Eigen Function Expansion and Applications.
[Pages:6]Eigen Function Expansion and Applications.
In this worksheet we will use the eigenfunction expansion to solve nonhomogeneous equation.
a/ The theory. b/ Example1: Solving the Euler equation in two ways. c/ Example2: non uniqueness of solution.
The idea is simple:
a/ The theory: Let L be a regular Sturm-Liouville operator on an interval (a, b) together with regular boundary conditions. That is, (note the minus sign in front of the 2nd derivative)
L( u ) = - x p( x ) x u + q( x ) u Let f be a function in L2(a, b). We consider the problem
Lu = f
We denote by n, ( n, x ) the eigenvalues and corresponding normalized eigenfunctions of L.
L( ( n, x ) ) = n ( n, x )
(1)
We know that { ( n, x ) } makes a complete orthonormal set in L2(a, b). We write f as
f( x ) = fn ( n, x )
(2)
n=1
where fn is the fourier coefficients of f with respect to the base { ( n, x ) }.
Let's assume that the solution u of the nonhomogeneous equation can be expressed as
u( x ) = un ( n, x )
(5)
n=1
We want to compare the coefficients un in (5) with those fn in (2).
Formally, we apply the operator L to the infinite sum (5) to compute L( u ).
Lu(
x
)
=
Ln
=
1
un
(
n,
x
)
=
n
=
1
un
L(
(
n,
x
)
)
=
n
=
1
un
n
(
n,
x
)
Using (1) and the equation (3) we have
Page 1
un n ( n, x ) = fn ( n, x )
n=1
n=1
We then equalize the coefficients of
( n, x ) on both sides and find
un =
fn n
.
Using
this
in
(5)
we
have
(see the note that follows)
u( x ) =
fn ( n, x )
n=1
n
(6)
Important: If n = 0 for some n, that is if 0 is an eigenvalue for L then the above makes no sense
(dividing by 0) unless fn = 0 accordingly. We can see this from the relation un n = fn . Thus, we must
have
fn
=
b
f(
x
)
(
n,
x
)
dx
=
0.
In
other
words,
a
The right hand side f must be orthogonal to all eigenfunctions which correspond to the 0
eigenvalue, in order the equation Lu = f has a solution.
Moreover, if this is the case then the corresponding un can be arbitrary constant. This also says that the problem Lu = f does not have an unique solution.
In fact, let us denote the eigenfunctions corresponding to the zero eigenvalue by ( 1, x ),..., ( m, x ) and the eigenfunctions corresponding to the nonzero eigenvalues by ( 1, x ),...,( n, x ),... Then the solution to Lu = f is not unique and given by
u(
x
)
=
i
m = 1
Ci
(
i,
x
)
+
n
=
1
fn
( n, n
x
)
where the constants { Ci } are arbitrary.
EXAMPLE 1: Consider the Euler operator with Neumann conditions [Sturm-Liouville type for p( x ) = 1 , q( x ) = 0, w( x ) = 1] over the interval I = { x | 0 < x < b }. The boundary conditions are type 2 at both end points.
Euler differential operator
Boundary conditions
2 L( y ) = x x y( x )
yx( 0 ) = 0 and yx( b ) = 0 a/ Can we find the Green function following the formula? b/ Find the expansion of the Green function in terms of the eigenfunctions (see (7).
Page 2
c/ What is the condition should be imposed on f such that the equation L( u ) = f has a solution?
d/ Let b=1, and f( x ) = x -
1 2
sin(
x
).
Solve
Lu
=
f
using the undetermined constant
method and the series method.
Compare by graphing.
f/ let b = 1 and f be the piecewisedefined function
-1 1
1 - x 0 and x - 0
2 1
- x 0 and x - 1 0 2
Solve the problem L( u ) = f using eigenfunction series (6). Compare the results by graphing.
Solution:
a/ Green function: To compute the Green function according to the text, we have to find two
fundamental solutions u1,u2 of Lu = 0 which satisfy the BC at the ends respectively. Since the general
solution of Lu = 0 is
> restart:u:=x->C1 + C2*x;
u := x C1 + C2 x We determine u1 and u2 by finding the constants to match the BC. At x = 0, we have > subs(x=0,diff(u(x),x)=0);
So we can take u1( x ) = 1. At x = b we have > subs(x=b,diff(u(x),x)=0);
C2 = 0
C2 = 0 Again, u2 must be a constant. However, u1,u2 cannot make two linear independent solutions. Hence, the formula for the Green function cannot be applied here! b/ Eigenfunction expansion for G: We need to determine the eigenvalues and eigenfunctions of the Euler differential operator with Neumann BC. As an exercise, you should check that:
1
0 = 0 is an eigenvalue with the normalized eigenfunction ( 0, x ) =
. b
In fact, this causes the nonexistence of two linear independent solutions.
The other eigenvalues and corresponding normalized eigenfunctions are > lambda[n]:=n^2*Pi^2/b^2;
n2 2 n := b2 > phi:=(n,x)->sqrt(2/b)*cos(n*Pi*x/b);
:= ( n, x )
2
1 b
cos n
x b
Thus the partial eigen-expansion for G is given by (excluding the first eigenfunction ( 0, x ))
> G_Pseries:=(m,x,z)->sum(phi(n,x)*phi(n,z)/lambda[n],n=1..m);
m ( n, x ) ( n, z )
G_Pseries := ( m, x, z )
n=1
n
Page 3
Let's take m = 2 and plot the graph of this partial sum. > b:=1:eval(G_Pseries(2,x,z));
cos( x ) cos( z ) 1 cos( 2 x ) cos( 2 z )
2
2
+ 2
2
> plot3d(G_Pseries(20,x,z),x=0..1,z=0..1);
c/ Solvability condition: Since the 0 is an eigenvalue, in order to the problem Lu = f has a solution
1 we must have that f is orthogonal to the eigenfunction ( 0, x ) = to this eigenvalue 0. That is,
b > Int(f(x)*1/sqrt(b),x=0..b)=0;
1
f( x ) dx = 0
0
d/ Solve the equation by undetermined constant method: The right hand side f( x ) = x -
1 2
sin(
x
).
We check the solvability condition
> f:=x->(x-1/2)*sin(Pi*x); eval(int(f(x),x=0..1));
f := x
x -
1 2
sin(
x
)
0 We looking for a particular solution of the form: up( x ) = ( A x + B ) sin( x ) + ( C x + D ) cos( x ) where A, B are constants to be determined. > up:=x->(A1*x+B1)*sin(Pi*x)+(C1*x+D1)*cos(Pi*x);
Page 4
up := x ( A1 x + B1 ) sin( x ) + ( C1 x + D1 ) cos( x ) Substitute this into the equation Lu = f, we have > diff(up(x),x,x) = (x-1/2)*sin(Pi*x);
collect(%,[sin(Pi*x),cos(Pi*x)]);
2 A1 cos( x ) - ( A1 x + B1 ) sin( x ) 2 - 2 C1 sin( x ) - ( C1 x + D1 ) cos( x ) 2 =
x
-
1 2
sin(
x
)
( - ( A1 x +
B1 ) 2 -
2 C1 ) sin( x ) +
( 2 A1 -
( C1 x +
D1 ) 2 ) cos( x ) = x -
1 2
sin(
x
)
Compare the coefficients, we infer that
> A1:=-1/Pi^2; B1:=1/(2*Pi^2); C1:=0; D1:=-2/Pi^3;
1 A1 := - 2
11 B1 := 2 2
C1 := 0
1 D1 := - 2 3 > diff(up(x),x,x)=f(x); #check against the equation
- -
x 2 +
1 2
1 2
sin( x )
2
=
x
-
1 2
sin(
x
)
So, a particular solution is ( A1 x + B1 ) sin( x ) + D1 cos( x ) and the general solution is then given
by
> u:=x->C3 + C2*x + up(x);
u := x C3 + C2 x + up( x ) We need to find the constants to match the BC. > subs(x=0,diff(u(x),x))=0; subs(x=1,diff(u(x),x))=0;
sin( 0 ) 1 cos( 0 )
C2 +
2
+ 2
= 0
sin( ) 1 cos( )
C2 +
2
- 2
=0
1 Thus, the solution is (C2 = - and C1 is arbitrary).
2 > u:=x->-1/(2*Pi)*x + up(x);
1x u := x - 2 + up( x ) > F[n]:=int(f(x)*phi(n,x),x=0..1);
Page 5
1 2 ( - 4 sin( n ) n - cos( n ) + cos( n ) n2 + - n2 )
Fn := - 2
2 ( 1 + n )2 ( - 1 + n )2
> F[1]:=int(f(x)*phi(1,x),x=0..1);
12 F1 := - 4
Fn Then the "Fourier" coefficients for u is given by
n > U[n]:= -F[n]/lambda[n]; U[1]:= -F[1]/subs(n=1,lambda[n]);
1 2 ( - 4 sin( n ) n - cos( n ) + cos( n ) n2 + - n2 )
Un := 2
4 ( 1 + n )2 ( - 1 + n )2 n2
12 U1 := 4 3 Thus, the (partial) "Fourier" series for u is > uP_series:=(m,x)-> U[1]*phi(1,x) + sum(U[n]*phi(n,x),n=2..m);
uP_series := ( m, x )
U1 ( 1, x ) +
n
m =
2
Un
(
n,
x
)
> eval(uP_series(3,x));
1 cos( x ) 1 cos( 3 x )
2
3
- 36
3
Finally, we plot the result given by the integral and the partial sum above in the same graph for a comparison. It can be seen that, even with one term (m = 1), the partial sum approximates well the solution. > plot([u(x),uP_series(3,x)],x=0..1,color=[red,blue],title="Green_so
l: RED, Partial Sum: BLUE");
Page 6
One can see that the two graphs are away from each other by a constant distance. This tells us that u( x ) and the solution given by the series are different by just a constant. This is in accordance with what we know in the case the homogeneous problem has nonzero solution so that the nonhomogeneous equation exhibits nonuniqueness. f/ Define new f: > f:=x->PIECEWISE([-1, -x uP_series:=(m,x)-> sum(U[n]*phi(n,x),n=1..m);
m
uP_series := ( m, x ) Un ( n, x )
n=1
Let's compute few terms of this series
> eval(uP_series(3,x));
cos( x ) 4 cos( 3 x )
4
3
- 27
3
We plot the series with 50 (!) terms
> plot(uP_series(50,x),x=0..1);
To check against the equation we can plot the 2nd derivative of the partial series and the function f( x ). We can see that they are pretty close! > plot([diff(uP_series(50,x),x,x),f(x)],x=0..1,color=[red,blue],titl
e="2nd derivative: RED, f(x): BLUE",discont=true);
Page 8
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