eigen problem for direct scattering method

eigen problem for direct scattering method

Consider the KdV equation $$u_{t}+6uu_{x}+u_{xxx}=0$$ with initial
condition $$u(x,0)= \begin{cases} 1 &\text{if } x \in [-1,0] ,\\ 0 &\text
{if } x \in (-\infty,-1)\cup(0,\infty),\end{cases}$$.
The associated initial state scattering problem is governed by the
eigenvalue problem: $$v_{xx}+u(x,0)v=\lambda v$$. Prove that each $\lambda
\in (-\infty,0)$ is an eigenvalue for smooth $v$.
My approach: Let $-\lambda = k^2$ and $-\lambda+1= l^2$ where $k>0,l>0$.
The eigen problem becomes ODEs $$v= \begin{cases} -l^2v &\text{if } x \in
[-1,0] ,\\ -k^2v &\text {if } x \in
(-\infty,-1)\cup(0,\infty),\end{cases}$$ which implies $$v= \begin{cases}
A\cos(kx)+B\sin(kx) &\text{if } x \in (-\infty,-1) ,\\ C\cos(lx)+D\sin(lx)
&\text {if } x \in [-1,0],\\E\cos(kx)+F\sin(kx) &\text{if} x \in
(0,\infty),\end{cases}$$. In which $A,B,C,D,E,F$ are constants.
Since $v$ is continuously differentiable, so
$v(-1^{+})=v(-1^{-}),v(0^{+})=v(0^{-}),v'(-1^{+})=v'(-1^{-}),v'(0^{+})=v'(0^{-})$.
(Note: the general solution to the ODEs shows that $v$ is bounded, so
there is no extra boundness condition imposed here). So one has $6$
variables in $4$ linear equations. By formulating the augemented matrix
for the corresponding homogeneous system one can see there are infinitely
many non-trivial solutions $(A,B,C,D,E,F)$.
Since $\lambda<0$ is arbitrary, so the assertion is proved.
Is my approach a correct way ?