Before learning about Sturm’s oscillation and separation theorems, one should be aware of ordinary differential equations (ODE), first and second-order ODE, and homogeneous second order linear DEs. A basic understanding of these terms is necessary to learn Sturm’s oscillation and separation theorem.

Read:

• Ordinary differential equations
• System of linear First-order ODE
• First-order differential equations
• Homogeneous differential equations
• Solution of a differential equation

Sturm Oscillation Theorem

We can define the Sturm oscillation theorem in two ways. They are:

Statement 1:

The function F_n has q − 1 number of roots in the open interval (a, b) precisely.

Statement 2:

Suppose p and q are two integers such that p ≤ q and consider a set of coefficients, a_p, a_p+1, a_p+2, . . . , a_q so that not all of them are equal to 0, then the function

Sturm Separation Theorem

In mathematics, the Sturm separation theorem in ordinary differential equations illustrates the location of roots of solutions of homogeneous second-order linear ODE. The theorem says that given two linearly independent solutions of the second-order homogeneous linear differential equation, the zeros of the two solutions are alternatives to each other.

Statement:

Suppose u(x) and v(x) are a pair of linearly independent solutions of a homogeneous second-order linear ODE of the form y′′ + q(x)y = 0 such that;

(i) The zeroes of the non-trivial solution of the above equation are isolated.

(ii) Let x₁ and x₂ be the two consecutive roots (or zeroes) of u(x) then v(x) has exactly one root in (x₁, x₂).

Sturm Separation Theorem Proof

Proof for (i): The zeroes of the non-trivial solution of the above equation are isolated.

Let us assume that the zero sets of a solution have a limit point. One can establish that the solution is trivial by satisfying an initial value problem with zero initial values.

Proof for (ii): Let x₁ and x₂ be the two consecutive roots (or zeroes) of u(x), then v(x) has exactly one root in (x₁, x₂)

Let us assume that v(x) does not contain a root in the interval (x₁, x₂).

So, we can say that v(x) will not include a root in the closed interval [x₁, x₂].

Since u(x) and v(x) are a basic pair that is linearly independent, we can define a function g(x) in the interval [x₁, x₂].

Therefore by Rolle’s theorem, we can say that there exists a root c in (x₁, x₂) that means x₁ < c < x₂ such that g′(c) = 0.

This implies that Wronskian is the root at c, and this is not feasible as u(x) and v(x) are an essential equivalent and linearly independent pair.

Therefore, v(x) has at least one root in the interval (x₁, x₂).

Hence, we can conclude that v(x) does not have more than one root in the open interval (x₁, x₂).

If the above statement is false, there would be at least two v(x) roots in the open interval (x₁, x₂).

Without losing generality, let us assume two consecutive roots of v(x) be x₃ and x₄ in the interval (x₁, x₂), such that x₂ > x₄ > x₃ > x₁ or we can express x₁ < x₃ < x₄ < x₂.

Hence, by previous statements, u(x) will include a zero in an open interval (x₃, x₄).

This contradicts that x₁ and x₂ are two consecutive roots of u(x).

Therefore, v(x) has exactly one root in (x₁, x₂).

Hence proved.

Sturm Separation Theorem Solved Problem

Question:

Check the Sturm separation theorem for the given differential equation.

(d²y/dx²) – (dy/dx) = 0

(or) y′′ − y = 0

Solution:

Given differential equation is:

y′′ − y = 0

This is the second order linear ordinary differential equation.

Let us find the solution set of this differential equation.

As we know, the solution of the given differential equation is of the form y = e^rx.

So, r².e^rx – e^rx = 0

e^rx(r² – 1) = 0

r² – 1 = 0

r² = 1

So, r = 1, -1

Therefore, the solutions are e^x and e^-x.

Let u(x) = e^x and v(x) = e^-x.

(i) u(x) = e^x and v(x) = e^-x are two linearly independent solutions, and none of these have roots.

(ii) F₁(x) = sinh x and F₂(x) = cosh x are two linearly independent solutions.

Also, F₁(x) has one root, and F₂(x) has no root on R.