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\author{Gabriel C. Drummond-Cole}
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\title{Math Physics\\November 9, 2006}
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All right, so I want to finish this up today, so I'll try to buzz along.  So remember we've been talking about electrostatics on $N=\eE^3.$  I turned this into a PDE course.  If you've ever taken a physics class, electromagentism, it always turns into a PDE class.  There are a couple of things you do initially.  You want to solve Poisson's equation for the electric potential $\phi\in \Omega^0(N).$  This satisfies $\Delta \phi=*\rho$ for electric charge density $\rho\in \Omega^3_c(N).$

Along the way, we've used Green's second formula to prove this lemma:
\begin{lemma}
Let $U$ be a connected open set with bounded closure; two continuous functions $a,b$ from $\bar{U}$ harmonic on $U, \Delta a=\Delta b=0.$  If these guys satisfy the same boundary conditions, then they are equal on $U.$\end{lemma}  This gives us solutions to the Dirichlet boundary value problem.

For fixed $f\in C^\infty \boundary U)$ solve $\Delta\phi=0$ on $U$ with $\phi|_{\boundary U}=f.$

Such a solution, if it exists, is unique.

How do we solve this?  So we want $\Delta\phi=0$ in $U.$  We won't have any charge sitting in $U.$  Then the charge is sitting outside of $U$  In order to find the field inside $U$ we have to solve the Dirichlet problem.  The field $\phi$ inside $U$ doesn't, only depends on the charge distribution outside $U$ up to the extent that it determines $\phi_0.$

We've proved uniqueness in this roundabout way, using this lemma.  The method also gives us solutions, so the method also implies uniqueness, and is a more ``physical'' approach.  So that's what we'll get going on now.  Recall Green's first formula, that for $a,b\in \Omega^0(U)$ we have $$\langle da,db\rangle_U\int_{\boundary U} a*db -\int_U a\wedge *\Delta b.$$

So we're going to use Green's first formula.  Denote $C^\infty(U)$ as continuous functions on $\bar{U}$ that are smooth on $U.$  I'll call $C_0^\infty(U)$ the elements $a$ of $C^\infty(U)$ such that $a|_{\boundary U}=0.$  Then $Harm(U)$ are elments $b\in C^\infty(U)$ that are harmonic on the interior.

I chose $a$ and $b$ to satisfy these conditions, and then Green's first formula tells us that these are orthogonal to one another.  Then  $dC_0^\infty\perp dHarm$ with respect to $\langle\ ,\ \rangle_U$

In particular, if $u\in C^\infty_0\cap Harm,$ then $u$ must be zero.  Green's first formula implies more.  That is, $dC^\infty=dC_0^\infty \oplus dHarm.$  So what does this decomposition imply for us?  For any $f\in C^\infty,$ we can write $df=da+db$ uniquely for $a\in C^\infty_0$ and $b\in Harm.$  Then there is a projection operator $dC^\infty \to dC^\infty_0$ with $\pi(df)=da$ and $(1-\pi)(df)=db.$

This gives us a way to solve the Dirichlet boundary value problem.  Let me remind you what we want.  We want $\Phi$ smooth on the interior and continuous on the closure, such that $\Phi|_{\boundary U}=\phi$ and $\Delta\Phi=0.$  Take any $f\in C^\infty$ with $f|\boundary U=\phi.$  Take $\Phi$ as $b$ up to a constant.  With a very little bit of functional analysis we were able to solve this problem.  To solve Poisson's equation you need more functional analysis.

One thing I want to point out is these minimization principles.  You know that harmonic functions should minimize distance of one-forms.  What does this have to do with harmonic forms with the boundary conditions here.  Notice that $$||df||^2_U=||da||_U^2+||d\Phi||_U^2\ge ||\d\Phi||^2_U.$$  Thus $\Phi$ minimizes the function $f\to ||df||^2_U$ among $f$ that restrict to certain boundary values.  This is step with what we know.

I'll just point out, we'll see that the modulus squared can be interpreted physically, so that $\frac{1}{2}||d\Phi||^2=\frac{1}{2}||E||^2,$ so in this sense what happens is that $\Phi$ minimizes the modulus squared, the energy.

Define $K(N\times N)\backslash \Delta\to \rR$ via $K(x,y)=\frac{1}{4\pi}\frac{1}{||x-y||}$ for $x,y\in N.$  I can write this for any two elements of $N$ but it's only defined off the diagonal.  For now fix $x,$ and note that $*dK(x,\cdot)=\frac{1}{4\pi}\beta_x,$ where this is the volume form for the two-form centered at $x.$

[Is that a $\beta.$]

Yeah, $\beta,$ ahh, $\beta.$

[What's the joke?]

It's from the Simpsons.  Everyone leaves and Snake, he robs the house, and he's stealing their VCR and they have a $\beta$

[--a betamax--]

and he says [laughing] ``ahh, beta.''

So onyway.  The identity we derived last time was $$
a(x)=\frac{1}{4\pi}\int_{\boundary U}(a\beta_a + \frac{*da}{r_x})-\frac{1}{4\pi}\int_U\frac{*\Delta a}{r_x}.$$
So this is
$$
\int_{\boundary U} (a*dK(x,\cdot)+K(x,\cdot)*da)-\int_U*K(x,\cdot)\Delta a.$$

From the Dirichlet boundary problem, for fixed $x\in U,$ the function $K(x,\cdot)$ is smooth on the boundary so I know there exists a unique harmonic function $h(x,\cdot)$ such that $h_U(x,y)=-K(x,y)$ for all $y.$
\begin{lemma}
\begin{enumerate}\item The function $h_U:\bar{U}\times \bar{U}\backslash\Delta \to \rR$ is continuous and differentiable in $y$ on $U.$
\item For fixed $x,\ \Delta_y h_u=0.$
\item G_u(x,y)=K(x,y)+h_U(x,y)=0$ for all $y\in \boundary U.$
\end{enumerate}
\end{lemma}
This is not hard to prove, but we'll see if we have time later.

I want to say some nice things about $G.$  The second one, 
\begin{lemma} $G(x,y)=G(y,x).$  \end{lemma}Take a ball of radius $\epsilon$ centered at both $x$ and $y.$  Apply Green's second formula to $U$ minus these balls.  The functions we consider are $a=G(x,\cdot)$ and $b=G(y,\cdot).$  Now, both $a$ and $b$ are harmonic on the domain, and both vanish on th boundary $\boundary U.$  So we get 
$$\int_{\boundary B_\epsilon(x)} G(x,\cdot)*dG(y,\cdot)+\int_{\boundary B_\epsilon(y)}G(x,\cdot)*dG(y,\cdot)=
\int_{\boundary B_\epsilon(x)} G(y,\cdot)*dG(x,\cdot)+\int_{\boundary B_\epsilon(y)}G(y,\cdot)*dG(x,\cdot).
$$
We claim that the left hand side goes to $G(x,y)$ as $\epsilon\to 0$ and by symmetry the right hand side goes to $G(y,x).$

The second term of the left hand side, plugging in, is $\int_{\boundary B_\epsilon(y)}G(x,\cdot)*dK(y,\cdot)+\int_{\boundary B_\epsilon(y)}G(x,\cdot)*dh(y,\cdot).$  The second term here goes to zero, since these functions are both smooth on the whole ball.  Then the first half gives $G(x,y)$ because $dK(y,\cdot)$ is $\beta_y.$

What about the other half? You can imagine that we're going to expand a $G$ in terms of $h$ and $K.$  You get $$\int_{\boundary B_\epsilon(x)}K(x,\cdot)*dG(y,\cdot)+\int_{\boundary B_\epsilon(x)}h(x,\cdot)*dG(y,\cdot).$$ The first term here is zero for the same reason as before.  The first term is $\frac{1}{4\pi\epsilon}\int_{\boundary B_\epsilon(x)}*dG(y,\cdot),$ which is $$\frac{1}{4\pi\epsilon}\int_{B_\epsilon(x)}d*d G(y,\cdot).$$ 

Let's just kill this bird.  $G$ has these nice properties, it's harmonic one variable at a time, and so on.  For any $a\in C^\infty(U),$ apply Green's second formula again, wher $b=G(x,\cdot).$  We'll do this on the domain $U\backslash B_\epsilon(x).$  So we have $$\int_{\boundary U}a*dG(x,\cdot)-\int_{\boundary B_\epsilon(x)} a*dG(x,\cdot)+\int_{\boundary B_\epsilon(x)}G(x,\cdot)*da=-\int_U G(x,\cdot)*\Delta a.$$

So expanding the left hand side I get
$$
\int_{\boundary B_\epsilon(x)} K(x,\cdot)*da+\int_{\boundary B_\epsilon(x)}h(x,\cdot)*da=\frac{1}{4\pi\epsilon}\int_\int_{\boundary B_\epsilon(x)}*da+\int_{\boundary B_\epsilon(x)}h(x,\cdot)*da.$$
The first term is of order $\epsilon$ and the second of order $\epsilon^2,$ so both of thsee go to zero.  The other terms, well, we also have the $*dG$ terms, which gives us $a(x),$ and we're left with for any interior smooth, closure continuous function $$a(x)=\int_U G(x,\cdot )*\Delta a+\int_{\boundary U}a*dG(x,\cdot).$$
Using Green's functions, we can find the value of a smooth function at any point by looking at its Laplacian and its boundary values.  So if $\Delta\phi=\rho_0,$ and $\phi|_{\boundary U}=0,$ then $\phi(x)=\int_U *G(x,y)\rho_0(y).$

If you have $G,$ you also have a way to solve the boundary value problem, so that if $\phi|_{\boundary U}=f$ and $\Delta\phi=0$ then $\phi(x)=\int_{\boundary U} f(y)*dG(x,y).$
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