My problem concerns proving the existence of solutions to a set of linear PDEs on a semi-infinite strip. We have 2 variables, r which takes the range (h, infinity), h > 0, and xi which takes the range [-1, 1].
The issue is that I wish to solve the Cauchy problem near the boundary r = h, but on r = h the PDEs change type: here they are parabolic, and for r > h, they are elliptic. This is causing the problem that if I try to put the equations into canonical form, the variable that r is transformed to becomes singular on r = h.
What can be done here? The way it seems, if I use r = h for the Cauchy curve, and hence have a parabolic equation on the boundary, how can I define a solution arbitrarily near r = h if suddenly the type changes, for *any* r > h? Or am I thinking about it wrongly?
Thanks in advance for any pointers.
(EXTRA INFO: The equation in w(r, xi) is of the form
r^2*mu*(d^2 w/dr^2) + r^2*(d mu/dr)*(dw/dr) = f(xi)*(d^2 w/d xi^2)
with f(xi) >0 for all xi in (-1 ,1), and where mu(r) is a monotonically increasing one-to-one function such that mu(h) = 0 and where all d's are partial (except the one attached to mu). Transforming this to canonical form, we find we must use a variable like
alpha(r) = integral (from R = h to r) of [1/(R*sqrt(mu(R))) ] dR,
which clearly has problems with the lower limit - unless r = h in which case alpha(h) = 0, another thing which is slightly baffling me!)
If you seek a smooth solution as r -> h, then I suspect only one initial condition can be imposed. Let's consider a simplified problem, an ODE for w(r):
d/dr ((r-h) dw/dr) = 0,
where mu(r)=r-h for this discussion. The solution is of the form
w(r) = A log|r-h| + B
for some constants A,B. If we seek smooth solutions, then A = 0, so
w(r) = B,
which means we need dw/dr(0) = 0 for a smooth solution to exist.
First you should set up the Cauchy (or boundary value) problem. This set up depends on the function mu(r). If you want to consider singular solutions you may set up
weighted Cauchy problem. To understand details consider the simple case: f=0, mu=b(r-h)^k. In these case you can find the exact solutions and figure out what kind of Cauchy problem provides the existence of a solution.
Sub case 1. k=1. (see comments of Ravi Shankar)
Sub case 2. 0
Gro: Thank you. It will take me a while to think about and respond to all of your answer. First though - why do you think that substitution is better? Firstly, my alpha changes the equations to canonical form, and secondly, your alpha still has the problem of having a singular limit as r=h. Can you explain please?
Also - what do you mean by 'the exact solutions' - do you mean the series expansions near r=h? I have derived these boundary conditions already, there is freedom of choice of the value of w at r=h for all xi, with w'(h) determined; and freedom of choice of both w and w' at infinity.
Hi Erik. Your substitution making the term next to u_{rr} equal 1. Similar simplification one can make by dividing the equation over r^2\mu. Using the substitution mention by me, one can get rid of the first order terms. But which substitution to use depends on what method you are using to prove the existence. So it good to try them all and see which one works better. By exact solution I mean representation by integrals, which could be helpful if you prove the existence using iterations. I think power series expansion could work better, but under some analyticity conditions and restriction on class of solutions that are interesting from the physical point of view.
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