Question

Let $$ f^{\prime \prime}(x)+p(x) f(x)=0 \quad \text { and } \quad g^{\prime \prime}(x)+p(x) g(x)=0, \quad a

Short answer

Question: Prove that if the Wronskian \(W=f'g-fg'\) of two functions \(f\) and \(g\) is a non-zero constant on an interval \((a, b)\), and if \(f(x)\) has two zeros in this interval, then \(g(x)\) must have a zero between these two zeros of \(f(x)\). Answer: If the Wronskian is a non-zero constant, we can apply the Intermediate Value Theorem to a function \(h(x)=\frac{f(x)}{g(x)}\) derived from the given functions. Evaluating \(h(x)\) at the zeros of \(f(x)\) yields \(h(x_1)=h(x_2)=0\). Since the derivative \(h'(x)\) only changes sign when \(g(x)\) changes sign, there must exist a point \(c\in(x_1, x_2)\) such that \(g(c)=0\), proving that \(g(x)\) has a zero between the two zeros of \(f(x)\).

Step-by-step solution

Computing the derivative of the Wronskian W

We have \(W=f'g-fg'\). We want to find its derivative with respect to \(x\): $$ W'=\frac{d}{dx}(f'g-fg')=(f''g+f'g')-(f'g'+fg'') $$

Using the given differential equations

We can write the given equations as \(f''=-p(x)f\) and \(g''=-p(x)g\). We now substitute this into the expression for \(W'\): $$ W'=(-p(x)f\cdot g+f'g')-(f'g'+(-p(x)g\cdot f)) $$

Simplifying and showing W' is 0

We can now simplify the expression: $$ W'=(-p(x)fg+f'g')-(f'g'+(-p(x)gf))=0 $$ Since \(W'=0\), it implies that the Wronskian \(W\) is constant on \((a, b)\). #part b#:

Using the given conditions

We are given that \(W\neq0\), \(f(x_1)=f(x_2)=0\), and \(a<x_1<x_2<b\). Our goal is to show that there exists a \(c\in(x_1, x_2)\) such that \(g(c)=0\).

Defining the ratio of f to g

Define the function \(h(x)=\frac{f(x)}{g(x)}\). We want to find the value of \(h'(x)\) to apply the Intermediate Value Theorem. By quotient rule, we have: $$ h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} $$ Notice that \(h'(x)=-\frac{W}{g(x)^2}\).

Applying the Intermediate Value Theorem

Since \(h'(x)=-\frac{W}{g(x)^2}\), and we know that \(W\neq 0\), the sign of \(h'(x)\) is determined by the sign of \(g(x)^2\). Because \(g(x)^2\) is always non-negative, then \(h'(x)\) is either \(0\) or has the opposite sign of \(W\). Now, evaluate the function \(h(x)\) at the points \(x_1\) and \(x_2\), i.e., find \(h(x_1)\) and \(h(x_2)\). Note that \(f(x_1)=f(x_2)=0\), so we have: $$ h(x_1)=\frac{f(x_1)}{g(x_1)}=0 $$ and $$ h(x_2)=\frac{f(x_2)}{g(x_2)}=0 $$ Between \(x_1\) and \(x_2\), either \(h'(x)\) must be \(0\) or change signs. Since \(h'(x)\) changes signs only when \(g(x)\) changes signs, there must exist a point \(c\in(x_1, x_2)\) such that \(g(c)=0\). This completes the proof.

Key concepts

Real Analysis

Real Analysis is a branch of mathematics that deals with the theory and application of functions of real numbers. It is primarily concerned with the properties of real-valued functions, sequences, and series. In the context of differential equations, Real Analysis provides the rigor behind the concepts such as limits, continuity, and differentiability, which are fundamental to understanding the behavior of solutions to differential equations.

For instance, the concept of a derivative emanates from Real Analysis. Understanding how derivatives behave and how we manipulate them is key to solving and analyzing differential equations. The derivations often involve real-valued functions, ensuring the solutions are consistent with the real-world phenomena they are expected to model.

In the given exercise, we use Real Analysis principles to compute the derivative of the Wronskian and prove its constancy on an interval. This involves the application of derivatives, function manipulation, and ultimately relies on the structure provided by Real Analysis to ensure that the functions and operations we discuss are well-defined and adhere to the expected properties of real-valued functions.

Differential Equations

Differential Equations are mathematical equations that involve functions and their derivatives. They describe various phenomena where there is a rate of change, such as in physics, economics, and biology. An essential type of differential equation is the second-order linear homogeneous equation, which appears in the original problem provided.

In the context of our exercise, we deal with two functions, denoted as f and g, and their relation through a second-order linear homogeneous differential equation with a variable coefficient p(x). Solving such equations often involves proving certain properties of their solutions, like the constancy of the Wronskian, which gives insight into the behavior of these functions and their interrelations.

The process used to verify that the Wronskian is constant entails initially computing the derivative of the Wronskian and then using the differential equations provided to simplify this derivative. This simplicity comes from the assumption that the solutions f and g are continuously differentiable over the interval (a, b), thereby satisfying the differential equations over the entire interval. These steps deeply involve the application of differential equations theory and demonstrate how solutions to these equations portray a fundamental relationship that is persistent throughout the interval.

Intermediate Value Theorem

The Intermediate Value Theorem is a key concept in calculus, a part of Real Analysis, which states that if a function is continuous on a closed interval \[a, b\], and takes different signs at the endpoints of the interval, then it must have a root in \(a, b\). This theorem is extensively used to prove the existence of solutions in a given range without necessarily finding the exact value of the solutions.

In the exercise improvement advice, applying the Intermediate Value Theorem is crucial for proving part (b). Here, we create a new function h(x) that is determined by the ratio of the two functions f(x) and g(x). If f and g are continuous, so too is h, and we can apply the Intermediate Value Theorem to h(x) to infer the existence of a root.

By examining h'(x) and W, and acknowledging that W ≠ 0, it is surmised that there must be a c in \(x_1, x_2\) where g(c) = 0. This conclusion is drawn without directly solving the equation g(x) = 0, which showcases the power of the Intermediate Value Theorem in confirming the presence of a root based on conditions at the boundaries and the continuity of g(x).