Verify whether the trivial solution of a system of…

Verify whether the trivial solution of a system of differential equations is stable in the sense of Lyapunov, given that the general solution of the system has the form $$x=\frac{c_1-c_2 t}{1+t^2},\qquad y=(c_1 t^3+c_2)\,e^{-t}.$$

Step 1 · Identification of the trivial solution

Setting $x\equiv0$, $y\equiv0$ requires $c_1-c_2t=0$ for all $t$ (hence $c_1=c_2=0$), consistent with $(c_1t^3+c_2)e^{-t}\equiv0$.

Step 2 · Definition of Lyapunov stability

$x=0,y=0$ is stable if for every $\varepsilon>0$ and $t_0$ there is $\delta>0$ such that $|x(t_0)|^2+|y(t_0)|^2<\delta^2$ implies $|x(t)|^2+|y(t)|^2<\varepsilon^2$ for all $t\geq t_0$.

Step 3 · Initial conditions at $t=t_0$

Solving the linear system for $c_1,c_2$ in terms of $x_0=x(t_0)$, $y_0=y(t_0)$ (via Cramer's rule; the coefficient determinant $\Delta=\tfrac{e^{-t_0}(1+t_0^4)}{1+t_0^2}\neq0$ for finite $t_0$) shows $c_1,c_2$ depend continuously on $(x_0,y_0)$, vanishing when $(x_0,y_0)=(0,0)$: $|c_1|+|c_2|\leq K(t_0)(|x_0|+|y_0|)$.

Step 4 · Estimation of $|x(t)|$ for $t\geq t_0\geq0$

Using $\tfrac{|t|}{1+t^2}\leq\tfrac12$ and $\tfrac{1}{1+t^2}\leq1$: $$|x(t)|\leq|c_1|+\tfrac{|c_2|}{2}\leq|c_1|+|c_2|.$$

Step 5 · Estimation of $|y(t)|$ for $t\geq t_0\geq0$

The function $h(t)=t^3e^{-t}$ has maximum $27e^{-3}$ at $t=3$, so $$|y(t)|\leq27e^{-3}|c_1|+|c_2|.$$

Step 6 · Combined bound

$$\sqrt{x(t)^2+y(t)^2}\leq\bigl(1+27e^{-3}\bigr)(|c_1|+|c_2|) =M_1(|c_1|+|c_2|)\leq\sqrt2\,M_1\sqrt{c_1^2+c_2^2}.$$

Step 7 · Bounding $c_1,c_2$ in terms of initial data

Combining with Section~3: $\sqrt{x(t)^2+y(t)^2}\leq\sqrt2\,M_1K(t_0) \sqrt{x_0^2+y_0^2}$ for all $t\geq0$.

Step 8 · Verification of Lyapunov stability

Let $C(t_0)=\sqrt2\,M_1K(t_0)$. Given $\varepsilon>0$, choose $\delta=\varepsilon/C(t_0)$; then $\sqrt{x_0^2+y_0^2}<\delta$ implies $\sqrt{x(t)^2+y(t)^2}<\varepsilon$ for all $t\geq t_0$: Lyapunov stability holds.

Step 9 · Checking asymptotic stability

As $t\to+\infty$: $x(t)=\tfrac{c_1-c_2t}{1+t^2}\to0$ (since $t/(1+t^2)\to0$), and $y(t)=(c_1t^3+c_2)e^{-t}\to0$ (exponential decay dominates polynomial growth). So $\sqrt{x(t)^2+y(t)^2}\to0$ for every $(c_1,c_2)$: the trivial solution is also asymptotically stable.

Step 10 · Uniformity with respect to $t_0$

Since $K(t_0)$ grows with $t_0$ (through the factor $t_0^3$), the required $\delta$ shrinks as $t_0\to+\infty$: the solution is stable but not uniformly stable.

Step 11 · Summary of estimates

Step 12 · Conclusion

$$\boxed{ \text{The trivial solution }x\equiv0,\;y\equiv0 \text{ is \textbf{asymptotically stable} in the sense of Lyapunov.} }$$ For any fixed $t_0$, the bounds $|x(t)|\leq|c_1|+\tfrac{|c_2|}{2}$ and $|y(t)|\leq27e^{-3}|c_1|+|c_2|$ (for $t\geq0$) show that small initial data give small solutions for all $t\geq t_0$ (Lyapunov stability), while $x(t)\to0$ and $y(t)\to0$ as $t\to+\infty$ for every $(c_1,c_2)$ confirms asymptotic stability.

Summary Table

Additional Notes

• Because an explicit general solution is already available, no linearisation or eigenvalue argument is needed: stability is read off directly from how each solution component behaves as a function of the initial data and of time.