Nice Properties of Sturm-Liouville Equations

So in my previous post I brought up Sturm-Liouville Equations and mentioned they had nice properties to them.  Also recall that I said you can define the differential operator L[y] anyway you like to.  In this post L[y]=(py')'+qy'.  Which is the Sturm-Liouville definition.

The first important thing to talk about is Lagrange's Identity. This theorem states..

Let u and v be functions having continuous second derivatives on the interval [a,b]. Then,

Where W[u,v] is the Wronskian of the functions u and v.


The value of p in this theorem is always constant.  This can be proved by using Abels Identity on the right hand side of the equation.
To prove this we will  basically use the definition of L[y], the product rule and add and subtract pu'v' simultaneously.

If we integrate both sides we obtain what is known as Green's Formula. 

It can be shown (try it!)  that if the functions u and v satisfy the boundary conditions to the Sturm-Liouville equation, the right hand side is equal to zero.  If we define an inner product as..

Then Greens formula (When it equals zero!) takes the form..

Any operator that satisfies this expression is called a selfadjoint operator.

Okay now on to our next important property of Sturm-Liouville Equations. 

 The Eigenvalues and Eigenfunctions that arise will always be real. 

To prove this we will start out with the Sturm Liouville Equation and take the complex conjugate of both sides. (r(x) is defined to always be real.)

It can be shown that the complex eigenfunction also satisfies the boundary conditions, therefore the complex value of lambda is also an eigenvalue to the problem.  If we take the equation (Before we took the complex conjugate of both sides) and multiplied both sides by the complex eigenfunction and then integrate...

Because both real and complex values of lambda are eigenvalues that give rise to eigenfunctions that satisfy the initial conditions, we can use Greens Formula to obtain..

Therefore we can equate the two expressions and divide out the integral...

Therefore, the only way this can be true is if lambda is real.

If all the eigenvalues are just scalar multiples of one another, then the values are said to be simple.  This leads us to our third important fact.

All eigen values of the regular Sturm-Liouville Eigenvalue problem are simple.

This one isn't that bad to prove.  All you have to do is assume that you have two eigenfunctions corresponding to the same eigenvalue that satisfy the initial conditions.  (The two functions will be Psi and Phi.)

Notice that we solved for the derivatives.  Recall that if the Wronskian is equal to zero, then the two functions are linearly dependent.  This is what we get.

Therefore, the two functions are just scalar multiples of one another.  This means that we essentially only get one eigenfunction for each eigenvalue.

On to the last nice property of Sturm-Liouville Equations!  

All eigenfunctions of these equations will be orthogonal on the interval [a.b].

To prove this, we again assume we have two eigenfunctions except these functions correspond to different  values of lambda.  All we do here is use Greens Formula. (Mu corresponds to another eigenvalue.)

Since lambda does not equal Mu ever, we can conclude that the integral must equal Mu.  Since the integral is the definition of orthogonality, the two eigenfunctions with respect to the weight function r(x), must be orthogonal.