BVP Green's Functions¶

Copyright (C) 2026 Andreas Kloeckner

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In [2]:
import numpy as np
import numpy.linalg as la
import sympy as sp
import matplotlib.pyplot as plt
In [3]:
x = sp.Symbol("x")
z = sp.Symbol("z")

Let's consider the ODE system $$ \boldsymbol y'(x) = \begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \boldsymbol y(x) + \begin{bmatrix} 3\\ 1 \end{bmatrix} $$ i.e. the harmonic oscillator with boundary conditions $$ \begin{align*} y_1(0) &= 2\\ y_1'(\pi/2) - 3 y_1(\pi/2) &= -4\\ \end{align*} $$

Set up system data and the fundamental solution matrix $\boldsymbol Y$:

In [4]:
A = sp.Matrix([[0, 1], [-1, 0]])
rhs = sp.Matrix([[3, 1]]).T
a = sp.sympify(0)
b = sp.pi/2
B_a = sp.Matrix([[1, 0], [0, 0]])
B_b = sp.Matrix([[0, 0], [-3, 1]])
c = sp.Matrix([[2, -4]]).T

xmesh = np.linspace(float(a), float(b), 1000)

solutions = [sp.sin(x), sp.cos(x)]
Y = sp.Matrix([
   [sol.diff(x, num_derivs) for sol in solutions]
   for num_derivs in [0, 1]
])
Y
Out[4]:
$\displaystyle \left[\begin{matrix}\sin{\left(x \right)} & \cos{\left(x \right)}\\\cos{\left(x \right)} & - \sin{\left(x \right)}\end{matrix}\right]$

Check that the columns satisfies the homogeneous ODE:

In [5]:
Y.diff(x) - A @ Y
Out[5]:
$\displaystyle \left[\begin{matrix}0 & 0\\0 & 0\end{matrix}\right]$

Solving the volume-homogeneous/"boundary-only" BVP¶

Plug in the BCs.

In [6]:
B_a @ Y.subs(x, a)
Out[6]:
$\displaystyle \left[\begin{matrix}0 & 1\\0 & 0\end{matrix}\right]$
In [7]:
B_b @ Y.subs(x, b)
Out[7]:
$\displaystyle \left[\begin{matrix}0 & 0\\-3 & -1\end{matrix}\right]$

What do the matrix entries of $\boldsymbol Q$ mean?

In [8]:
Q = B_a @ Y.subs(x, a) + B_b @ Y.subs(x, b)
Q
Out[8]:
$\displaystyle \left[\begin{matrix}0 & 1\\-3 & -1\end{matrix}\right]$

What does $\boldsymbol Q^{-1} \boldsymbol c$ mean?

In [9]:
Q.inv() @ c
Out[9]:
$\displaystyle \left[\begin{matrix}\frac{2}{3}\\2\end{matrix}\right]$
In [10]:
Phi = Y @ Q.inv()
Phi
Out[10]:
$\displaystyle \left[\begin{matrix}- \frac{\sin{\left(x \right)}}{3} + \cos{\left(x \right)} & - \frac{\sin{\left(x \right)}}{3}\\- \sin{\left(x \right)} - \frac{\cos{\left(x \right)}}{3} & - \frac{\cos{\left(x \right)}}{3}\end{matrix}\right]$

Observe:

In [112]:
B_a @ Phi.subs(x, a) + B_b @ Phi.subs(x, b)
Out[112]:
$\displaystyle \left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$

Specifically:

In [115]:
B_a @ Phi.subs(x, a)
Out[115]:
$\displaystyle \left[\begin{matrix}1 & 0\\0 & 0\end{matrix}\right]$
In [114]:
B_b @ Phi.subs(x, b)
Out[114]:
$\displaystyle \left[\begin{matrix}0 & 0\\0 & 1\end{matrix}\right]$

Check $\boldsymbol y_B = \boldsymbol \Phi \boldsymbol c$ satisfies the BCs:

In [11]:
yB = Phi @ c
yB
Out[11]:
$\displaystyle \left[\begin{matrix}\frac{2 \sin{\left(x \right)}}{3} + 2 \cos{\left(x \right)}\\- 2 \sin{\left(x \right)} + \frac{2 \cos{\left(x \right)}}{3}\end{matrix}\right]$
In [12]:
B_a @ yB.subs(x, a)
Out[12]:
$\displaystyle \left[\begin{matrix}2\\0\end{matrix}\right]$
In [13]:
B_b @ yB.subs(x, b)
Out[13]:
$\displaystyle \left[\begin{matrix}0\\-4\end{matrix}\right]$

Plot it:

In [14]:
plt.plot(xmesh, sp.lambdify(x, yB[0], "numpy")(xmesh))
Out[14]:
[<matplotlib.lines.Line2D at 0x7feee2576ba0>]
No description has been provided for this image

Solving the boundary-homogeneous/"volume-only" BVP¶

Set up the two pieces of the Green's function:

In [104]:
Gleft = Phi @ B_a @ Phi.subs(x, a) @ Phi.subs(x, z).inv()
Gright = - Phi @ B_b @ Phi.subs(x, b) @ Phi.subs(x, z).inv()
Gleft
Out[104]:
$\displaystyle \left[\begin{matrix}\frac{\left(- \frac{\sin{\left(x \right)}}{3} + \cos{\left(x \right)}\right) \left(- 3 \sin^{2}{\left(z \right)} - \sin{\left(z \right)} \cos{\left(z \right)} + 3\right)}{- \sin{\left(z \right)} + 3 \cos{\left(z \right)}} & - \left(- \frac{\sin{\left(x \right)}}{3} + \cos{\left(x \right)}\right) \sin{\left(z \right)}\\\frac{\left(- \sin{\left(x \right)} - \frac{\cos{\left(x \right)}}{3}\right) \left(- 3 \sin^{2}{\left(z \right)} - \sin{\left(z \right)} \cos{\left(z \right)} + 3\right)}{- \sin{\left(z \right)} + 3 \cos{\left(z \right)}} & - \left(- \sin{\left(x \right)} - \frac{\cos{\left(x \right)}}{3}\right) \sin{\left(z \right)}\end{matrix}\right]$

Define an overall function and plot:

In [105]:
G0 = sp.Piecewise(
    (Gleft[0, 0], z < x),
    (Gright[0, 0], z > x),
)
plt.plot(xmesh, sp.lambdify(x, G0.subs(z, 0.9), "numpy")(xmesh))
G0
Out[105]:
$\displaystyle \begin{cases} \frac{\left(- \frac{\sin{\left(x \right)}}{3} + \cos{\left(x \right)}\right) \left(- 3 \sin^{2}{\left(z \right)} - \sin{\left(z \right)} \cos{\left(z \right)} + 3\right)}{- \sin{\left(z \right)} + 3 \cos{\left(z \right)}} & \text{for}\: x > z \\\frac{\left(- 3 \sin{\left(z \right)} - \cos{\left(z \right)}\right) \sin{\left(x \right)}}{3} & \text{for}\: x < z \end{cases}$
No description has been provided for this image

Verify that G satisfies the volume-homogeneous ODE:

In [106]:
sp.simplify(Gleft.diff(x) - A @ Gleft)
Out[106]:
$\displaystyle \left[\begin{matrix}0 & 0\\0 & 0\end{matrix}\right]$
In [107]:
sp.simplify(Gright.diff(x) - A @ Gright)
Out[107]:
$\displaystyle \left[\begin{matrix}0 & 0\\0 & 0\end{matrix}\right]$

Verify that G satisfies the boundary conditions.

  • Note that $x= a<z$, so that we use the $x < z$ Gright branch.
In [108]:
B_a @ Gright.subs(x, a)
Out[108]:
$\displaystyle \left[\begin{matrix}0 & 0\\0 & 0\end{matrix}\right]$
In [109]:
B_b @ Gleft.subs(x, b)
Out[109]:
$\displaystyle \left[\begin{matrix}0 & 0\\0 & 0\end{matrix}\right]$

Verify the jump condition:

In [101]:
sp.simplify(Gleft.subs(x, z) - Gright.subs(x, z))
Out[101]:
$\displaystyle \left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$

Put together the volume solution:

In [86]:
yV =  (
    sp.integrate(Gleft @ rhs.subs(x, z), (z, a, x))
    + sp.integrate(Gright @ rhs.subs(x, z), (z, x, b))
)
yV
Out[86]:
$\displaystyle \left[\begin{matrix}- \frac{\left(- 3 \sin{\left(x \right)} - \cos{\left(x \right)}\right) \sin{\left(x \right)}}{3} - \left(- \sin{\left(x \right)} + 3 \cos{\left(x \right)}\right) \sin{\left(x \right)} + \frac{2 \left(- \sin{\left(x \right)} + 3 \cos{\left(x \right)}\right) \tan{\left(\frac{x}{2} \right)}}{\tan^{2}{\left(\frac{x}{2} \right)} + 1} - \left(\frac{\sin{\left(x \right)}}{3} - \cos{\left(x \right)}\right) \cos{\left(x \right)} - \frac{5 \sin{\left(x \right)}}{3} - \cos{\left(x \right)}\\- \frac{\left(- 3 \sin{\left(x \right)} - \cos{\left(x \right)}\right) \cos{\left(x \right)}}{3} + \frac{2 \left(- 3 \sin{\left(x \right)} - \cos{\left(x \right)}\right) \tan{\left(\frac{x}{2} \right)}}{\tan^{2}{\left(\frac{x}{2} \right)} + 1} - \left(- \sin{\left(x \right)} + 3 \cos{\left(x \right)}\right) \cos{\left(x \right)} - \left(\sin{\left(x \right)} + \frac{\cos{\left(x \right)}}{3}\right) \cos{\left(x \right)} + \sin{\left(x \right)} - \frac{5 \cos{\left(x \right)}}{3}\end{matrix}\right]$

Verify the ODE:

In [87]:
sp.simplify(yV.diff(x) - A @ yV - rhs)
Out[87]:
$\displaystyle \left[\begin{matrix}\left(2 \left(\sin{\left(x \right)} - 3 \cos{\left(x \right)}\right) \tan^{2}{\left(\frac{x}{2} \right)} + \frac{- 2 \sin{\left(x \right)} + \sin{\left(2 x \right)} + 6 \cos{\left(x \right)} - 3 \cos{\left(2 x \right)} - 3}{\cos{\left(x \right)} + 1}\right) \cos^{2}{\left(\frac{x}{2} \right)}\\\left(2 \left(3 \sin{\left(x \right)} + \cos{\left(x \right)}\right) \tan^{2}{\left(\frac{x}{2} \right)} + \frac{- 6 \sin{\left(x \right)} + 3 \sin{\left(2 x \right)} - 2 \cos{\left(x \right)} + \cos{\left(2 x \right)} + 1}{\cos{\left(x \right)} + 1}\right) \cos^{2}{\left(\frac{x}{2} \right)}\end{matrix}\right]$

OK then... Verify the ODE numerically:

In [88]:
sp.simplify(yV.diff(x) - A @ yV - rhs).subs(x, 0.2).evalf()
Out[88]:
$\displaystyle \left[\begin{matrix}6.86973595315652 \cdot 10^{-17}\\-7.55670954847217 \cdot 10^{-17}\end{matrix}\right]$

Plot it:

In [89]:
plt.plot(xmesh, sp.lambdify(x, yV[0], "numpy")(xmesh))
Out[89]:
[<matplotlib.lines.Line2D at 0x7feee13f0ad0>]
No description has been provided for this image