# Eigenfunction expansion method in PDE solutions

1. Oct 3, 2006

### pivoxa15

How does this method work? What are the mathematical ideas behind this method? Unlike separation of variables techniques, where things can be worked out from first principles, this method of solving ODE seems to find the right formulas and apply which I feel uncomfortable about.

2. Oct 3, 2006

### HallsofIvy

Staff Emeritus
Essentially, you are saying that (for self-adjoint differential operators) the eigenvectors form a basis for the linear space of all solutions to the differential equation. Typically, the eigen-functions are chosen so as to form an ortho-normal (or at least orthogonal) basis.

3. Oct 3, 2006

### pivoxa15

What are examples of eigenvectors that you talk about in a DE? Could you give an example?

Why does it only work for homogeous BC? Is it because otherwise you won't get orthogonality?

4. Oct 3, 2006

### Rach3

Consider the differential operator $${d^2}/{d x ^2}$$ acting on the vector space of (twice-diffentiable, continuous) functions on [0, Pi] which satisfy f(0) = f(Pi) = 0. Then Sin(n x) is an eigenfunction for positive integer n:
$$\frac{d ^2}{d x ^2} \sin n x = -n^2 \sin (x)$$

Note this is a vector space: linear combinations of functions satisfying f(0)=f(Pi)=0 themselves satisfy that equation.

You don't get a vector space, because a sum of two solutions to an inhomogenous equation is not a solution.

e.g., solutions x1, x2 to the equation Lx=f,
L(x1+x2) = Lx1 + Lx2 = 2f != f

Last edited by a moderator: Oct 3, 2006
5. Oct 3, 2006

### Rach3

$$\frac{d ^2}{d x ^2} \sin n x = -n^2 \sin (x)$$

In case you ask: the linear differential operator d^2/dx^2 acts on the eigenfunction sin(nx), and yields the eigenfunction again scaled by an eigenvalue (-n^2).

The reason we're looking at integers n in sin(nx) is because of our boundary condition, f(Pi)=0. For nonintegeral a, sin(ax) does not satisfy this and is not (by definition) a member of the vector space.

6. Oct 3, 2006

### Rach3

What have you learned so far, and what in particular don't you get?

Also, have you studied much linear algebra?

7. Oct 4, 2006

### pivoxa15

I have done a first course in linear algebra so I can understand your language. I noticed you haven't mentioned eigenvectors. In this case, would the eigenvectors not be vectors but eigenfunctions? Because the space is a function space although functions can be part of a vector space because a vector space is the most general space in linear algebra?

But we are dealing with functions not vectors so only the word eigenfunctions should be used?

Last edited: Oct 4, 2006
8. Oct 4, 2006

### pivoxa15

What do you mean by self-adjoint differential operators? How can you know when one is one? I see the definition in here
http://en.wikipedia.org/wiki/Differential_operator
especially,
<u,Tv>=<T*u,v>
T is a self adjoint operator if and only if T=T*
so <u,Tv>=<Tu,v>

If T is the second derivative operator than it would matter what u and v are to qualify whether T is self adjoint or not wouldn't it? Because if u and v were sin and cosine functions than T would be self adjoint but if they were some other functions like u=x, v=x^3 than T being the second derivative, wouldn't be self adjoint. This seem a bit strange that T is dependent on u and v? Why? Does that make T, that is the second derivative operator self adjoint or not?

Are self-adjoint differential operators related to self adjoint matrices in which case the solution space produces vectors which are orthogonal to each other. So a similar analogy can be drawn with functions

Last edited: Oct 4, 2006
9. Oct 4, 2006

### pivoxa15

I think you should have put in a boundary value for f to make it more clear.

I.e. say f(L)=1 is a boundary condition. If x1 and x2 are solutions than x1+x2=x3 is not a solution because x1(L)+x2(L)=x3(L)=1+1=2!=1 which means x3(x) is not a solution to the differential equation because it does not satisfy the boundary condition. Therefore in general the addition of two functions is not a solution of the original DE because it would not satisfy the boundary condition(s).

Another reason why nonhomogenous BC won't work is because the solutions won't be orthogonal functions which could be problematic for one because we can't use the fourier series to construct our infinite solution in that the constants for each n in that series cannot be determined. Two it could have something to do with what HallsofIvy was talking about in terms of selfadjoint differential operators.

Last edited: Oct 4, 2006
10. Oct 4, 2006

### HallsofIvy

Staff Emeritus
In a function space the "usual" inner product is
$$\int f(x)g(x)dx[/itex] where the integral is over whatever set the functions are defined on, so saying a differential operator, L, is self adjoint means [tex]\intL(f(x))g(x)dx= \intf(x)L(x)dx[/itex] In particular, if L is a "Sturm-Liouville" operator: [tex]\frac{d }{dx}\left(p(x)\frac{dy}{dx}\right)+ q(x)y$$
with boundary conditions y(a)= y(b)= 0, then
$$\int_a^b \left(\frac{d }{dx}\left(p(x)\frac{df}{dx}\right)+ q(x)f\right)g(x)dx= \int_a^b \frac{d }{dx}\left(p(x)\frac{df}{dx}\right)g(x)dx+ \intq(x)f(x)g(x)dx$$
That second integral is clearly "symmetric" in f and g. Do the first integral by parts, letting u= g(x),
$$dv= \frac{d }{dx}\left(q(x)\frac{df}{dx}\right)dx$$
so that
$$du= \frac{dg}{dx}dx$$
and
$$v= p(x)\frac{df}{dx}[/itex] Since f and g are 0 at both ends, that integral is [tex]-\int_a^bp(x)\frac{dg}{dx}\frac{df}{dx}dx$$
Now "reverse" it. Do another integration by parts, letting
$$u= p(x)\frac{dg}{dx}$$
and
$$dv= \frac{df}{dx}$$
and its easy to see that we have just swapped f and g.

No, that is not correct. For an operator to be self adjoint, <Tu,v>= <u, Tv> must be true for all u and v in the space. If you mean here that T= d2/dx2 then it is self adjoint (it is a Sturm-Liouville operator above with p(x)= 1, q(x)= 0). However, note that the "vector space" on which we are working must be the set of infinitely differentiable functions, that are equal to 0 at two specified points. That lets out your "x" and "x3" examples.

Yes, both matrices and linear differential operators are linear transformations on vector spaces and all properties of self adjoint linear operators apply to them as well. (Linear differential operators, however, are necessarily operators on infinite dimensional vector spaces.)

11. Oct 5, 2006

### pivoxa15

The first condition is imposed because of the self adjoint differential operator (this operator comes naturally to solve solutions of PDE) which by definition can only operate on infinitely differentiable functions (which includes x and x^3)

The second condition is imposed because the solutions of the differential equation must satisfy the two BC and we only want to include these functions in our vector space (these functions also satisfy the conditions of a vector space but many others that don't satisfy homogenous BC don't).

However it is the second condition which disallow x and x^3.

Would you say that d2/dx2 is a self adjoint operator if and only if it is a Sturm-Liouville operator?

All Correct?

$$\intL(f(x))g(x)dx= \intf(x)L(x)dx$$
Are you representing the operator as L(x) or L? Is the first (x) a function?

Last edited: Oct 5, 2006
12. Oct 5, 2006

### HallsofIvy

Staff Emeritus
Sorry my Latex got messed up. It should have been
$$\int L(f(x))g(x)dx= \int f(x)L(g(x))dx$$

(I forgot the spaces after \int.)

13. Oct 5, 2006

### HallsofIvy

Staff Emeritus
?? $\frac{d }{dx^2}$ is both self adjoint anda Sturm-Liouville operator- is no "if and only if" about it. A general second order differential operator is Sturm-Liouville, then it is self adjoint. I think the other way is also true but I'm not sure. Of course, self adjoint is defined for operators in any inner-product space while "Sturm-Liouville" applies only to second order differential operators.

14. Oct 5, 2006

### pivoxa15

Actually, I think there are other self adjoint differential operators, i.e. The Hamiltonian operator. Or is the Hamiltonian operator a branch of the Sturm-Liouville operator. I wonder if the Sturm-Liouville operator is the most general second order differential operator? If so than the 'the other way' is also true.

Returning to the more general question, how do we go from a PDE to the method of eigenfunction expansion? This process for me is a bit unclear. For example, say we have a 2D Poisson equation (equaling a constant C). To work out by separation of variables is not possible, so we have to use eigenfunction expansion. But how do we justify which eigenfunction to use? Is it only based on the BC, i.e. If Dirichlet than sine eigenfunctions?

Last edited: Oct 6, 2006
15. Oct 6, 2006

### HallsofIvy

Staff Emeritus
No, the Sturm-Liouville operators are not the "most general second order differential operators". For example,
[tex]\frac{dy}{dx^2}+ x\frac{dy}{dx}+ y[/itex] is a second order differential operator but is not a Sturm-Liouville operator. But, of course, that example is not self-adjoint.

16. Oct 6, 2006

### pivoxa15

Yes. I made mistake, I meant whether the Sturm-Liouville operators were the most general self adjoint differential operators.