Subject: Re: Help with PDE? 
From: Robt Bryant  (bryant@math.duke.edu)
Date: 1995/04/17
Newsgroup: sci.math.research 

Dear Jason Smith,

        You wrote:

%> I need an analytical solution to the following PDE.
%>
%> d^2 V / dx^2 + d^2 V / dz^2 = a(z) V / (x^2 + b^2)
%>
%> where V=V(x,z), b is a constant and a(z) is
%> a given function.  Can anyone help, please?

        N.B.:  In the following discussion, I have 
substituted a `$y$' where you have written `$z$', mainly 
because it fits better notationally with my usage.  I trust 
that this won't confuse you.  Also, I have decided to write 
in proper tex source code so that you (and others) can
typeset this for greater ease of reading.  Thus, your 
original equation is now written in the form
$$
 {{\partial^2 V}\over{\partial x^2}}
+{{\partial^2 V}\over{\partial y^2}}
= {{a(y)}\over{(x^2+b^2)}}\, V\,.
$$

        The answer depends on what you mean by the phrase 
`analytical solution'.  If you mean `explicit formula' in 
some sense, then you still have to put some bounds on what 
you mean by `explicit formula'.  For example, consider the
case where $a\equiv 0$.  Then the equation for $V$ is just
Laplace's equation and the `explicit solution' people 
usually mean is that a solution~$V$ must be of the form 
$$
V(x,y) = \Re\bigl( H(x+ i\,y) \bigr)
$$
where~$H$ is a (possibly multiple-valued) holomorphic 
function of the variable~$z=x+iy$ on the domain of~$V$
and $\Re(z)$ means the real part of~$z$..  
This is a good thing to say, but not always so helpful.  For 
example, this certainly allows you to construct lots of local
solutions, but it doesn't give you a particularly nice way 
to solve boundary value problems, at least not directly.

        Still, you may ask if your original equation admits 
a solution in terms of a holomorphic function.  And the answer 
is that it sometimes does.  For example, when~$k\ge0$ is an 
integer, consider the equation
$$
 {{\partial^2 V}\over{\partial x^2}}
+{{\partial^2 V}\over{\partial y^2}}
= {{k(k{+}1)}\over{x^2}}\, V\,,
$$
i.e., your equation when $b=0$ and $a(y)\equiv k(k{+}1)$. 
It can be shown that the general solution in any simply
connected domain in the half-plane $x>0$ can be written in
the form
$$
V(x,y) = x^{k+1}\,
\left(
{{\partial^2}\over{\partial x^2}}
+{{\partial^2}\over{\partial y^2}}
\right)^k
\left( {{\Re\bigl(H(x+iy)\bigr)}\over{x}}
\right)
$$
where~$H$ is a holomorphic function in the domain of~$V$.
Moreover, it can be shown that when $b=0$ and $a$ is a
constant, then the only times when there is a general 
solution to your equation of the form 
$$
V(x,y) = 
K\bigl(x+iy, H(x+iy), H'(x+iy), \ldots,H^{(m)}(x+iy)\bigr)
$$
where $m\ge0$ is some integer and~$K$ is some specified function 
(not assumed to be holomorphic) on some domain in~${\bf C}^{m+2}$ 
are exactly when $a = k(k{+}1)$ for some non-negative integer~$k$, 
in which case, one can take $m=k$ and no lower value of~$m$ will do.

        In fact, one can considerably generalize what is allowable 
as a `general solution' and the non-existence result still holds 
in this case.

        The deepest work in this area is nearly 100 years old
now.  While some results were known for equations of the form
$$
 {{\partial^2 V}\over{\partial x^2}}
+{{\partial^2 V}\over{\partial y^2}}
= \lambda(x,y)\, V\,,
$$
(the so-called `Moutard equations') as far back as the late 
18th century, the subject really got going with the work of
Darboux starting around 1870.  Darboux developed an algorithm
(the `method of Darboux') for testing equations far more 
general than this for the existence of `general solutions'.
It was developed by Goursat and then further developed by
Moutard and others.  Basically, there is a sequence~$I_k$ 
($k\ge 0$) of invariants (expressible as differential 
polynomials in~$\lambda$) with the property that there is
a `general solution' of the above type (or even a much more 
general type, after the work of Lie and Goursat) if and only
if $I_k$ vanishes identically for some~$k$.  I refer you 
to Volume~$2$ of Darboux' monumental {\it Th\'eorie 
G\'en\'erale des Surfaces} for details, especially 
Chapitres~II--IX. 

        In your particular case, I don't know if there are
any choices of $a$ and $b$ other than the ones that I have 
already mentioned for which Darboux' Method  works.  I would 
have to compute the sequence of invariants in order to check 
this.  I thought I might have time to do this over the weekend, 
but I'm afraid that I just haven't found the time.
I figured that you would rather have the part of the solution
that I do know and the points to the literature now rather
than wait for a fuller explanation and calculation that I
might not have time to do before I leave for the summer.

        I hope that this has been helpful.

        Yours,

        Robert Bryant


\bye