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&#039;&#039;+ \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&gt; 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}