v2: Added section 3.
Some prerequisites: vector calculus, Heaviside step function, Dirac delta function
In electrostatics, the presence of electric charge influences the value of electric field through Gauss’ law
where is electric field, is charge density, and is permittivity of free space. It is often convenient to introduce a scalar potential via
The Gauss’ law then becomes
In order to solve this equation, one considers Green’s function such that
where A solution [In fact, in order to solve for Green’s function, one has to also specify boundary conditions. So in our case, it is possible that there are also other solutions.] is
where This then allows us to obtain the scalar potential
whose form looks very familiar to any physicist.
What I want to discuss in this post is not on electrostatics, but on Green’s function of Laplacian operator. I will quote the Green’s functions and discuss a few alternative ways to show that they are indeed Green’s functions of the Laplacian operator. The main purpose is just to do it for fun, not necessarily following any mainstream presentation. I might even discuss some heuristic approaches.
2. Green’s function of 1D Laplacian
Simply by looking at this Green’s function, one can be convinced because it is not a smooth function. While it can be differentiated twice, the second derivative is infinite at and hence is not well-defined. This behaviour seems to get along well with Dirac’s delta function on the RHS of eq.(7).
2.1. Using Heaviside step function
The behaviour of a discontinuous function can be well captured by Heaviside step function, which is defined as
Its plot looks like a step:
The Heaviside step function can be used in order to simplify the writing of the definition of a piecewise function. For example, suppose we have a function
This function can be written using only one line as
So we can write
The derivative of Heaviside step function is given by Dirac delta function:
So from this result, we can easily see that eq.(7) is indeed satisfied.
2.2. Replacing by the limit of smooth function sequence
The presence of Dirac delta function might not be so friendly for many, especially mathematicians. So instead of dealing directly with Dirac delta function, one can also consider it as a sequence of functions. To make it even more convenient, one can demand that functions in the sequence are all smooth. So in this way, since the RHS of eq.(7) is replaced by a sequence of smooth functions, one has to also replace the LHS of eq.(7) in the similar manner. This amounts to replacing by a sequence of smooth functions.
A simple choice is given by
Clearly, the limit as of this sequence is So
By moving into the parenthesis, the expression in the parenthesis then becomes and hence we recover eq.(7).
Note that during the derivation, we have not been careful with moving around the limit In general, the order of the operations matter, especially when infinity is involved. In fact, we should verify that the operations we have made are allowed, or else we should have kept the order of the operations at all times. Well, we can consider ourselves lucky as physicists as we are leaving all these worries to mathematicians… [Actually, I am not yet able to tackle these issues nor see the need for doing this any soon. So readers interested in learning these delicate matters are advised to look elsewhere.]
2.3. Replacing by the limit of smooth function sequence (II)
Let us consider an alternative sequence which represents
For convenience, call
But the the limit of the quantity in the bracket on the LHS is just So we recover eq.(7).
2.4. Replacing by the limit of smooth function sequence (III)
The alternative possibilities seem endless. As yet another example, let us consider a sequence
It reaches as That is
The figure below illustrates that this is really the case.
I leave the verification that this case satisfies eq.(7) as an exercise for interested readers. But to ease off the calculations, I give results of derivatives:
3. Green’s function for 2D Laplacian
where is 2D Dirac’s delta function satisfying
where A contour plot is given below. The darker the colour the higher the value of Green’s function.
3.1. Using divergence theorem
It is clear that Green’s function eq.(36) is well-defined whenever So by excluding the point it should satisfy
This can easily be shown to be the case. For this we make use of the intermediate results
and similar ones for derivatives wrt
The analysis however is invalid for For this case, let us use divergence theorem. Consider a closed surface which encloses the volume Divergence theorem states that
where is a unit vector pointing out from each point on the surface.
In 2d, a closed surface is in fact a closed loop, whereas the enclosed volume is in fact the region inside the closed loop. The idea is to consider a small loop enclosing the origin. In particular, let us consider a circle of radius centred at So A sketch is given by the figure below.
Let us consider an integral
Let us compute
By integrating eq.(33) over the area enclosed by a small circle centred at the LHS agrees with the LHS of eq.(44). By using the property of Dirac’s delta function, the RHS from the two equations can also be shown to agree. So we have finished showing that eq.(36) satisfies eq.(33).
3.2. Working in polar coordinates with cut-off
NB: This method is still not fully justified. Proceed with care.
In polar coordinates, eq.(33) becomes
Let us cut the solution eq.(36) off for That is, we consider
On the other hand, the same operation on RHS of eq.(50) gives
as required. Note that the second term on the penultimate step vanishes because as
4. Green’s function for 3D Laplacian
[To be continued…]