Question

If \(0

Short answer

Based on the given second order homogeneous linear differential equation, we applied the Sturm-Picone Comparison Theorem to help us derive an inequality related to the distance between successive zeros of its solutions. We solved two auxiliary equations corresponding to the lower and upper bounds of the continuous function q(x), and by using the boundary conditions and Sturm-Picone Comparison Theorem, we derived the inequality: $$\frac{\pi}{\sqrt{M}}<x_{2}-x_{1}<\frac{\pi}{\sqrt{m}}$$ This result concludes our proof that the distance between successive zeros of a solution to the given linear differential equation is bounded by these inequalities.

Step-by-step solution

Introduce the Sturm-Picone Comparison Theorem

Sturm-Picone Comparison Theorem states that, if \(y_1(x)\) and \(y_2(x)\) are two linearly independent solutions of the given differential equation, and \(z(x)\) is a solution of the following auxiliary equation: \(z''(x) + Mz(x) = 0\), and \(z(x_1) = 0\), then the following inequality holds: $$y_1(x_2)z(x_1) - y_1(x_1)z(x_2) > 0$$

Find the solution of the auxiliary equation

Next, we need to find the solution of the auxiliary equation: $$z''(x) + Mz(x) = 0$$ This equation is a simple harmonic oscillator equation, whose general solution is given by: $$z(x) = a\cos(\sqrt{M} x) + b\sin(\sqrt{M} x)$$

Apply boundary conditions and find z(x)

To find the specific solution \(z(x)\) satisfying our boundary condition \(z(x_1)=0\), we plug it into the expression of \(z(x)\) and get: $$ 0 = a\cos(\sqrt{M} x_1) + b\sin(\sqrt{M} x_1)$$ Consider the case when the sine term is nonzero at \(x_1\). Therefore, let $$a\cos(\sqrt{M} x_1)=0$$ which implies that $$a=0$$. Now we have: $$z(x) = b\sin(\sqrt{M} x)$$

Evaluate Sturm-Picone Comparison Theorem expression

Now, evaluate the Sturm-Picone Comparison Theorem expression for our specific equation using the found \(z(x)\): $$y_1(x_2)z(x_1) - y_1(x_1)z(x_2) = y_1(x_2) \cdot 0 - y_1(x_1) b\sin(\sqrt{M} x_2)$$ We know that \(y_1(x_1)=0\), so the above expression simplifies to: $$y_1(x_2)z(x_1) - y_1(x_1)z(x_2) = -y_1(x_1) b\sin(\sqrt{M} x_2)$$ This should be greater than 0. Since \(y_1(x_1) = 0\), we must have that: $$\sin(\sqrt{M} x_2) > 0$$ It implies that \(\sqrt{M}(x_2-x_1) > \pi\). And we obtain: $$x_2 - x_1 > \frac{\pi}{\sqrt{M}}$$

Apply Sturm-Picone Comparison Theorem again

Now, consider a second auxiliary equation: \(w''(x) + mw(x) = 0\). Analogously, we can derive that: $$x_2 - x_1 < \frac{\pi}{\sqrt{m}}$$

Present the final result

Combining the results from Steps 4 and 5, we arrive at the final inequality: $$\frac{\pi}{\sqrt{M}}<x_{2}-x_{1}<\frac{\pi}{\sqrt{m}}$$ which proves the given exercise statement.

Key concepts

Sturm-Picone Comparison Theorem

The Sturm-Picone Comparison Theorem is an essential tool in analyzing solutions of second-order linear differential equations. This theorem helps us compare different solutions by considering their oscillatory properties. It shows relations between specific points, called zeros, where solutions repeatedly switch their sign.
In the context of our exercise, we use the Sturm-Picone Comparison Theorem to make conclusions about the intervals between successive zeros of solutions to certain differential equations. This is pivotal as it establishes bounds which are key to proving inequalities involving the differences between these zeros.

• This theorem is particularly useful when dealing with differential equations like our given expression, where the solutions might not be easily solved exactly.
• It enables one to derive inequalities without having to solve the problem explicitly, providing a broader understanding of the behavior of differential equation solutions.

harmonic oscillator

A harmonic oscillator in the realm of differential equations refers to a system that experiences harmonic motion described by a second-order linear differential equation. In mathematics, these equations often help model oscillatory systems such as springs and pendulums.
The solution of the harmonic oscillator equation, like the one we dealt with in the exercise, is typically a combination of sine and cosine functions. These functions capture the repetitive or cyclic nature of motion that solutions to these equations depict.

• This is the particular equation used to define the auxiliary function in the step-by-step solution: \(z''(x) + Mz(x) = 0\).
• Its general solution is given by \(z(x) = a\cos(\sqrt{M} x) + b\sin(\sqrt{M} x)\), illustrating how oscillations inherently lead to sine and cosine terms.

boundary conditions

Boundary conditions are crucial in restricting the general solutions of differential equations to a specific context which suits the problem at hand. They specify the values a solution must take on the boundaries of the domain, thus "anchoring" the equation to known points.
For the exercise, we have a boundary condition \(z(x_1)=0\). Applying this boundary condition refines our understanding of the harmonic functions describing the system by reducing the constants involved, to fit within the required framework.

• They ensure that the solution is tailored to satisfy the particular physical, geometric, or other constraints imposed by a specific problem.
• Applying boundary conditions can simplify differential equations significantly, allowing us to find particular solutions that are well-suited to the problem context.

differential equation solutions

Solutions to differential equations hold the key to understanding the behavior of dynamic systems. They encompass a broad range of techniques to facilitate different scenarios and equation types.
For the exercise presented, finding solutions involves working through auxiliary equations such as \(z''(x) + Mz(x) = 0\) and confirming the constraints that solutions satisfy. Solving these equations often provides oscillatory solutions expressed with trigonometric functions.

• The general solution, as encountered with the harmonic oscillator component, provides a template that gets refined through boundary conditions, ensuring the solution's suitability for specific cases.
• Understanding the bounds \(\frac{\pi}{\sqrt{M}}