Question

If the characteristic polynomial of \(f^{\prime \prime}+a f^{\prime}+b f=0\) has real roots, show that \(f=0\) is the only solution satisfying \(f(0)=0=f(1)\).

Short answer

The only solution satisfying the conditions is \(f(x) = 0\).

Step-by-step solution

Determine the Characteristic Equation

Given the differential equation \(f'' + a f' + b f = 0\), the characteristic polynomial is obtained by replacing \(f\) with \(e^{\lambda x}\). This gives the equation \(\lambda^2 + a \lambda + b = 0\).

Analyze the Conditions for Real Roots

The roots of the characteristic equation \(\lambda^2 + a \lambda + b = 0\) are real if its discriminant \(a^2 - 4b\) is greater than or equal to zero. This means that \((\lambda_1, \lambda_2)\) are real numbers.

Consider the Form of the General Solution

When the roots are real and distinct, i.e., \(\lambda_1 eq \lambda_2\), the general solution of the differential equation is \(f(x) = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_2 x}\). If the roots are real and equal, \(\lambda_1 = \lambda_2 = \lambda\), the solution is \(f(x) = (C_1 + C_2 x) e^{\lambda x}\).

Apply Initial Conditions

To meet the conditions \(f(0) = 0\) and \(f(1) = 0\), substitute into the general solution form. For distinct roots, \(f(0) = C_1 + C_2 = 0\) and \(f(1) = C_1 e^{\lambda_1} + C_2 e^{\lambda_2} = 0\). Equivalently for repeated roots, \(C_1 = 0\) or both equations become zero.

Show Zero Solution with Real Distinct Roots

With distinct roots, from \(C_1 + C_2 = 0\) and \(C_1 e^{\lambda_1} + C_2 e^{\lambda_2} = 0\), you express \(C_2 = -C_1\) and substitute to get \(C_1 (e^{\lambda_1} - e^{\lambda_2}) = 0\). If \(e^{\lambda_1} eq e^{\lambda_2}\), then \(C_1 = 0\). Thus \(C_1 = C_2 = 0\) and \(f(x) = 0\).

Show Zero Solution with Repeated Roots

For repeated roots, the conditions \(f(x) = (C_1 + C_2 x)e^{\lambda x}\) yield \(f(0) = C_1 = 0\) and \(f(1) = C_2 e^{\lambda} = 0\). Since \(e^{\lambda} eq 0\), \(C_2 = 0\) must hold. This gives \(f(x) = 0\).

Key concepts

Characteristic Polynomial

The characteristic polynomial is a central concept when working with linear differential equations like \( f'' + a f' + b f = 0 \). Essentially, it helps us understand the behavior of the solutions to the differential equation by transforming it into an algebraic problem. The process involves replacing the function \( f \) with an exponential function \( e^{\lambda x} \). This substitution leads to the characteristic polynomial, which in this case is \( \lambda^2 + a \lambda + b = 0 \).

Why is the polynomial important? It tells us about the type of solutions we can expect. These solutions are related to the roots of the polynomial:

• Distinct Real Roots: The general solution will be a combination of exponential functions of these roots.
• Repeated Roots: The solution will be slightly different, incorporating the variable \( x \) itself multiplied by an exponential.

By analyzing the characteristic polynomial, we can decide on the correct form of the general solution, making it a powerful tool in solving differential equations.

Real Roots

Real roots of a characteristic polynomial are significant because they determine the real-valued behavior of solutions to a differential equation. To determine if the roots are real, we use the discriminant of the quadratic equation \( \lambda^2 + a \lambda + b = 0 \). The discriminant is given by \( a^2 - 4b \).

If \( a^2 - 4b \geq 0 \), the roots \( \lambda_1 \) and \( \lambda_2 \) are real. Let's see what this implies:

• Distinct Real Roots: If the roots are not equal, the solution comprises two distinct exponential terms: \( f(x) = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_2 x} \).
• Repeated Real Roots: If the roots are the same, \( \lambda_1 = \lambda_2 = \lambda \), the solution takes the form \( f(x) = (C_1 + C_2 x) e^{\lambda x} \).

When dealing with boundary conditions, knowing the type of roots greatly assists in finding the unique solution to the problem at hand.

Initial Conditions

Initial conditions play a crucial role in determining the specific solution to a differential equation from the set of possible general solutions.

In our example, we are given the conditions \( f(0) = 0 \) and \( f(1) = 0 \). These translate into specific equations for the constants \( C_1 \) and \( C_2 \) in the general solutions. Here's a step-by-step guide on how these conditions narrow down the exact solution:

• For Distinct Real Roots: Substituting \( f(0) = C_1 + C_2 = 0 \), we find \( C_2 = -C_1 \). Applying \( f(1) = C_1 e^{\lambda_1} + C_2 e^{\lambda_2} = 0 \), and using the relation \( C_2 = -C_1 \), we determine that \( C_1 \) must be zero unless \( e^{\lambda_1} = e^{\lambda_2} \), which is generally not true. Thus \( C_1 = C_2 = 0 \), making \( f(x) = 0 \).
• For Repeated Real Roots: Given \( f(x) = (C_1 + C_2 x) e^{\lambda x} \), applying \( f(0) = C_1 = 0 \) and \( f(1) = C_2 e^{\lambda} = 0 \) leads to \( C_2 = 0 \), since \( e^{\lambda} eq 0 \). Therefore, \( f(x) = 0 \).

Using initial conditions, we find not just any solution, but the specific one that satisfies these given constraints, which in this challenge demonstrates the zero solution \( f(x) = 0 \).