Existence of a fixed point of a nonlinear integral operator

Question

This is an elaboration on my post on MO.

In the paper "A Boundary Value Problem Associated with the Second Painlevé Transcendent and the Korteweg-de-Vries Equation" by Hastings and McLeod, the authors study the ODE $$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}-x y=2y |y|^\alpha , \quad -\infty<x<\infty, \quad \alpha>0. $$ In their studies, they formulate the following integral equation [Section 2] $$y_k(x)=k \text{Ai}(x)+2\int_x^\infty \left\{ \text{Ai}(x) \text{Bi}(t)-\text{Bi}(x) \text{Ai}(t) \right\} y_k(t) \left| y_k(t) \right|^\alpha \mathrm{d} t, $$ and write that the equation "can be solved (uniquely) by iteration, and this gives both $y_k$ and its continuous dependence on $k$.

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I think that as it stands, the integral equation is erroneous, and should be corrected to $$y_k(x)=k \text{Ai}(x)+2 \pi \int_x^\infty \left\{ \text{Ai}(x) \text{Bi}(t)-\text{Bi}(x) \text{Ai}(t) \right\} y_k(t) \left| y_k(t) \right|^\alpha \mathrm{d} t. $$

I was trying to make the "by iteration" part more rigorous. I think that the formal way to prove that solutions exist and depend continuously over the parameter $k$ is to consider the right hand side of the integral equation as an integral operator, and then show that it is a contraction mapping under an appropriate choice of a metric.

So, define $$\Phi_k:X_k \to X_k $$ by $$\Phi_k[y](x)=k \text{Ai}(x)+2 \pi \int_x^\infty \left\{ \text{Ai}(x) \text{Bi}(t)-\text{Bi}(x) \text{Ai}(t) \right\} y(t) \left| y(t) \right|^\alpha \mathrm{d} t $$

We need set-up the domain $X_k$ in a way which makes $\Phi_k$ well defined, and map $X_k$ into itself. Moreover, we need to define a metric $d_k:X_k \times X_k \to [0,+\infty)$ for which $(X_k,d_k)$ is complete, and $\Phi_k$ is a contraction.

In order for $\Phi_k[y]$ to exist as a function, a natural requirement is

• $y$ is continuously defined over some neighbourhood of $+\infty$, that is an interval of the form $(x_0,\infty)$.

In fact since $$ \begin{align} \operatorname{Ai}(t)& \sim \frac{\mathrm{e}^{-\frac{2}{3} t^{3/2}}}{2 \pi^{1/2} t^{1/4}} \quad \to +\infty \\ \operatorname{Bi}(t)& \sim \frac{\mathrm{e}^{\frac{2}{3} t^{3/2}}}{ \pi^{1/2} t^{1/4}} \quad t \to +\infty \end{align} $$ it is natural to add a growth restriction in order to guarantee the improper integrals converge:

• There exists a constant $A>0$ such that $$\limsup_{t \to \infty} \left| \frac{y(t)}{\operatorname{Ai}(t)} \right| \leq A .$$

In order to get some insights on choosing the metric, I've considered differences of the form $$\Phi_k[y_1](x)-\Phi_k[y_2](x)=2 \pi \int_x^\infty \left\{ \text{Ai}(x) \text{Bi}(t)-\text{Bi}(x) \text{Ai}(t) \right\} \left( y_1(t) \left| y_1(t) \right|^\alpha- y_2(t) \left| y_2(t) \right|^\alpha \right) \mathrm{d} t. $$

I'm pretty much stuck here. Any advice on what other properties I should place on $y \in X_k$, as well choosing a metric on $X_k$ would be appreciated. Also, there is the continuity in $k$ part, which I don't know how to tackle. Thank you.

Answer 1

An important point mentioned in the paper is that the solution only exists for sufficiently large $x$. I suggest the norm $$||y|| = \sup_{x\geq x_0} |y(x)/\operatorname{Ai}(x)|$$ where $x_0$ is to be chosen later (depending on $k$ and $\alpha$, and positive to avoid any Airy function zeroes).

In other words $y/\operatorname{Ai}(x)$ is a bounded continuous function in the range $x\geq x_0$, with the sup norm. The space of bounded continuous functions is complete under the sup norm, so the Banach contraction mapping theorem saying that if $\Phi$ is a contraction under this norm, then a solution to $\Phi(y)=y$ exists.

As for continuity in $k$: temporarily writing $\Phi_k$ to show dependence on $k$, if furthermore $||\Phi_k(y)||\leq c||y||$ for some $c<1$ uniformly for $k\in[k_0,k_1]$, then we can apply the contraction mapping theorem on solutions for all $k\in [k_0,k_1]$ simultaneuously, giving a contraction on a space of bounded continuous maps $[k_0,k_1]\times \mathbb R_{\geq x_0}\to\mathbb R$, where a family of maps $y=(y_k)_{k\in[k_0,k_1]}$ are sent to the family of maps $(\Phi_k(y_k))_{k\in[k_0,k_1]}$, and norm $||y-y'||'=\sup_{k\in[k_0,k_1]}||y_k-y'_k||$.

We want to obtain the following bound for $x\geq x_0$. $$|\Phi_k[y_1](x)-\Phi_k[y_2](x)|\leq c\operatorname{Ai}(x)||y_1-y_2||.$$

The left-hand-side is \begin{align*} 2\pi \int_x^\infty \left\{ \text{Ai}(x) \text{Bi}(t)-\text{Bi}(x) \text{Ai}(t) \right\} \left( y_1(t) \left| y_1(t) \right|^\alpha- y_2(t) \left| y_2(t) \right|^\alpha \right) \mathrm{d} t\\ \leq 4\pi\operatorname{Ai}(x) \int_x^\infty \exp(-\alpha\tfrac 2 3 t^{3/2}) \left| \hat y_1(t) \left| \hat y_1(t) \right|^\alpha- \hat y_2(t) \left| \hat y_2(t) \right|^\alpha \right| | \mathrm{d} t \end{align*} where $\hat y(t)$ means $y(t)/\operatorname{Ai}(t)$. Using $|a|a|^\alpha - b|b|^\alpha|\leq (\alpha+1)|a - b|\cdot(|a|^\alpha + |b|^\alpha)$ I think it is clear that $x_0$ can be chosen to be $O(|k|)$ to get the expression above to be less than $\tfrac 1 2\operatorname{Ai}(x)||y_1-y_2||$ for all $x\geq x_0$ and $||y_1||,||y_2||\leq |2k|$.