f_{{1}} \left( t \right) \epsilon $

Now look for a solution of the form

$x \left( t \right) =x_{{0}} \left( t \right) +x_{{1}} \left( t \right) \epsilon $

Replacing this solution in the equation we have

${\frac {d^{2}}{d{t}^{2}}}x_{{0}} \left( t \right) + \left( {\frac {d^{ 2}}{d{t}^{2}}}x_{{1}} \left( t \right) \right) \epsilon+\beta\, \left( {\frac {d}{dt}}x_{{0}} \left( t \right) + \left( {\frac {d}{dt }}x_{{1}} \left( t \right) \right) \epsilon \right) +\epsilon\,{ {\rm e}^{-\lambda\, \left( x_{{0}} \left( t \right) +x_{{1}} \left( t \right) \epsilon \right) }}=f_{{0}} \left( t \right) +f_{{1}} \left( t \right) \epsilon $

expanding the exponential we obtain

${\frac {d^{2}}{d{t}^{2}}}x_{{0}} \left( t \right) +\beta\,{\frac {d}{d t}}x_{{0}} \left( t \right) + \left( {\frac {d^{2}}{d{t}^{2}}}x_{{1}} \left( t \right) +\beta\,{\frac {d}{dt}}x_{{1}} \left( t \right) +{ {\rm e}^{-\lambda\,x_{{0}} \left( t \right) }} \right) \epsilon=f_{{0} } \left( t \right) +f_{{1}} \left( t \right) \epsilon $

Then we deduce that

${\frac {d^{2}}{d{t}^{2}}}x_{{0}} \left( t \right) +\beta\,{\frac {d}{d t}}x_{{0}} \left( t \right) =f_{{0}} \left( t \right) $

and

${\frac {d^{2}}{d{t}^{2}}}x_{{1}} \left( t \right) +\beta\,{\frac {d}{d t}}x_{{1}} \left( t \right) +{{\rm e}^{-\lambda\,x_{{0}} \left( t \right) }}=f_{{1}} \left( t \right) $

The solution at zero order is

$x_{{0}} \left( t \right) =\int _{0}^{t}\! \left( \int _{0}^{\tau}\!f_{ {0}} \left( \sigma \right) {{\rm e}^{\beta\,\sigma}}{d\sigma}+C_{{1}} \right) {{\rm e}^{-\beta\,\tau}}{d\tau}+C_{{2}} $

Replacing this solution in the equation at first order we obtain a linear equation which can be solved formally as

$x_{{1}} \left( t \right) =\int _{0}^{t}\! \left( \int _{0}^{\tau}\!(f_{ {1}} \left( \sigma \right) - {{\rm e}^{-\lambda\,x_{{0}} \left( \sigma \right) }} ){{\rm e}^{\beta\,\sigma}}{d\sigma}+C_{{3}} \right) {{\rm e}^{-\beta\,\tau}}{d\tau}+C_{{4}}$

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  • $\begingroup$ See update3 please $\endgroup$ – Jackson Hart Dec 10 '14 at 21:14
  • $\begingroup$ See update 4 too please $\endgroup$ – Jackson Hart Dec 10 '14 at 22:58
  • $\begingroup$ Can the last equation I have me solved with undetermined coeffs? $\endgroup$ – Jackson Hart Dec 11 '14 at 1:59

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