04 - Comparison Principle

Definition

Let $g:{\mathbb{R}}^2\to \mathbb{R}$ be continuously differentiable
Let $u(x)$ and $v(x)$ be differentiable functions which satisfy

$$u^{\prime }(x)\le g(x,u(x))\text{ and }{v}^{\prime }(x)\ge g(x,v(x))$$

on some interval $I$

If $u(x_0)\le v(x_0)$ for some $x_0\in I$ then

$$u(x)\le v(x)\text{ for all }x\in I,x\ge x_0$$

If $u(x_0)\ge v(x_0)$ for some $x_0\in I$ then

$$u(x)\ge v(x)\text{ for all }x\in I,x\le x_0$$

Proof

Since $g$ is continuously differentiable use MVT to check that $h\to {\mathbb{R}}^3\to \mathbb{R}$ of 3 variables
which is defined by difference quotient

$$h(x,y_1,y_2):=\{\begin{array}{ll}\frac{g(x,y_1)-g(x,y_2)}{y_1-y_2}& \text{ if }{y}_1\ne y_2\\ {\partial }_{y}g(x,y_1)& \text{ if }{y}_1=y_2\end{array}$$

is continuous

As both functions $u(x)$ and $v(x)$ are continuous on interval $I$ then composition

$$a(x):=h(x,u(x),v(x))\text{ is also continuous}$$

Crucially the function is chosen so that we can write

$$g(x,v(x))-g(x,u(x))=a(x)(v(x)-u(x))$$

From assumptions of theorem then we get $w(x):=v(x)-u(x)$ satisfying

$$w^{\prime }(x)=v^{\prime }(x)-u^{\prime }(x)\ge g(x,v(x))-g(x,u(x))=a(x)w(x)$$

Setting $A(x)={\int }_{x_0}^{x}a(t)dt$ to get function $A^{\prime }(x)=a(x)$ hence

$$(e^{-A(x)}w(x){)}^{\prime }=e^{-A(x)}(w^{\prime }(x)-A^{\prime }(x)w(x))$$

Hence

$$e^{-A(x)}w(x)\text{ is non-decreasing}$$

If $u(x_0)\le v(x_0)$ then

$$e^{-A(x_0)}w(x_0)=w(x_0)\ge 0$$

Hence for $x\ge x_0$ we have $e^{-A(x)}w(x)\ge 0$ and $w(x)\ge 0$ so

$$u(x)\le v(x)$$

On the other hand if $u(x_0)\ge v(x_0)$ for $x\le x_0$ then

$$e^{-A(x)}w(x)\le w(x_0)\le 0$$

Hence

$$u(x)\ge v(x)$$