Question

For each of the following initial value problems use the results of Problem 28 to find the differential equation satisfied by \(Y(s)=\mathcal{L}[\phi(t)\\},\) where \(y=\phi(t)\) is the solution of the given initial value problem. \(\begin{array}{ll}{\text { (a) } y^{\prime \prime}-t y=0 ;} & {y(0)=1, \quad y^{\prime}(0)=0 \text { (Airy's equation) }} \\ {\text { (b) }\left(1-t^{2}\right) y^{\prime \prime}-2 t y^{\prime}+\alpha(\alpha+1) y=0 ;} & {y(0)=0, \quad y^{\prime}(0)=1 \text { (Legendre's equation) }}\end{array}\) Note that the differential equation for \(Y(s)\) is of first order in part (a), but of second order in part (b). This is duc to the fact that \(t\) appcars at most to the first power in the equation of part (a), whereas it appears to the second power in that of part (b). This illustrates that the Laplace transform is not often useful in solving differential equations with variable coefficients, unless all the coefficients are at most linear functions of the independent variable.

Short answer

In this exercise, we found the differential equations satisfied by the Laplace transform of the solution Y(s) for two initial value problems: Airy's equation and Legendre's equation. For Airy's equation, Part (a): The resulting differential equation for Y(s) is: $$s^2Y(s) - s - \frac{d}{ds}Y(s) = 0$$ For Legendre's equation, Part (b): The resulting differential equation for Y(s) is: $$(s^2 + \alpha(\alpha+1))Y(s) + 1 - 2\frac{d}{ds}Y(s)= 0$$

Step-by-step solution

Part (a) - Airy's Equation:

Given the Airy's equation and initial conditions: \(y^{\prime\prime} - ty = 0\), \(y(0)=1\), and \(y^{\prime}(0)=0\). To find the Laplace transform of the equation, we can follow these steps: 1. Take the Laplace transform of each term in the given equation, resulting in: \(\mathcal{L}[y^{\prime\prime}](s) - \mathcal{L}[ty](s) = 0\) 2. Use properties of the Laplace transform to evaluate each term. For the first term, we get: \(s^2Y(s) - sy(0) - y^{\prime}(0) = s^2Y(s) - s\) 3. For the second term, use integration by parts: \(\int_0^\infty te^{-st}y(t)dt = -\frac{d}{ds}\int_0^\infty e^{-st}y(t)dt = -\frac{d}{ds}Y(s)\) 4. Now the Laplace-transformed equation looks like: \(s^2Y(s) - s + \frac{d}{ds}Y(s) = 0\) 5. Apply the given initial conditions \(y(0)=1\) and \(y^{\prime}(0)=0\), we get \(s^2Y(s) - s - \frac{d}{ds}Y(s) = 0\) So, the differential equation satisfied by \(Y(s)\) for part (a) is: $$s^2Y(s) - s - \frac{d}{ds}Y(s) = 0$$

Part (b) - Legendre's Equation:

Given the Legendre's equation and initial conditions: \((1-t^2)y^{\prime\prime} - 2ty^{\prime} + \alpha(\alpha+1)y = 0\), \(y(0)=0\), and \(y^{\prime}(0)=1\). To find the Laplace transform of the equation, we can follow these steps: 1. Take the Laplace transform of each term in the given equation, resulting in: \(\mathcal{L}[(1-t^2)y^{\prime \prime}](s) - \mathcal{L}[2ty^{\prime}](s) + \mathcal{L}[\alpha(\alpha+1)y](s) = 0\) 2. Use properties of the Laplace transform to evaluate each term. For the first term, apply theorem related to higher-order derivatives of t-transform: \(\int_0^\infty (1-t^2)e^{-st}y^{\prime\prime}(t)dt = s^2Y(s) - sy(0) + y^{\prime}(0) = s^2Y(s) + 1\) 3. For the second term, use integration by parts: \(\int_0^\infty 2te^{-st}y^{\prime}(t)dt = -2\frac{d}{ds}\int_0^\infty e^{-st}y(t)dt = -2\frac{d}{ds}Y(s)\) 4. For the third term, we obtain: \(\alpha(\alpha+1)\int_0^\infty e^{-st}y(t)dt = \alpha(\alpha+1)Y(s)\) 5. Now the Laplace-transformed equation looks like: \(s^2Y(s) + 1 - 2\frac{d}{ds}Y(s) + \alpha(\alpha+1)Y(s) = 0\) 6. Apply the given initial conditions \(y(0)=0\) and \(y^{\prime}(0)=1\), we get \(s^2Y(s) + 1 - 2\frac{d}{ds}Y(s) + \alpha(\alpha+1)Y(s) = 0\) So, the differential equation satisfied by \(Y(s)\) for part (b) is: $$(s^2 + \alpha(\alpha+1))Y(s) + 1 - 2\frac{d}{ds}Y(s)= 0$$

Key concepts

Airy's Equation

Airy's equation, represented by the differential equation \(y'' - ty = 0\), is a classic example of a linear differential equation with a variable coefficient. This equation appears in various physical contexts, such as quantum mechanics and optics, representing scenarios where the potential varies linearly with distance. In the context of the Laplace transform, the initial value problem for Airy's equation is transformed into an algebraic equation in terms of \(Y(s)\), the Laplace transform of the solution \(y(t)\).

The use of Laplace transforms to solve Airy's equation leverages specific properties of the transform, such as the ability to deal with derivatives (\(\mathcal{L}[y'']\)) and the multiplication of the function by a variable (\(\mathcal{L}[ty]\)). When solving the transformed equation, it is crucial to apply initial conditions to obtain a unique solution.

Legendre's Equation

Legendre’s equation is another form of differential equation which is critical in mathematical physics, especially in solving problems with spherical symmetry. The general form is \( (1 - t^2) y'' - 2ty' + \alpha(\alpha + 1) y = 0 \), where \( \alpha \) is a constant, representing the degree of the associated Legendre polynomials.

Through the Laplace transform, Legendre’s equation is manipulated from a second-order variable-coefficient ordinary differential equation into a more tractable form. Here, the critical insight is to remember that solving Legendre's equation involves not just the Laplace transform, but also an astute application of initial conditions, which in this case define the behavior of the solution at the origin (\( t = 0 \)). These initial conditions are crucial to finding an explicit solution for \( Y(s) \) after performing the Laplace transform.

Initial Value Problems

Initial value problems (IVPs) are a fundamental type of problem in differential equations. An IVP involves finding a function \( y(t) \) that satisfies a differential equation along with specified values, called the initial conditions, at a given starting time. These initial conditions, typically given as \( y(0) \) and \( y'(0) \) for first-order and second-order IVPs respectively, help determine a unique solution to the differential equation.

In the case of Laplace transformation, initial value problems are particularly convenient, as the transform turns differential equations into algebraic ones, making it easier to solve for \( Y(s) \) and thus for \( y(t) \) indirectly. Integrating the Laplace transform into the solution process allows us to encapsulate initial conditions directly into our transformed equations.

Variable Coefficient Differential Equations

Variable coefficient differential equations are those in which the coefficients of the differential terms are not constant but vary with the independent variable (often time or position). Solutions to these equations are less straightforward due to the variable nature, which affects standard solution techniques like separation of variables or integrating factors.

As highlighted in the exercise, variable coefficient differential equations present challenges when applying the Laplace transform, as it is most effective when the coefficients are at most linear functions of the independent variable. This limitation occurs because the Laplace transform linearly handles the derivatives, but non-linear variations in coefficients introduce complexity that the transform alone cannot simplify. Nonetheless, for equations with coefficients that are at most linear in \( t \), such as Airy's equation, the Laplace transform still offers a powerful tool for solving these IVPs.