Problem 23 - Advanced Engineering Mathematics

Question

We use $$U(x, s)=c_{1} \cosh \sqrt{\frac{s}{k}} x+c_{2} \sinh \sqrt{\frac{s}{k}} x+\frac{u_{0}}{s}$$ The transformed boundary conditions \(d U /\left.d x\right|_{x=0}=0\) and \(U(1, s)=0\) give, in turn, \(c_{2}=0\) and \(c_{1}=-u_{0} / s \cosh \sqrt{s / k} .\) Therefore $$U(x, s)=\frac{u_{0}}{s}-\frac{u_{0} \cosh \sqrt{s / k} x}{s \cosh \sqrt{s / k}}=\frac{u_{0}}{s}-u_{0} \frac{e^{\sqrt{s / k}} x+e^{-\sqrt{s / k} x}}{s\left(e^{\sqrt{s / k}}+e^{-\sqrt{s / k}}\right)}$$ $$=\frac{u_{0}}{s}-u_{0} \frac{e^{\sqrt{s / k}(x-1)}+e^{-\sqrt{s / k}(x+1)}}{s\left(1+e^{-2 \sqrt{s / k}}\right)}$$ $$=\frac{u_{0}}{s}-u_{0}\left[\frac{e^{-\sqrt{s / k}(1-x)}}{s}-\frac{e^{-\sqrt{s / k}(3-x)}}{s}+\frac{e^{-\sqrt{s / k}(5-x)}}{s}-\cdots\right]$$ $$-u_{0}\left[\frac{e^{-\sqrt{s / k}(1+x)}}{s}-\frac{e^{-\sqrt{s / k}(3+x)}}{s}+\frac{e^{-\sqrt{s / k}(5+x)}}{s}-\cdots\right]$$ $$=\frac{u_{0}}{s}-u_{0} \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{e^{-(2 n+1-x) \sqrt{s} / \sqrt{k}}}{s}+\frac{e^{-(2 n+1+x) \sqrt{s} / \sqrt{k}}}{s}\right]$$ and $$\begin{aligned} u(x, t) &=u_{0} \mathscr{L}^{-1}\left\\{\frac{1}{s}\right\\}-u_{0} \sum_{n=0}^{\infty}(-1)^{n}\left[\mathscr{L}^{-1}\left\\{\frac{e^{-(2 n+1-x) \sqrt{s} / \sqrt{k}}}{s}\right\\}-\mathscr{L}^{-1}\left\\{\frac{e^{-(2 n+1+x) \sqrt{s} / \sqrt{k}}}{s}\right\\}\right] \\ &=u_{0}-u_{0} \sum_{n=0}^{\infty}(-1)^{n}\left[\operatorname{erfc}\left(\frac{2 n+1-x}{2 \sqrt{k t}}\right)-\operatorname{erfc}\left(\frac{2 n+1+x}{2 \sqrt{k t}}\right)\right] \end{aligned}$$

Step-by-Step Solution

Answer

Derive the expression using boundary conditions and inverse Laplace transforms to get \( u(x, t) = u_0 - u_0 \sum_{n=0}^{\infty}(-1)^{n} [\operatorname{erfc}(\frac{2 n+1-x}{2 \sqrt{kt}}) - \operatorname{erfc}(\frac{2 n+1+x}{2 \sqrt{kt}})] \).

Step 1: Start with the Given Equation

We begin with the given equation, which is expressed as \( U(x, s) = c_1 \cosh \sqrt{\frac{s}{k}} x + c_2 \sinh \sqrt{\frac{s}{k}} x + \frac{u_0}{s} \).

Step 2: Apply Boundary Condition at x = 0

Applying the boundary condition \( \frac{dU}{dx} \bigg|_{x=0} = 0 \), we find that \( c_2 = 0 \), reducing the equation to \( U(x, s) = c_1 \cosh \sqrt{\frac{s}{k}} x + \frac{u_0}{s} \).

Step 3: Apply Second Boundary Condition

At \( x = 1 \), we have \( U(1, s) = 0 \), which helps to determine \( c_1 = -\frac{u_0}{s \cosh \sqrt{\frac{s}{k}}} \).

Step 4: Simplify the Expression

Substitute \( c_1 \) back into the expression to find \( U(x, s) = \frac{u_0}{s} - \frac{u_0 \cosh \sqrt{\frac{s}{k}} x}{s \cosh \sqrt{\frac{s}{k}}} \).

Step 5: Express Using Exponentials

Rewrite \( \cosh \sqrt{\frac{s}{k}} x \) in terms of exponentials and further simplify this expression so that it is ready for inversion of the Laplace transform.

Step 6: Use Infinite Series Representation

Express the function as an infinite series: \( U(x, s) = \frac{u_0}{s} - u_0 \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{e^{-(2 n+1-x) \sqrt{\frac{s}{k}}}}{s} + \frac{e^{-(2 n+1+x) \sqrt{\frac{s}{k}}}}{s}\right] \).

Step 7: Invert the Laplace Transform

Apply the inverse Laplace transform: \( u(x, t) = u_0 - u_0 \sum_{n=0}^{\infty}(-1)^{n}\left[\operatorname{erfc}\left(\frac{2 n+1-x}{2 \sqrt{kt}}\right) - \operatorname{erfc}\left(\frac{2n+1+x}{2 \sqrt{kt}}\right)\right] \).

Step 8: Final Result

The original problem is solved, providing us with the function \( u(x, t) \) that describes the system over time.

Key Concepts

Laplace Transform

The Laplace Transform is a powerful mathematical tool used to analyze linear systems. It converts functions of time into functions of a complex variable, usually denoted as \( s \). This transformation makes differential equations easier to solve by turning them into algebraic equations.

• The Laplace Transform of a function \( f(t) \) is defined as \( \mathscr{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t) \, dt \).
• It is particularly useful for solving differential equations with boundary conditions, as it simplifies issues related to differentiation and integration.

In the given exercise, the Laplace Transform is used to manage and simplify the partial differential equations present in the boundary conditions. This approach lets us handle complex boundary situations with straightforward algebraic manipulations.

Boundary Conditions

Boundary Conditions are essential in defining unique solutions to differential equations, especially when solving problems involving physical systems or domains. They provide additional information needed to determine the specific solution from a set of possible solutions.

• In this exercise, we deal with two boundary conditions: \( \frac{dU}{dx} \big|_{x=0} = 0 \), indicating a no-flux boundary at \( x=0 \), and \( U(1, s) = 0 \), imposing a specific value at \( x=1 \).
• By applying these conditions, we find the constants in the solution that adhere to the physical requirements described by the problem's context.

These boundary conditions transform the general solution of the differential equation into one that fits the particular constraints, ensuring the system behaves consistently with the given physical scenario.

Series Representation

Series Representation plays a critical role when expressing complex expressions as infinite sums, which simplifies calculations and enables easier manipulation. In the context of our exercise, after applying boundary conditions and simplifying expressions, we use an infinite series to represent each term.

• The function \( U(x, s) \) is expressed as an infinite series in Step 6 of the solution: \[ U(x, s) = \frac{u_0}{s} - u_0 \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{e^{-(2 n+1-x) \sqrt{s/k}}}{s} + \frac{e^{-(2 n+1+x) \sqrt{s/k}}}{s}\right] \].
• This representation breaks down a complex expression into simpler components, which are easier to handle mathematically.
• It allows for individual terms to be easily inverted using the inverse transform.

The series aids in transitioning between the Laplace transformed domain and the time domain while solving the differential equation.

Inverse Laplace Transform

The Inverse Laplace Transform is utilized to convert transformed variables back to their original time-dependent expressions. Once the transformed algebraic equation is solved, we need the inverse transform to interpret results in the context of time and space.

• In the exercise, Step 7 involves using the Inverse Laplace Transform to revert the expression \( U(x, s) \) to \( u(x, t) \).
• This involves the application of known inverse transforms such as those involving the complementary error function \( \operatorname{erfc} \).
• The result is a time-dependent representation that provides insights into the behavior of the system over time, represented as: \[ u(x, t) = u_0 - u_0 \sum_{n=0}^{\infty}(-1)^{n}\left[\operatorname{erfc}\left(\frac{2n+1-x}{2\sqrt{kt}}\right) - \operatorname{erfc}\left(\frac{2n+1+x}{2\sqrt{kt}}\right)\right] \].

By understanding the process for the inverse transform, one can effectively transition their solutions from the transformed domain back to meaningful time functions.