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MathGroup Archive 1992

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Solving eigenvalue problems for ode's

  • To: mathgroup at yoda.physics.unc.edu
  • Subject: Solving eigenvalue problems for ode's
  • From: news at murdoch.acc.virginia.edu
  • Date: Thu, 4 Jun 92 16:25:49 EDT

I posted this problem earlier this week but I got a few responses 
asking for more detail, so here it is.

This is the an aerospace problem for supersonic flow over a simply 
supported flat plate.  I have the following governing pde and 
boundary conditions:
	
	d*W""""(x,t) + m*W''(x,t) + a*W"(x,t) - b*W'(x,t) = 0
								(1)
	W(0,t)=W(L,t)=W""(0,t)=W""(L,t) = 0, 0<= x <= L, t>0
	
	where
		" - denotes differentiation w.r.t. x
		' - denotes differentiation w.r.t. t
		W - transverse motion of plate
	and a, b, d, m are positive constants.

Since this is a linear pde, I would like to find the normal modes 
of (1).  I can then expand W(x,t) and then apply Galerkin's method 
to reduce the problem to a set of equations.

Applying the separation of variables technique, we assume that
W(x,t)=F(x)*G(t).  Subbing into (1) gives

	d*F''''     a*F'       m*G''     b*g'
	-------  +  ----   = - -----  +  ---- = v
	   F         F           G        G

where and ' denotes differentiation and v is a constant. The sign of
v can be less than, equal to, or greater than zero.  The normal modes 
are represented by F(x).  Rearranging gives the classic eigenvalue 
problem;

	d*F''' + a*F' - v*F = 0
								(2)
	F(0)=F(L)=F''(0)=F''(L) = 0
and
	m*G'' - b*G' + v*G = 0.					(3)

The constants a and b are related to the the air speed.  If a=b=0, 
then we have a simply supported flat plate and solving (2) gives 
the normal modes as K*sin(n*pi*x/L), K is a constant.  The eigenvalues 
are v=(n*pi/L)^4,  n=0,1,-1,2,-2,...
(This problem was been solved many times.)

Howevwer, for a and b nonzero, the solution is not as easy.  I figure 
let Mathematica handle it.  Use DSolve to solve (2) for any 3 of the
4 boundary conditions.  Solve the resulting expression for v subject
to the 4th boundary condition. This fails because applying either of
the conditions F''(0)=0 or F''(L)=0 to (2) gives the trivial solution 
of zero.  

For v>0, The Mathematica command stream looks like:

	DSolve[{d F'''' + a F' - v F == 0, F[0] == 0, F[L] == 0,
		F''[0] == 0}, F, x]

or

	DSolve[{d F'''' + a F' - v F == 0, F[0] == 0, F[L] == 0,
		F''[L] == 0}, F, x]

which both return 

	{}.

But if I evaluate

	DSolve[{d F'''' + a F' - v F == 0, F[0] == 0, F[L] == 0}, F, x]

I get a expression as expected but how do I apply the 2 remaining
boundary conditions.

The constant v can be negative or zero also.  Of course, I must 
consider the possibility that the solution may not exist at all.  

I hope this clarifies the problem.  Please e-mail response to me.

Keith





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