Problem 5 - Advanced Engineering Mathematics

Question

$$\begin{aligned} c_{i} &=\frac{2 \alpha_{i}^{2}}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} \int_{0}^{2} x J_{0}\left(\alpha_{i} x\right) d x \\ &=\frac{2 \alpha_{i}^{2}}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} \cdot \frac{1}{\alpha_{i}^{2}} \int_{0}^{2 \alpha_{i}} t J_{0}(t) d t \\ &=\frac{2}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} \int_{0}^{2 \alpha_{i}} \frac{d}{d t}\left[t J_{1}(t)\right] d t \\ &=\left.\frac{2}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} t J_{1}(t)\right|_{0} ^{2 \alpha_{i}} \\ &=\frac{4 \alpha_{i} J_{1}\left(2 \alpha_{i}\right)}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} \end{aligned}.$$ Thus $$f(x)=4 \sum_{i=1}^{\infty} \frac{\alpha_{i} J_{1}\left(2 \alpha_{i}\right)}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} J_{0}\left(\alpha_{i} x\right).$$

Step-by-Step Solution

Answer

The solution for \( c_i \) simplifies to \( \frac{4\alpha_i J_1(2\alpha_i)}{(4\alpha_i^2 + 1)J_0^2(2\alpha_i)} \) and is used to express \( f(x) \).

Step 1: Factor Out Constants

The given equation starts with the expression \( c_i = \frac{2\alpha_i^2}{(4\alpha_i^2 + 1)J_0^2(2\alpha_i)} \int_{0}^{2} x J_0(\alpha_i x) \, dx \). Notice that the constant \( \frac{2\alpha_i^2}{(4\alpha_i^2 + 1)J_0^2(2\alpha_i)} \) can be factored out from the integral as it does not depend on \( x \).

Step 2: Change of Variables

Perform a change of variables in the integral to simplify it. Let \( t = \alpha_i x \), then \( dt = \alpha_i \, dx \) and the limits change from \( 0 \) to \( 2\alpha_i \). So, \( \int_{0}^{2} x J_0(\alpha_i x) \, dx = \frac{1}{\alpha_i^2} \int_{0}^{2\alpha_i} t J_0(t) \, dt \).

Step 3: Apply Integration by Parts

To solve the integral \( \int t J_0(t) \, dt \), we use integration by parts. Let \( u = t \) and \( dv = J_0(t) \, dt \). Then \( du = dt \) and \( v = J_1(t) \) using the fact that the derivative of \( J_0(t) \) is \( -J_1(t) \). The integral becomes \( \int t J_0(t) \, dt = [t J_1(t)] - \int J_1(t) \, dt \).

Step 4: Simplify the Integral Result

Continuing from Step 3, \([t J_1(t)] - \int J_1(t) \, dt \) reduces to \([t J_1(t)]_{0}^{2\alpha_i} - 0 \) because \( \int J_1(t) \, dt \) equals zero when evaluated at the precise limits provided in the problem statement.

Step 5: Evaluate the Boundary Terms

Evaluate \( [t J_1(t)]_{0}^{2\alpha_i} \). This evaluates to \( 2\alpha_i J_1(2\alpha_i) - 0 \) when plugging in the limits.

Step 6: Combine Back with Original Factor

Now replace the integral in the original constant expression so that \( c_i = \frac{2}{(4\alpha_i^2 + 1)J_0^2(2\alpha_i)} \cdot 2\alpha_i J_1(2\alpha_i) \), which simplifies to \( \frac{4\alpha_i J_1(2\alpha_i)}{(4\alpha_i^2 + 1)J_0^2(2\alpha_i)} \).

Step 7: Solution for the Given Function

Finally utilize this simplified expression for \( c_i \) in the provided series to express \( f(x) \) as \( 4 \sum_{i=1}^{\infty} \frac{\alpha_i J_1(2\alpha_i)}{(4\alpha_i^2 + 1)J_0^2(2\alpha_i)} J_0(\alpha_i x) \).

Key Concepts

Integration by Parts

Integration by parts is a handy technique to solve integrals where direct integration is not straightforward. It's based on the product rule for differentiation and is particularly useful when dealing with products of functions, such as in our exercise where we have the integral \( \int t J_0(t) \ dt \). The idea is to transform the problem into a potentially easier integral to solve.

The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

Here's a step-by-step breakdown:

• Identify parts: Pick \( u \) and \( dv \) from the integral. In our case, \( u = t \) and \( dv = J_0(t) \, dt \).
• Differentiation and integration: Find \( du = dt \) and integrate to get \( v = J_1(t) \).
• Substitute and simplify: Use the formula to get \( [t J_1(t)] - \int J_1(t) \, dt \).

This approach helps break down complex integrals into simpler parts, making them manageable to solve.

Change of Variables

When confronted with an integral that seems a bit tricky, changing variables can make the problem much simpler. This technique, known as a U-substitution in some contexts, helps transform the integral into a more familiar or simpler form.

In this problem, we perform a change of variables by setting \( t = \alpha_i x \), which means that \( dt = \alpha_i \, dx \). This substitution not only simplifies the function inside the integral but also stretches or compresses the limits of integration. It transforms our original integral from:

• \( \int_{0}^{2} x J_0(\alpha_i x) \, dx \)

to:

• \( \frac{1}{\alpha_i^2} \int_{0}^{2\alpha_i} t J_0(t) \, dt \).

The change of variables is a powerful tool to untangle complicated integrands by rewriting them in a simpler form, often leading to easier integration paths.

Series Solutions

Series solutions are a way to express complex functions or solutions as infinite sums. This is particularly valuable when dealing with differential equations or functions like Bessel functions, which arise from physical problems.

In our context, we have expressed the entire function \( f(x) \) as a series:

• \( f(x) = 4 \sum_{i=1}^{\infty} \frac{\alpha_i J_1(2\alpha_i)}{(4\alpha_i^2 + 1)J_0^2(2\alpha_i)} J_0(\alpha_i x) \)

Each term in the series involves computations around Bessel functions and specific coefficients \( c_i \), which account for the individual harmonics or modes of the function.

Series solutions offer a way to approximate functions through partial sums, where more terms yield a closer approximation. This method is critical in physics and engineering, enabling the modeling of waves or heat distributions, among other phenomena.