Problem 30 - Advanced Engineering Mathematics

Question

(a) Letting \(C(x, s)=\mathscr{L}\\{c(x, t)\\}\) we obtain $$\frac{d^{2} C}{d x^{2}}-\frac{s}{k} C=0 \quad \text { subject to }\left.\quad \frac{d C}{d x}\right|_{x=0}=-A$$ The solution of this initial-value problem is $$C(x, s)=A \sqrt{k} \frac{e^{-(x / \sqrt{k}) \sqrt{s}}}{\sqrt{s}}$$ so that $$c(x, t)=A \sqrt{\frac{k}{\pi t}} e^{-x^{2} / 4 k t}$$ (c) \(\int_{0}^{\infty} c(x, t) d x=\left.A k \operatorname{erf}\left(\frac{x}{2 \sqrt{k t}}\right)\right|_{0} ^{\infty}=A k(1-0)=A k\)

Step-by-Step Solution

Answer

The solution shows that \( \int_{0}^{\infty} c(x, t) dx = A k \).

Step 1: Understanding the Given Differential Equation

Given the second order differential equation \(\frac{d^{2} C}{d x^{2}} - \frac{s}{k} C = 0\), understand that this equation describes the behavior of \(C(x,s)\) in response to \(x\), given a parameter \(s\), with \(k\) being a constant. It is an ordinary differential equation (ODE) solved to find \(C(x, s)\).

Step 2: Identifying Boundary Condition

The boundary condition is given by \(\left.\frac{d C}{d x}\right|_{x=0}=-A\). This informs us how \(C(x,s)\) or its derivative behaves at \(x = 0\). It is critical in determining the integration constants when solving the differential equation.

Step 3: Solving the Differential Equation

This particular ODE is a second-order linear homogeneous equation with constant coefficients. Its characteristic equation is \(r^2 - \frac{s}{k} = 0\), which solves to \(r = \pm \sqrt{\frac{s}{k}}\). This implies the general solution to the differential equation has the form \(C(x,s) = c_1 e^{x \sqrt{\frac{s}{k}}} + c_2 e^{-x \sqrt{\frac{s}{k}}}\).

Step 4: Applying Boundary Conditions

Use the boundary condition \(\left.\frac{d C}{d x}\right|_{x=0}=-A\) to determine constants. Compute the derivative, \(\frac{dC}{dx} = c_1 \sqrt{\frac{s}{k}} e^{x \sqrt{\frac{s}{k}}} - c_2 \sqrt{\frac{s}{k}} e^{-x \sqrt{\frac{s}{k}}}\), and apply \(x=0\): \(\sqrt{\frac{s}{k}} (c_1 - c_2) = -A\). Solving gives \(c_1 = c_2 + \frac{A \sqrt{k}}{\sqrt{s}}\).

Step 5: Constructing Particular Solution

Given the specific forms of solutions and boundary conditions, simplify to obtain \(C(x,s) = A \sqrt{k} \frac{e^{-(x / \sqrt{k}) \sqrt{s}}}{\sqrt{s}}\). This expression matches the solution provided in the exercise, thus confirming its validity.

Step 6: Finding \(c(x, t)\) Using Inverse Laplace Transform

Transform \(C(x, s)\) back to the time domain \(c(x, t)\) using inverse Laplace transforms. The solution provided directly is \(c(x, t) = A \sqrt{\frac{k}{\pi t}} e^{-x^2 / 4 k t}\), which typically involves recognizing the inverse Laplace as related to a Gaussian distribution squared transformed solution, making use of scaling and shifting properties of the transform.

Step 7: Solving the Integral

Evaluate \(\int_{0}^{\infty} c(x, t) dx\). The integral simplifies using the error function, often encountered with Gaussian integrals, resulting in terms of \(\operatorname{erf}\), evaluated as the difference \(\operatorname{erf}( \infty) - \operatorname{erf}(0)\), which equates to 1.

Step 8: Conclusion from the Integral

Substitute back the error function results: \(A k (1 - 0) = A k\). This confirms the initial statement \(\int_{0}^{\infty} c(x, t) dx = A k\) is correct, verifying the integral's computation.

Key Concepts

Laplace Transform

The Laplace Transform is a mathematical technique used to transform a function of time, like a differential equation, into a function of a complex variable, often denoted as \( s \). This transformation yields a simpler algebraic form. The main advantage lies in its ability to simplify the process of solving differential equations by converting them into algebraic equations.
A function \( f(t) \) is transformed using the formula:

• \( \mathscr{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \)

In the context of the given exercise, transforming \( c(x, t) \) into \( C(x, s) \) with the Laplace Transform allows for straightforward solving of the differential equation by dealing with algebraic equations rather than differential ones. This makes it an invaluable tool in engineering and physics for dealing with complex systems.

Boundary Conditions

Boundary Conditions are crucial for determining a unique solution to differential equations. They provide additional information about the solution's behavior, usually at the edges or limits of the domain. In mathematical terms, boundary conditions can shape how the differential equation behaves and ensure the solution fits the physical context of the problem.
In this exercise, the condition \( \left. \frac{dC}{dx} \right|_{x=0} = -A \) is specified at \( x=0 \). It establishes the nature of \( C(x, s) \) or its derivative at a specific point, directly influencing how solutions integrate and involved constants are determined. Usually, such conditions describe how the system behaves at the system's start (like temperature, pressure, or concentration changes) and are essential to accurately simulate real-life scenarios.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) involve functions of a single independent variable and their derivatives. These equations are imperative in modeling continuous systems across various scientific and engineering fields. ODEs can describe a range of natural phenomena including motion, growth, decay, and wave propagation.
The ODE given in the exercise is \( \frac{d^{2} C}{d x^{2}} - \frac{s}{k} C = 0 \). This second-order ODE is solved by first finding its characteristic equation, yielding solutions in exponential forms. These solutions reflect the intrinsic dynamics governed by the differential terms of the equation, crucial for modeling how systems evolve over space or time.
Solving these equations often requires strategies beyond simple algebraic calculations, necessitating tools like Laplace Transforms to make them manageable.

Inverse Laplace Transform

The Inverse Laplace Transform is used to revert a transformed function back to its original time-domain form. After working in the \( s \)-domain to simplify problem-solving, we often need to convert results back to function of time or space variables. The inverse transform is essential in obtaining the real-world solution from the transformed equations.
Mathematically, this involves retrieving \( f(t) \) from \( F(s) \), often requiring complex contour integrals or matching results to known Laplace pairs.
In our problem, the inverse Laplace is used to transform \( C(x, s) \) back to time domain to get \( c(x, t) \). The result reflects on the nature of solutions and typically involves identifying inverse pairs from transformation tables. This specific process bridges the gap between the simplified math world and tangible problem-solving in time-domain modeling.