May 7, 1996
This worksheet does some of the calculations of the take-home part of
the final exam for Math 312.
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When looking for the eigenvalues of this problem, I looked at the
equation u^'' = -L^2u, and then wrote down solutions of
the form a*sin(Lx) + b*cos(Lx). The eigenvalues, L, with the given
boundary conditions were then seen to satisfy the equation
tan(L*pi) = 3L/(1-2L^2), with
eigenfunctions equal to sin(Lx) - Lcos(Lx).
It soon became apparent that these weren't all of the
eigenfunctions. So I went back to the beginning, and said let's also
look for solutions of the form u^'' = L^2u. This leads to solutions
of the form u = a*e^(Lx) +b*e^(-Lx). Imposing the
boundary conditions leads to the equations
a=-b*(1-L)/(1+L) and
(1+2*L)(1+L)=(1-L)(1-2*L)e^(2*L*pi).
We do not need to look for complex eigenvalues, since this is a
self-adjoing problem and the eigenvalues of such problems
are always real. It is an easy calculation to show that zero is also
not an eigenvalue. The only solution of u^'' = 0 which
satisfies the boundary conditions is the function which is
identically zero.
-------------------------------------------------------------------------------
> f1:=(1+2*x)*(1+x)/((1-x)*(1-2*x)) - exp(2*x*Pi);
(1 + 2 x) (1 + x)
f1 := ----------------- - exp(2 x Pi)
(1 - x) (1 - 2 x)
Examining this function we soon realize that it has four real zeros:
two negative and two positive. The two positive ones
will not give us any new eigenfunctions, so we just use the two
negative values to generate eigenfunctions.
-------------------------------------------------------------------------------
> plot(f1,x=-1.5..1.5,y=-0.5..0.5);
** Maple V Graphics **
-------------------------------------------------------------------------------
> lambda[1]:=fsolve(f1,x=-2..-1);lambda[2]:=fsolve(f1,x=-0.5..0);
lambda[1] := -1.010406664
lambda[2] := -.2065109110
Note: the eigenvalues associated with these numbers are these numbers
squared. That is, the eigenvalues are positive in
contrast to the situation with the trigonometric eigenfunctions.
The following Maple code defines the first two eigenfunction.
> i:='i':for i from 1 to 2 do
> phi[i]:=lambda[i]*cosh(lambda[i]*x) - sinh(lambda[i]*x); od;i:='i':
phi[1] := - 1.010406664 cosh(1.010406664 x) + sinh(1.010406664
x)
phi[2] := - .2065109110 cosh(.2065109110 x) + sinh(.2065109110
x)
-------------------------------------------------------------------------------
> plot({seq(phi[i],i=1..2)},x=0..Pi,y=-1..8);
** Maple V Graphics **
Actually plot these eigenfunctions. Note that the first eigenfunction
has not zeros on the interval [0,pi], while the second eigenfunction
has exactly one zero on this interval.
-------------------------------------------------------------------------------
> f2:=tan(x*Pi)-3*x/(1-2*x^2);
x
f2 := tan(x Pi) - 3 --------
2
1 - 2 x
-------------------------------------------------------------------------------
> plot(f2,x=0..6,y=-2..2);
** Maple V Graphics **
-------------------------------------------------------------------------------
> i:='i':for i from 3 to 10 do
-------------------------------------------------------------------------------
> lambda[i]:= fsolve(f2,x=i-2+0.5..i-1+0.5)
> od;
lambda[3] := 1.745608956
lambda[4] := 2.836576436
lambda[5] := 3.878880392
lambda[6] := 4.903619978
lambda[7] := 5.919912466
lambda[8] := 6.931470762
lambda[9] := 7.940103067
lambda[10] := 8.946798427
Remember that the eigenvalues associated with these number are the
negatives of their squares.
-------------------------------------------------------------------------------
> i:='i':for i from 3 to 10 do
> phi[i]:=sin(lambda[i]*x) -lambda[i]*cos(lambda[i]*x);od;
phi[3] := sin(1.745608956 x) - 1.745608956 cos(1.745608956
x)
phi[4] := sin(2.836576436 x) - 2.836576436 cos(2.836576436
x)
phi[5] := sin(3.878880392 x) - 3.878880392 cos(3.878880392
x)
phi[6] := sin(4.903619978 x) - 4.903619978 cos(4.903619978
x)
phi[7] := sin(5.919912466 x) - 5.919912466 cos(5.919912466
x)
phi[8] := sin(6.931470762 x) - 6.931470762 cos(6.931470762
x)
phi[9] := sin(7.940103067 x) - 7.940103067 cos(7.940103067
x)
phi[10] := sin(8.946798427 x) - 8.946798427 cos(8.946798427
x)
-------------------------------------------------------------------------------
> plot({seq(phi[i],i=1..5)},x=0..Pi,y=-4..4);
** Maple V Graphics **
The above graphs are the plots of the first five
eigenfunctions. Notice that the first eigenfunction has no zeros on
the
interval [0,pi], the second eigenfunction has one zero, the third,
has two zeros, etc. This is not a coincidence. It is a fact that
the nth eigenfunction of self adjoint second order differential
operators, with appropriate boundary conditions has exactly
n-1 zeros in the interval.
-------------------------------------------------------------------------------
> i:='i':for i from 1 to 10 do
> Int(sin(x)*phi[i],x=0..Pi)/Int(phi[i]^2,x=0..Pi);a[i]:=evalf(");
> od:i:='i':
-------------------------------------------------------------------------------
The following Maple code defines the partial sum of the solution to
the partial differential equation, which consists of the first 10
terms.
>
i:='i':upart:=sum(a[i]*phi[i]*exp(lambda[i]^2*t),i=1..2)+sum(a[i]*phi[i]*exp(-lambda[i]^2*t),i=3..10);
upart := - 1.079105861
( - 1.010406664 cosh(1.010406664 x) + sinh(1.010406664 x))
exp(1.020921627 t)
+ 1.512299999
( - .2065109110 cosh(.2065109110 x) + sinh(.2065109110 x))
exp(.04264675636 t)
+ .3238372277
(sin(1.745608956 x) - 1.745608956 cos(1.745608956 x)) exp( -
3.047150627 t)
- .001324927234
(sin(2.836576436 x) - 2.836576436 cos(2.836576436 x)) exp( -
8.046165877 t)
+ .02287068329
(sin(3.878880392 x) - 3.878880392 cos(3.878880392 x)) exp( -
15.04571310 t)
- .00008447335488
(sin(4.903619978 x) - 4.903619978 cos(4.903619978 x)) exp( -
24.04548889 t)
+ .006258056095
(sin(5.919912466 x) - 5.919912466 cos(5.919912466 x)) exp( -
35.04536361 t)
- .00001493870789
(sin(6.931470762 x) - 6.931470762 cos(6.931470762 x)) exp( -
48.04528692 t)
+ .002570896202
(sin(7.940103067 x) - 7.940103067 cos(7.940103067 x)) exp( -
63.04523671 t)
-5
- .4167256030*10
(sin(8.946798427 x) - 8.946798427 cos(8.946798427 x)) exp( -
80.04520209 t)
Note: the exponential term is positive for the first two summands,
and negative for the last ones. Thus, as t tends to infinity
our solution, depending upon the initial condition, can also become
infinite. If the first two terms in the eigenfunction
expansion of the initial condition are not present, then the
associated solution to the partial differential equation will tend to
zero as t tends to infinity.
-------------------------------------------------------------------------------
> plot(subs(t=2,upart),x=0..Pi);
** Maple V Graphics **
> evalf(subs({x=1,t=2},upart));
3.157656557
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The rest of the take home exam is routine to work out.