Problem 7 - Advanced Engineering Mathematics

Question

We use $$U(x, s)=c_{1} \cosh \frac{s}{a} x+c_{2} \sinh \frac{s}{a} x$$ Now \(U(0, s)=0\) implies \(c_{1}=0,\) so \(U(x, s)=c_{2} \sinh (s / a) x .\) The condition \(E d U /\left.d x\right|_{x=L}=F_{0}\) then yields \(c_{2}=F_{0} a / E s \cosh (s / a) L \) and so $$U(x, s)=\frac{a F_{0}}{E s} \frac{\sinh (s / a) x}{\cosh (s / a) L}=\frac{a F_{0}}{E s} \frac{e^{(s / a) x}-e^{-(s / a) x}}{e^{(s / a) L}+e^{-(s / a) L}}$$ $$=\frac{a F_{0}}{E s} \frac{e^{(s / a)(x-L)}-e^{-(s / a)(x+L)}}{1+e^{-2 s L / a}}$$ $$=\frac{a F_{0}}{E}\left[\frac{e^{-(s / a)(L-x)}}{s}-\frac{e^{-(s / a)(3 L-x)}}{s}+\frac{e^{-(s / a)(5 L-x)}}{s}-\cdots\right]$$ $$-\frac{a F_{0}}{E}\left[\frac{e^{-(s / a)(L+x)}}{s}-\frac{e^{-(s / a)(3 L+x)}}{s}+\frac{e^{-(s / a)(5 L+x)}}{s}-\cdots\right]$$ $$=\frac{a F_{0}}{E} \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{e^{-(s / a)(2 n L+L-x)}}{s}-\frac{e^{-(s / a)(2 n L+L+x)}}{s}\right]$$ and $$u(x, t)=\frac{a F_{0}}{E} \sum_{n=0}^{\infty}(-1)^{n}\left[\mathscr{L}-1\left\\{\frac{e^{-(s / a)(2 n L+L-x)}}{s}\right\\}-\mathscr{L}-1\left\\{\frac{e^{-(s / a)(2 n L+L+x)}}{s}\right\\}\right]$$ $$=\frac{a F_{0}}{E} \sum_{n=0}^{\infty}(-1)^{n}\left[\left(t-\frac{2 n L+L-x}{a}\right) \mathscr{U}\left(t-\frac{2 n L+L-x}{a}\right)\right.$$ $$\left.-\left(t-\frac{2 n L+L+x}{a}\right)^{\prime} U\left(t-\frac{2 n L+L+x}{a}\right)\right]$$

Step-by-Step Solution

Answer

The function \( U(x, s) \) is expressed with \( c_2 \) found using boundary conditions, leading to an infinite series representation for the solution \( u(x, t) \).

Step 1: Understanding Initial Condition

We start with the function \( U(x, s) = c_1 \cosh \left(\frac{s}{a} x\right) + c_2 \sinh \left(\frac{s}{a} x\right) \). The condition \( U(0, s) = 0 \) implies that \( c_1 = 0 \), because the hyperbolic cosine term must disappear, reducing \( U(x, s) \) to \( c_2 \sinh \left(\frac{s}{a} x\right) \).

Step 2: Applying the Derivative Condition

Given \( \frac{dU}{dx} \bigg|_{x=L} = \frac{F_0}{E} \), we differentiate \( U(x, s) = c_2 \sinh \left( \frac{s}{a} x \right) \) to find \( c_2 \). The derivative is \( U'(x, s) = c_2 \frac{s}{a} \cosh \left( \frac{s}{a} x \right) \). Set this equal to \( \frac{F_0}{E} \) at \( x = L \), solve for \( c_2 = \frac{F_0 a}{E s \cosh \left( \frac{s}{a} L \right)} \).

Step 3: Substituting for \( c_2 \)

Substitute \( c_2 \) back into \( U(x, s) \):\[U(x, s) = \frac{F_0 a}{E s} \frac{\sinh \left( \frac{s}{a} x \right)}{\cosh \left( \frac{s}{a} L \right)}\] This is a simplified expression for the function \( U(x, s) \).

Step 4: Rewriting \( \sinh \) and \( \cosh \)

Express the \( \sinh \) and \( \cosh \) functions in terms of exponential functions:\[U(x, s) = \frac{F_0 a}{E s} \frac{e^{\frac{s}{a} x} - e^{-\frac{s}{a} x}}{e^{\frac{s}{a} L} + e^{-\frac{s}{a} L}}\] Then write it as:\[U(x, s) = \frac{F_0 a}{E s} \frac{e^{\frac{s}{a}(x-L)} - e^{-\frac{s}{a}(x+L)}}{1 + e^{-2 \frac{s}{a} L}}\]

Step 5: Infinite Series Representation

Using the geometric series and symmetry property, this equation can be rewritten:\[U(x, s) = \frac{a F_0}{E} \sum_{n=0}^{\infty} (-1)^n \left[ \frac{e^{-\frac{s}{a}(2nL + L - x)}}{s} - \frac{e^{-\frac{s}{a}(2nL + L + x)}}{s} \right]\] This representation helps simplify subsequent steps.

Step 6: Finding \( u(x, t) \)

The final step is inverting the Laplace transform for each term in the series. We have\[u(x, t) = \frac{a F_0}{E} \sum_{n=0}^{\infty} (-1)^n \left[ \left(t - \frac{2nL + L - x}{a}\right) \mathscr{U}\left(t - \frac{2nL + L - x}{a}\right) - \left(t - \frac{2nL + L + x}{a}\right) \mathscr{U}\left(t - \frac{2nL + L + x}{a}\right) \right]\] where \( \mathscr{U} \) is the unit step function.

Key Concepts

Laplace Transform

The Laplace Transform is a powerful mathematical tool used to transform a function of time into a function of a complex variable, typically denoted as \( s \). This transformation is particularly useful in solving differential equations, such as partial differential equations (PDEs), by converting them into algebraic equations that are often easier to solve.
The process involves integrating a given time-domain function, \( f(t) \), multiplied by \( e^{-st} \) over the interval from 0 to infinity, yielding the Laplace-transformed function, \( F(s) \): \[ F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \]Some advantages of using the Laplace Transform include its ability to:

• Simplify complex differential equations.
• Handle piecewise and discontinuous functions.
• Provide insight into the stability and behavior of systems.

In the context of our exercise, the Laplace Transform is utilized to find the solution \( u(x, t) \) by inverting each term in the series representation found earlier. This involves integrating back from the \( s \)-domain to the \( t \)-domain, essentially performing the opposite of the Laplace transformation. This inversion is realized through sophisticated mathematical techniques, often involving tables of common transforms and properties. Understanding these transformations is crucial for working with PDEs and boundary value problems.

Hyperbolic Functions

Hyperbolic functions, including \( \sinh(x) \) and \( \cosh(x) \), are analogues of trigonometric functions for hyperbolic geometry. They are used in this exercise to express the solution to the partial differential equation.
Hyperbolic sine \( \sinh(x) \) and cosine \( \cosh(x) \) are defined using exponential functions:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

These functions have properties similar to sine and cosine in trigonometry, like addition formulas and derivatives, yet they are distinguished by their unique definitions that involve exponential growth.

In the given problem, these hyperbolic functions enable the formulation of the solution \( U(x, s) = c_2 \sinh\left(\frac{s}{a} x\right) \) because their combinations satisfy boundary conditions much like sines and cosines satisfy particular conditions in traditional differential equations.
By rewriting \( \sinh \) and \( \cosh \) in terms of exponential functions, the solution can be manipulated into a format that's more conducive to further analysis, such as inverse transformations or additional algebraic manipulation. Recognizing the role of hyperbolic functions in solving and simplifying these equations is key in many areas of engineering and physics, especially in problems involving hyperbolic PDEs.

Boundary Conditions

Boundary conditions are essential constraints in solving differential equations, especially PDEs, as they specify the behavior of a solution at the boundaries of the domain. They can take various forms depending on the physical context of the problem.
In our exercise, two boundary conditions play a crucial role:

• The initial condition \( U(0, s) = 0 \) ensures that the hyperbolic cosine term \( c_1 \cosh\left(\frac{s}{a} x\right) \) vanishes, simplifying the expression to \( c_2 \sinh\left(\frac{s}{a} x\right) \).
• The derivative condition \( \frac{dU}{dx} \big|_{x=L} = \frac{F_0}{E} \) introduces an additional relationship necessary to determine the unknown constant \( c_2 \).

Boundary conditions can be expressed as fixed (Dirichlet), where the function is specified at the boundaries, or as derivative conditions (Neumann), where the rate of change is specified. The exercise deals with both by leveraging these constraints to determine the functional form of the solution that fits the given conditions.

Understanding how to apply and interpret these conditions is vital. It provides the context needed to solve PDEs accurately, ensuring that the solution is not just mathematically correct but also physically meaningful, allowing it to describe actual scenarios encountered in science and engineering.