Problem 3 - Advanced Engineering Mathematics

Question

The solution of $$a^{2} \frac{d^{2} U}{d x^{2}}-s^{2} U=0$$ is in this case $$U(x, s)=c_{1} e^{-(x / a) s}+c_{2} e^{(x / a) s}$$ since \(\lim _{x \rightarrow \infty} u(x, t)=0\) we have \(\lim _{x \rightarrow \infty} U(x, s)=0 .\) Thus \(c_{2}=0\) and $$U(x, s)=c_{1} e^{-(x / a) s}$$ If \(\mathscr{L}\\{u(0, t)\\}=\mathscr{L}\\{f(t)\\}=F(s)\) then \(U(0, s)=F(s) .\) From this we have \(c_{1}=F(s)\) and $$U(x, s)=F(s) e^{-(x / a) s}$$ Hence, by the second translation theorem, $$u(x, t)=f\left(t-\frac{x}{a}\right) \mathscr{U}\left(t-\frac{x}{a}\right)$$

Step-by-Step Solution

Answer

The solution is \(u(x, t) = f(t - \frac{x}{a}) \mathscr{U}(t - \frac{x}{a})\).

Step 1: Identify the Differential Equation

We start with the differential equation, given as \(a^{2} \frac{d^{2} U}{d x^{2}}-s^{2} U=0\). Our task is to find a solution for \(U(x, s)\).

Step 2: Consider the Solution Involving Exponentials

The solution for this type of differential equation generally takes the form \(U(x, s) = C_1 e^{-\frac{x}{a}s} + C_2 e^{\frac{x}{a}s}\). Exponential functions are solutions because they satisfy the form of the differential equation given. Here, \(C_1\) and \(C_2\) are constants to be determined.

Step 3: Apply Boundary Conditions to Eliminate \(C_2\)

Given the condition \(\lim_{x \rightarrow \infty} U(x, s) = 0\), we know that the term \(e^{\frac{x}{a}s}\) grows exponentially when \(x\) tends to infinity, and so it can't satisfy the boundary condition. This means \(C_2 = 0\), simplifying the solution to \(U(x, s) = C_1 e^{-\frac{x}{a}s}\).

Step 4: Determine \(C_1\) Using Initial Conditions

We are given \(\mathscr{L}\{u(0, t)\} = \mathscr{L}\{f(t)\} = F(s)\), which implies \(U(0, s) = F(s)\). Substituting \(x=0\) in the solution \(U(x, s) = C_1 e^{-\frac{x}{a}s}\) gives \(C_1 = F(s)\).

Step 5: Construct the Final Solution for \(U(x, s)\)

Substituting \(C_1 = F(s)\) back into our solution, we get \(U(x, s) = F(s) e^{-\frac{x}{a}s}\).

Step 6: Apply Inverse Laplace Transform Using the Second Translation Theorem

By the second translation theorem, if \(U(x, s) = F(s) e^{-\frac{x}{a}s}\), then \(u(x, t) = f(t - \frac{x}{a}) \mathscr{U}(t - \frac{x}{a})\). Here, \(\mathscr{U}\) is the unit step function, indicating a shift in the function \(f\).

Key Concepts

Differential Equations

Understanding differential equations is essential for solving many problems in physics and engineering. They are equations that involve an unknown function and its derivatives. In our case, the differential equation is \(a^{2} \frac{d^{2} U}{d x^{2}} - s^{2} U = 0\). This represents a second-order linear differential equation, where \(U\) is a function of \(x\) and \(s\).

Differential equations are used to describe various phenomena such as heat, sound, elasticity, and quantum mechanics. Solving them involves finding a function that satisfies the equation for all values involved, like \(U(x, s)\) in our example. Here, the presence of \(e^{\frac{x}{a}s}\) and \(e^{\frac{-x}{a}s}\) hints at a solution involving exponential functions.

The general solution form \(U(x, s) = C_1 e^{\frac{-x}{a}s} + C_2 e^{\frac{x}{a}s}\) arises because exponential functions are inherently compatible solutions for linear differential equations with constant coefficients.

Boundary Conditions

Boundary conditions are crucial in determining the specific solution to a differential equation. They define the behavior of the solution at the boundaries of the domain. In our situation, we have the boundary condition \(\lim_{x \rightarrow \infty} U(x, s) = 0\).

This particular boundary condition implies that as \(x\) becomes very large, \(U(x, s)\) should approach zero. In practical terms, it helps eliminate non-physical solutions. Here, it gets rid of the \(C_2 e^{\frac{x}{a}s}\) part of the solution since this term grows with \(x\) and cannot logically tend to zero as \(x\) goes to infinity.

By applying this boundary condition, we effectively narrow down our solution to \(U(x, s) = C_1 e^{\frac{-x}{a}s}\), making it physically meaningful for the problem at hand.

Inverse Laplace Transform

The inverse Laplace transform is the process used to retrieve a time-domain function from its Laplace-transformed counterpart. It is especially useful in solving differential equations when we work with variables in the frequency domain. In our example, we start with \(U(x, s)\) and want to find \(u(x, t)\).

The inverse Laplace transform relies on known pairs of transforms or using complex inversion formulae to return to the original function. In this exercise, it involves computing the inverse of expressions that are modified by operations like multiplication or translation. This allows us to represent the original function in terms and functions of time, transitioning from \(s\) to \(t\).

In practice, software tools or tables of transforms are often used to streamline this process, making it easier to handle more complex transforms.

Second Translation Theorem

The second translation theorem, or shifting theorem, is a powerful tool in the context of Laplace transforms. It states that a shift in the time domain corresponds to a multiplication by an exponential factor in the Laplace domain. This is particularly useful for functions that have been shifted through time.

For our exercise, once we have \(U(x, s) = F(s) e^{-\frac{x}{a}s}\), the second translation theorem allows us to conclude \(u(x, t) = f\left(t - \frac{x}{a}\right) \mathscr{U}\left(t - \frac{x}{a}\right)\).

Here, the unit step function \(\mathscr{U}\) plays a critical role. It implies that \(f(t)\) is shifted by \(\frac{x}{a}\) time units. This translation aligns the frequency domain manipulations with real-time physical phenomena, providing a clear and robust interpretation of the system's response to initial conditions.