Population dynamics, square root and zero derivative Hy everybody ! 
I'm studying population dynamics for my calculus exam, and I don't understand something that seems really easy, so I thought you might be able to help me out ;)
Here's the thing. I have this differential equation $\frac{dN}{dt} = \sqrt{N}$.
Our book makes us realize that both $N(t) = 0$ and $N(t) = \frac{t^2}{4}$ are solution, which makes sense so far, simply by replacing in the original equation.
Now suppose we start at $N(0) = 0$. How can $N$ start growing like $\frac{t^2}{4}$ if it's derivative is $0$ at $t = 0$ ? Because zero derivative should mean no growth, so $\sqrt{N}$ should remain zero, which means still no growth, and so on. My brain is melting right now.
 A: Zero derivative at one point does not mean no growth at other points. 
For example the  derivative of function $$N(t)=\frac {t^2}{4}$$ is $$N'(t)=t/2$$ That is the derivative is zero at $t=0$ but it is not zero at other points. The point of the problem is that in this case the solution is not unique.  
A: Apart from these two solutions you can also combine them to solutions that are zero up to some point $t\le c$ and then follow the shifted quadratic function, $N(t)=\frac14(t-c)^2$ for $t>c$.
The irritation you get is based on the mis-understanding of what it means that a derivative is zero. It just means that $N(h)=o(h)$ for $h\approx 0$. 
If the differential equation were Lipschitz-continuous, small deviations of this size in any consistent integration method advancing with step size $h$ add up to an error that is still small for the whole solution. In other words, the Euler method and every other consistent method converges towards the unique exact solution. This would still be true if one were to introduce artificial random perturbations of size $h^2$.
In this case however, a small initial error gets magnified into a different solution. If one takes the Euler method with random perturbations, $N_{k+1}=N_k+hf(N_k)+h^2r_k$, the numerical solution will rapidly move away from the unstable zero solution and follow closely the other $N(t)=\frac14(t-c)_+^2$, with a higher probability for $c$ being close to zero.

def Euler(f,t):
    y = np.zeros_like(t);
    for k in range(1,len(t)):
        h=t[k]-t[k-1];
        y[k]=y[k-1]+h*f(y[k-1])+h**2*(np.random.random()-0.5)
    return y

t = np.linspace(0,2,41);
plt.subplot(2,1,1); plt.title("Non-Lipschitz: $y'=\sqrt{|y|}$")
for k in range(15): y=Euler(lambda y: abs(y)**0.5, t); plt.plot(t,y);
plt.grid(); 
plt.subplot(2,1,2); plt.title("Lipschitz: $y'=0.5y+0.3$")
for k in range(15): y=Euler(lambda y: 0.5*y+0.3, t); plt.plot(t,y);
plt.grid(); plt.show()

A: You are correct. There are indeed two solutions. If the population is zero it will stay zero. The only way it will grow is if the initial conditions are positive. However, the population zero solution is "unstable" in the sense that even if the initial condition is slightly positive, the $t^2$ solution will kick in and push the population away from zero.
