Problem 15 - Advanced Engineering Mathematics

Question

We use $$U(x, s)=c_{1} e^{-\sqrt{s} x}+c_{2} e^{\sqrt{s} x}$$ The condition \(\lim _{x \rightarrow \infty} u(x, t)=0\) implies \(\lim _{x \rightarrow \infty} U(x, s)=0,\) so we define \(c_{2}=0 .\) Hence $$U(x, s)=c_{1} e^{-\sqrt{s} x}$$ The transform of \(u(0, t)=f(t)\) is \(U(0, s)=F(s) .\) Therefore $$U(x, s)=F(s) e^{-\sqrt{s} x}$$ and $$u(x, t)=\mathscr{L}^{-1}\left\\{F(s) e^{-x \sqrt{s}}\right\\}=\frac{x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{f(t-\tau) e^{-x^{2} / 4 \tau}}{\tau^{3 / 2}} d \tau$$

Step-by-Step Solution

Answer

The condition leads to $c_{2} = 0$, and the solution becomes $u(x, t) = \\frac{x}{2 \\sqrt{\\pi}} \\int_{0}^{t} \\frac{f(t-\\tau) e^{-x^{2} / 4 \\tau}}{\\tau^{3 / 2}} d \\tau$.

Step 1: Understand the given function U(x, s)

The function given is \(U(x, s) = c_{1} e^{-\sqrt{s} x} + c_{2} e^{\sqrt{s} x}\). It is a solution form based on certain conditions.

Step 2: Apply the boundary condition

Given that \(\lim _{x \ ightarrow \infty} U(x, s) = 0\), this implies the second term, \(c_{2} e^{\sqrt{s} x}\), must vanish as \(x \ ightarrow \infty\) because it grows exponentially. Hence setting \(c_{2} = 0\), simplifies to \(U(x, s) = c_{1} e^{-\sqrt{s} x}\).

Step 3: Implement initial condition transformation

The transform given by \(u(0, t) = f(t)\) implies that \(U(0, s) = F(s)\) where \(F(s)\) is the Laplace transform of \(f(t)\). Use this to express \(U(x, s) = F(s) e^{-\sqrt{s} x}\).

Step 4: Apply inverse Laplace transform

To find \(u(x, t)\) in terms of \(t\), we need to apply the inverse Laplace transform \(\mathscr{L}^{-1}\) on \(F(s) e^{-x \sqrt{s}}\). This results in the integral expression for \(u(x, t)\): \(u(x, t) = \frac{x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{f(t-\tau) e^{-x^{2} / 4 \tau}}{\tau^{3 / 2}} d \tau\).

Key Concepts

Boundary Conditions

Boundary conditions are essential in boundary value problems. These conditions help specify the solution of a differential equation. They usually define the behavior of the solution at specific points in space or time. In the context of Laplace transforms, boundary conditions ensure we find the right solution that fits the situation.
In our case, the boundary condition given by \(\lim _{x \to \infty} U(x, s) = 0\) implies that as x becomes very large, the solution must approach zero. This condition forces the coefficient \(c_{2}\) to be zero. This is because the term \(c_{2} e^{\sqrt{s} x}\) grows exponentially with x, making the solution diverge as x increases.
By eliminating this divergent term, we ensure the resulting function \(U(x, s) = c_{1} e^{-\sqrt{s} x}\) satisfies the condition, decaying to zero as x becomes infinitely large. Thus, the boundary condition is satisfied, ensuring a meaningful solution.

Inverse Laplace Transform

The inverse Laplace transform is crucial in shifting a function from the frequency domain back to the time domain. This operation allows us to derive the time-based function from its Laplace-transformed counterpart.
In our exercise, the function \(U(x, s) = F(s) e^{-\sqrt{s} x}\) is transformed back to the time domain using the inverse Laplace transform. This gives us the solution \(u(x, t)\) in its integral form.
The integral resulting from the inverse Laplace transform here is:

• \(u(x, t) = \frac{x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{f(t-\tau) e^{-x^{2} / 4 \tau}}{\tau^{3 / 2}} d \tau\) is obtained.

This form showcases how \(u(x, t)\) depends on the past values of a given function \(f(t)\) and evolves over time.

Integral Transform

An integral transform is a mathematical operation that converts a function into an integral through integration. The Laplace transform is a well-known integral transform used for solving differential equations by transforming them into more manageable algebraic equations.
Specifically, the Laplace transform converts a time-domain function, such as \(f(t)\), into a complex frequency domain representation, \(F(s)\). This process is based on the integral:

• \[ F(s) = \int_0^\infty e^{-st} f(t) \, dt. \]

In our exercise, the Laplace transform helps in expressing the boundary problem in terms of polynomials, easing the process of finding solutions for differential equations. Using the inverse later, we retrieve the original time-domain function in a comprehensible form.

Exponential Functions

Exponential functions are pivotal in this Laplace transform problem. These functions are recognizable by their constant base being raised to a variable exponent, expressed generally as \(e^{bx}\). They are fundamental in modeling growth and decay processes.
In the original exercise, the solution \(U(x, s) = c_{1} e^{-\sqrt{s} x}\) forms an exponential function. The negative exponent ensures that the function decays as x increases, fulfilling the boundary condition requirement that the solution approaches zero at infinity.

• The term \(e^{-\sqrt{s} x}\) naturally models a decay process where \(\sqrt{s}\) governs the rate of decay as x changes.
• Conversely, the term \(c_{2} e^{\sqrt{s} x}\) would model an exponential growth, hence it is set to zero under the boundary condition.

By understanding this behavior, we ensure solutions fit real-world decay scenarios, like cooling or voltage dissipation, accurately with time.