In the case that we have a two-point boundary value problem, *i.e.*, when the boundary conditions are fixed on the extremes of the interval, we have a two-sided Green’s function, also called Green’s function, only.

We consider the second-order ordinary differential equation in the Sturm-Liouville form

\frac{\mathrm{d}}{\mathrm{d}x}[p(x)\frac{\mathrm{d}}{\mathrm{d}x}y(x)]+q(x)y(x)=-f(x)

for {x}_{0}<x<{x}_{1} and the boundary conditions y({x}_{0})=0=y({x}_{1}), and comparing this problem with the corresponding one-sided Green’s function, we have put -f(x) in the place of f(x) for convenience only.

The two-sided Green’s function, denoted by {\mathcal{G}}_{b}(x|\xi ) for each *ξ*, satisfies the following problem consisting of homogeneous ordinary differential equation

\frac{\mathrm{d}}{\mathrm{d}x}[p(x)\frac{\mathrm{d}}{\mathrm{d}x}{\mathcal{G}}_{b}(x|\xi )]+q(x){\mathcal{G}}_{b}(x|\xi )=0

for x\ne \xi and the homogeneous boundary conditions

{\mathcal{G}}_{b}({x}_{0}|\xi )=0={\mathcal{G}}_{b}({x}_{1}|\xi ).

This Green’s function must satisfy also the continuity

{\mathcal{G}}_{b}({\xi}^{+}|\xi )={\mathcal{G}}_{b}({\xi}^{-}|\xi )

and a jump discontinuity on the first derivative [1]

\frac{\mathrm{d}}{\mathrm{d}x}{\mathcal{G}}_{b}(x|\xi ){|}_{x={\xi}^{+}}-\frac{\mathrm{d}}{\mathrm{d}x}{\mathcal{G}}_{b}(x|\xi ){|}_{x={\xi}^{-}}=-\frac{1}{p(\xi )}\cdot

The solution y(x) can be interrelated as a displacement and be given by

y(x)={\int}_{{x}_{0}}^{{x}_{1}}{\mathcal{G}}_{b}(x|\xi )f(\xi )\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi ,

where f(\xi ) is a force per unit length. {\mathcal{G}}_{b}(x|\xi ) is the displacement at *x* due to a force of unit magnitude concentrated at *ξ*. In this case, we have {\mathcal{G}}_{b}(x|\xi )={\mathcal{G}}_{b}(\xi |x), the so-called reciprocity law.

### 3.1 Constant coefficients

As an example, we discuss the following ordinary differential equation:

\frac{{\mathrm{d}}^{2}}{\mathrm{d}{x}^{2}}y(x)+2b\frac{\mathrm{d}}{\mathrm{d}x}y(x)+({a}^{2}+{b}^{2})y(x)=-f(x)

with the homogeneous boundary conditions y({x}_{0})=0=y({x}_{1}).

First, we construct the corresponding Green’s function, as we have seen before. Two linearly independent solutions of the homogeneous ordinary differential equation are

{y}_{1}(x)=\frac{{\text{e}}^{-bx}}{a}sinax\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{y}_{2}(x)=\frac{{\text{e}}^{-bx}}{a}cosax.

Imposing that {y}_{1}(x) satisfies {y}_{1}({x}_{0})=0, we can write

{y}_{1}(x)=\frac{A(\xi )}{a}{\text{e}}^{-bx}sin[a(x-{x}_{0})].

On the other hand, to satisfy {y}_{2}({x}_{1})=0, we put

{y}_{2}(x)=\frac{B(\xi )}{a}{\text{e}}^{-bx}sin[a(x-{x}_{1})],

where A(\xi ) and B(\xi ) must be determined.

Using the continuity at x=\xi and the jump discontinuity of the first derivative, we obtain a system of two algebraic equations involving A(\xi ) and B(\xi ) whose solution is

A(\xi )=\frac{sin[a({x}_{1}-\xi )]}{sin[a({x}_{1}-{x}_{0})]}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}B(\xi )=-\frac{sin[a(\xi -{x}_{0})]}{sin[a({x}_{1}-{x}_{0})]}\cdot

Thus, the Green’s function is given by the following expression:

{\mathcal{G}}_{b}(x|\xi )=\frac{exp[-b(x+\xi )]}{asin[a({x}_{1}-{x}_{0})]}\{\begin{array}{c}sin[a(x-{x}_{0})]sin[a({x}_{1}-\xi )],\phantom{\rule{1em}{0ex}}{x}_{0}<x<\xi ,\hfill \\ sin[a(\xi -{x}_{0})]sin[a({x}_{1}-x)],\phantom{\rule{1em}{0ex}}\xi <x<{x}_{1}.\hfill \end{array}

Using this Green’s function, the solution of the boundary value problem can be written as follows:

y(x)={\int}_{{x}_{0}}^{{x}_{1}}{\mathcal{G}}_{b}(x|\xi )f(\xi )\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \cdot

### 3.2 Laplace transform

As we have already said, the Laplace transform converts the differential equation with constant coefficients into an algebraic equation whose solution is

F(s)=\frac{sy(0)+{y}^{\prime}(0)}{{(s+b)}^{2}+{a}^{2}}-\frac{G(s)}{{(s+b)}^{2}+{a}^{2}},

where F(s) and G(s) are the corresponding Laplace transforms of y(x) and f(x), respectively.

Using the inverse Laplace transform and the convolution product, we get

\begin{array}{rcl}y(x)& =& y(0){\text{e}}^{-bx}\frac{cosax}{a}+{y}^{\prime}(0){\text{e}}^{-bx}\frac{sinax}{a}\\ -\frac{1}{a}{\int}_{0}^{x}{\text{e}}^{-b(x-\xi )}sin[a(x-\xi )]f(\xi )\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi .\end{array}

Thus, substituting the boundary conditions, we obtain two algebraic equations, a system involving y(0) and {y}^{\prime}(0). Solving this algebraic system, we obtain

\begin{array}{rcl}y(x)& =& \frac{-1/a}{sin[a({x}_{1}-{x}_{0})]}{\int}_{{x}_{0}}^{x}{\text{e}}^{-b(x+t)}sin[a({x}_{1}-x)]sin[a({x}_{0}-t)]f(t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\\ -\frac{1/a}{sin[a({x}_{1}-{x}_{0})]}{\int}_{x}^{{x}_{1}}{\text{e}}^{-b(x+t)}sin[a({x}_{1}-t)]sin[a({x}_{0}-x)]f(t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\end{array}

which can be rewritten as follows:

y(x)={\int}_{{x}_{0}}^{{x}_{1}}{\mathcal{G}}_{b}(x|t)f(t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t,

where the Green’s function is given by

{\mathcal{G}}_{b}(t|x)=\frac{exp[-b(x+t)]}{asin[a({x}_{1}-{x}_{0})]}\{\begin{array}{cc}sin[a({x}_{1}-x)]sin[a(t-{x}_{0})],\hfill & {x}_{0}<t<x,\hfill \\ sin[a({x}_{1}-t)]sin[a(x-{x}_{0})],\hfill & x<t<{x}_{1},\hfill \end{array}

which is the same expression as that obtained in Section 3.1.

At this point we conclude that for a problem involving initial conditions, the Laplace integral transform is more convenient since y(0) and {y}^{\prime}(0) are known. On the other hand, *i.e.*, for a problem involving boundary conditions, the Sturm-Liouville, as opposed to the Laplace integral transform, is more convenient in the sense that the calculation is much more simple.