# Fractional Laplacian problem on half-line

Is it possible to obtain an explicit solution for the following fractional problem on the half-line? $$(-\Delta)^\alpha u(x) + M u'(x) + K u(x) + C = 0 \quad \text{ in } (0,\infty)$$ $$u(x) = a, \quad u'(x) = b \text{ in } (-\infty, 0]$$ where $$M,K,C,a,b$$ are constants and $$(-\Delta)^\alpha$$ is the Fractional Laplacian.

• What would $u(x) = a$ and $u'(x) = b$ for $x < 0$ mean if $b \ne 0$? Did you mean $u(x) = b x + a$, or something similar? Jan 29 at 0:13
• @MateuszKwaśnicki I wanted to give the analogous of the condition $u(x) = a$, $u'(x) = b$ that you would impose at $x=0$ for a classical ODE, but now I see the problem. What non-zero boundary conditions can we prescribe? Jan 29 at 8:25
• Uh, there is a dozen of (non-equivalent) ways to restrict the fractional Laplace operator to a domain. The answer likely depends on what you are trying to model using this non-local equation. Is this coming from a "real-life" application, or you are trying to solve some abstract problem? Jan 29 at 9:14
• @MateuszKwaśnicki I would like to use the one in the link (singular integral formulation) with b.c. in $(-\infty,0]$ Jan 29 at 10:54
• Got that. But for some reason you refer the local problem with boundary condition $u'(0)=a$ and $u'(0)=b$. This has no obvious counterpart for non-local problem with an exterior condition. For example, you could require $u(x) \sim a x^{\alpha-1} + b x^\alpha$ as $x \to 0^+$ (and $u(x) = 0$ if $x \leqslant 0$). Or $u(x) = a e^{b x}$ for $x \leqslant 0$. The answer really depends on what your equation is meant to model. Jan 29 at 13:40

This is just an extended comment, where I list what I know about the question, but it does not actually answer the question. However, it suggests that the general answer might be complicated.

Suppose that we consider zero "exterior" condition: $$u(x) = 0$$ when $$x \leqslant 0$$ (I think I have very little to say in the more general case).

• The equation $$(-\Delta)^\alpha u = 0$$ (i.e. $$M = K = C = 0$$) admits two linearly independent positive solutions: $$u(x) = x^\alpha$$ and $$x^{\alpha - 1}$$. This is a rather old result: more general case has been studied by Silverstein in 1980, https://www.jstor.org/stable/2243167.

• The equation $$(-\Delta)^\alpha u + M u' = 0$$ (i.e. $$K = C = 0$$, $$M \ne 0$$) is also covered by Silverstein. There should be an "explicit" expression (involving some integrals): roughly speaking, one has to identify the appropriate Wiener–Hopf factor and invert the Laplace transform.

• The equation $$(-\Delta)^\alpha u + C = 0$$ (i.e. $$M = K = 0$$, $$C \ne 0$$) has no solutions. Off the top of my head, I do not have a reference, but the argument is based on the following fact: the same equation in $$(0, R)$$ has a solution $$c_\alpha C (x - R x)_+^\alpha$$, which goes to infinity as $$R \to \infty$$ — and there is no way to compensate that by imposing appropriate behaviour for $$x > R$$.

• The eigenvalue problem $$(-\Delta)^\alpha u + K u = 0$$ (i.e. $$M = C = 0$$, $$K \ne 0$$) has an "explicit" solution if $$K < 0$$, which behaves as $$\sin(|K|^{1/\alpha} x + \tfrac{(1 - \alpha) \pi}{4})$$ away from the boundary, but the formula is somewhat complicated. See Example 6.1 in my paper from 2011, https://doi.org/10.4064/sm206-3-2. Another formula involves the double sine function, see my paper with Alexey Kuznetsov from 2018, https://doi.org/10.1214/18-EJP134. The method

• The same equation for $$K > 0$$ (or, more generally, complex $$K \notin (-\infty, 0]$$) is somewhat simpler and, if I remember correctly, it is also addressed in the above paper with Alexey Kuznetsov.

• The equation $$(-\Delta)^\alpha u + M u' + K u = 0$$ can be dealt with by similar methods. This is not (yet) written anywhere, I suppose, but some preliminary work is in my preprint from 2018, https://arxiv.org/abs/1811.06617.

• My (vague) impression is that if $$C \ne 0$$ and $$M < 0$$, there will be no solution, while if $$C \ne 0$$ and $$M > 0$$, there should be a solution that could possibly be found again by applying similar methods.

(I am sorry if the above looks like I was just advertising my own work. This is not what I intended; it just so happened that I had been working quite a while on particular variants of your equation.)

• Thank you. In all these cases, why don't we prescribe also a condition on the first derivative $u'(x) = 0$ if $x \le 0$? Isn't it needed for the problem to be well-posed (as it would be for the case with the $u''$ instead of the fractional Laplacian)? Jan 29 at 19:25
• I may be missing something: for the local problem, one usually prescribes the values of $u(0)$ and $u'(0)$ to have a unique solution; the non-local one requires the exterior condition on $u$ over $(-\infty, 0]$ (plus continuity at zero and some sort of boundedness at infinity). The derivative $u'$ over $(-\infty, 0)$ is just the derivative of the exterior condition, there is no further degree of freedom here. Also, the right-sided derivative at zero will typically fail to exist: $u(x)$ behaves as $p + q x^\alpha$ as $x \to 0^+$. Jan 29 at 22:40
• Thanks! I missed what happens in the case $(-\Delta)^\alpha u(x) + M u'(x) + C = 0$? Jan 31 at 18:42
• If $M < 0$ and $\alpha > \tfrac12$, there should be a solution growing linearly at infinity. I do not have a proof, but a probabilistic argument is pretty clear: this is the equation for the mean time it takes the corresponding Lévy process to hit the negative half-axis. If $\alpha > \tfrac12$, in the long run the "drift term" $M u'$ wins with the "stable process" corresponding to $(-\Delta)^\alpha$, so the time should be roughly proportional to the distance. If $M > 0$, I expect no solution, and if $m < 0$ and $\alpha < \tfrac12$ I guess there is no solution, too, but this is even more vague. Jan 31 at 19:41