Problem 26 - Advanced Engineering Mathematics

Question

(a) We use $$U(x, s)=c_{1} e^{-(s / a) x}+c_{2} e^{(s / a) x}+\frac{v_{0}^{2} F_{0}}{\left(a^{2}-v_{0}^{2}\right) s^{2}} e^{-\left(s / v_{0}\right) x}$$ The condition \(\lim _{x \rightarrow \infty} u(x, t)=0\) implies \(\lim _{x \rightarrow \infty} U(x, s)=0,\) so we must define \(c_{2}=0 .\) Consequently $$U(x, s)=c_{1} e^{-(s / a) x}+\frac{v_{0}^{2} F_{0}}{\left(a^{2}-v_{0}^{2}\right) s^{2}} e^{-\left(s / v_{0}\right) x}$$ The remaining boundary condition transforms into \(U(0, s)=0 .\) From this we find $$c_{1}=-v_{0}^{2} F_{0} /\left(a^{2}-v_{0}^{2}\right) s^{2}$$ Therefore, by the second translation theorem $$U(x, s)=-\frac{v_{0}^{2} F_{0}}{\left(a^{2}-v_{0}^{2}\right) s^{2}} e^{-(s / a) x}+\frac{v_{0}^{2} F_{0}}{\left(a^{2}-v_{0}^{2}\right) s^{2}} e^{-\left(s / v_{0}\right) x}$$ and $$\begin{aligned} u(x, t) &=\frac{v_{0}^{2} F_{0}}{a^{2}-v_{0}^{2}}\left[\mathscr{L}^{-1}\left\\{\frac{e^{-\left(x / v_{0}\right) s}}{s^{2}}\right\\}-\mathscr{L}^{-1}\left\\{\frac{e^{-(x / a) s}}{s^{2}}\right\\}\right] \\ &=\frac{v_{0}^{2} F_{0}}{a^{2}-v_{0}^{2}}\left[\left(t-\frac{x}{v_{0}}\right)^{2} U\left(t-\frac{x}{v_{0}}\right)-\left(t-\frac{x}{a}\right)^{2} U\left(t-\frac{x}{a}\right)\right] \end{aligned}$$ In the case when \(v_{0}=a\) the solution of the transformed equation is $$U(x, s)=c_{1} e^{-(s / a) x}+c_{2} e^{(s / a) x}-\frac{F_{0}}{2 a s} x e^{-(s / a) x}$$ The usual analysis then leads to \(c_{1}=0\) and \(c_{2}=0 .\) Therefore $$U(x, s)=-\frac{F_{0}}{2 a s} x e^{-(s / a) x}$$ and $$u(x, t)=-\frac{x F_{0}}{2 a} \mathscr{L}-1\left\\{\frac{e^{-(x / a) s}}{s}\right\\}=-\frac{x F_{0}}{2 a} \mathscr{U}\left(t-\frac{x}{a}\right)$$

Step-by-Step Solution

Answer

The function \(u(x, t)\) is defined by inverse transforms; different forms apply when \(v_0 = a\) and \(v_0 \neq a\).

Step 1: Understand Given Expression

We start with the expression for \(U(x, s)\):\[U(x, s) = c_1 e^{-(s / a) x} + c_2 e^{(s / a) x} + \frac{v_0^2 F_0}{(a^2 - v_0^2) s^2} e^{-(s / v_0) x}\].We're told to find conditions such that \(\lim_{x \to \infty} U(x, s) = 0\).

Step 2: Apply Boundary Condition

Given \(\lim_{x \to \infty} U(x, s) = 0\), the term that grows with \(x\), \(c_2 e^{(s / a) x}\), must vanish. This implies setting \(c_2 = 0\).

Step 3: Update Expression with c2 = 0

Substitute \(c_2 = 0\) into the expression:\[U(x, s) = c_1 e^{-(s / a) x} + \frac{v_0^2 F_0}{(a^2 - v_0^2) s^2} e^{-(s / v_0) x}\].

Step 4: Apply U(0, s) = 0 Condition

We now use the boundary condition \(U(0, s) = 0\):\[c_1 + \frac{v_0^2 F_0}{(a^2 - v_0^2) s^2} = 0\], which results in \(c_1 = -\frac{v_0^2 F_0}{(a^2 - v_0^2) s^2}\).

Step 5: Substitute c1 and Simplify

Replacing \(c_1\), the expression becomes:\[U(x, s) = -\frac{v_0^2 F_0}{(a^2 - v_0^2) s^2} e^{-(s / a) x} + \frac{v_0^2 F_0}{(a^2 - v_0^2) s^2} e^{-(s / v_0) x}\].

Step 6: Compute Inverse Laplace Transform

The goal is to find \(u(x, t)\) via inverse Laplace transform:\[u(x, t) = \frac{v_0^2 F_0}{a^2 - v_0^2}\left[i\mathscr{L}^{-1}\left(\frac{e^{-(x / v_0) s}}{s^2}\right)-\mathscr{L}^{-1}\left(\frac{e^{-(x / a) s}}{s^2}\right)\right]\].

Step 7: Solve Using Translation Theorem

Using the second translation theorem of inverse Laplace transform:\[\mathscr{L}^{-1}\left(\frac{e^{-(x / v_0) s}}{s^2}\right) = \left(t - \frac{x}{v_0}\right) U\left(t - \frac{x}{v_0}\right)\].Similarly,\[\mathscr{L}^{-1}\left(\frac{e^{-(x / a) s}}{s^2}\right) = \left(t - \frac{x}{a}\right) U\left(t - \frac{x}{a}\right)\].

Step 8: Final Solution for General Case

Combine the results to express \(u(x, t)\):\[u(x, t) = \frac{v_0^2 F_0}{a^2 - v_0^2}\left[\left(t-\frac{x}{v_0}\right)^2 U\left(t-\frac{x}{v_0}\right) - \left(t-\frac{x}{a}\right)^2 U\left(t-\frac{x}{a}\right)\right]\].

Step 9: Special Case v0 = a

For \(v_0 = a\), simplify \(U(x, s)\) further:\[U(x, s) = c_1 e^{-(s / a) x} + c_2 e^{(s / a) x} - \frac{F_0}{2a s} x e^{-(s / a) x}\].Boundary conditions lead to \(c_1 = 0\) and \(c_2 = 0\).

Step 10: Final Expression for v0 = a

With \(c_1 = 0\), \(c_2 = 0\), we find:\[U(x, s) = - \frac{F_0}{2 a s} x e^{-(s / a) x}\].

Step 11: Inverse Transform for v0 = a

Compute inverse Laplace transform:\[u(x, t) = -\frac{x F_0}{2 a} \mathscr{L}^{-1}\left(\frac{e^{-(x / a) s}}{s}\right) = -\frac{x F_0}{2 a} U\left(t - \frac{x}{a}\right)\].

Key Concepts

Boundary Conditions

Boundary conditions are essential criteria used to determine specific solutions of a differential equation. They indicate how the solution behaves at the boundaries of the domain. Boundary conditions help streamline complex equations into practical and solvable problems.

In our problem, we come across a boundary condition for the function \( U(x, s) \). We need to ensure that it approaches zero as \( x \) approaches infinity, \( \lim_{x \to \infty} U(x, s) = 0 \). This ensures the stability of the system as it extends towards infinity.

Applying the condition helps us zero out any terms like \( c_2 e^{(s / a) x} \), which grow unbounded as \( x \) increases. This step simplifies the function, leading to a manageable expression for further calculations.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the context of Laplace Transforms, these equations model dynamic systems influenced by isolated or continuous inputs.

The provided exercise deals with linear combinations of exponential terms that naturally arise in the solution to linear differential equations with constant coefficients. It's important to manipulate these differential equations correctly to accommodate the given conditions.

For instance, when we set a boundary condition like \( U(0, s) = 0 \), we solve for constants \( c_1 \) and \( c_2 \) involved in the differential equation solution. By substituting and rearranging terms based on these boundary conditions, we derive expression solutions consistent with the original differential equation.

Inverse Laplace Transform

The inverse Laplace transform is a technique used to convert a function back from its Laplace-transformed state to its original time-domain expression. This is particularly useful for analyzing time-dependent behaviors of dynamical systems described by differential equations.

In our exercise, \( u(x, t) \) is obtained by taking the inverse Laplace transform of \( U(x, s) \). The process involves employing the second translation theorem, a valuable tool that facilitates the evaluation of shifted Laplace transforms.

The theorem helps translate expressions like \( \frac{e^{-(x / v_0) s}}{s^2} \) into their time-domain counterparts, allowing us to write compact solutions for \( u(x, t) \) related to system behaviors over time. This reverse transformation effectively completes the solution process initiated in the frequency domain.

Exponential Functions

Exponential functions are an essential class of functions in mathematics. They exhibit rapid growth or decay, which is pivotal in solutions involving Laplace Transforms.

In this exercise, exponential functions form a fundamental part of the solution \( U(x, s) \). Terms like \( e^{-(s / a) x} \) or \( e^{-(s / v_0) x} \) describe processes that evolve smoothly or sharply depending on parameters \( s \), \( a \), and \( v_0 \).

Proper control and simplification of exponential terms ensure that the solutions remain finite and stable as systems, like the ones described here, extend in space or time. Understanding how these terms behave with conditions such as \( \lim_{x \to \infty} U(x, s) = 0 \) is vital to deriving meaningful, applicable results in physical applications.