This might be an easy question in the theory of ordinary differential equation. But since I know very little about it, I posted it here and hope to get some answers or references.

Consider the equation $$x\phi'(x)-c\phi(x)=xf(x), x>0$$ where $c$ is some constant.

My question is that under what (growth) conditions, the solution $\phi(x)$ has the following property: the limit $$\lim_{x\to 0}x^{-c}\phi(x)$$ exists and nonzero?

Thank you very much for your help and sorry if this question is not appropriate here.

  • $\begingroup$ Robert answered it, but in general, when faced with a problem of this type one can define, say, $a(x)=x^{-c}\phi(x)$, so $\phi(x)=x^c a(x)$, and then substitute into the differential equation for $\phi$, obtaining a differential equation for $a$, which you can use to understand the asymptotics of $a$. $\endgroup$ – Will Sawin Jan 9 '12 at 23:00

The solution to your linear differential equation is $\phi(x) = x^c F(x)$ where $F(x)$ is any antiderivative of $f(x) x^{-c}$. In order for some solution to have $\lim_{x \to 0} x^{-c} \phi(x)$ exist, the necessary and sufficient condition is that the improper integral $\int_0^\varepsilon f(x)\ x^{-c}\ dx$ converges for some $\varepsilon > 0$. Then the limit will exist for all the solutions. However, the limit will be $0$ for one of those solutions, namely $\phi(x) = x^c \int_0^x f(t)\ t^{-c}\ dt$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.