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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 $x \in [0,\infty)$?

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  • $\begingroup$ Presumably $u_1(0) \geq u_2(0)$ and $\alpha \geq 0$. Otherwise I don't see how the result could be true, even in the $C^1$ case. $\endgroup$
    – Paul Bryan
    Apr 18, 2017 at 3:20
  • $\begingroup$ I'm not so familiar with first order viscosity equations. The second order case is described in the delightfully written "User's guide to viscosity solutions of second order partial differential equations" (arxiv.org/abs/math/9207212). The difficulty faced there is that one must work with sub- and super-jets which is overcome by a semi-convexity regularisation. This is quadratic and I don't know if the analogous requirements are met for first order regularisations. See section 3 for the details. $\endgroup$
    – Paul Bryan
    Apr 18, 2017 at 3:25
  • $\begingroup$ Getting carried away here! See also mathoverflow.net/questions/253300/… and note the comment about non-decreasing $H$ which corresponds to $\alpha \geq 0$ here. $\endgroup$
    – Paul Bryan
    Apr 18, 2017 at 3:26
  • $\begingroup$ @PaulBryan: yes, I assume $u_1(0) = u_2(0)$. However, I need the result to be true for all $\alpha \in \mathbb{R}$ and I hope it is. A proof seems to be sketched in php.math.unifi.it/users/cime/Courses/2011/04/201142-Notes.pdf (page 22, section 5.2 Several variations), but I cannot complete all the missing details. Any ideas? $\endgroup$
    – user102027
    Apr 18, 2017 at 9:28
  • $\begingroup$ You don't have the time variable. The general rule is without the time variable $H$ needs to be monotone increasing in the $u$ variable and hence $\alpha \geq 0$. When you also have $t$, adding an exponential term $e^{-\lambda t}$ allows you to compenste for a lack of monotonicity of $H$. But you don't have this option available to you. $\endgroup$
    – Paul Bryan
    Apr 18, 2017 at 10:45

2 Answers 2

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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.

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  • $\begingroup$ Very good! This solution probably would have been obvious had I been teaching a course on ODE's right now. Standard technique for first order ODE's: make the LHS a total derivative. $\endgroup$
    – Paul Bryan
    Apr 20, 2017 at 6:25
  • $\begingroup$ By the way, the quadratic regularisation I mentioned is only needed if you want to compare two viscosity sub/super solutions. If one of them is smooth, then it works just as you wrote. $\endgroup$
    – Paul Bryan
    Apr 20, 2017 at 6:26
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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.

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  • $\begingroup$ I updated the answer with a counter example. $\endgroup$
    – Paul Bryan
    Apr 19, 2017 at 1:59
  • $\begingroup$ The solution is not greater than the subsolution! $0 \geq -1/2x + x^2$ is not true for sufficiently large $x$. That's why it's a counter example. In other words, your claim is false for $\alpha < 0$. $\endgroup$
    – Paul Bryan
    Apr 19, 2017 at 2:27
  • $\begingroup$ However your $u_2$ is not a subsolution, I think. $\endgroup$
    – user102027
    Apr 19, 2017 at 9:49
  • $\begingroup$ Yes. Sorry! Carried an extra two. $\endgroup$
    – Paul Bryan
    Apr 20, 2017 at 6:19

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