Goulet Dave ACM101c 5/6/05
Inner Products and Self-Adjointness
If we have an inner product and a linear operator L, then the adjoint of L, if it exists, is defined as the operator which satisfies XLf, g\ = Xf, L* g\ (1)
for all f and g in the space of interest. If L=L* then the operator is said to be self-adjoint. Self-adjoint operators have lots of nice properties which we'll see some of shortly. Suppose we have a given set of functions and two inner products defined on those functions
b Xf, g\ = f HxL g HxL ,, x a b a
where the overbar denotes complex conjugation. Suppose further that these functions are eigenfunctions of some Sturm-Liouville eigenvalue problem. Ly + l w y = Hpy'L ' + q y + l w y = 0 (3)
Xf, g\w = f HxL g HxL w HxL ,, x
(2)
with some given boundary conditions we won't specify yet. Consider any two of these eigenfunctions, f(x) and g(x). Integration by parts gives XL f, g\ - Xf, L g\ = Hg p f ' - ' p fLx=b - Hg p f ' - ' p fLx=a g g (4)
Notice that if the boundary term stuff appearing on the right side of the equation vanishes, then the Sturm-Liouville operator is self adjoint. Example u '' + l u = 0 u H0L = u H1L u ' H0L = u' H1L
(5)
Suppose y and z are twice differentiable and that y satisfies the boundary conditions. XLy, z\ = y '' ,, x = y ' H0L z H0L - y ' H1L z H1L - y ' ' ,, x = z z
1 1 0 0 1
We see that if z satisfies the same boundary conditions then XLy,z\=Xy,Lz\. Hence the operator Ly=y'' is self-adjoint over the space of all functions on [0,1] which are square integrable, twice differentiable, and satisfy periodic boundary conditions. Example y '' + y' + l y = 0 y H0L = 0 y H1L = 0
y ' H0L z H0L - y ' H1L H1L - y H0L z ' H0L + y H1L z ' H1L + y z '' ,, x = z y ' H0L Hz H0L - H1LL + y H0L Hz ' H1L - ' H0LL + Xy, Lz\ z z
0
(6)
(7)
2 Suppose y and z are twice differentiable and that y satisfies the boundary conditions XLy, z\ = Hy '' + y 'L z ,, x = y ' H0L z H0L - y ' H1L z H1L - Hy ' + yL z ' ,, x =
1 1
We see from this that the operator is not self-adjoint. The adjoint problem is y '' - y ' + l y = 0 y H0L = 0 y H1L = 0
y ' H0L z H0L - y ' H1L H1L + y Hz '' - z 'L ,, x z
1 0
0
0
(8)
(9)
Example
Suppose y and z are twice differentiable and that y satisfies the boundary conditions. As in the last two examples we use integration by parts XLy, z\ = y ' H0L Hz H0L - H1LL + y H1L z ' H1L + Xy, Lz\ z y '' + l y = 0 y ' H1L = 0 y H0L = y H1L (11)
y '' + l y = 0 y H0L = 0 y ' H0L = y' H1L
(10)
We see that if z satisfies the same boundary conditions, then the boundary terms won't vanish and the operator isn't self-adjoint. Hence the adjoint problem is
(12)
So while the differential operator itself is formally self-adjoint, the operator formed from the combination of differential and boundary operators isn't self-adjoint.
Sturm-Liouville Eigenvalue Problems
A second order ODE is in Sturm-Liouville form if it is written as Hp HxL y 'L ' + q HxL y + l r HxL y = 0 axb f y '' + g y ' + h y = 0 can be put into Sturm-Liouville form by the use of an integration factor
g g h (15) IY f ,,x y'M ' + Y f ,,x y = 0 f As explained above, the choice of boundary conditions is what makes the Sturm Liouville operator self-adjoint or not. Each of the following types of boundary conditions lead to self-adjoint Sturm-Liouville problems.
(13)
A general second order ODE of the form
(14)
We'll have a regular Sturm-Liouville eigenvalue problem if the boundary conditions are of the form b1 y HaL + b2 y' HaL = 0 b3 y HbL + b4 y ' HbL = 0 (16)
3 We'll have a singular Sturm-Liouville eigenvalue problem if p(x) vanishes at an endpoint and we have conditions of the form: p HaL = 0 y HaL < and or p HbL = 0 y HbL <
(17)
We'll have a periodic Sturm-Liouville eigenvalue problem if we're given boundary conditions of the form y HaL = y HbL y ' HaL = y ' HbL (18)
Example
The boundary conditions are periodic, but as it stands this isn't a periodic S-L eigenvalue problem. To frame it as such we need to put the ODE into S-L form. 1 1 J y'N' + l 2 y = 0 x x The weight function is w(x)=1/x2 . So the inner product that we'll use is Xf, g\ =
2 1
x y'' - y ' + ly = 0 y H1L = y H2L y ' H1L = y' H2L
(19)
(20)
Self-Adjoint Operators
f HxL g HxL ,, x x2
(21)
If we have a complex inner product space with the inner products Xu,v\ and Xu, v\w defined above and linear operator L which is a self -adjoint with respect to the unweighted inner product, then there are some easy to derive properties of the eigenvalues and eigenfunctions of L. Here are two very important ones
The eigenvalues are real Suppose Lu=lwu. By the properties of inner products: l Xu, u\w = Xlu, u\w (22)
By definition of the weighted inner product and the equation u satisfies: Xlu, u\w = XLu, u\ Xu, Lu\ = Xu, lu\w (23)
By self-adjointness:
By definition of the weighted inner product and the equation u satisfies: (25)
XLu, u\ = Xu, Lu\
(24)
By the properties of inner products: Xu, lu\w = l Xu, u\w (26)
4 Hence Because u is a non trivial function, Xu, u\w is the square of the norm of u which is non-zero. So we have proved that l=l , i.e. l is real. Suppose we have two eigenvalues l g with corresponding eigenfunctions u and v, Lu=lwu and Lv=gwv By the properties of inner products l Xu, v\w = Xlu, v\w XLu, v\ = Xu, Lv\ Xlu, v\w = XLu, v\ Xu, Lv\ = Xu, gv\w (28) l Xu, u\w = l Xu, u\w (27)
The eigenfunctions are orthogonal
By definition of the weighted inner product and the equation u satisfies: (29)
By self-adjointness:
(30)
By definition of the weighted inner product and the equation v satisfies: (31)
By the properties of inner products and the fact that the eigenvalues are real Xu, gv\w = g Xu, v\w (32)
Hence
This result can be generalized a bit. Suppose that we've solved Lu = lwu +boundary conditions (34)
Since l g it must be that Hu, vLw =0 i.e. u and v are orthogonal.
l Xu, v\w = g Xu, v\w
(33)
and L* v = lwv +corresponding adjoint boundary conditions Then l Xu, v\w = Xlu, v\w = XLu, v\ = Xu, L* v\ = Xu, gv\w = g Xu, v\w Xu, v\w = 0 (35)
(36)
If l then the eigenfunctions for these two problems are orthogonal. g
(37)
Other Theorems for Regular S-L Eigenvalues Problems See your class notes or a text for theorems on and proofs of the following facts The eigenfunctions are unique. There are an infinite number of eigenfunctions There are an infinite number of eigenvalues with no accumulation point The eigenfunctions form a complete basis
5 This last property, together with the orthogonality condition I derived above, is extremely useful in the context of separation of variables and eigenfunction expansions of solutions to PDE. See the class notes or a text to learn which of these results hold for irregular and periodic S-L problems.
Green's Functions
The General Abstract Notion of a Green's Function
Let L be some linear differential operator which also incorporates information about boundary/initial conditions. Let f be some function describing inhomgeneities in possibly both the differential equation and the initial/boundary conditions. Suppose we want to solve Lu = f (38)
Now, let d represent delta functions in the equation and in the initial/boundary data. And suppose we have already solved L* G = d
*
(39)
Where L is the adjoint operator and adjoint boundary/initial conditions. Then formally we have the following string of equalities u = Xd, u\ = XL* G, u\ = XG, Lu\ = XG, f\ (40)
In this abstract framework it didn't matter if we were solving an ODE or PDE. It didn't matter if the equations were inhomogeneous or if the initial/boundary data was inhomogeneous. The basic principle of superposition should yield the desired result in any case (if applied correctly!). You don't need to understand this discussion too deeply. The main points you should take away are: (i) For ODE or PDE the essence of using Green's functions to solve inhomogeneous problems is superposition (ii) You can put the d into the DE and then use that Green's function to solve an inhomogeneous DE (iii) You can put the d into the initial/boundary data and then use that Green's function to solve a DE with inhomogeneous initial/boundary data Example We can use the solution of Gt = Gxx + d Hx - xL G Hx, 0L = 0 G H0, tL = G H1, tL = 0
(41)
To solve
Example
We can use the solution of
yt = yxx + f HxL y Hx, 0L = 0 y H0, tL = y H1, tL = 0
(42)
To solve
Gt = Gxx + d Ht - xL G Hx, 0L = 0 G H0, tL = G H1, tL = 0
(43)
6 yt = yxx + f HtL y Hx, 0L = 0 y H0, tL = y H1, tL = 0
(44)
Example
We can use the solution of
To solve
Gt = Gxx G Hx, 0L = d Hx - xL G H0, tL = G H1, tL = 0 yt = yxx y Hx, 0L = f HxL y H0, tL = y H1, tL = 0 Gt = Gxx G Hx, 0L = 0 G H0, tL = 0 G H1, tL = d Ht - xL yt = yxx y Hx, 0L = 0 y H0, tL = 0 y H1, tL = f HtL
(45)
(46)
Example
We can use the solution of
(47)
To solve
(48)
Eigenfunction Expansion of Green's Functions
Suppose we've already solved an eigenvalue problem for the operator L Lu + lwu = 0 with some boundary conditions. We found the eigenfunctions and the eigenvalues 8ln , yn HxL< (50)
*
(49)
Suppose that we've also solved the eigenvalue problem for the adjoint operator L with corresponding adjoint boundary conditions. L* u + gwu = 0 to find the eigenfunctions and eigenvalues 8gn , zn HxL< (52) (51)
With an orthogonality condition (see previous section) Hyn , zm Lw = 0 n m Nn n = m (53)
Now suppose we want to solve the Greens function problem LG = d Hx - xL (54)
7 Suppose we know that the eigenfunctions form a complete basis. We'll look for a solution of the form G = , an yn HxL
n
(55)
Orthogonality gives us
We compute the inner product of both sides of the DE with an eigenfunction to find HLG, zn L = HG, L* zn L = HG, -gn w zn L = -gn HG, zn Lw Hd Hx - xL, zn Lw = zn HxL w HxL HG, zn Lw zn HxL w an HxL = = -gn Nn Nn (57)
HG, zn Lw = Nn an
(56)
Putting this together gives
(58)
So the solution is
Example
Because the operator is self-adjoint we solve the eigenvalue problem y '' + g y = 0 y H0L = y H1L = 0
G '' = d Hx - xL G H0L = G H1L = 0
zn HxL yn HxL G = -w HxL ,, gn Nn
n
(59)
(60)
(61)
We then find
Example
The corresponding eigenvalue problem isn't self-adjoint. y '' + y' + l y = 0 y H0L = y H1L = 0 (64)
G '' + G' = d Hx - xL G H0L = G H1L = 0
Sin Hn p xL Sin Hn p xL -2 G = , n2 p2 n=1
(62)
(63)
We discovered this in a previous section. The eigenfunctions are yn = -x2 Sin Hn p xL H2 p nL2 + 1 ln = 4 Previously we found that the adjoint problem is y '' - y ' + l y = 0 y H0L = y H1L = 0
(65)
(66)
This has eigenfunctions
8 zn = x2 Sin Hn p xL H2 p nL2 - 1 gn = 4 We have the orthogonality condition
(67)
So the Green's function is
Green's Functions and Inhomogeneous Boundary Conditions.
We know a good deal about Green's functions for self-adjoint problems, but when the boundary conditions are inhomogeneous, the theorems don't directly apply since the problem isn't self-adjoint. There are two ways to fix this Suppose we have already solved LG = d Hx - xL BG = 0 Lu = f Bu = k How can we use G to help solve this? (70)
Sin Hn p xL Sin Hn p xL G Hx xL = 8 Hx-xL2 , 2 1 - H2 p nL n=1
1 0 m n Xyn , zm \ = x2 Sin Hn p xL -x2 Sin Hm p xL ,, x = 12 m = n 0
(68)
(69)
Where L is self adjoint and B represents some linear homogeneous boundary conditions that we discussed above. Now suppose we want to solve (71)
Method 1
Consider two problems Lv = f Bv = 0 Lw = 0 Bw = k (72) (73)
The first can be solved using G. The second can be solved using any of the methods you know for homogeneous ODE boundary value problems. After solving these both, form the sum u=v+w. By linearity, this solves Lu = f Bu = k (74)
Method 2
Sometimes changing dependent variables can transform Lu = f Bu = k into a problem like (75)
9 Lv = g Bv = 0 Which can then be solved using G. Example (76)
Make the change of dependent variable z=y-x (78)
y '' + y = f HxL y H0L = 0 y H1L = 1
(77)
This new variable satisfies
Example
Make the change of variables
ut - uxx = 0 u Hx, 0L = f HxL u H0, tL = g HtL u H1, tL = h HtL
z '' + z = f HxL - x z H0L = 0 z H1L = 0
(79)
(80)
Define F(x,t)=(g'(t)-h'(t))x-g'(t) and k(x)=f(x)+(g(0)-h(0))x-g(0). The variable v now satisfies vt - vxx = F Hx, tL v Hx, 0L = k HxL v H0, tL = 0 v H1, tL = 0
v = u + Hg HtL - h HtLL x - g HtL
(81)
(82)
Example's Of Building Green's Functions
Example Solve y '' + f HxL = 0 y H0L = y HLL = 0
(83)
We do this by building a Green's function G '' = -d Hx - xL G H0L = G HLL = 0 G Hx xL = (84)
Since the forcing is zero for all x except x=x, the solution will be of the form a+bx 0 x < x c+dx x < x L (85)
We choose the coefficients to satisfy the two boundary conditions
10 0 = G H0 xL = a 0 = G HL xL = c + d L G Hx xL = (86)
This gives
The Green's function will be continuous with a discontinuous first derivative. The magnitude of the jump is given by Gx Hx+ xL - Gx Hx- xL = lim
e0 x+e x-e
bx 0x<x d Hx - LL x < x L
(87)
This condition together with the continuity condition gives d - b = -1 b x = d Hx - LL G Hx xL =
Gxx Hx xL ,, x = lim
e0
x+e
x-e
-d Hx - xL ,, x = -1
(88)
(89)
1 HL - xL x 0 x < x L
Solving these we find
Note the symmetry in x and x. This is generally true of self adjoint operators (think of the analogy to self adjoint real matrices). Example y '' + l2 y = f HxL y H0L = y ' H1L = 0 l 0
1 HL - xL x x < x L L
(90)
(91)
We'll solve this by building a Green's function G '' + l2 G = d Hx - xL G H0L = G ' H1L = 0 (92)
We do this in two different ways. Method 1: The Usual Approach The Green's function will be of the form G= A Sin Hl xL + B Cos Hl xL 0 x < x C Sin Hl xL + D Cos Hl xL x < x 1
(93)
To determine the constants, we first fit the boundary conditions B=0 C Cos HlL - D Sin HlL = 0 G= (94)
These give
Continuity and the jump condition on the derivative at x=x require A Sin Hl xL = E Cos Hl Hx - 1LL E l Sin Hl Hx - 1LL + A l Cos Hl xL = -1 (96)
A Sin Hl xL 0x<x E Cos Hl Hx - 1LL x < x 1
(95)
We conclude
11
- Cos Hl Hx-1LL Sin xL Hl x < x 1 l Cos l -Cos Hl Hx-1LL Sin HlxL 0 x < x l Cos l
G=
(97)
What happens if l makes the denominator zero? Can we still build a Green's function? Can we still solve the original inhomogeneous problem? If so,will the solution be unique? Method 2:Eigenfunction Expansion Let's first solve an eigenvalue problem and use the eigenfunctions as a complete basis with which to build our Green's function. z '' + l2 z = 0 z H0L = z ' H1L = 0 r2 + l2 = 0 (98)
Since this is a constant coefficient linear ODE it will have at least one solution of the form r x . Plugging this in (99)
So the general solution is of the form z = A l x + B -l x The boundary conditions give A+B=0 A l l - B l -l = 0 Systems of equations only have non-trivial solutions when the determinant of the corresponding matrix is zero 1 1 0 = l l -l -l = l H--l - l L (102) (101) (100)
l=0 is one solution, other solutions of this are found by rearranging 2 l = -1 = p H2 n+1L (103)
a quick check shows that l=0 gives z=0 which is trivial. The other values of l give ln = H2 n + 1L p 2 zn = Sin Hln xL
n=0
(104)
Since the eigenfunctions are complete in the space of piecewise continuous functions, we can look for a solution of the form G = , an Sin Hln xL
1 0
(105)
Since the eigenfunctions come from a regular S-L eigenvalue problem we have an orthogonality condition 2 Sin Hln xL Sin Hlm xL ,, x =
1 0
0 n m 1 n=m
(106)
So we have
Multiplying both sides of the ODE by one of the eigenfunctions and integrating gives
1 2 1 Sin Hln xL G '' ,, x + l an = Sin Hln xL 2 0
an = 2 Sin Hln xL G HxL ,, x
(107)
(108)
Integra...