Euclidean Ordinary Differential Equations with Smooth Dependence: Examples

The curve $x(t)=x_0+tv$ satisfies $x(0)=x_0$ and $x^{\mathrm{\prime }}(t)=v$ for every $t$, so [given] it solves the ODE.

By [L1], the local solution through $x_0$ is unique, so the displayed [L1, step 1.1] affine curve is the solution and the time-$t$ map is translation by $tv$.

Differentiating the displayed formula gives [given] $x^{\mathrm{\prime }}(t)=(e^t{x}_1^0,-e^{-t}{x}_2^0)=Ax(t)$, and $x(0)=(x_1^0,x_2^0)$. So it is a solution.

The matrix $\mathrm{\Phi }$ satisfies ${\mathrm{\Phi }}^{\mathrm{\prime }}(t)=A\mathrm{\Phi }(t)$ and $\mathrm{\Phi }(0)=I_2$, so by [F1, L1, L2, step 1.1] [F1] and [L1] it is the fundamental matrix of this system. Its determinant is $e^t{e}^{-t}=1$, which is consistent with [L2].

Therefore the linear system has the stated solution operator and [step 2.1] fundamental matrix.

Setting $p=q^{\mathrm{\prime }}$ turns $q^{\mathrm{\prime }\mathrm{\prime }}+q=0$ into the displayed first-order system, and [given] conversely differentiating the first equation and substituting the second recovers $q^{\mathrm{\prime }\mathrm{\prime }}+q=0$.

Differentiating the displayed sine-cosine formulas gives [L1, L2, step 1.1] $q^{\mathrm{\prime }}=p$ and $p^{\mathrm{\prime }}=-q$, so they solve the first-order system with $(q(0),p(0))=(q_0,p_0)$. By [L1] the solution is unique, and [L2] identifies the system as the oscillator written in first-order form.

Therefore the harmonic oscillator fits the first-order smooth-ODE framework [step 2.1] exactly as claimed.

The support of $V(x)=\rho (x)v$ is contained in the compact support of $\rho$, [given] and $V$ is smooth because it is a scalar multiple of the constant vector $v$ by a smooth scalar function.

Therefore [L1] applies and makes every maximal trajectory of $V$ global. [L1, step 1.1] Outside the support of $\rho$, the field vanishes and the solution is locally constant, which is consistent with that completeness conclusion.

The curve $x(t)=e^{\lambda t}{x}_0$ satisfies $x(0)=x_0$ and [given] $x^{\mathrm{\prime }}(t)=\lambda e^{\lambda t}{x}_0=\lambda x(t)$, so it solves the ODE.

Differentiating the explicit formula gives [L1, L2, step 1.1] ${\mathrm{\partial }}_{x_0}x(t;\lambda ,x_0)=e^{\lambda t}$ and ${\mathrm{\partial }}_{\lambda }x(t;\lambda ,x_0)=t{e}^{\lambda t}{x}_0$, and higher derivatives are again polynomial multiples of $e^{\lambda t}$. Thus the solution depends smoothly on both data variables, exactly as [L1] and [L2] predict.

So this ODE is a concrete instance of smooth dependence on initial data and [step 2.1] parameters.

The augmented system has $s(t)=t+t_0$ because $s^{\mathrm{\prime }}=1$ and $s(0)=t_0$. Then [given] $x^{\mathrm{\prime }}=s(t)x=(t+t_0)x$, so solving this linear scalar equation gives the displayed exponential formula.

Writing $\tau =t+t_0$ turns the displayed solution into [L1, step 1.1] $x(\tau )=x_0{e}^{({\tau }^2-t_0^2)\mathrm{/}2}$, which is exactly the solution of the original nonautonomous equation. This is the concrete reduction promised by [L1].