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?

  • 1
    $\begingroup$ The problem is ill-posed. The system of ODEs has a unique solution (at least locally due to locally Lipschitz flow) that does not satisfy the algebraic constraint . $\endgroup$ – RTJ Dec 9 '15 at 18:00

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 $$ N=0,\quad S^2=\frac{4}{49}. $$


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