A way of solving this PDE? (Other than method of characteristics)

Question

I've been trying to work with this coupled PDE for some time. Here, $P = P(k,t)$

$$\frac{\partial{P}}{\partial t}-\frac{\sin(k)}{t}\frac{\partial{P}}{\partial k}=\left(\cos(k)-1\right){P}$$

To solve this PDE, I looked at the ODE

$$\frac{dP}{dt}=\left(\cos(k)-1\right)P$$

We know that

$$\frac{dP}{dt}=\frac{\partial P}{\partial t} + \frac{dk}{dt}\frac{\partial P}{\partial k}$$

So we can see that whenever

$$\frac{dk}{dt}=\frac{-\sin(k)}{t}$$

the solution to the ODE gives us a solution to the PDE.

But is there another way to get a more general solution than this?

Thanks for the help :)

Answer 1

I wrote some notes on the solution of Poses by characteristics: http://hyperkahler.co.uk/wp-content/uploads/2013/06/characteristics.pdf

Answer 2

$$\frac{\partial{P}}{\partial t}-\frac{\sin(k)}{t}\frac{\partial{P}}{\partial k}=\left(\cos(k)-1\right){P}$$ $$\frac{dt}{1}=-t\frac{dk}{\sin(k)}=\frac{dP}{(\cos(k)-1)P}$$ From $\frac{dt}{1}=-t\frac{dk}{\sin(k)}$ : $$t\tan(k/2)=c_1$$ From $-t\frac{dk}{\sin(k)}=\frac{dP}{(\cos(k)-1)P}=-\frac{c_1}{\tan(k/2)}\frac{dk}{\sin(k)}$

$\frac{dP}{P}=-c_1\frac{\cos(k)-1}{\tan(k/2)\sin(k)}dk=c_1dk$

$Pe^{-c_1k}=c_2$ $$Pe^{-t\tan(k/2)k}=c_2$$ The general solution of the PDE is : $$P(k,t)=e^{t\tan(k/2)k}F\Big(t\tan(k/2)\Big)$$ where $F$ in an arbitrary function (to be determined if some sufficient boundary conditions are specified).