Question

Suppose \(P_{0}, P_{1},\) and \(P_{2}\) are continuous on \((a, b)\) and let \(x_{0}\) be in \((a, b) .\) Show that if either of the following statements is true then \(P_{0}(x)=0\) for some \(x\) in \((a, b)\). (a) The initial value problem $$P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=0, \quad y\left(x_{0}\right)=k_{0}, \quad y^{\prime}\left(x_{0}\right)=k_{1}$$ has more than one solution on \((a, b)\). (b) The initial value problem $$P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=0, \quad y\left(x_{0}\right)=0, \quad y^{\prime}\left(x_{0}\right)=0$$ has a nontrivial solution on \((a, b)\).

Short answer

In summary, we have analyzed the two initial value problems (a) and (b) separately and used their given conditions to show that a nontrivial solution or a unique solution exists on the interval \((a, b)\). We used Wronskian, Abel's formula, and some auxiliary functions to establish that if either of these initial value problems holds, then there must exist some \(x\) in \((a, b)\) for which \(P_0(x) = 0\). The claim has been proven in both cases.

Step-by-step solution

Initial Value Problem (a)

First, we look at the initial value problem (a). Given that it has more than one solution on \((a, b)\), let \(y_1(x)\) and \(y_2(x)\) be two different solutions to the problem. Now, since \(y_1(x)\) and \(y_2(x)\) both satisfy the given differential equation and initial conditions, their linear combination, \(y(x) = c_1 y_1(x) + c_2 y_2(x)\), must also satisfy the differential equation for some non-zero constants \(c_1\) and \(c_2\). Moreover, since \(y_1(x_0) = y_2(x_0) = k_0\), we have \(y(x_0) = k_0\) and \(y'(x_0) = k_1\).

Wronskian of \(y_1(x)\) and \(y_2(x)\)

Next, we will compute the Wronskian \(W(x) = y_1(x)y_2'(x) - y_1'(x)y_2(x)\) for the solutions \(y_1(x)\) and \(y_2(x)\). Since \(y_1(x)\) and \(y_2(x)\) are linearly independent, their Wronskian \(W(x)\) must be non-zero.

Using Abel's Formula

Now, we will use Abel's formula to relate the Wronskian to the coefficients of the differential equation: \(W'(x) = -\frac{P_1(x)}{P_0(x)}W(x)\). Since \(W(x)\) is nonzero for all \(x \in (a, b)\), we can integrate both sides with respect to \(x\): \(\int \frac{W'(x)}{W(x)} dx = -\int \frac{P_1(x)}{P_0(x)} dx\).

Conclusion for Problem (a)

Since \(y(x)\) is nontrivial, there exists an \(x_1 \in (a, b)\) such that \(y(x_1) = 0\). Define \(z(x) = y'(x) + \int \frac{P_1(x)}{P_0(x)} y(x) dx\). Integrate the expression obtained in the previous step again, and we have \(y'(x) = \int \frac{P_1(x)}{P_0(x)}y(x) dx\). Now plugging in \(x_1\), we get: \(z(x_1) = y'(x_1) + \int \frac{P_1(x_1)}{P_0(x_1)} y(x_1) dx = y'(x_1) = 0\). Since \(y(x_1) = 0\) and \(z(x_1) = 0\), we can conclude that \(P_0(x_1) = 0\) as desired. This proves the claim in case (a).

Initial Value Problem (b)

Now, we look at the initial value problem (b). We are given that it has a nontrivial solution on \((a, b)\). Let \(y(x)\) be that solution.

Wronskian of \(y(x)\) and \(1\)

We will compute the Wronskian \(W_2(x) = y'(x) - 0 \cdot 1\) for the function \(y(x)\) and the constant function \(1\). Since \(y(x)\) is a nontrivial solution, the Wronskian \(W_2(x)\) must be non-zero.

Using Abel's Formula and Conclusion for Problem (b)

As in case (a), we can use Abel's formula: \(W_2'(x) = -\frac{P_1(x)}{P_0(x)}W_2(x)\). However, in this case, the Wronskian \(W_2(x) = y'(x)\). Since \(y(x)\) is a nontrivial solution, there must exist an \(x_2 \in (a, b)\) such that \(y(x_2) = 0\), and so, \(y'(x_2) = W_2(x_2) = 0\). Therefore, \(P_0(x_2) = 0\), proving the claim in case (b).

Key concepts

Initial Value Problem

Understanding an *initial value problem* involves finding a function that not only satisfies a given differential equation but also meets specific initial conditions. Usually, these problems appear in the form of finding a solution to an ordinary differential equation, like:

• \(P_0(x) y'' + P_1(x) y' + P_2(x) y = 0\)
• With initial conditions \(y(x_0) = k_0\) and \(y'(x_0) = k_1\).

To solve such problems, one must ensure that the solution respects the initial values at the given point \(x_0\). This implies that the function not only solves the equation, but also initiates from a specified state. Initial value problems can have either a unique solution or multiple solutions, and the properties of these solutions reveal much about the coefficients \(P_0(x), P_1(x),\) and \(P_2(x)\).
Particularly, if more than one solution exists, it suggests the possible zero of the leading coefficient \(P_0(x)\) within the interval, indicating a malfunction in the fundamental existence and uniqueness solutions theorem.

Continuous Functions

Continuous functions are key players in mathematics, especially when analyzing differential equations. A function \(f(x)\) is called continuous over an interval \((a, b)\) if, for every point \(x\) in this interval, the function values vary smoothly without abrupt changes or gaps. In simpler words, you can draw the curve of the function with a single stroke, without lifting your pencil.

• Continuity is crucial when dealing with differential equations because it guarantees the integration and differentiation can be performed smoothly.
• Without continuity over an interval, solutions to an equation may not exist or be well-defined.

In differential equations like the one in our exercise, where functions \(P_0(x), P_1(x),\) and \(P_2(x)\) are continuous over \((a, b)\), we ensure that the operations and resulting behaviors of the differential equation remain predictable and workable within that interval.
Discontinuities could lead to undefined or unpredictable behavior, hence, continuous functions play a critical role in ensuring the integrity of solutions.

Wronskian

The *Wronskian* is a special determinant used in the study of differential equations to decide whether a set of solutions is linearly independent. Suppose you have two functions \(y_1(x)\) and \(y_2(x)\), the Wronskian \(W(x)\) is calculated as:\[ W(x) = \begin{vmatrix} y_1(x) & y_2(x) \ y'_1(x) & y'_2(x) \end{vmatrix} = y_1(x)y'_2(x) - y'_1(x)y_2(x) \]The Wronskian has several key properties:

• If \(W(x) eq 0\) for at least one point within the interval, then \(y_1(x)\) and \(y_2(x)\) are linearly independent.
• Conversely, if \(W(x) = 0\) at every point in \((a, b)\), the functions are linearly dependent.

In the context of the exercise, the Wronskian is used to determine whether the solutions of the initial value problem are unique or not. A non-zero Wronskian suggests that the solutions are indeed unique unless \(P_0(x)\) is zero somewhere in \((a, b)\), implying possible multiple solutions. Therefore, finding the Wronskian serves as a handy tool in comprehending solution behaviors of differential equations.