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I have a differential equation of the form $\ddot{x} = c\cdot x \cdot a(t)$ where $a(t)$ contains an explicit time dependence. The explicit time dependence is too complex so that the differential equation cannot be solved analytically.

Now my question is if, as an approximation, I can assume $a(t)$ to be constant for solving the differential equation ($a(t) \rightarrow a$) and then plug this explicit time dependence back into the solution ($a \rightarrow a(t)$)? If so, what conditions need to be satisfied (e.g. the variation of $a(t)$ needs to be small compared to the variation of $x(t)$)? Or does this work only for specific kinds of differential equations?

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  • $\begingroup$ Maybe you could provide the expression for $a(t)$? In general the approximation you suggest would be poor, but for some cases it could work $\endgroup$ – Yuriy S Oct 27 '18 at 14:51
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    $\begingroup$ A WKB-like approach gives $$x(t)=(ca(t))^{-1/4}\exp\left(\pm\int_0^t (ca(s))^{1/2}\, ds\right)$$ as a first approximation of basis functions. $\endgroup$ – Lutz Lehmann Oct 27 '18 at 14:52
  • $\begingroup$ @YuriyS For example $a(t) = \exp(-t^2/(2\sigma^2))$. But I'd like to have a more general overview of what are the deciding criteria for $a(t)$ for which that approach can be applied. $\endgroup$ – a_guest Oct 27 '18 at 21:48
  • $\begingroup$ @a_guest, you'll have to specify some properties of $a(t)$. In your example you have a "bump" which goes to $0$ as $t \to \infty$, so it will affect the solution very much, as for large times $x(t)$ becomes a linear or constant function, while for $t \to 0$ it has the form $A \cosh \sqrt{ca}t +B \sinh \sqrt{ca} t$. For other $a(t)$ you will have a very different behavior. In your case if you simply "plug in" $a(t \to \infty)$ you are going to have $x=A=\text{const}$ which is only a particular solution of equation $\ddot{x}=0$. So I wouldn't use this kind of approximation for large times. $\endgroup$ – Yuriy S Oct 27 '18 at 21:57
  • $\begingroup$ See the perturbation method I suggest in my answer, which is more justified than the "plugging back" approach. Or try WKB as LutzL suggests. But be careful to check that the assumptions for both methods are satisfied for your case $\endgroup$ – Yuriy S Oct 27 '18 at 23:05
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A lot of things depend on the properties of $a(t)$, but I will mention some general methods I would use.

For $t \to 0$ the approach is obvious: search for the series solution. If we can represent:

$$x(t)=c_0+c_1 t+c_2 t^2+ \cdots$$

$$a(t)=a_0+a_1 t+a_2 t^2+ \cdots$$

We have:

$$2c_2=c a_0 c_0$$

$$6c_3=c (a_1c_0+ a_0c_1)$$

And so on. With $c_0,c_1$ determined by the initial conditions.

For large times the approach will have to be different. Maybe we can represent $a(t)$ as a series in $1/t$, in which case it will be easy to find the asymptotic solution as a series same as above.

If not, we can try to use various other methods. Like WKB method, mentioned by @LutzL. In fact, you can scroll to an example describing a general equation very similar to your case.

Another approach to try would be using perturbation theory. (We denote $ca(t)=b(t)$ for simplicity.)

Assuming that variable part of $b(t)$ is small, that is, it can be represented as a series in some small parameter $\beta \ll 1$, like this:

$$b(t)=B+\beta b_1(t)+ \beta^2 b_2(t) +\dots$$

$$B=\text{const}$$

Going back to the original equation, we can try to also represent $x(t)$ this way:

$$x(t)=x_0(t)+\beta x_1(t)+ \beta^2 x_2(t) +\dots$$

Then we have:

$$\ddot{x}=b(t) x$$

$$(\ddot{x_0}+\beta \ddot{x_1}+ \beta^2 \ddot{x_2} +\dots)=(B+\beta b_1(t)+ \beta^2 b_2(t) +\dots)(x_0+\beta x_1+ \beta^2 x_2 +\dots)$$

Matching the powers of $\beta$, we have:

$$\ddot{x_0}=B x_0$$

$$\ddot{x_1}=B x_1+b_1(t)x_0(t)$$

$$\ddot{x_2}=B x_2+b_1(t)x_1(t)+b_2(t)x_0(t)$$

And so on. For $x_0$ we obviously have the general solution:

$$x_0(t)=C_1 e^{\sqrt{B} ~t}+C_2 e^{-\sqrt{B} ~t}$$

The other equations all have the same form:

$$\ddot{x_n}=B x_n+f(t)$$

with $f(t)$ a known function. This is a linear inhomogeneous equation and can be solved exactly or approximately by the usual methods. We only need to find a particular solution, because we know the general solution has the same form as $x_0$.

We can use the variable coefficient method, searching for $x_{np}$ as:

$$x_{np}=C(t) e^{\sqrt{B}t}$$

Which brings us to:

$$(\ddot{C}+\sqrt{B} ~\dot{C})e^{\sqrt{B}t}=f(t)$$

$$\dot{C}=D(t)$$

$$\dot{D}+\sqrt{B}~ D=e^{-\sqrt{B}t}f(t)$$

This is a 1st order inhomogeneous equation which is solved in quadratures.

This is a little involved in general, but allows us to find as many orders of approximation as we want.

A different form of the equation might help. Let us change the function:

$$x(t)=e^{y(t)}$$

Then we have:

$$\ddot{y}+\dot{y}^2=b(t)$$

Now introducing another function:

$$z(t)=\dot{y}(t)$$

we have a nonlinear 1st order ODE:

$$\dot{z}+z^2=b(t)$$

This is a Riccati equation. This equation has a number of nice properties, for example if we know any particular solution, we can find the general one.

Riccati equation has also been extensively studied by many researchers for the last 200 years or so, which is a good reason to search online for the methods relevant to your case.

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  • $\begingroup$ Thanks for all the information, I'll need some time to digest it. But as far as I see, for the example of $c=-1\, , \; a=\exp(-t^2)$, neither of the approaches is really applicable. I'd like to consider the whole range $t \in [0, \infty)$ so a series expansion won't work. For WKB to first order I obtain $[\cos(...) + \sin(...)]\exp(t^2/4)$ which doesn't seem correct due to the $\exp$ t-dependence (I'd rather expect something like a linear dependence). Perturbation approach doesn't seem to work for $\exp(-t^2)$ and since I don't know any particular solution I can't use Riccati equation either. $\endgroup$ – a_guest Oct 29 '18 at 16:16
  • $\begingroup$ For $\ddot x +e^{-t^2}x=0$ select some small $ϵ$, for instance $ϵ=0.01$, then for $t>\sqrt{-\lnϵ}\approx 2.146$ you get in a first approximation $\ddot x=0$, which is also the observed behavior of the numerical solutions, some arc and then linear behavior. $\endgroup$ – Lutz Lehmann Oct 29 '18 at 17:00
  • $\begingroup$ @a_guest, you can't make a single workable approximation for the whole range, not in general. You have to deal with different ranges in different ways, that's unavoidable $\endgroup$ – Yuriy S Oct 29 '18 at 17:28

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