Differential Equations - Sturm-Liouville Examples

The following will just be some examples of putting equations in Sturm-Liouville form.

Lets put the following into Sturm-Liouville form

So now we must find out what r, q and P are.

 

Then we can solve for r q and P

 

Then plug this into Sturm-Liouville form

Back to Sturm-Liouville

Differential Equations - Sturm-Liouville Form

The following will be on transforming an equation into Sturm-Liouville form which is useful in finding the eigenvalues/eigenfunctions of a differential equation within a given boundary. It does this by creating a Hermitian operator.

• Operator - a series of "operations" that act upon a function, sort of like a set of instructions. This can be anything from adding and subtracting certain parts of the function to taking the derivative, or integral. This maps a function from one vector space to another, and is not necessarily one to one.
• Hermitian Operator - an operator that is self adjoint, which means that if you put the operator into matrix form to act upon an equation, that matrix is the same when you take the conjugate transpose of it.
• Conjugate Transpose - Taking the transpose of a matrix, which means to switch the indices from (1,2) to (2,1) for the entire matrix. Then conjugate any terms that have imaginary components, which means switch the signs of the imaginary terms.

Sturm Liouville Form

Given a second order differential operator where D², and D represent partial derivatives acting on some variable

We will first have this operator act on the following equation where the operator takes the place of the second order derivative acting on y where lambda is an eigenvalue, explained in a separate section.

giving us the following

We can then rearrange this by dividing through by a₂(x)

Now we multiply through by e to the integral of a₁ over a₂ and replace the operator with just y'' and y' for simplicity

Now we are going to once again regroup this into

by taking advantage of the product rule for taking derivatives, which makes the above two parts equivalent. Now we are going to define the parts of this equation by functions r, q, and p.

Then we plug into this form.

Differential Equations - Power Series

In order to solve a differential equation by the use of power series, we simply find the coefficients then attempt to match the created series to a maclaurin or taylor expansion of a known function. Let us try solving for the solution of the following differential equation.

y'' - xy' + y = 0

We substitue in for the differentials with a power series that represent the differential.

Which will give us the following equation

Now we set y to 0 to solve around the origin, which is an ordinary point in this case, as well as changing the indices from m to k. For the first term, we set 

m - 2 = k

For the second and third terms we just set

m = k

Giving us the equation

Now we need to match our indices so we'll write the first term where k = 0, then write the combined sum starting from k = 1.

Which we can then solve for the recursion relation.

Now that we have the recursion relation we can then solve for the first few coefficient terms in the solution.

C₂ = -1/2 C₀
C₃ = 0
C₄ = 1/12 C_{2 =} -1/24 C₀

Which gives us the following as the solution to this power series problem

Differential Equations - Euler Cauchy Equation

The euler cauchy equation is a method for solving differential equation with variable coefficients but only in a specific form. The form is that the equation must have the variable match the order of the differential next to it, so for example x³y'''. We then plug this into a polynomial and solve in order to find the coefficients to the general solution. Depending on the solutions of the roots of the polynomial we get different general solutions. Lets find the solution to the following equation.

Now lets take the information we need from this and plug it into the following equation which is the characteristic equation for an Euler Cauchy equation thats second order.

Where we get b and c knot from

So this means that if x² has some coefficient or didn't match this form, we'd have to adjust the equation to make the first term x². In this case though its fine, and we just have to find b(x) and c(x), which are

b(x) = -3

c(x) = 4

which means that our equation will be 

Which after factoring becomes 

So r is equal to 2 twice, which means its a repeated result, one of the possibilities. Now the general solution to the Euler Cauchy equation is x to the n where n is the solution, but in this case we have a repeated result. This means we need to add an additional factor of ln(x) into our solution giving us the following.

Which is our final result assuming there's no initial conditions.

Differential Equations - Constant Coefficients

This will go over the process of solving simple differential equations with constant coefficients by using the characteristic equation. The basic idea is to replace the order of the differential with a polynomial that has the same power and coefficients as the differential. The solution of this polynomial, depending on the sign or type of number the solution is gives us the coefficients to the general solutions we already know for this type of differential equation. Lets go over the following example

We can replace this with

r³ - 3 r² + 6 r - 18 = 0

We then solve this equation in order to find the coefficients to a general solution.

r² (r - 3) + 6( r - 3) = 0
(r - 3) ( r² + 6) = 0
r = 3, +- (6)^1/2i

So now that we solved this equation, we can note that theres a 3. This means that the coefficient of an exponential is 3

y = c₁e^3x

The complex solutions result in a different way, resulting in sin and cosine solutions. This can be proven using complex analysis and substituting in using euler's equation.

Resulting in a final combined solution of this

Which is in a form we can use in order to completely solve if we have initial conditions to find the coefficients c₁, c₂ and c₃. Lets use the example that at y(0) = 1, y'(0) = 1, y''(0) = 1. This would mean that

After taking the derivative, then the double derivative, and plugging in for x = 0. This gives us a system of equations which we can solve by taking the reduced row echelon form of it.

Giving us a solution for all the coefficients and a final answer of