Editing Green's function (section)

==Example==
Find the Green function for the following problem, whose [[Green's function number]] is X11:
<math display="block">\begin{align}
 Lu & = u'' + k^2 u = f(x) \\
 u(0)& = 0, \quad u{\left(\tfrac{\pi}{2k}\right)} = 0.
\end{align}</math>

'''First step:''' The Green's function for the linear operator at hand is defined as the solution to
{{NumBlk||<math display="block">G''(x,s) + k^2 G(x,s) = \delta(x-s). </math>|Eq. {{EquationRef|<nowiki>*</nowiki>}}}}

If <math>x\ne s</math>, then the delta function gives zero, and the general solution is
<math display="block">G(x,s)=c_1 \cos kx+c_2 \sin kx.</math>

For <math>x < s</math>, the boundary condition at <math>x=0</math> implies
<math display="block">G(0,s)=c_1 \cdot 1+c_2 \cdot 0=0, \quad c_1 = 0</math>
if <math>x < s</math> and <math>s \ne \tfrac{\pi}{2k}</math>.

For <math>x>s</math>, the boundary condition at <math>x = \tfrac{\pi}{2k}</math> implies
<math display="block">G{\left(\tfrac{\pi}{2k},s\right)} = c_3 \cdot 0+c_4 \cdot 1=0, \quad c_4 = 0 </math>

The equation of <math>G(0,s) = 0</math> is skipped for similar reasons.

To summarize the results thus far:
<math display="block">G(x,s) = \begin{cases}
 c_2 \sin kx, & \text{for } x < s, \\[0.4ex]
 c_3 \cos kx, & \text{for } s < x.
\end{cases}</math>

'''Second step:''' The next task is to determine <math>c_2</math> and {{Nowrap|<math>c_3</math>.}}

Ensuring continuity in the Green's function at <math>x = s</math> implies
<math display="block">c_2 \sin ks=c_3 \cos ks</math>

One can ensure proper discontinuity in the first derivative by integrating the defining differential equation (i.e., {{EquationNote|*|Eq. *}}) from <math>x=s-\varepsilon</math> to <math>x=s+\varepsilon</math> and taking the limit as <math>\varepsilon</math> goes to zero. Note that we only integrate the second derivative as the remaining term will be continuous by construction.
<math display="block">c_3 \cdot (-k \sin ks)-c_2 \cdot (k \cos ks)=1</math>

The two (dis)continuity equations can be solved for <math>c_2</math> and <math>c_3</math> to obtain
<math display="block">c_2 = -\frac{\cos ks}{k} \quad;\quad c_3 = -\frac{\sin ks}{k}</math>

So Green's function for this problem is:
<math display="block">G(x,s) = \begin{cases}
 -\frac{\cos ks}{k} \sin kx, & x<s, \\
 -\frac{\sin ks}{k} \cos kx, & s<x.
\end{cases}</math>