Consider \[ x''+x'+x^3=0 \] and \[ F(x,x')=(x+x')^2+(x')^2+x^4. \]
- Show that \(F\) decreases along trajectories of the differential equation.
- Show that every solution satisfies \[ (x(t),x'(t))\longrightarrow(0,0) \qquad(t\to\infty). \]
Solution
Set \[ y(t)\coloneqq x'(t), \qquad E(t)\coloneqq F(x(t),y(t)). \] Then the differential equation is \[ x'=y, \qquad y'=-y-x^3. \]
1. Along every solution,\[ E'(t)=-2x(t)^4-2y(t)^2\leq0. \]
Differentiating \[ E=(x+y)^2+y^2+x^4 \] and using \(x'=y\) and \(y'=-y-x^3\) gives \[ \begin{aligned} E' &=2(x+y)(x'+y')+2yy'+4x^3x'\\ &=2(x+y)(-x^3)+2y(-y-x^3)+4x^3y\\ &=-2x^4-2y^2. \end{aligned} \] Thus \(E\) is nonincreasing, and it decreases strictly away from \((x,y)=(0,0)\).
2. Every solution is defined for all \(t\geq0\), and both \(x(t)\) and \(y(t)\) are bounded on \([0,\infty)\).
By step 1, \[ 0\leq E(t)\leq E(0). \] Since \[ y(t)^2+x(t)^4\leq E(t), \] both \(x(t)\) and \(y(t)\) remain bounded as long as the solution exists. The vector field \[ (x,y)\longmapsto (y,-y-x^3) \] is locally Lipschitz. A maximal solution that remains in a bounded subset of \(\RR^2\) cannot terminate at a finite positive time, so the solution extends to all \(t\geq0\).
3. One has\[ \int_0^\infty x(t)^4\,dt<\infty, \qquad \int_0^\infty y(t)^2\,dt<\infty. \]
The nonincreasing function \(E\) is bounded below by \(0\), so \(E(t)\) has a finite limit \(E_\infty\geq0\). Integrating step 1 from \(0\) to \(T\) gives \[ 2\int_0^T\bigl(x(t)^4+y(t)^2\bigr)\,dt =E(0)-E(T) \leq E(0). \] Letting \(T\to\infty\) proves both integrability statements.
4. Every solution satisfies\[ (x(t),x'(t))\longrightarrow\boxed{(0,0)} \qquad(t\to\infty). \]
By step 2, \(y=x'\) is bounded, so \(x\) is uniformly continuous on \([0,\infty)\). Also \[ y'=-y-x^3 \] is bounded because both \(x\) and \(y\) are bounded; hence \(y\) is uniformly continuous as well.
3. Therefore \(x(t)\to0\).
The same argument applied to the uniformly continuous function \(y\), using the finiteness of \(\int_0^\infty y(t)^2\,dt\) from step 3, gives \(y(t)\to0\). Since \(y=x'\), the asserted limit follows.
5. Q.E.D.
Step 1 proves part 1, and step 4 proves part 2.