Singular points of non-linear ODE

Question

EDIT: Sorry i messed up, I forgot a minus sign in front of the left hand side. I added it now.

I am not sure how to proceed with this. Given this non-linear ODE$$\partial_{t}u(t,x)=-\cot(t)\left[\frac{1}{8u(t,x)}\left(6u(t,x)^2-4A(x)u(t,x)+B(x)\right)\right]$$ fot $t\in(0,\pi)$, can I calculate the behavior of $u(t,x)$ for $t\to0,\pi$?

Answer 1

Treat the equation as an ODE in $t$ (which it is): the equation is first order and separable, so can be solved explicitly by integration. You obtain the equation $$ \sin(t) = f(u), $$ with $f(u)$ a function involving arctan and exp. This equality holds for all $t$ (and all $x$), so taking $t \to 0,\pi$ makes the left hand side zero. So, $u$ will converge to the zeroes of $f$ when $t \to 0,\pi$. So, all you have to do is find $f$, and determine its zeroes.

Answer 2

$$\partial_{t}u(t,x)=\cot(t)\left[\frac{1}{8u(t,x)}\left(6u(t,x)^2-4A(x)u(t,x)cot(t+B(x)\right)\right]$$ Since there is no differential with respect to $x$ the equation is an ODE with the only variable $t$ and where $x$ can be considered as a parameter. $$\frac{du}{dt}=\frac18 \cot(t)\left(6u-4A+\frac{B}{u}\right)$$ Let $z=\ln(\sin(t))\quad;\quad dz=\cot(t)dt$ $$\frac{du}{dz}=\frac18 \left( 6u-4A+\frac{B}{u}\right)$$ This is a separable ODE. $$z=8\int \frac{du}{6u-4A+\frac{B}{u}}$$ $$z=\frac32\ln(6u^2-4Au+B)-\frac{4A}{3\sqrt{\frac32 B-A^2} }\tan^{-1}\left(\frac{A-3u}{\sqrt{\frac32 B-A^2}} \right)+C$$ The solution on the form of implicit equation is : $$\ln(\sin(t))=\frac32\ln(6u^2-4Au+B)-\frac{4A}{3\sqrt{\frac32 B-A^2} }\tan^{-1}\left(\frac{A-3u}{\sqrt{\frac32 B-A^2}} \right)+C$$ $A,B,C$ are functions of $x$ : $$\ln(\sin(t))=\frac32\ln\left(6u(x,t)^2-4A(x)u(x,t)+B(x)\right)-\frac{4A(x)}{3\sqrt{\frac32 B(x)-A(x)^2} }\tan^{-1}\left(\frac{A(x)-3u(x,t)}{\sqrt{\frac32 B(x)-A(x)^2}} \right)+C(x)$$ $${{\sin(t)=} =F(x)\bigg(6u(x,t)^2-4A(x)u(x,t)+B(x)\bigg)^{3/2} {\exp\left(-\frac{4A(x)}{3\sqrt{\frac32 B(x)-A(x)^2} }\tan^{-1}\left(\frac{A(x)-3u(x,t)}{\sqrt{\frac32 B(x)-A(x)^2}} \right) \right)}}$$ $F(x)$ is an arbitrary function to be determined according to some boundary condition.

Now, coming back to the original question, answering about the behaviour of $u(x,t)$ for $t\to0,\pi$ seems problematical if no boundary condition is specified.

An hypothetical answer for any boundary condition might be $$6u(x,0)^2-4A(x)u(x,0)+B(x)\sim 0$$ $$u(x,0)\sim \frac{2A(x)\pm\sqrt{4A(x)^2-6 B(x)}}{6}$$

NOTE :

This result is suggested by inspection of the original equation where $\cot(t)$ tends to infinity for $t\to0$. Then, most likely $6u(t,x)^2-4A(x)u(t,x)+B(x)\to 0$.