I want to find functions $N:\mathbb{R} \to \mathbb{R}$ and $S:\mathbb{R} \to \mathbb{R}$ satisfying the following:
$0 = -6N(x)^2 + 15 S(x)^2$
$\frac{dN}{dx} = \frac{12}{49} - 3 S(x)^2$
$\frac{dS}{dx} = 3 N(x) S(x)$
.....subject to some initial condition $N(0)=n_0$.
The way I've thought about doing this is plugging the constraint $S(x)^2 = \frac{6}{15} N(x)^2$ into the ODE for $\frac{dN}{dx}$. This gives me the IVP:
$\frac{dN}{dx} = \frac{12}{49} - \frac{6}{5} N(x)^2$ with $N(0)=n_{0}$
The solution to this IVP is:
$N(x) = \frac{\sqrt{10}}{7} \tanh\left( \frac{6\sqrt{10}}{35}x+\tanh^{-1}\left( \frac{7}{\sqrt{10}} n_{0} \right) \right)$
I have checked this pretty thoroughly and this works out well.
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Now what comes up next is what is confusing me. Should I now plug $N(x)$ into the ODE for $\frac{dS}{dx}$? This would give me the ODE:
$\frac{dS}{dx} = 3 N(x) S(x) = \frac{3\sqrt{10}}{7} \tanh\left( \frac{6\sqrt{10}}{35}x+\tanh^{-1}\left( \frac{7}{\sqrt{10}} n_{0} \right) \right) S(x)$
Furthermore, because of the constraint $S(x)^2 = \frac{6}{15} N(x)^2$, the initial condition for this function must be $S(0) = s_{0} = \pm \sqrt{\frac{6}{15}} |n_{0}|$. Solving this next IVP gives the function: $S(x) = \pm \sqrt{\frac{2}{5}} |n_{0}| \left(1-\left(\frac{7}{\sqrt{10}}n\right)^2\right)^{\frac{5}{4}} \left[ \cosh \left( \frac{6\sqrt{10}}{35}x+\tanh^{-1}\left( \frac{7}{\sqrt{10}} n_{0} \right) \right) \right]^{\frac{5}{2}}$
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So all seems well and I'm happy to have found solutions to the DEs...
HOWEVER, when I plug $S(x)$ and $N(x)$ into the constraint condition, I find that the constraint equation is NOT satisfied! Meaning, $-6N(x)^2 + 15 S(x)^2 \neq 0$.
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What is going on? Since I used the constraint condition to solve the ODEs, I would think that the solutions should be automatically satisfying the constraint condition. Why isn't this the case?
The system is overdetermined. Two differential equations, an algebraic equation and one initial condition.
Divide the differential equations: $$ \frac{N'}{S'}=\frac{12/49-3\,S^2}{3\,N\,S}.\tag1 $$ Differentiating the algebraic relation we get $$ 12\,N\,N'=30\,S\,S'\implies \frac{N'}{S'}=\frac{5\,S}{2\,N}.\tag2 $$ From (1) and (2) $$ \frac{12/49-3\,S^2}{3\,N\,S}=\frac{5\,S}{2\,N}\implies S^2=\frac{8}{343}. $$ This implies that $N$ and $S$ are constants, but the constant solution of the system is $$ N=0,\quad S^2=\frac{4}{49}. $$