Consider \[ x'=\frac{x^3-x}{1+e^x}. \]
- Find all constant solutions.
- Discuss \(\lim_{t\to\infty}x(t)\) for the solution satisfying \(x(0)=1/2\).
Solution
Put \[ F(x)\coloneqq\frac{x^3-x}{1+e^x} = \frac{x(x-1)(x+1)}{1+e^x}. \]
1. The constant solutions are exactly\[ \boxed{x(t)\equiv-1,\qquad x(t)\equiv0,\qquad x(t)\equiv1.} \]
A constant function \(x(t)\equiv c\) solves the equation exactly when \(F(c)=0\). Since \(1+e^c>0\) for every real \(c\), this is equivalent to \[ c(c-1)(c+1)=0, \] whose solutions are \(c=-1,0,1\).
2. On the interval \((0,1)\),\[ F(x)<0. \]
For \(0<x<1\), one has \[ x>0, \qquad x-1<0, \qquad x+1>0, \qquad 1+e^x>0. \] The displayed factorization of \(F\) therefore gives \(F(x)<0\).
3. The solution with \(x(0)=1/2\) satisfies\[ 0<x(t)\le\frac12 \]for every \(t\ge0\) in its maximal forward interval, and it is strictlydecreasing there.
The vector field \(F\) is smooth, so the initial-value problem has locally unique solutions. While the solution lies in \((0,1)\), step 2 gives \(x'(t)<0\), hence it decreases from \(1/2\).
It cannot reach \(0\) at a finite time. If \(x(t_0)=0\), then both the given solution and the constant solution \(x\equiv0\) solve the same initial-value problem at time \(t_0\). Uniqueness would force them to agree, contradicting \(x(0)=1/2\). Thus \(x(t)>0\), and consequently it remains in \((0,1/2]\) and is strictly decreasing.
4. The solution from step 3 exists for every \(t\ge0\).
By step 3, its graph remains in the compact interval \([0,1/2]\). The smooth vector field \(F\) is bounded and locally Lipschitz on a neighborhood of that interval. Therefore a finite maximal forward endpoint would admit continuation of the solution, contradicting maximality. Hence the forward interval is \([0,\infty)\).
5. The limit\[ L\coloneqq\lim_{t\to\infty}x(t) \]exists and satisfies \(F(L)=0\).
3. Since \(F(x)\le0\) on \([0,1/2]\), the only remainingpossibility is \(F(L)=0\).
6. For the solution with \(x(0)=1/2\),\[ \boxed{\lim_{t\to\infty}x(t)=0.} \]
By step 5, the limit is a zero of \(F\) in \([0,1/2]\). Step 1 shows that the only such zero is \(0\).
7. Q.E.D.
Step 1 answers part (1), and step 6 answers part (2).