Vrftsjtryk

Show that $frac{du}{dt}$=$u^{a}$ where $u(0)=0$ has two soultions for $0<a<1$ and one solution for $a=0,1$

I have tried integrating this by saying
$intfrac{du}{u^{a}}$=$int{1}dt$
which gave me the horrible solution of

$$frac{u(t)}{(u(t))^{a}}=(t+c)(1-a)$$ and assuming that is correct how do I proceed to show that the solutions depend on $a$?

The first step in solving the problem is to note that $u(t)=0$ is always a solution of the ODE Cauchy problem under analysis, i.e. of
$$
begin{cases}
dfrac{mathrm{d}u(t)}{mathrm{d}t}=u^a(t) \
\
u(0)equiv u_0=0
end{cases}quad0le ale 1.tag{1}label{1}
$$
The second step is trying to find another solution of the problem eqref{1} in the "critical exponent" case, i.e. $0< a<1$. As the already noted by the OP, the best way to do so is perhaps to use the standard Barrow's formula ([1], §1.5 p. 19),
$$
t-t_0=intlimits_{u_0}^{u(t)}frac{mathrm{d}xi}{v(xi)}tag{2a}label{2a}
$$
which, for our problem eqref{1}, takes the form:
$$
t=intlimits_{0}^{u(t)}frac{mathrm{d}xi}{xi^a}tag{2b}label{2b}
$$
Now, it's easy to see that the integral at the right side of eqref{2b} is not defined for $a=1$ (and thus in this case it does not represent a solution for eqref{1}), while for $aneq 1$ we have
$$
u(t)=
begin{cases}
0& a=0\
left(dfrac{t}{1-a}right)^frac{1}{1-a} &0<a<1
end{cases}tag{3}label{3}
$$
Now we have reached our goal: in step 1 and step 2 we have shown that eqref{1} has one solution $u(t)=0$ if $a=0,1$, and has at least two solutions, given by the following formula
$$
u(t)=
begin{cases}
0\
left(dfrac{t}{1-a}right)^frac{1}{1-a}
end{cases}text{ for } 0<a<1
$$

Final note: for $a=0,1$ it can be proved that the shown solutions are de facto unique. The related existence and uniqueness theorem ([1], §2.2 p. 36), says that if $v:mathbb{R}tomathbb{R}$ is continuously differentiable, then the solution of the given Cauchy problem exists, is unique and is given by formula eqref{2a} if $v(u_0)neq 0$, or by
$$
quad u(t)=u_0=mathrm{const.}:text{ if }v(u_0)=0tag{4}label{4}
$$
(solutions of type $eqref{4}$ are commonly called equilibrium points). Thus, in the class of Cauchy problems to which eqref{1} belongs, a general method to determine the non uniqueness of the solution is to check if $v(u)notin C^1$ and, in case of affirmative answer, see if there are equilibrium points and other, non constant solutions, that have the same equilibrium point as initial value.

[1] Vladimir Igorevic Arnol'd, "Ordinary differential equations", various editions from MIT Press and from Springer-Verlag, MR1162307 Zbl 0744.34001.