How to solve the ODE: $L(\varphi)=\ddot \varphi - (n-2) \tanh t \dot \varphi + n\varphi\frac{1}{\cosh^2 t }=0$, where $\sinh t=\frac{e^t-e^{-t}}{2}$, $\cosh t=\frac{e^t+e^{-t}}{2}$, $\tanh t=\frac{\sinh t}{\cosh t}$. The answer is $L(\varphi)=0$ has two linearly independent solutions $\varphi_1(t) = \tanh t$ and $\varphi_2(t) = e^{(n-2)|t|}(1 + O(e^{t}))$ as $t \rightarrow -\infty$. (for e.g., $\varphi_2(t) = \tanh t \int_{c}^t \frac{\cosh^n \tau}{\sinh^2 \tau} d\tau$ for some $c < 0$.) But, how can I find the two linearly independent solutions?
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ I don't quite understand the question. You have already given explicit formulas for $\varphi_1(t)$ and $\varphi_2(t)$, which are solutions and are independent. Sounds like you already have an answer. $\endgroup$– Igor KhavkineCommented Apr 15, 2023 at 13:08
-
$\begingroup$ Sorry, this is my fault! The solutions are given from some papers, I don't list it here. My question is how to find the two soutions like $\varphi_1$, $\varphi_2$ in details. This is shown in that paper, but not given the detials. $\endgroup$– Davidi ConeCommented Apr 15, 2023 at 14:52
-
$\begingroup$ This looks very much like the Legendre differential equation. Try to make $t=\tanh(x)$ $\endgroup$– Claude LeiboviciCommented Apr 18, 2023 at 5:46
-
$\begingroup$ Could you list the paper you are referring to, please? $\endgroup$– Thomas BenjaminCommented May 21, 2023 at 0:58
-
$\begingroup$ It was stated in the paer：The axisymmetric $\sigma_k$-Nirenberg problem, in the page 57. $\endgroup$– Davidi ConeCommented May 22, 2023 at 1:09
Add a comment
|