Question

Consider the Sturm-Liouville problem $$ -\left[p(x) y^{\prime}\right]^{\prime}+q(x) y=\lambda r(x) y $$ $$ a_{1} y(0)+a_{2} y^{\prime}(0)=0, \quad b_{1} y(1)+b_{2} y^{\prime}(1)=0 $$ where \(p, q,\) and \(r\) satisfy the conditions stated in the text. (a) Show that if \(\lambda\) is an eigenvalue and \(\phi\) a corresponding eigenfunction, then $$ \lambda \int_{0}^{1} r \phi^{2} d x=\int_{0}^{1}\left(p \phi^{2}+q \phi^{2}\right) d x+\frac{b_{1}}{b_{2}} p(1) \phi^{2}(1)-\frac{a_{1}}{a_{2}} p(0) \phi^{2}(0) $$ provided that \(a_{2} \neq 0\) and \(b_{2} \neq 0 .\) How must this result be modified if \(a_{2}=0\) or \(b_{2}=0\) ? (b) Show that if \(q(x) \geq 0\) and if \(b_{1} / b_{2}\) and \(-a_{1} / a_{2}\) are nonnegative, then the eigenvalue \(\lambda\) is nonnegative. (c) Under the conditions of part (b) show that the eigenvalue \(\lambda\) is strictly positive unless \(q(x)=0\) for each \(x\) in \(0 \leq x \leq 1\) and also \(a_{1}=b_{1}=0\)

Short answer

In summary, we solved the given Sturm-Liouville problem by following these steps: 1. We substituted the eigenvalue and associated eigenfunction into the given Sturm-Liouville equation and integrated over the defined boundary. 2. We applied integration by parts and used the given boundary conditions to simplify the equation. In part (a), we derived the resulting equation: $$\lambda\int_{0}^{1} r\phi^{2} d x=\int_{0}^{1}\left(p \phi^{2}\right)' d x+\frac{b_{1}}{b_{2}} p(1) \phi^{2}(1)-\frac{a_{1}}{a_{2}} p(0) \phi^{2}(0)$$ In part (b), we showed that the eigenvalue λ is nonnegative, given the provided conditions. In part (c), we further demonstrated that under specific conditions, λ must be strictly positive.

Step-by-step solution

Multiply Sturm-Liouville equation by \(\phi(x)\) and integrate over \(x\) from 0 to 1.

Suppose that \(\lambda\) is an eigenvalue and \(\phi\) is its associated eigenfunction, then we multiply the Sturm-Liouville equation by \(\phi(x)\) and integrate over \(x\) from 0 to 1: \begin{align*} \int_{0}^{1}\left[-\left[p(x) \phi'(x)\right]' + q(x)\phi(x)\right]\phi(x) dx = \lambda\int_{0}^{1} r(x)\phi(x)^2 dx \end{align*}

Apply Integration by Parts to \(-\left[ p(x) \phi'(x) \right]'\) term

Integration by parts allows us to simplify the integration. So we have: $$-\int_{0}^{1}\left[p(x)\phi'(x)\right]' \phi(x) dx= -\left[p(x)\phi'(x) \phi(x)\right]\Bigr |_0^1+ \int_0^1 p(x) \phi'(x) \phi'(x) dx$$

Substitute boundary conditions

Now, we can substitute the boundary conditions \(a_{1} \phi(0)+a_{2} \phi'(0)=0\) and \(b_{1} \phi(1)+b_{2} \phi'(1)=0\) in the terms resulted from the integration by parts in the above step. $$= -\left[b_1 \phi(1) + b_2 \phi'(1)\right]\phi(1) + \left[a_1\phi(0)+a_2\phi'(0)\right]\phi(0) + \int_0^1 p(x) \phi'(x)\phi'(x) dx$$ $$= -\frac{b_{1}}{b_{2}} p(1) \phi^2(1)+\frac{a_{1}}{a_{2}} p(0) \phi^2(0) + \int_0^1 p(x) \phi'(x)\phi'(x) dx$$

Combine integrals and simplify

Now, we can combine the resulting terms from Step 3 with the equation given in Step 1. \begin{align*} & -\frac{b_{1}}{b_{2}} p(1) \phi^{2}(1)+\frac{a_{1}}{a_{2}} p(0) \phi^{2}(0) + \int_0^1 p(x) \phi'(x)\phi'(x) dx + \int_{0}^{1} q(x)\phi(x)^2 dx \\ &= \lambda\int_{0}^{1} r(x)\phi(x)^2 dx \end{align*} or $$\lambda\int_{0}^{1} r\phi^{2} d x=\int_{0}^{1}\left(p \phi^{2}\right)' d x+\frac{b_{1}}{b_{2}} p(1) \phi^{2}(1)-\frac{a_{1}}{a_{2}} p(0) \phi^{2}(0)$$ #Part B#

Show that λ is nonnegative

It is given that \(q(x) \geq 0\) and \(-a_1/a_2\), \(b_1/b_2\) are nonnegative. Therefore, from the equation in part A, we have: $$\lambda \int_{0}^{1} r\phi^{2} dx \geq 0$$ Since \(r(x)\) and \(\phi(x)^2\) are both nonnegative, their integral is nonnegative too. This implies that \(\lambda\) is nonnegative. #Part C#

Show that λ is strictly positive under given conditions

Let us consider the conditions given in the problem for part (c). We are given that \(q(x)=0\) for each \(x\) in \(0 \leq x \leq 1\) and \(a_{1}=b_{1}=0\). Under these conditions, from the equation we derived in part A: $$\lambda \int_{0}^{1} r\phi^{2} dx = \int_{0}^{1} p(x) \phi'(x) \phi'(x) dx$$ If \(\lambda=0\), then \(\int_{0}^{1} p(x) \phi'(x) \phi'(x) dx = 0\). But since \(p(x)>0\) and \(\phi'(x)\) is non-zero for the corresponding eigenfunction, the integral must be positive, which is a contradiction. Therefore, λ must be strictly positive.

Key concepts

Eigenvalue

Eigenvalues are crucial in understanding the solutions for differential equations like the Sturm-Liouville problem. In simple terms, an eigenvalue, often denoted as \( \lambda \), is a special number that allows a differential equation to have solutions, called eigenfunctions, that satisfy given conditions. It acts as a "scaling factor" for these solutions.

In our specific problem, eigenvalues are linked to the expression \(-[p(x) y']'+q(x) y=\lambda r(x) y\). They determine how the function \( y(x) \) behaves under this equation. Identifying eigenvalues involves setting up a boundary value problem, where solutions exist only for specific values of \( \lambda \).

For instance, when \( \lambda = 0 \), the equation reduces its complexity drastically, a crucial step in understanding the nature of eigenvalues for solving the problem.

Eigenfunction

Eigenfunctions accompany eigenvalues in forming solutions for differential equations. In the context of the Sturm-Liouville problem, an eigenfunction is the function \( \phi(x) \) that satisfies both the differential equation and the boundary conditions for a given eigenvalue \( \lambda \).

Think of it as a "shape" or "mode" that the solution takes, specific to the eigenvalue. For each eigenvalue, there might be one or more eigenfunctions that fit the bill.

• Eigenfunctions are often normalized for simplicity, meaning that they are adjusted to a standard form.
• They must satisfy the conditions like \( a_1 y(0) + a_2 y'(0) = 0 \) and \( b_1 y(1) + b_2 y'(1) = 0 \), ensuring the solution is valid throughout the interval in question.

Identifying the correct eigenfunction is essential for solving and understanding system dynamics modeled by the Sturm-Liouville problems.

Boundary Conditions

Boundary conditions are stipulations set at the edges of the interval on which the differential problem is defined. They help ensure that the solutions, eigenfunctions in this case, conform to necessary physical or theoretical constraints.

For the Sturm-Liouville problem we're solving, the boundary conditions are:

• At \( x = 0 \): \( a_1 y(0) + a_2 y'(0) = 0 \)
• At \( x = 1 \): \( b_1 y(1) + b_2 y'(1) = 0 \)

These conditions ensure the solutions are valid across the entire interval \([0,1]\). They define the behavior of the solution at the interval's endpoints. Adjusting these conditions (e.g., setting \( a_2 = 0 \) or \( b_2 = 0 \)) results in modifying how the problem is solved and potentially changes the set of eigenvalues and eigenfunctions.

Integration by Parts

Integration by parts is a mathematical technique employed to simplify and solve integrals, especially involving products of functions. It transforms the product of a derivative and another function into simpler integrals.

In the Sturm-Liouville problem, integration by parts is applied to the term \(-[p(x)\phi'(x)]'\) in the process of deducing properties of eigenvalues and eigenfunctions. The formula used is:

\[ \int u \, dv = uv - \int v \, du \]

Here, functions \( u \) and \( dv \) are carefully chosen based on the problem, leading to a simpler integral or an expression that's easier to manage. This method allows us to substitute boundary conditions effectively and is pivotal in deriving expressions that relate eigenvalues and their associated eigenfunctions.

For instance, it aids in isolating and simplifying terms in our equation like \(-[p(x)\phi'(x)]' \phi(x)\), streamlining the path to find eigenvalues.