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G Some Tools of Vector Analysis
the ‘effective force’:
∇
2 φ = f ,
(G.45)
This relation is a special case of a linear operator acting on an unknown function y
giving a known function f :
Ly = f .
(G.46)
In fact, even Newton’s 2nd law can be put in this form, with y being the coordinate
of a particle (as a function of time) and f being the force on the particle, and
L = md 2 /dt 2 . A formal solution to Eq. (G.46) can be found by the following idea.
The ‘forcing term’ can be thought of as a combinations of impulses, each of short
duration. If f does not depend on y, the equation is linear in the unknown y, so the
general solution can be expressed as the sum of impulse solutions. Let the impulse
be expressed as f = δ, where δ is a function which vanishes outside a small region
in the space {x} over which L operates, and δ is taken to have unit total impulse, i.e.
δ(x)dx = 1. For application to Newton’s law, x = t. The solution to the impulse
problem is simpler than the arbitrary-force case. Let G(x, x ) be a solution to the
impulse equation:
LG(x, x
) = δ(x − x
) ,
(G.47)
where the region of non-vanishing impulse can be varied by changing x . 7 Then the
general solution for arbitrary force is
y(x) = y o (x) +
G(x, x
)f (x
)dx
,
(G.48)
where y 0 (x) is the solution to the homogeneous equation: Ly 0 = 0. That this is the
solution to the inhomogeneous equation (G.46) can be seen by operating on the left
and right by L.
For the Poisson equation (G.45), one can verify that the Green’s function in an
unbounded region is 8
G P (x, x
) = −
1
4π
1
|r − r |
,
(G.49)
7 G is called the Green’s function, or the ‘kernel’ of the equation. The impulse δ(x − x ) is call
the Dirac delta. Note: We do not assume that the impulse time goes to zero, so we do not have to
introduce a singular Dirac delta. Our δ is non-zero over a finite region, but whose width is much
smaller than the extent of variations of y.
8 For those who know some electrostatics, you may recognize this Green’s function as proportional
to the electric potential for a point charge. Such a point charge acts as the ‘impulse’ for the equation
for the electric potential ∇ 2 V = −4πρ q .
G Some Tools of Vector Analysis
the ‘effective force’:
∇
2 φ = f ,
(G.45)
This relation is a special case of a linear operator acting on an unknown function y
giving a known function f :
Ly = f .
(G.46)
In fact, even Newton’s 2nd law can be put in this form, with y being the coordinate
of a particle (as a function of time) and f being the force on the particle, and
L = md 2 /dt 2 . A formal solution to Eq. (G.46) can be found by the following idea.
The ‘forcing term’ can be thought of as a combinations of impulses, each of short
duration. If f does not depend on y, the equation is linear in the unknown y, so the
general solution can be expressed as the sum of impulse solutions. Let the impulse
be expressed as f = δ, where δ is a function which vanishes outside a small region
in the space {x} over which L operates, and δ is taken to have unit total impulse, i.e.
δ(x)dx = 1. For application to Newton’s law, x = t. The solution to the impulse
problem is simpler than the arbitrary-force case. Let G(x, x ) be a solution to the
impulse equation:
LG(x, x
) = δ(x − x
) ,
(G.47)
where the region of non-vanishing impulse can be varied by changing x . 7 Then the
general solution for arbitrary force is
y(x) = y o (x) +
G(x, x
)f (x
)dx
,
(G.48)
where y 0 (x) is the solution to the homogeneous equation: Ly 0 = 0. That this is the
solution to the inhomogeneous equation (G.46) can be seen by operating on the left
and right by L.
For the Poisson equation (G.45), one can verify that the Green’s function in an
unbounded region is 8
G P (x, x
) = −
1
4π
1
|r − r |
,
(G.49)
7 G is called the Green’s function, or the ‘kernel’ of the equation. The impulse δ(x − x ) is call
the Dirac delta. Note: We do not assume that the impulse time goes to zero, so we do not have to
introduce a singular Dirac delta. Our δ is non-zero over a finite region, but whose width is much
smaller than the extent of variations of y.
8 For those who know some electrostatics, you may recognize this Green’s function as proportional
to the electric potential for a point charge. Such a point charge acts as the ‘impulse’ for the equation
for the electric potential ∇ 2 V = −4πρ q .