I am reading a paper that deals with the solution of the Sturm-Liouville problem:

$u''(t) + \rho (t) u + \lambda ^{-1}u= -f $

$ u(0)=u(1)=0$

For $\rho(t) \leq 0 $. First it is solved the problem:

$u''(t) + \rho (t) u= -f $

$u(0)=u(1)=0 \hspace{1cm}$ (1)

Picking two arbitrary linearly independent solutions $u_1,u_2, u_1(0)=0$ $u_2(1)=0$ and using Variation of constants method to obtain a particular solution $u_p$. Then, by imposing to the general solution $u=u_p +a u_1 +b u_2$ the boundary conditions of the original (1) problem, it is found the solution that the solution to (1) is written in integral form as $ \int_{0}^{1} k(t,s)f(s)ds$. So it is defined the operator $K$:

$$K:L^2([0,1]) \hspace{1cm} \longrightarrow \hspace{1cm} L^2([0,1])$$ $$ \hspace{4cm} f \hspace{1cm} \longrightarrow \hspace{1cm} u(t)= \int_{0}^{1} k(t,s)f(s)ds$$

Where u is the solution to the ODE $u'' + \rho u=f$ (With the boundary conditions $u(0)=u(1)=0$) and $k(t,s)$ is Green's Function:

$$k(t,s):=\left\{\begin{matrix} - \frac{u_2(t)u_1(s)}{W(0)}& s \leq t \\ -\frac{u_2(s)u_1(t)}{W(0)}& t\leq s \end{matrix}\right. $$

I understand the last part of the paper which uses spectral theorem for compact self-adjoint operators to solve the initial problem. But I have a few questions:

  1. It is shown that the operator $K$ is injective, so that for a given $f \in L^2([0,1])$ there is a unique solution $u \in L^2([0,1])$. Is this necessary? Couldn't it be shown using the fact that two different $\hat{u_1}, \hat{u_2}$ lineally independent solutions yield the same Green's function as $u_1, u_2$?

  2. Also, it is shown that if $f \in C([0,1])$ then $u \in C^2([0,1])$. This is done by derivating u in its integral form two times. Again, Is this necessary? As u verifies $u'' + \rho u=f$ then $u''$ is also continuous so that $u \in C^2([0,1])$ as long as f is continuous. Am I missing something?

  • $\begingroup$ But the operator is defined after the explicit solution has been found, and sends every $f \in L^2([0,1])$ to the solution $u$ of (1) $\endgroup$
    – D1X
    Jan 12, 2016 at 14:20
  • $\begingroup$ what do you mean with "two linearly independent solutions of (1) give the same green's function" ? $\endgroup$
    – reuns
    Jan 12, 2016 at 14:29
  • $\begingroup$ and first we solve $u'' + \rho u = 0$ not $u'' + \rho u = -f$ $\endgroup$
    – reuns
    Jan 12, 2016 at 14:30
  • $\begingroup$ It is easy to show that two different $\hat{u_1}, \hat{u_2}$ lineally independent solutions yield the same Green's function as $u_1, u_2$. Also yes, of course, to solve a second order linear ode you first solve the homogeneous one. What's your point? $\endgroup$
    – D1X
    Jan 12, 2016 at 15:07

1 Answer 1


In the first set of equations, I think you meant $$ u''(t)+\rho(t)u(t) + \frac{1}{\lambda}u(t) = -f(t) $$ instead of $$ u''(t)+\rho(t)u(t) + \frac{1}{\lambda} = -f(t). $$ That's the typical way such problems are defined.

For the second set of equations $$ \begin{array}{c} u''(t)+\rho(t)u(t) = -f(t),\\ u(0)=0,\;\;\; u(1)=0, \end{array} \;\;\;\; (\dagger) $$ the solution may not exist for all $f$ and may not be unique if it does exist. It depends on $\rho$. For example, suppose $\rho(t)=\pi^2$. Then $u(t)=\sin(\pi t)$ is a solution of the homogeneous equation $$ u''(t)+\pi^2u(t) = 0, \\ u(0)=0,\;\;\; u(1)=1. $$ Therefore if you have a solution $u_0$ of the inhomogeneous system $(\dagger)$ with $\rho(t)=\pi^2$, then $u_0+C\sin(\pi t)$ is also a solution of the inhomogenous equation for any constant $C$. So there have to be conditions on $\rho$.

  • $\begingroup$ $\rho(t) \leq 0$ Also yes, it is defined as you say. It was a typo. $\endgroup$
    – D1X
    Jan 12, 2016 at 17:36
  • $\begingroup$ If $\rho \le 0$, then the homogenous system has only the $0$ solution, which implies that the inhomogenous has a unique solution. That's because $u''+\rho u=0,\;u(0)=u(1)=1$, then you can assume $u$ is real, and you obtain $0=\int_{0}^{1}u''u+\rho u^2 dt = u'u|_0^1-\int_{0}^{1}u'^2dt+\int_0^1\rho u^2=\int_{0}^{1}-u'^2+\rho u^2 dt$, which implies $u=0$ because $\rho \le 0$. $\endgroup$ Jan 12, 2016 at 17:44
  • $\begingroup$ Then I think I am missinterpreting something. The paper first considers the problem (Sturm-Liouville) for any $\rho \in C([0,1])$, then says it is assumed that $\rho \leq 0$ but at the step of solving (1). So I wonder now Why is this problem solved at the first place... $\endgroup$
    – D1X
    Jan 12, 2016 at 19:32
  • $\begingroup$ And to my first question, it is indeed not necessary to show that $K$ is injective, Right? $\endgroup$
    – D1X
    Jan 12, 2016 at 19:47
  • $\begingroup$ @D1X : Uniqueness of solutions requires injectivity. Arguing that a particular type of construction is unique is not quite the same as saying there is only one solution. You need to say a little more if you are going to try that approach. $\endgroup$ Jan 12, 2016 at 20:23

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