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Robert Israel
Robert Israel
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There are really two separate cases: $$\eqalign{a' + 1 &= +\sqrt{1 - k^2 a^2}\cr a' + 1 &= -\sqrt{1 - k^2 a^2}\cr} $$ You could switch continuously from one to the other only on the curve $1 - k^2 a^2 = 0$ (but at almost all points on that curve the trajectories will be either entering or leaving the forbidden region $1 - k^2 a^2 < 0$).

Let's consider the $+$ case, with an initial condition $a(x_0) = a_0$ where $x_0 > 0$ and $0 < a_0 k(x_0) < 1$. For $x > x_0$ we will have $0 < a(x) < a_0 < 1/k(x)$$0 < a(x) < a_0 < 1/k(x_0)$, and $$ 1 - \dfrac{k^2 a^2}{2} > \sqrt{1-k^2 a^2} > 1 - k^2 a^2 $$ so $$-\dfrac{k^2}{2} > \dfrac{a'}{a^2} = - \left(\frac{1}{a}\right)' > -k^2$$ Thus for $x > x_0$ $$ - 2 \tanh(x) + 2 \tanh(x_0) > \dfrac{1}{a(x_0)} - \dfrac{1}{a(x)} > -4 \tanh(x) + 4 \tanh(x_0) $$ Of course, as $x \to +\infty$ we have $\tanh(x) \to 1$, and we find that the limit of $a(x)$ must be nonzero.

There are really two separate cases: $$\eqalign{a' + 1 &= +\sqrt{1 - k^2 a^2}\cr a' + 1 &= -\sqrt{1 - k^2 a^2}\cr} $$ You could switch continuously from one to the other only on the curve $1 - k^2 a^2 = 0$ (but at almost all points on that curve the trajectories will be either entering or leaving the forbidden region $1 - k^2 a^2 < 0$).

Let's consider the $+$ case, with an initial condition $a(x_0) = a_0$ where $x_0 > 0$ and $0 < a_0 k(x_0) < 1$. For $x > x_0$ we will have $0 < a(x) < a_0 < 1/k(x)$, and $$ 1 - \dfrac{k^2 a^2}{2} > \sqrt{1-k^2 a^2} > 1 - k^2 a^2 $$ so $$-\dfrac{k^2}{2} > \dfrac{a'}{a^2} = - \left(\frac{1}{a}\right)' > -k^2$$ Thus for $x > x_0$ $$ - 2 \tanh(x) + 2 \tanh(x_0) > \dfrac{1}{a(x_0)} - \dfrac{1}{a(x)} > -4 \tanh(x) + 4 \tanh(x_0) $$ Of course, as $x \to +\infty$ we have $\tanh(x) \to 1$, and we find that the limit of $a(x)$ must be nonzero.

There are really two separate cases: $$\eqalign{a' + 1 &= +\sqrt{1 - k^2 a^2}\cr a' + 1 &= -\sqrt{1 - k^2 a^2}\cr} $$ You could switch continuously from one to the other only on the curve $1 - k^2 a^2 = 0$ (but at almost all points on that curve the trajectories will be either entering or leaving the forbidden region $1 - k^2 a^2 < 0$).

Let's consider the $+$ case, with an initial condition $a(x_0) = a_0$ where $x_0 > 0$ and $0 < a_0 k(x_0) < 1$. For $x > x_0$ we will have $0 < a(x) < a_0 < 1/k(x_0)$, and $$ 1 - \dfrac{k^2 a^2}{2} > \sqrt{1-k^2 a^2} > 1 - k^2 a^2 $$ so $$-\dfrac{k^2}{2} > \dfrac{a'}{a^2} = - \left(\frac{1}{a}\right)' > -k^2$$ Thus for $x > x_0$ $$ - 2 \tanh(x) + 2 \tanh(x_0) > \dfrac{1}{a(x_0)} - \dfrac{1}{a(x)} > -4 \tanh(x) + 4 \tanh(x_0) $$ Of course, as $x \to +\infty$ we have $\tanh(x) \to 1$, and we find that the limit of $a(x)$ must be nonzero.

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Robert Israel
Robert Israel
  • 54.2k
  • 1
  • 76
  • 152

There are really two separate cases: $$\eqalign{a' + 1 &= +\sqrt{1 - k^2 a^2}\cr a' + 1 &= -\sqrt{1 - k^2 a^2}\cr} $$ You could switch continuously from one to the other only on the curve $1 - k^2 a^2 = 0$ (but at almost all points on that curve the trajectories will be either entering or leaving the forbidden region $1 - k^2 a^2 < 0$).

Let's consider the $+$ case, with an initial condition $a(x_0) = a_0$ where $x_0 > 0$ and $0 < a_0 k(x_0) < 1$. For $x > x_0$ we will have $0 < a(x) < a_0 < 1/k(x)$, and $$ 1 - \dfrac{k^2 a^2}{2} > \sqrt{1-k^2 a^2} > 1 - k^2 a^2 $$ so $$-\dfrac{k^2}{2} > \dfrac{a'}{a^2} = - \left(\frac{1}{a}\right)' > -k^2$$ Thus for $x > x_0$ $$ - 2 \tanh(x) + 2 \tanh(x_0) > \dfrac{1}{a(x_0)} - \dfrac{1}{a(x)} > -4 \tanh(x) + 4 \tanh(x_0) $$ Of course, as $x \to +\infty$ we have $\tanh(x) \to 1$, and we find that the limit of $a(x)$ must be nonzero.