# Different integral equations on solution of Sturm-Liouville equation

Consider a differential equation $$u''(x) + \lambda^2 u(x) = q(x)u(x), \;\;\; x>0, \; \Im\lambda \geqslant 0, \; \lambda \neq 0$$ with boundary condition $u(0)=0$. Here potential $q(x)$ is continuous and $q(x) \in L_1\left([0,\infty],1+x\right)$. I solved a problem $$\left\{ \begin{array}{c} v''(x) + \lambda^2 v(x) = 0, \\ v(0) = 0, \; v'(0) = 1. \end{array} \right.$$ The solution is $v(x) = \frac{\sin(\lambda x)}{\lambda}$. Then for $L = \frac{d^2}{dx^2}+\lambda^2$ we have $L(v(x)\chi(x)) = \delta(x)$. Then $$u(x) = (v\chi)*(qu)(x)=\int\limits_{0}^{x}\frac{\sin \lambda(x-t)}{\lambda}q(t)u(t)dt$$ satisfies a differential equation $Lu(x) = q(x)u(x)$. But I was told that there is an another integral equation on $u$ (maybe, different from the above $u$): $$u(x) = e^{i \lambda x}+\int\limits_{x}^{\infty} \frac{\sin \lambda(t-x)}{\lambda} q(t)u(t)dt.$$ Please, help me to obtain it.

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I found a way of receiving such solutions. Let $u_0(x)$ be a solution of homogenous system $Lu_0 = 0$. We can choose such $u_0(x)$ with appropriate asymptotics at zero or at infinity. Next, if we want to obtain a solution of system $Lu = qu$ with asymptotics of $u_0(x)$ at zero (or at infinity) we can define $u(x)$ ($\tilde{u}(x)$) by the integral equation $$u(x) = u_0(x)+\int\limits_{0}^{x} \frac{\sin \lambda(x-t)}{\lambda} q(t) u(t) \, dt \\ \left( \tilde{u}(x) = u_0(x)+\int\limits_{x}^{\infty} \frac{\sin \lambda(t-x)}{\lambda} q(t) \tilde{u}(t) \, dt \right)$$ Then $$Lu = Lu_0 + L( v \chi* qu) = 0 + L(v\chi)*qu = 0 + \delta * qu = qu \\ \left( L\tilde{u} = Lu_0 + L(v(-x)\chi(-x)*q\tilde{u}) = 0 + \delta*q\tilde{u} = q\tilde{u} \right)$$ and $u$ ($\tilde{u})$ has the same asymptotics at zero (infinity) as $u_0$. Particularly, if $u_0(x) = 0$ then $$u(x) = \int\limits_{0}^{x} \frac{\sin \lambda(x-t)}{\lambda} q(t) u(t) \, dt$$ solves $Lu = qu$ with initial condition $u(0)=0$ and if $u_0(x) = e^{i \lambda x}$ then $$\tilde{u}(x) = e^{i \lambda x} + \int\limits_{x}^{\infty} \frac{ \sin \lambda(t-x)}{\lambda} q(t) \tilde{u}(t) \, dt$$ solves $L\tilde{u} = q\tilde{u}$ with asympotics $\tilde{u} \sim e^{i \lambda x}$ when $x \to + \infty$.