How Prove that $f$ is unique Assume that there exsits a smooth positive function $f$ on $(0,1)$ satisfying the differential equation
$$-f''-\dfrac{f'}{r}+\dfrac{f}{r^2}=f(1-f^2)$$
together with with boundary conditions $f(0)=0$ and $f(1)=1$.
Prove that $f$ is
unique.

the book give follow solution:Let $f_{1}$ and $f_{2}$ be two positive functions
  satisfying the hypotheses. Dividing the differential equation by $f$ and subtracting
  the corresponding equations, we obtain 
  $$-f''_{1}-\dfrac{f'_{1}}{r}+\dfrac{f_{1}}{r^2}=f_{1}(1-f^2_{1})$$
  $$-f''_{2}-\dfrac{f'_{2}}{r}+\dfrac{f_{2}}{r^2}=f_{2}(1-f^2_{2})$$
  $$\Longrightarrow -\dfrac{f''_{1}}{f_{1}}+\dfrac{f''_{2}}{f_{2}}-\dfrac{1}{r}\left(\dfrac{f'_{1}}{f_{1}}-\dfrac{f'_{2}}{f_{2}}\right)=-(f^2_{1}-f^2_{2})\tag 1$$

Multiplying the above equality by $r(f^2_{1}-f^2_{2})$ and integrating over $(0,1)$ yields
$$\int_{0}^{1}\left(f'_{1}-\dfrac{f_{2}}{f_{1}}f'_{2}\right)^2rdr+\int_{0}^{1}\left(f'_{2}-\dfrac{f_{1}}{f_{2}}f'_{1}\right)^2rdr=-\int_{0}^{1}(f^2_{1}-f^2_{2})^2rdr\tag2$$
Therefore $f_{1}=f_{2}$
My Question: $(1)\Longrightarrow (2)$,I can't understand How get it? Thank you
because when $(1)$ multiplying $r(f^2_{1}-f^2_{2})$ and integrating over $(0,1)$ then
$$\int_{0}^{1}(f^2_{1}-f^2_{2})\left(-\dfrac{f''_{1}}{f_{1}}+\dfrac{f''_{2}}{f_{2}}\right)rdr-\int_{0}^{1}(f^2_{1}-f^2_{2})\left((\dfrac{f'_{1}}{f_{1}}-\dfrac{f'_{2}}{f_{2}}\right)dr=-\int_{0}^{1}(f^2_{1}-f^2_{2})^2rdr$$
so we only prove this
$$\int_{0}^{1}(f^2_{1}-f^2_{2})\left(-\dfrac{f''_{1}}{f_{1}}+\dfrac{f''_{2}}{f_{2}}\right)rdr-\int_{0}^{1}(f^2_{1}-f^2_{2})\left((\dfrac{f'_{1}}{f_{1}}-\dfrac{f'_{2}}{f_{2}}\right)dr=\int_{0}^{1}\left(f'_{1}-\dfrac{f_{2}}{f_{1}}f'_{2}\right)^2rdr+\int_{0}^{1}\left(f'_{2}-\dfrac{f_{1}}{f_{2}}f'_{1}\right)^2rdr$$
maybe use integration by parts ? But I can't.Thank you
 A: Partial proof of uniqueness: The differential equation reads
$$\tag{1} \frac{(rf^{\prime})^{\prime}}{f}-\frac{1}{r}~=~r(f^2-1), \qquad r~\in~ ]0,1] ,\qquad f~>~0. $$
For two solutions $f_1$ and $f_2$ we derive
$$\tag{2} \frac{(rf_1^{\prime})^{\prime}}{f_1}-\frac{(rf_2^{\prime})^{\prime}}{f_2} ~\stackrel{(1)}{=}~r(f_1^2-f_2^2). $$
Next
$$
\frac{d}{dr}\left[ \left(\frac{ rf_1^{\prime}}{f_1}-\frac{rf_2^{\prime}}{f_2} 
\right)(f_1^2-f_2^2)\right] $$
$$~=~\left(\frac{(rf_1^{\prime})^{\prime}}{f_1}-\frac{(rf_2^{\prime})^{\prime}}{f_2}\right)(f_1^2-f_2^2)
+rf_1^{\prime} \frac{d}{dr}\left(\frac{f_1^2-f_2^2}{f_1}\right)
+rf_2^{\prime} \frac{d}{dr}\left(\frac{f_2^2-f_1^2}{f_2}\right) $$
$$~\stackrel{(2)}{=}~r(f_1^2-f_2^2)^2
+rf_1^{\prime} \left(f_1^{\prime}-\frac{2f_2f_2^{\prime}}{f_1}+\frac{f_2^2f_1^{\prime}}{f_1^2}\right)
+rf_2^{\prime} \left(f_2^{\prime}-\frac{2f_1f_1^{\prime}}{f_2}+\frac{f_1^2f_2^{\prime}}{f_2^2}\right) $$
$$~=~r(f_1^2-f_2^2)^2
+r \left(f_1^{\prime}-\frac{f_2f_2^{\prime}}{f_1}\right)^2
+r \left(f_2^{\prime}-\frac{f_1f_1^{\prime}}{f_2}\right)^2 ~\geq ~0.\tag{3}$$
The idea is now to integrate both sides of eq. (3) from $r=0$ to $r=1$ to conclude that $f_1=f_2$. Here we will definitely need the upper boundary condition $f_1(1)=f_2(1)$. Also there remains to argue (which we haven't rigorously done) that the lower limit $r\to 0^{+}$ is well-behaved, which presumable involves the boundary condition $f(0)=0$. 
