Dynamical systems proof that $ f(t)$ is less than or equal to $g(t)$

Suppose that $f'(t) \le F(f(t),t)$ where $F$ is continuously differentiable. If $g(t)$ is a solution to the equation $g'(t) = F(g(t),t)$ and $g(a) = f(a)$, then prove that $f(t) \le g(t)$ for all $t \ge a$

Solution 1:

The ODE $y'(t)=F(y(t),t)$ has a parametrized general solution or flow $\phi(t;t_0,x_0)$. The exact solution of the claim is thus obtained as $g(t)=\phi(t;a,f(a))$. Further properties of the flow are \begin{align} \text{ODE: } && \partial_1\phi(t;t_0,x_0)&=F(\phi(t;t_0,x_0),t),\\ \text{composition: } && \phi(t;t_0,x_0)&=\phi(t;s,\phi(s;t_0,x_0)),\\ \text{$s$-derivative: } && 0&=\partial_2\phi(t;s,\phi(s;t_0,x_0))+\partial_3\phi(t;s,\phi(s;t_0,x_0))F(\phi(s;t_0,x_0),s),\\ \text{set $s=t_0$: } && \partial_2\phi(t;t_0,x_0)&=-\partial_3\phi(t;t_0,x_0)F(x_0,t_0). \end{align} where $\partial_k\phi$ is the partial derivative for the $k$th argument.

Lemma: $\phi$ is monotonously increasing in its third argument, if $x_0<y_0$ then $\phi(t;t_0,x_0)<\phi(t;t_0,y_0)$ for all times.

Prf.: If solutions start at different values at the same time, then their curves can not cross, as at the crossing point they would be solutions of the same IVP and thus identical by the uniqueness theorem.

Cor.: Thus $\partial_3\phi(t;t_0,x_0)\ge 0$.

Now consider $$ h(t)=\phi(a;t,f(t)) $$ which projects back the graph of $f$ along the flow trajectories of $F$ to the initial values at $t=a$. Then \begin{align} h'(t)&=\partial_2\phi(a;t,f(t))+\partial_3\phi(a;t,f(t))f'(t) \\ &=\partial_3\phi(a;t,f(t))[-F(f(t),t)+f'(t)] \\ &\le 0 \end{align}

Thus because of $h(a)=f(a)$ we get $h(t)\le h(a)=f(a)$ for $t\ge a$ and in the grand conclusion $$ f(t)=\phi(t;a,h(t))\le \phi(t;a,f(a))=g(t), $$ as was claimed.