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.

Sturm-Liouville Boundary Value Problem

So this seems like a good time to bring up Sturm Liouville Boundary Value Problems.  In my previous post I derived the heat equation and specifically said that we will treat all the physical parameters such as thermal conductivity as uniform throughout the wire as a simplifying assumption.  But what if we choose not to do this?  Well then, picking up the equation from my previous post, we will have..

The last two terms on the right hand side combined are G(x,t) from the previous post.  The logic behind these terms is that the heat generated inside the wire is proportional to the temperature itself with a proportionality constant q(x), and all other types of heat sources are given by h(x,t).

Now in the previous post, we said that the boundary conditions at the ends of the wire could either be kept at zero temperature or insulated and kept at a constant temperature.  What can also happen is the wire could be partially insulated, we would therefore need to combine the two boundary value problems as a generalization.

These conditions are called separated boundary conditions since one set of boundary conditions does not relate in any way to the other.  We still need an initial heat distribution of u(x,0)=f(x).

We will now use the method of separation of variables to separate the equation in terms of t and x.  We need to assume that the solution is of the form u(x,t)=X(x)T(t).  I always thought the word "assume" was kind of misleading when learning things like this.  I always thought, "Why would you assume that?"  Instead of thinking of it as an assumption, think of it as an educated guess that we are rolling with until we get nonsense, at that point that will tell us our guess is wrong.  So, plugging this in and separating the variables we get..

We need to make a logical leap.  What exactly are we saying here?  We are saying that two functions of two different variables are equal to each other at every point of x and t imaginable.  The only way this could happen is if they are both equal to a constant value.  We will denote this value with the Greek letter lambda.  Setting each side equal to lambda, we obtain.. (We will only focus on the function of x as this is the most important.)

This equation and the boundary conditions is what is known as a Sturm Liouville boundary value problem.  Notice we did a little change of notation and called the function y instead of X.  The partial derivatives also changed into normal derivatives because the function is only of one variable.  These types of functions are very important in solving the heat equation and any other type of partial differential equation where they show up.  If one does a little expanding, you can see that the Sturm-Liouville problem is actually a normal second order differential equation.

So the obvious question comes to mind, does this mean that all second order homogeneous differential equations can be put into Sturm Liouville form?  The answer to that is yes!  In much the same way you use an integrating factor to solve first order differential equations, you use an integrating factor to combine the first two terms of the differential equation.

The Sturm Liouville problem  is what is known as an eigenvalue problem.  An eigenvalue problem says that if I define an operator L[y], what values of lambda does L[y]=λy.  The operator L is called a differential operator and can be defined any way you like.  In the case of the Sturm Liouville problem L is defined as..

The r(x) in the problem is a weight function and sometimes may also be thrown in.  The function r(x) in our case will be 1.  So we can write the Sturm Loiuville problem as L[y]=λy.  To those of you who have had linear algebra, eigenvalue problems come up as Ax=λx.  Where A is the operator in this case.  

It is important to note that you must have boundary values or some sort of initial condition to be able to find the eigenvalues.

At this point, I could delve into an entire discussion on corresponding eigenfunctions and ultimately the solution to the eigenvalue problem, and I will probably do this in the future, but at the moment I just don't have the time.

One thing I want to mention before I sign off, the Sturm Liouville eigenvalue problem shown has a lot of very special properties that make it nice to deal with.  I will again go into the details in another post. Until then, see yah!