Solve second order ode with Green function

[the king]

I had made a post in the past about the same problem and unfortunately I wasn't clear enough
so I am trying again.

I am studying an article and there I found without any proof that the solution of:
Let $g \in \mathbb{C}$ and let $u:(0,\infty)\to \mathbb{C}$
$$ -u''+\lambda^2u=f\,\, on \,(0,\infty);\quad u(0)=g,\quad \lim_{x\to\infty}u(x)=0,$$
is given by using a Green's function and it is the following:

For $$z\in (0,\infty),$$
$u(z)=\frac{1}{2\lambda^2}\Bigl(e^{-\lambda z}\int_{0}^{\lambda z}e^yf(y/\lambda)dy+e^{\lambda z}\int_{\lambda z}^{\infty}e^{-y}f(y/\lambda)dy- e^{-\lambda z}\int_{0}^{\infty}e^{-y}f(y/\lambda)dy\Bigr)+ge^{-\lambda z}.$

I would be very grateful if someone explain me how can green's function be find for this problem, because I am interested to find a respectively solution for the problem,

$$ -u''+\lambda^2u'=f\,\, on \,(0,\infty);\quad u(0)=g,\quad \lim_{x\to\infty}u(x)=0.$$

 

[HallsofIvy]

Do you not have the definition of the "Green's function"? That's really all you need.

The Green's function for linear differential operator L(f), on $(0, \infty)$ is the function G(x,t) such that
L(G(x,t))= 0 for every t.
G(x,t) is continuous for all x, t.
The derivative of G(x, t), with respect to x, is continuous for x< t and continuous for x> t but has a step discontinuity of 1 at x= t.
G(0, t)= 0 and $\lim_{x\to\infty} G(x, t)= 0$.

Okay, $-G''+ \lambda G= 0$ so that $G(x, t)= A(t)e^{\lambda x}+n B(t)e^{-\lambda x}$ for x< t, and $G(x, t)= C(T)e^{\lambda x}+ D(t)e^{-\lambda x}$ for x> t.
(We can, without loss of generality, assume that $\lambda\ge 0$.)


Obviously, x= 0< t so $G(0, t)= A(t)+ B(t)= 0$: $G(x, t)= A(t)(e^{\lambda x}- e^{-\lambda x}$ for x< t.

Obviously, $x= \infty> t$ so $\lim_{x\to\infty} G(x,t)= \lim_{x\to\infty} (C(t)e^{\lambda x}+ D(t)e^{-\lambda t}$. And $e^{\lambda x}$ will go to infinity so we must have C(t)= 0 for all t: $G(x, t)= D(t)e^{-\lambda x}$ for x> t.

The fact that the function must be continuous at x= t means that we must have
$A(t)(e^{\lambda t}- e^{-\lambda t}= D(t)e^{-\lambda t}$

The fact that the derivative, with respect to x, must have a step discontinuity at x= t means that we must have
$\lambda A(t)(e^{-\lambda t}+ e^{-\lambda t})+ \lambda D(t)e^{-\lambda t}= 1$

Those last two equations allow you to solve for A(t) and D(t) as functions of t.

 

[the king]

Ok I tried to solve it but obviously somewhere I have done a mistake.

First we need to find Green's function for the above problem.

Integration by parts:
$$
\int_{0}^\infty\bigl(-u''(z)+\lambda^2u'(z)\bigr)G(z,\zeta)d\zeta=\int_{0}^\infty-u''(z)G(z,\zeta)d\zeta+\int_{0}^\infty-u''(z)G(z,\zeta)d\zeta
$$
Now,
\begin{align*}
\int_{0}^\infty-u''(z)G(z,\zeta)d\zeta&=-u'(z)G(z)\biggr|_0^\infty-\int_{0}^\infty-u'(z)G_z(z,\zeta)d\zeta\\
&=-u'(z)G(z)\biggr|_0^\infty+u(z)G_z(z)\biggr|_0^\infty-\int_{0}^\infty-G_{zz}(z,\zeta)d\zeta,
\end{align*}
and
\begin{align*}
\int_{0}^\infty\lambda^2 u'(z)G(z,\zeta)d\zeta&=\lambda^2 u(z)G(z,\zeta)\biggr|_0^\infty+\int_{0}^\infty-\lambda^2G_z(z,\zeta)u(z)d\zeta.\\
\end{align*}
By equation (\ref{ode}) we get
\begin{align*}
\int_{0}^\infty G(z,\zeta)f(\zeta)d\zeta=-&u'(z)G(z)\biggr|_0^\infty+u(z)G_z(z)\biggr|_0^\infty +\lambda^2 u(z)G(z,\zeta)\biggr|_0^\infty\\
&+\int_{0}^\infty \bigl(-G_{zz}(z,\zeta)-\lambda^2G(z,\zeta)\bigr)d\zeta.
\end{align*}
\underline{Boundary conditions}:
$$
-u'(z)G(z,\zeta)\biggr|_0^\infty=-u'(\infty)G(\infty,\zeta)+u'(0)G(0,\zeta),
$$
so we enforce,
$$
G(0,\zeta)=G(\infty,\zeta)=0,
$$
$$
-u(z)G_z(z,\zeta)\biggr|_0^\infty=-u(\infty)G_z(\infty,\zeta)+u(0)G_z(0,\zeta)=-gG_z(0,\zeta),
$$
$$
\lambda^2 u(z)G(z,\zeta)\biggr|_0^\infty=\lambda^2 u(\infty)G(\infty,\zeta)-\lambda^2 u(0)G(0,\zeta)=-\lambda^2 gG_z(0,\zeta).
$$
Furthermore, we want
$$
\int_0^\infty \bigl(-G_{\zeta \zeta}(z,\zeta)-\lambda^2G_\zeta(z,\zeta) u(\zeta)\bigr )d\zeta=\int_0^\infty \delta(\zeta-z)u(\zeta)d\zeta=u(z)
$$
So we have
$$
\begin{cases}
G''+ \lambda^2 G'= 0,\\
G(0,\zeta)=G(\infty,\zeta)=0,
\end{cases}
$$
so, for $\mu=\lambda^2$,
\begin{equation}
G(z,\zeta)=
\begin{cases}
A(\zeta)+B(\zeta)e^{-\mu z},\quad z<\zeta\\
C(\zeta)+D(\zeta)e^{-\mu z},\quad z>\zeta\\
\end{cases}
\end{equation}
For $z=0<\zeta$, $G(0,\zeta)=A(\zeta)+B(\zeta)=0$ and for $z=\infty>\zeta, \,\lim_{z \to \infty} G(z,\zeta)=C(\zeta)+D(\zeta)e^{-\mu z}=0$, so $C(\zeta)=0$. The fact that the function must be continuous at $z=\zeta$, means that we must have $D(\zeta)=A(\zeta)(e^{\mu \zeta}-1)$.
The fact that the derivative, with respect to $z$, must have a step discontinuity at $z=\zeta$, means that we must have
$\lim_{z \to \zeta}G_z(z,\zeta)\bigr|_{z^-}^{z^+}=1,$ therefore $-\mu D(\zeta)e^{-\mu \zeta}-\mu A(\zeta)e^{-\mu \zeta}=1$ and so $A(\zeta)={-1}/{\mu}$ and $D(\zeta)=\frac{-(e^{\mu \zeta}-1)}{\mu}.$ Thus Green's function is
$$
G(z,\zeta)=\begin{cases}
\frac{-1}{\mu}(1-e^{-\mu z}),\quad 0<z<\zeta,\\
-\frac{(e^{\mu\zeta}-1)}{\mu}e^{-\mu z},\quad 0<\zeta<z,\\
\end{cases}
$$
Therefore the solution is given by:
\begin{equation}
\begin{aligned}
u(z)=\frac{-1}{\mu}\int_{0}^{z}(1-e^{-\mu \zeta})f(\zeta)d\zeta-\frac{e^{-\mu z}}{\mu}\int_{ z}^\infty (e^{\mu\zeta}-1)f(\zeta)d\zeta+ge^{-\mu z}
\end{aligned}
\end{equation}