How to integrate when zero is in denominator?

Question

I have this equation:

$$(p_0 p_1)'=\dfrac{4}{r^2-1}p_0'$$

where $'$ is derivative $\dfrac{d}{dz}$.

I need to solve it (find $p_1$) for case when $r=1$ (I already know $p_0$), and I don't know what to do because in that case I will have zero in denominator. I can make this shape:

$$(r^2-1)(p_0 p_1)'=4p_0'$$ where I will have zero on left side and just $p_0'$ on the right side, but this equation is part of the system, where I already found solution for $p_0$.

Can you tell me please do you see any solution of this? $$ $$ Before upper equation it was necessary to solve this one:

$$4\beta\dfrac{\partial^2 u}{\partial r^2} + \dfrac{4\beta}{r}\dfrac{\partial u}{\partial r}=\dfrac{\partial p_1}{\partial z}$$

where $\beta$ is constant, $p=f(z)$, $r$ is radial coordinate, $z$ is longitudinal coordinate and $u=u(r,z)$ and with additional conditions:

$r=0: \dfrac{\partial u}{\partial r}=0$;

$r=1: u=0$.

I got solution: $$ u(r,z)=\frac1{16\beta}\,(r^2-1)\,\frac{dp_1}{dz}(z). $$ from this equation is my part in upper equation $(r^2-1)$.

Because first equation is probably not correct because zero in denominator, can you tell me please did I make mistake in solution for $u(r,z)$?

Answer 1

When $r=1$, the equation $$(p_0p_1)' = \frac{4}{r^2-1}p_0'$$ is meaningless because the right side of the equation is not defined.

The equation $$(r^2-1)(p_0p_1)' = 4p_0'$$ is still meaningfull, however it cannot be used to calculate $p_1$ because the equation is true for any $p_1'$, as long as $p_0$ is a constant function.

Therefore, from only the details written in the question, calculting $p_1$ is impossible.

Answer 2

Supposing that your equation $\quad (p_0 p_1)'=\dfrac{4}{r^2-1}p_0'\quad$ is correct :

In case of $r\neq\pm1$ $$p_0 p_1=\int\dfrac{4}{r^2-1}p_0'dr+c_1$$ $$p_1(r)=\frac{1}{p_0}\int\dfrac{4}{r^2-1}p_0'dr+\frac{c_1}{p_0}$$ Then, study the limit for $r\to 0$.

Depending on the form of the function $p_0(r)$ , possibly the limit might be finite and then $p_1(0)$ might be significant.