Prove that the condition lim (t→+∞)∈t_0^t a(s) ds<+∞ is…

Prove that the condition $$\lim_{t\to+\infty}\int_0^t a(s)\,ds<+\infty$$ is necessary and sufficient for the trivial solution of the equation $$\frac{dx}{dt}=a(t)\,x,$$ where $a(t)$ is a continuous function, to be stable in the sense of Lyapunov.

Step 1 · General solution

Separating variables in $\dot x=a(t)x$ gives $$x(t)=x_0\,\exp\!\left(\int_0^t a(s)\,ds\right),$$ where $x_0=x(0)$. The trivial solution corresponds to $x_0=0$.

Step 2 · Notation

Define $A(t)=\int_0^t a(s)\,ds$, so $x(t)=x_0e^{A(t)}$ and $|x(t)|=|x_0|e^{A(t)}$.

Step 3 · Definition of Lyapunov stability

$x\equiv0$ is stable if for every $\varepsilon>0$ there is $\delta>0$ such that $|x_0|<\delta\implies|x(t)|<\varepsilon$ for all $t\geq0$.

Step 4 · Sufficiency

Suppose $A(t)\leq M<+\infty$ for all $t\geq0$. Given $\varepsilon>0$, choose $\delta=\varepsilon e^{-M}$. Then $|x_0|<\delta$ gives $|x(t)|\leq|x_0|e^M<\varepsilon$ for all $t\geq0$.

Step 5 · Necessity

Suppose $\sup_{t\geq0}A(t)=+\infty$, so $A(t_n)\to+\infty$ for some $t_n\to+\infty$. If the trivial solution were stable with $\varepsilon=1$, some $\delta>0$ would give $\tfrac{\delta}{2}e^{A(t)}<1$ for all $t\geq0$ (taking $x_0=\delta/2$), i.e.\ $A(t)<\ln\tfrac2\delta$ for all $t$ — contradicting $A(t_n)\to+\infty$. Hence $\sup_{t\geq0}A(t)<+\infty$ is necessary.

Step 6 · Reformulation of the condition

The condition $\sup_{t\geq0}A(t)<+\infty$ is precisely the boundedness of $A(t)=\int_0^t a(s)\,ds$, matching the statement of the problem.

Step 7 · Asymptotic stability — remark

$|x(t)|\to0\iff A(t)\to-\infty\iff\int_0^{+\infty}a(s)\,ds=-\infty$. Otherwise: stable but not asymptotically if $A(t)$ is bounded but does not tend to $-\infty$; unstable if $\sup A(t)=+\infty$.

Step 8 · Illustrative examples

$a(t)=\sin t$: $A(t)=1-\cos t\in[0,2]$, stable, not asymptotic. $a(t)=-1$: $A(t)=-t\to-\infty$, asymptotically stable. $a(t)=t$: $A(t)=t^2/2\to+\infty$, unstable.

Step 9 · Complete statement of the theorem

Theorem. Let $a:[0,+\infty)\to\mathbb{R}$ be continuous. The trivial solution of $\dot x=a(t)x$ is Lyapunov stable if and only if $\sup_{t\geq0}\int_0^t a(s)\,ds<+\infty$.

Proof. Sufficiency and necessity are exactly the arguments of Sections~4 and~5 above, applied to $A(t)=\int_0^t a(s)\,ds$.

Step 10 · Conclusion

$$\boxed{ \text{The trivial solution of }\dot{x}=a(t)x\text{ is Lyapunov stable} \iff \sup_{t\geq0}\int_0^t a(s)\,ds<+\infty. }$$ The proof relies on the explicit solution $x(t)=x_0\exp(\int_0^t a(s)\,ds)$: boundedness above of the integral gives a uniform bound witnessing stability, while unboundedness forces unbounded growth for any non-trivial initial condition.

Summary Table

Additional Notes

• The argument hinges on relating the scalar equation to its associated planar system and translating the geometric property (centre, exactness, symmetry) into an algebraic condition on the trace and determinant of the corresponding matrix.