Problem 6 - Active Calculus

Question

Airy's equation \(^{2}\) $$ y^{\prime \prime}-x y=0 $$ can be used to model an undamped vibrating spring with spring constant \(x\) (note that \(y\) is an unknown function of \(x\) ). So the solution to this differential equation will tell us the behavior of a spring-mass system as the spring ages (like an automobile shock absorber). Assume that a solution \(y=f(x)\) has a Taylor series that can be written in the form $$ y=\sum_{k=0}^{\infty} a_{k} x^{k} $$ where the coefficients are undetermined. Our job is to find the coefficients. (a) Differentiate the series for \(y\) term by term to find the series for \(y^{\prime}\). Then repeat to find the series for \(y^{\prime \prime}\). (b) Substitute your results from part (a) into the Airy equation and show that we can write Equation \((8.6 .4)\) in the form$$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}-\sum_{k=0}^{\infty} a_{k} x^{k+1}=0 $$ (c) At this point, it would be convenient if we could combine the series on the left in \((8.6 .5)\), but one written with terms of the form \(x^{k-2}\) and the other with terms in the form \(x^{k+1}\). Explain why $$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}=\sum_{k=0}^{\infty}(k+1)(k+2) a_{k+2} x^{k} $$ (d) Now show that $$ \sum_{k=0}^{\infty} a_{k} x^{k+1}=\sum_{k=1}^{\infty} a_{k-1} x^{k} $$ (e) We can now substitute \((8.6 .6)\) and \((8.6 .7)\) into \((8.6 .5)\) to obtain $$ \sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^{n}-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0 $$ Combine the like powers of \(x\) in the two series to show that our solution must satisfy $$ 2 a_{2}+\sum_{k=1}^{\infty}\left[(k+1)(k+2) a_{k+2}-a_{k-1}\right] x^{k}=0 $$ (f) Use equation (8.6.9) to show the following: i. \(a_{3 k+2}=0\) for every positive integer \(k\), ii. \(a_{3 k}=\frac{1}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)} a_{0}\) for \(k \geq 1\), iii. \(a_{3 k+1}=\frac{1}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)} a_{1}\) for \(k \geq 1\). (g) Use the previous part to conclude that the general solution to the Airy equation \((8.6 .4)\) is $$ \begin{aligned} y=& a_{0}\left(1+\sum_{k=1}^{\infty} \frac{x^{3 k}}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)}\right) \\ &+a_{1}\left(x+\sum_{k=1}^{\infty} \frac{x^{3 k+1}}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)}\right) \end{aligned} $$ Any values for \(a_{0}\) and \(a_{1}\) then determine a specific solution that we can approximate as closely as we like using this series solution.

Step-by-Step Solution

Answer

Differentiate the Taylor series for y. Substitute into Airy's equation and reindex. Find the recurrence relations and solve for coefficients to write the general solution.

Step 1: Differentiate the Series for y

Start with the given Taylor series:\[ y = \sum_{k=0}^{\infty} a_k x^k \]Differentiate term by term:\[ y' = \sum_{k=0}^{\infty} a_k k x^{k-1} = \sum_{k=1}^{\infty} a_k k x^{k-1} \]Differentiate again to find the second derivative:\[ y'' = \sum_{k=1}^{\infty} a_k k (k-1) x^{k-2} = \sum_{k=2}^{\infty} a_k k (k-1) x^{k-2} \]

Step 2: Substitute into Airy's Equation

Substitute the derivatives into Airy's equation:\[ y'' - x y = 0 \]\[ \sum_{k=2}^{\infty} a_k k (k-1) x^{k-2} - x \sum_{k=0}^{\infty} a_k x^k = 0 \]Notice that multiplying the second sum by \( x \) shifts the index up by 1:\[ \sum_{k=2}^{\infty} a_k k (k-1) x^{k-2} - \sum_{k=0}^{\infty} a_k x^{k+1} = 0 \]

Step 3: Reindex the Series

Reindex the first series so that both series are in terms of \( x^k \):\[ \sum_{k=2}^{\infty} (k-1) k a_k x^{k-2} = \sum_{k=0}^{\infty} (k+1)(k+2) a_{k+2} x^k \]This is valid because shifting the index in a series does not change the function.

Step 4: Reindex the Second Series

Similarly, reindex the second series so that both series are in terms of \( x^k \):\[ \sum_{k=0}^{\infty} a_k x^{k+1} = \sum_{k=1}^{\infty} a_{k-1} x^k \]Combines both re-indexed series into the form seen in equation (8.6.6).

Step 5: Combine the Series

Combine the series into one expression:\[ \sum_{k=0}^{\infty} (k+1)(k+2) a_{k+2} x^k - \sum_{k=1}^{\infty} a_{k-1} x^k = 0 \]Separate the case when \( k = 0 \) from the rest:\[ (2) a_2 + \sum_{k=1}^{\infty} [(k+1)(k+2) a_{k+2} - a_{k-1}] x^k = 0 \]

Step 6: Find the Recurrence Relation

Using the result of step 5, set each coefficient of \( x^k \) for \( k \ge 1 \) to zero:\[ (k+1)(k+2) a_{k+2} - a_{k-1} = 0 \]Thus we get the recurrence relation:\[ a_{k+2} = \frac{a_{k-1}}{(k+1)(k+2)} \]

Step 7: Find Specific Coefficients

From the recurrence relation, determine specific coefficients:1. \( a_{3k+2} = 0 \) for all positive integers \( k \).2. \( a_{3k} = \frac{1}{(2)(3)(5)(6) \cdots (3k-1)(3k)} a_0 \).3. \((3)(4)(6)(7) \cdots (3k)(3k+1)} a_1 \).

Step 8: Write the General Solution

Combine the results to conclude that the general solution to the Airy equation is:\[ y = a_0 \left(1 + \sum_{k=1}^{\infty} \frac{x^{3k}}{(2)(3)(5)(6) \cdots (3k-1)(3k)} \right) + a_1 \left( x + \sum_{k=1}^{\infty} \frac{x^{3k+1}}{(3)(4)(6)(7) \cdots (3k)(3k+1)} \right) \]

Key Concepts

differential equations

A differential equation is an equation that involves an unknown function and its derivatives. They are essential in modeling various physical systems and phenomena. In the given problem, we encounter a second-order linear differential equation known as Airy's equation, which is used to analyze the behavior of a spring-mass system. It appears as:
\[ y'' - xy = 0 \]
This type of differential equation requires us to find an unknown function, y, that satisfies this relationship with respect to a variable, x. Differential equations like Airy's equation play a crucial role in understanding how systems evolve over time, especially in engineering and physics contexts.

Taylor series

The Taylor series is a way to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For a function y=f(x), the Taylor series centered at x=0 is written as:
\[ y = \sum_{k=0}^{\infty} a_k x^k \]
In the given exercise, we assume a solution y=f(x) for Airy's equation in the form of this Taylor series. The coefficients a_k are undetermined and need to be found to express the function. By differentiating this series term by term, we obtain the series for the first and second derivatives of y. This step is necessary to substitute these series into the original differential equation to match terms by their powers of x. This method helps break down complex equations into a form that can be handled term by term.

spring-mass system

A spring-mass system is a classic example in physics and engineering used to demonstrate oscillatory motion. It usually involves a mass attached to a spring, where the system's behavior can be described by differential equations. In the present context, the Airy's equation models an undamped spring-mass system. This means that the equation accounts for the spring's aging (varying spring constant x) but neglects any damping forces that would lead to energy loss over time.

Understanding this system helps in interpreting the physical implications of the solution. For instance, finding that specific coefficients (a_3k+2 = 0) tells us about the specific behaviors and states the spring-mass system can exhibit as the system evolves.

series solution

A series solution involves expressing the solution to a differential equation as an infinite series. In our exercise, we use a Taylor series expansion to represent the unknown function y in terms of x. This results in solving for the coefficients a_k in:
\[ y = \sum_{k=0}^{\infty} a_k x^k \]
To find these coefficients, we differentiate the series, substitute into the original differential equation, and reindex the series for comparison. The form of these coefficients informs the exact structure of the general solution. In this case, the solution to Airy's equation results in:
\[ y = a_0 \left(1 + \sum_{k=1}^{\infty} \frac{x^{3k}}{(2)(3)(5)(6) \cdots (3k-1)(3k)} \right) + a_1 \left( x + \sum_{k=1}^{\infty} \frac{x^{3k+1}}{(3)(4)(6)(7) \cdots (3k)(3k+1)} \right) \]
This form allows us to know exactly how the solution behaves and approximates the system's behavior as precisely as needed. Each term in the series provides more depth to the solution, making this method particularly powerful for solving complex differential equations.