# Fundamental Solution to Laplace Equation

1. Jul 8, 2012

### starzero

Many texts in deriving the fundamental solution of the Laplace equation in three dimensions start by noting that the since the Laplacian has radial symmetry that

Δu=δ(x)δ(y)δ(z)

That all that needs to be considered is

d^2u/dr^2 + 2/r du/dr = δ(r)

For r > 0 the solution given is

u= c1/r + c2

I have no trouble accepting the fact that this is a solution.

My question is by what method is the solution obtained ?

I thought to apply a Laplace transform, however no initial values for u or the derivative of u are given.

Last edited: Jul 8, 2012
2. Jul 9, 2012

### JJacquelin

Let U(s) be the Laplace transform of u(r)
What are the Laplace transforms of du/dr, of d²u/dr², and of δ(r) ?
if the initial values are not settled, settled them as parameters, for example u(0)=a and u'(0)=b.
Then, what is the Laplace transform of d²u/dr² + 2/r du/dr = δ(r) ?

3. Jul 9, 2012

### homeomorphic

I prefer to use a little more physical intuition. The physical meaning of the fundamental solution of the Laplace equation is that it is the potential that results from putting down a point charge. So, to find the solution when the Laplacian is equal to some function, you just interpret that function as charge density, so that you integrate the fundamental solution. The Laplacian is just the divergence of the gradient of the potential function. So, the key to understanding this is to figure out why the divergence is zero, except at that point where the charge is sitting. The gradient should be inverse square to make the flux going through two concentric spherical shells equal, since surface area is proportional to r^2 (there is no flux in the radial direction, since the gradient vector fields points outward). Then, you get zero divergence, if you think about taking a little cube sitting between two spherical shells. Inverse square gradient implies potential proportional to 1/r.

For a more detailed discussion of this, you might try Vladimir Arnold's book on PDE, where he also discusses the solution in higher dimensions, as well as in dimension 2. I find PDE to be a pretty dull subject without the aid of physical interpretations like these.

4. Jul 9, 2012

### andrien

just see,it is
d/dr(r^2*du/dr)=0 because delta is zero for r>0

5. Jul 9, 2012

### starzero

I fully agree with you here. It is the physical interpretations that make the subject come alive.

Following up with your physical explaination I have a follow up question …
In three dimensions the fundamental solution becomes 1/(4πr) completely in line with the reasoning that the inverse square potential should be proportional to 1/r.
In two dimensions however the fundamental solution becomes (1/(2π))log(r). Far from dropping of as an inverse square, the fundamental solution now becomes infinite. I have no physical interpretation of this but I sure would like to have one.
Does the laplace/poisson equation only apply to electrostatics problems in three dimensions?

Another comment ---
I do just see it. The problem is by what means or techniques does one obtain the solution.
I know that there have been instances where someone (more clever than I) can just look at a particular equation and solve it more with cleverness than with technique. Is this such an equation or is there a technique ?

I am still thinking about this. I’m still not sure about the initial conditions. Can we even talk about u(0) and u’(0) ? Like I said…still mulling it over.

Thanks Again.

6. Jul 9, 2012

### homeomorphic

Sort of. Electrostatics is sort of inherently 3-d because the world is 3-d. However, in the 2-d case, if you wanted to think in terms of E and M, you could consider the electric field generated by an infinitely long charged wire. Perpendicular cross sections then give you what you want. Also, you could think of a think slice of water, stuck between two sheets. If you pump fluid in at one point and look at the velocity field, you get the potential you want. Alternatively, you could take a 2-d slice of some material and stick a heat source at precisely one point. Then, the temperature solves the 2-d Laplace equation.

In this particular instance, it appears that nature provided the solution, as I said above, since a point charge solves the problem. Inverse square laws precede the Laplacian historically, so we already had the solution in our hands. All that was left was the observation that these very important potentials satisfied that equation.

7. Jul 9, 2012

### homeomorphic

To put it another way, as far as motivation is concerned, you shouldn't set out to solve the Laplace equation from the outset. It's more like a property that is satisfied by these familiar fields. Start with the fields, observe that they satisfy the equation, not the other way around. After all, physically, it's a natural condition. If you want, in fluids, it simply says the fluid is incompressible. There's no need to "come up with" a solution you already have.

8. Jul 9, 2012

### HallsofIvy

Staff Emeritus
If you multiply through by $r^2$ you get $r^2d^2u/dr^2+ 2r du/dr= r^2\delta(r)$. That is an "equipotential" or "Euler-type" equation. The substitution x= ln(r) will convert that to an equation with constant coefficients which are the simplest kind of equation. Or, as we "look for" a solution of the form $e^{ax}$ for linear d.e.s with constant coefficients, we can "look for" a solution of the form $e^{a ln(r)}= e^{ln(r^a)}= r^a$. If $u= r^a$, then $du/dr= a r^{a-1}$ and $d^2u/dr^2= a(a-1)r^{a-1}$. Putting those into the associated homogeneous equation, we have $a(a-1)r^a+ 2ar^a= (a(a-1)+ 2a)r^a= (a^2+ a)r^a= 0$. Since r is not always 0, we must have $a^2+ a= a(a+ 1)= 0$ so that a= 0 and a= -1 are roots. Those give $a^{0r}= 1$ and $a^{-r}$ as independent solution to the associated homogeneous equation. Since this is a linear second order equation, the general solution to the associated homogeous equation (NOT the entire equation) is $C+ Dr^{-1}$.