For $x'(t)= f(t, x(t)) + \int_0^t g(x(s))\, ds$ show that if $x_1(0) \leq x_2(0)$ that also $x_1(t) \leq x_2(t)$

Question

Note

The Question has changed significantly since it was posed, but it is now as follows:

Question

Suppose we have some integro or delayed differential equation: $$ x'(t)=f(t,x(t)) + \int_0^t g(x(s))\, ds + \sum_n h(x(t-a_n)), $$ where we assume that conditions are satisfied such that this equation always has a unique, continuous solution. let $x_1(t)$ and $x_2(t)$ be two solutions with some boundary conditions $x_1(0)=x_1$ and $x_2(0)=x_2$ such that $x_1 \leq x_2$. I would now like to show that we have for all $t: x_1(t) \leq x_2(t)$. What are some methods for showing this? I know of the following method:

• Suppose there is some $t>0$ such that $x_1(t) > x_2(t)$, then there exists a value $s \in [0,t]$ such that $x_1(s)=x_2(s)$ and for all $u\leq s: x_1(u) \leq x_2(u)$. If one can show under these conditions that $x_1'(t) = f(t,x_1(t)) \leq f(t,x_2(t)) = x_2'(t)$ then it follows that $x_1(t)$ stays below $x_2(t)$ for all $t$.
• Write $x(t) = \int_0^t f(s,x(s))\, ds$ and show that $\int_0^t f(s,x_1(s))\, ds \leq \int_0^t f(s,x_2(s))\, ds$ for all $t$.

The first method is easy to apply but too restrictive and the second method is not restrictive at all but is hard to apply.

My question is a reference or an explenation of other methods in case the first method is not applicable.

An Example

Suppose for example we have the following Delay Differential Equation: $$ \begin{cases} x'(t) &= x(t)-1 \mbox{ if } t \leq 1\\ x'(t) &= x(t-1) (x(t) - x(t-1)) \mbox{ if } t >1 \end{cases} $$ this delay differential equation satisfies the property that if we have $x_1,x_2 \in (0.8,1)$ with $x_1 \leq x_2$ and $x_1(t), x_2(t)$ are solutions satisfying $x_i(0) = x_i$ then for all $t: x_1(t) \leq x_2(t)$. I can check this numerically but not analytically.

Answer 1

This is wrong in general

The differential equation $$x^\prime = \vert x \vert^{\dfrac{1}{2}}$$

has $x_1(t)=0$ as a solution. Another one is $$x_2(t)=\begin{cases} 0& \text{for } t\le 1\\ \dfrac{(t-1)^2}{4} & \text{for } t>1 \end{cases}$$

You have $x_2(0) \le x_1(0)$. However $x_2(2)>x_1(2)$.

You need to add additional hypothesis’s. The ones of Picard–Lindelöf theorem are for example sufficient as under this theorem solutions are unique.

If unicity of solutions is assumed, then you can use a simple argument.

Suppose that $x_2(t_0) > x_1(t_0)$ and $x_1(t_1)=x_2(t_1)$ with $t_0<t_1$. Then $x_1, x_2$ are two solutions on $[0, t_1]$. Another one is $$x_3(t)=\begin{cases} x_2(t) & \text{for } t\le t_0\\ x_1(t) & \text{for } t \ge t_0 \end{cases}$$ in contradiction with unicity of IVP.

Answer 2

This is not true without extra conditions.

Let $f(t,x) = 2 \sqrt{|x|}$.

Then let $x_p(t) = 0$ for $t < p$ and $x_p(t) = (t-p)^2$ for $t \ge p$. It is straightforward to show that if $p \ge 0$ then $x_p$ is a solution to the problem $\dot{x} = f(t,x)$, $x(0) =0$.

Also, note that $x_z(t) = 0$ is a solution.

Then we see that $x_1(0) \le x_z(0)$ but it is not true that $x_1(t) \le x_z(t)$ for all $t \ge 0$.