Why do 2nd-order linear ODEs have at most two independent solutions?

[LFCFAN]

Homework Statement

Why does the following ODE ALWAYS have two linearly independent solutions?

x''(t) + a(t) x'(t) + b(t) x(t) = f(t)


The characteristic polynomial argument is not sufficient?

 

[HallsofIvy]

That's a linear equation. It's fairly easy to show that the set of all solutions to an nth order linear homogeneous (f(t)= 0) differential equation forms an n dimensional vector space with the usual addition of functions and multiplication by a number. With f(t) NOT 0, the solution set is a "linear manifold" (think of a plane that does NOT contain the origin) so there exist a function such that every solution can be written as a member of an n dimensional vector space plus that function. The two linearly independent solutions are the "basis vectors" of the two dimensional vector space.

 

[LFCFAN]

Any mathematical elaboration I can have?

 

[HallsofIvy]

Do you understand what a "vector space" is?

In order to show that a set, with a given "addition" and "scalar multiplication" is a vector space you must show three things:
1) It is non empty. You can do that by showing that it contains the "zero vector" (the additive identity). Here, since the addition is ordinary addition of functions, the additive identity is the 0 function: f(x)= 0 for all x. Show that y= 0 satisfies $$a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0$$ for any functions. (The superscript in parentheses indicates the derivative.)

2) It is "closed under scalar multiplication". To do that show that if y(t) satisfies $$a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0$$ then so does ay(t) for any number a. That's just a matter of factoring out a.

3) It is "closed under addition". If y= u(t) and y= v(t) both satisfy $$a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0$$ then so does y= u+ v. That can be done because the nth derivative is itself "linear"- the nth deriviative of u+ v is the nth derivative of u plus the nth derivative of v.

To show that the solution set of a linear homogeneous nth order equation is a vector space of dimension n, use the "existence and uniqueness theorem" to define the n functions, $u_1$ that satisfies the given equation with initial conditions $u_1(0)= 1$, $u_1'(0)= u_1''(0)= \cdot\cdot\cdot= u_1^{(n-1)}(0)= 0$, $u_2(0)= 0$, $u_2'(0)= 1$, $u_2''(0)= \cdot\cdot\cdot= u_2^{(n-1)}(0)= 0$, etc., with "$u_i(t)$" being the solution to the differential equation having its ith derivative equal to 1 at t= 0 and all other values 0.

Those are sometimes referred to as the "fundamental solutions" to the differential equation. They have this very nice property: if y(t) is any solution to the differential equation, let $A_1= y(0)$, $A_2= y'(0)$, ... , $A_{n}= y^{(n-1)}(0)$. Then it is easy to show that $y(t)= A_1u_1(t)+ A_2 u_2(t)+ \cdot\cdot\cdot+ A_{n}u_{n-1}(t)$. That is, any solution to the differential equation can be written as a linear combination of the "fundamental solutions" so they form a basis for the vector space of all solutions showing that the vector space has dimension n. Any solution can be written as a linear combination of those n independent solutions.

Now, for the non-homogeneous equation, let Y(t) be any function satisfying the equation $a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= f(x)$. Then it is easy to show that if y is any other function satisfying that equation, u(t)= y(t)-Y(t) satisfies the equation $a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0$. That is, that y(t)= u(t)+ Y(t), it is the sum of a function in the vector space of solutions to the homogeneous equation plus that one given solution to the entire equation.

 

[HalfLostAndSearching]

There are several reasons why 2nd-order linear ODEs have at most two independent solutions. One of the main reasons is due to the fundamental theorem of linear algebra, which states that a system of linear equations can have at most as many independent solutions as the number of unknowns. In the case of a 2nd-order linear ODE, there are two unknown functions (x(t) and x'(t)), and therefore, there can be at most two independent solutions.

Another reason is related to the concept of linear independence. In order for two solutions to be considered independent, they must not be a linear combination of each other. This means that there must be at least two distinct solutions that cannot be expressed in terms of each other. In the case of a 2nd-order linear ODE, the general solution can be written as a linear combination of two independent solutions, and therefore, there can be at most two independent solutions.

Furthermore, the characteristic polynomial argument is not sufficient because it only considers the roots of the characteristic polynomial, which determine the form of the general solution. However, it does not take into account the coefficients of the ODE, which can also affect the number of independent solutions. In some cases, the characteristic polynomial may have repeated roots, but the ODE can still have two independent solutions due to the coefficients.

In conclusion, the limitation of two independent solutions in 2nd-order linear ODEs is a fundamental property of linear systems and is determined by the number of unknowns and the concept of linear independence.