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Let $f : \Bbb{R} \to \Bbb{R}$ be twice differentiable with $f'' > 0$, and let $u_- > u_+$ be real numbers. Show that there exists a solution $\varphi(x)$ to the following differential equation: $$ -s\varphi' + f'(\varphi)\varphi' = \varphi'' \tag{1} $$ such that $\lim_{x \to \pm\infty} \varphi(x) = u_\pm$, and where $s = \frac{f(u_+) - f(u_-)}{u_+ - u_-}$.

My initial attempt is to observe that this DE can be nicely integrated to the following: $$ \varphi' = f(\varphi) - s\varphi + C \tag{2} $$ Thus, it suffices to show the existence of a solution for this DE instead, where we are free to choose $C$. I attempted to bring RHS over to LHS, which gives: $$ \int \frac{1}{f(\varphi) - s\varphi + C} \; \mathrm{d}\varphi = x + D $$ where $D \in \Bbb{R}$. Thus, if we define: $$ g(x) = \int \frac{1}{f(x) - sx + C} \; \mathrm{d}x $$ and assuming that $g$ is invertible, then $\varphi(x) = g^{-1}(x)$ would be a solution to $(2)$. However, there are a few issues in this approach that we need to tackle:

  1. The integral will not make sense if $f(\varphi) - s\varphi + C$ vanishes at some point in $\Bbb{R}$. As we are free to choose $C$, if we can show that $f(\varphi) - s\varphi$ is bounded from either above or below, then such a choice of $C$ will exist. I suspect we can use the convexity and the definition of $s$ to prove this, but my attempts are futile so far.
  2. Should the integral make sense, another problem is if $g$ is invertible. However, this should not be an issue as by FTOC: $$ g'(x) = \frac{1}{f(x) - sx + C} $$ so if the denominator does not vanish, $g'$ is continuous and so must be strictly positive or negative, hence $g$ is strictly monotone, thus invertible.
  3. The biggest issue here is that this definition does not guarantee the requirement of $\lim_{x \to \pm\infty} \varphi(x) = u_\pm$. I tried to manipulate the integral to fit this condition, but to no avail so far.

I also tried other approaches, such as using Picard's iteration, but as this problem is not really an IVP they have not been successful.

Any help is appreciated.

EditPiAf
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Clement Yung
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  • You need $g(u_\pm)=\pm\infty$. This is not possible when the integrand is regular at these points, thus the denominator needs to be zero there, $f(u_\pm)-su_\pm+C=0$. These conditions should give the values for $s$ as given, and for $C$. It still remains to address all the other raised concerns. – Lutz Lehmann Aug 15 '20 at 14:07
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    Or even earlier, $\lim_{x\to\pm\infty}\phi(x)=u_\pm$ for a smooth function $\phi$ implies that $\lim_{x\to\pm\infty}\phi'(x)=0$, which again in the integrated equation leads to $0=f(u_±)−su_±+C$. So that $f(u_+)-f(u_-)=s(u_+-u_-)$ and $u_+f(u_-)-u_-f(u_+)=C(u_+-u_-)$. – Lutz Lehmann Aug 15 '20 at 14:19
  • @LutzLehmann I see, so the $C$ here is actually determined by the condition and is not up to our choice. Any idea where the convexity comes into place? – Clement Yung Aug 15 '20 at 15:22
  • @EditPiAf thank you, that was very helpful. If you don't mind, I want to clarify two things: 1) How can we ensure that the definition of $v(s)$ makes sense (i.e. the integrand is indeed integrable, and there exists such a function satisfying the equality since it's implicitly defined), and 2) how did you ensure that $v$ is a smooth decreasing function from $u_l$ to $u_r$? – Clement Yung Aug 17 '20 at 03:56
  • You do not need the integral, and also not the derivatives at $u_\pm$. You only need that $f(φ)−sφ+C$ vanishes at $u_\pm$, is strictly negative inside the interval and positive outside, due to the convexity of $f$. The general facts on one-dimensional dynamics then give the claim. – Lutz Lehmann Aug 17 '20 at 20:08
  • @LutzLehmann Thank you, but I'm new to one-dimensional dynamics. Am I right to interpret your comment as "as long as $\varphi'$ vanishes $u_\pu$ and is strictly negative inside the interval and positive outside, a solution to $\varphi$ exists? If so, where can I refer for what you said as the "general facts"? – Clement Yung Aug 18 '20 at 02:15
  • $φ$ exists locally because the integrated DE is differentiable and a solution starting between stationary solutions is forever bounded by these stationary solutions, giving the global existence. The "general facts" are essentially that the roots of the right side are the only limit points and solutions starting between two roots are monotonous with the roots as limit points. – Lutz Lehmann Aug 18 '20 at 06:27
  • @LutzLehmann I got it, thank you! – Clement Yung Aug 18 '20 at 14:28

1 Answers1

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Using the limits at $\pm\infty$, we find $$ C = su_+ - f(u_+) = su_- - f(u_-) \, , $$ $$ \text{and}\qquad \varphi' = f(\varphi) - f(u_+) - s(\varphi - u_+) = f(\varphi) - f(u_-) - s(\varphi - u_-) \, , $$ see this exercise in Evans PDE. The strict convexity of $\varphi\mapsto \varphi'$ follows from the strict convexity $f''>0$ of $f$. This property yields $\varphi' < 0$ for $\varphi \in \left]u_+, u_-\right[$. Therefore, $\varphi$ is a smooth decreasing function, that decreases from $u_-$ to $u_+$. To investigate the stability of the equilibrium $\varphi = u_\pm$, we compute the sign of the derivative $d\varphi'/d\varphi = f'(\varphi) - s$ at equilibrium, which is negative at $\varphi = u_+$ and positive at $\varphi = u_-$ due to strict convexity. Therefore, $u_+$ is an attractive equilibrium and $u_-$ is a repulsive equilibrium. Since the rhs. of the above differential equation is non-singular and does not possess additional roots, any bounded solution will necessarily connect both values $u_\pm$ through a smooth decreasing function $\varphi$. The integrand in $$ x+D = \int_{u_+}^{u_-} \frac{\text d \varphi}{f(\varphi) - f(u_+) - s(\varphi - su_+)} $$ is singular at the bounds $\varphi = u_\pm$. Convergence of this improper integral follows from its asymptotic behaviour at the bounds.

EditPiAf
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