Viscosity solutions for $u'(x) + \alpha u(x) - f(x) = 0$: supersolutions dominate subsolutions

Question

Let $$u'(x) + \alpha u(x) - f(x) = 0,$$ with $x \in [0,\infty)$ and $\alpha \in \mathbb{R}$. Suppose $f \in C(\mathbb{R})$.

If

• $u_1$ is a viscosity supersolution (or a viscosity solution, or a $C^1(\mathbb{R})$ solution) of $u'(x) + \alpha u(x) - f(x) = 0$,

• $u_2$ is a viscosity subsolution of $u'(x) + \alpha u(x) - f(x) = 0$,

• $u_1(0)=u_2(0)$,

how do I prove that $$u_1(x) \ge u_2(x)$$ for all $x \in [0,\infty)$?

Answer 1

If you set $v(x)=e^{\alpha x}u(x)$ then

$$v'(x) - e^{\alpha x}f(x) = 0. \ \ \ \ \ \ \ \ \ \ (*)$$

This equation has no zeroth order term, so the maximum principle applies with no restrictions. This is basically the trick @PaulBryan mentioned in a comment.

This works in the viscosity sense as well. Indeed, suppose $u$ is a viscosity subsolution of

$$u' + \alpha u - f = 0.$$

Let $x_0 \in \mathbb{R}$ and let $\phi$ be a smooth function for which $\phi(x_0)=v(x_0)$ and $v(x) \leq \phi(x)$ for $x$ near $x_0.$ Then $\psi(x_0) = u(x_0)$ and $\psi(x) \leq u(x)$ for $x$ near $x_0$, where $\psi(x) = e^{-\alpha x}\phi(x)$. Therefore $u-\psi$ has a local max at $x_0$ and hence

$$\psi'(x_0) + \alpha \psi(x_0) - f(x_0) \leq 0,$$

and so

$$\phi'(x_0) - e^{\alpha x_0}f(x_0) \leq 0.$$

This verifies that $v$ is a viscosity subsolution of (*). The supersolution verification is similar.

EDIT: Let me add, after removing the zeroth order term, the comparison principle argument is quite standard in the theory of viscosity solutions. You can see the book by Bardi and Capuzzo-Dolcetta for instance. You should pose the problem on the bounded domain $[0,M]$ as @PaulBryan did for the classical argument.

Answer 2

Edit: The $\alpha < 0$ case is answered by @Jeff in another answer. That answer works for $\alpha \geq 0$ also.

In the $C^1$ case here is a proof when $\alpha \geq 0$:

By assumption $$ \begin{split} u_1'(x) + \alpha u_1(x) - f(x) &\geq 0 \\ u_2'(x) + \alpha u_2(x) - f(x) &\leq 0 \end{split} $$ and $$ u_2(0) \leq u_1(0). $$

Now we show that for any $M > 0$, $u_2 \leq u_1$ for $x \in [0, M]$. By continuity we may choose $x_0 \in [0, M]$ such that $$ u_1(x_0) - u_2(x_0) \leq u_1(x) - u_2(x) $$ for all $x_0 \in [0, M]$. In fact, $u_1, u_2$ need not be continuous - $u_1$ need only be lower semi-continuous and $u_2$ need only be upper semi-continuous.

So we just need to show that $u_1(x_0) - u_2(x_0) \geq 0$. If $x_0 = 0$, then we're done. So suppose $x_0 \in (0, M]$. Then we have (allowing for the possibility $x_0 = M$: $$ u_1'(x_0) - u_2'(x_0) \leq 0. $$

Now just use this inequality along the assumptions: $$ u_2'(x_0) + \alpha u_2(x_0) - f(x_0) \leq 0 \leq u_1'(x_0) + \alpha u_2(x_0) - f(x_0) \leq u_2'(x_0) + \alpha u_2(x_0) - f(x_0). $$ Cancelling we then have $$ \alpha u_2(x_0) \leq \alpha u_1(x_0) $$ which implies the desired conclusion $u_1(x_0) - u_2(x_0) \geq 0$ provided $\alpha \geq 0$.

EDIT The following is not a counterexample!

EDIT: Counterexample when $\alpha < 0$.

Take $\alpha = -1$ and $f \equiv 0$. Then the equation is $$ u' - u = 0. $$ Let $u_1(x) = 0$ and $u_2(x) = -\tfrac{1}{2} x + x^2$. Then you can easily check that $u_1(0) = u_2(0) = 0$, $u_1$ is a supersolution, and $u_2$ is a subsolution. But for large $x$, $u_2(x) > 0 = u_1(x)$.

To proceed in the viscosity case, the notes referred to in the comments seem to suggest that one can do a quadratic regularitaztion, considering the function $$ w(x, y) = u_1(x) - u_2(y) + C (x-y)^2 $$ but I have not checked the details.