Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $x \in \mathbb{R}$ be a variable and $c\in\mathbb{R}$ a parameter. Also, let $f(x,c)$ be a function dependent on $x$ and $c$. Furthermore, define a Differential Equation which is solved by the function $y(x)$:

$$\left(\frac{d}{dx}\right)^2 y(x) + f(x,c)y(x)=0$$

Suppose, the general Differential Equation is too hard to solve, but for practical reasons only the solution with $c\approx 0$ is required. Naively I would expand $f(x,c)$ around $c=0$ and then compute the approximate Differential Equation. To the order $O(c^1)$ for example I would have to solve the following (where a prime denotes a derivative in respect to $c$):

$$\left(\frac{d}{dx}\right)^2 y(x) + \big(f(x,0)+f'(x,0)~c + O(c^2)\big)y(x)=0$$

If $f(x,0)$ and $f'(x,0)$ turn out to be simple enough and if we neglect the $O(c^2)$ terms, this approximated equation might turn out to be solvable.

Does this give a mathematically valid approximation for the solution $y(x)$? Maybe there are some subtleties which I did not consider? For instance, I have the feeling that the solution $y(x)$ should also be considered as a function of $c$ (as in $y(x,c)$) and be somehow involved in the expansion. It would be nice if someone knowledgeable could shed some light onto this.

share|improve this question
1  
It would depend on the function $f$. –  Avi Steiner Mar 3 '13 at 16:36
    
@Kagaratsch If you stay around 0 for c, your solution method is correct. –  Occupy Gezi Mar 14 '13 at 23:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.