I was wondering what the solution to the Falkner Skan equation, $$f^{\prime\prime\prime}+ff^{\prime\prime}+\beta(1-f^{\prime 2})=0$$ looks like for negative values of $\beta$ and whether solutions for a particular values of $\beta$ are unique or if you could get multiple solutions for a given value of $\beta$?
The conditions for this equation are $f(0)=f^\prime(0)=0$ and $f^\prime(x)\rightarrow1$ as $x\rightarrow \infty$
This is not an answer, but a remark due to the discussion in the comments.
Our equation is autonomous, because it doesn't contain the independent variable $x$.
$$f^{\prime\prime\prime}+ff^{\prime\prime}+\beta(1-f^{\prime 2})=0$$
So we can consider another function:
$$f'=u(f)$$
$$f''=\frac{du}{dx}=\frac{du}{df}\frac{df}{dx}=u \cdot u'$$
$$f'''=\frac{d}{dx}(u \cdot u')=u \cdot \frac{d}{df}(u \cdot u')=u (u'^2+uu'')$$
So we obtain a second order equation:
$$u (u'^2+uu'')+f uu'+\beta(1-u^2)=0$$
The boundary conditions are another story. From the first two conditions we know that $u(0)=0$, as for another condition, we'll need to work on that.
Note that for $u \to 1$ we can consider the approximate equation:
$$u u''+u'(u'+f)=0 \\ (u u')'+f u'=0 $$
And for $u \to 0$ another approximate equation:
$$u^2 u''+u ~u'(u'+f)+\beta=0$$