Problem 19 - Advanced Engineering Mathematics

Question

Solve the given boundary-value problem by an appropriate integral transform. Make assumptions about boundedness where necessary. Consider the boundary-value problem $$ \begin{aligned} &\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial u}{\partial t}, x>0, t>0 \\\ &\left.\frac{\partial u}{\partial x}\right|_{x=0}=-1, \lim _{x \rightarrow \infty} u(x, t)=0, \quad t>0 \\ &u(x, 0)=0, \quad x>0 \end{aligned} $$ Solve the BVP using the Laplace transform.

Step-by-Step Solution

Answer

The solution is \( u(x,t) = \text{erfc} \left( \frac{x}{2\sqrt{t}} \right) \).

Step 1: Apply Laplace Transform on Time

Take the Laplace Transform of both sides of the differential equation with respect to the time variable \(t\). We denote the Laplace Transform of \(u(x,t)\) by \(U(x,s)\). \[\mathcal{L}\left\{ \frac{\partial^2 u}{\partial x^2} - \frac{\partial u}{\partial t} \right\} = \frac{\partial^2 U}{\partial x^2} - sU + u(x,0) = 0\]Given \(u(x,0) = 0\), this simplifies to:\[\frac{\partial^2 U}{\partial x^2} = sU\]

Step 2: Solve the Transformed Differential Equation

The equation \(\frac{\partial^2 U}{\partial x^2} = sU\) is a second-order linear differential equation which can be solved using exponential functions. The general solution is:\[U(x,s) = A(s)e^{\sqrt{s}x} + B(s)e^{-\sqrt{s}x}\]Where \(A(s)\) and \(B(s)\) are functions to be determined from boundary conditions.

Step 3: Apply Boundary Condition at Infinity

Use the boundary condition \(\lim_{x \to \infty} u(x,t) = 0\). This implies \(\lim_{x \to \infty} U(x,s) = 0\). Since \(e^{\sqrt{s}x}\) grows unbounded as \(x\to\infty\), \(A(s)\) must be zero:\[U(x,s) = B(s)e^{-\sqrt{s}x}\]

Step 4: Apply Boundary Condition at x=0

Utilize the boundary condition \(\left. \frac{\partial u}{\partial x} \right|_{x=0} = -1\). In the Laplace domain, this becomes:\[\left. \frac{\partial U}{\partial x} \right|_{x=0} = -\sqrt{s}B(s) = -1\]Solving for \(B(s)\), we get:\[B(s) = \frac{1}{\sqrt{s}}\]Thus, the solution in the Laplace domain is:\[U(x,s) = \frac{1}{\sqrt{s}}e^{-\sqrt{s}x}\]

Step 5: Inverse Laplace Transform to Find u(x,t)

Take the inverse Laplace Transform of \(U(x,s)\) to find \(u(x,t)\). The inverse Laplace Transform of \(\frac{1}{\sqrt{s}}e^{-\sqrt{s}x}\) can be identified as a known transform:\[u(x,t) = \text{erfc}\left( \frac{x}{2\sqrt{t}} \right)\]where \(\text{erfc}(z)\) is the complementary error function.

Key Concepts

Boundary-value problem

A boundary-value problem involves a differential equation accompanied by a set of constraints called boundary conditions. In our specific case, we're working with a boundary-value problem involving a partial differential equation (PDE). The boundary conditions here are specified for the exponential behavior of a function as a variable approaches a particular limit.
Boundary-value problems:

• Directly specify values or derivatives of the solution.
• Are integral in solving real-world physical systems where boundaries or endpoints influence behavior within the system.

In our exercise, the boundary conditions include:

• The derivative of the function with respect to space, at the position zero, is set to −1.
• The function itself approaches zero as the spatial variable approaches infinity.
• At the initial time point, the function starts from zero.

These conditions dictate how we solve the PDE using techniques like the Laplace Transform, ensuring we find a unique and accurate solution to the behavior of the system in the given domain.

Partial differential equations

Partial differential equations (PDEs) are equations that involve the partial derivatives of a function of multiple variables. They're essential in describing various phenomena like heat, waves, fluids, and more.

• PDEs involve rates of change and are used to describe continuous systems.
• They often require specific boundary and initial conditions to solve.

In our scenario, the given PDE is \(\frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial u}{\partial t}\), indicating how the function \(u(x,t)\) changes over both the spatial variable \(x\) and the time variable \(t\). This type of equation models the diffusion process, like heat conduction, where heat spreads out over time.The Laplace Transform is applied here to convert the PDE with respect to time into an ordinary differential equation (ODE), focusing only on the spatial variable. This simplifies things significantly, allowing us to solve it more straightforwardly by dealing with algebraic expressions rather than partial derivatives.Ultimately, solving the PDE is an essential step in answering how the physical system behaves over space and time under specific constraints.

Complementary error function

The complementary error function, denoted as \(\text{erfc}(z)\), is a mathematical function related to the error function (erf). It's extensively used in probability, statistics, and PDE solutions as it describes distributions.

• The complementary error function \(\text{erfc}(z)\) is defined as \(1 - \text{erf}(z)\).
• It plays a prime role in representing solutions to diffusion-type equations.

In the context of our boundary-value problem, the complementary error function emerges during the inverse Laplace Transform process. When solving the transformed equation in the solution step, the inverse transform yields an expression involving \(\text{erfc}\). This appearance of \(\text{erfc}\) aligns perfectly with expectations for a diffusion equation under given initial and boundary conditions.The friendly nature of \(\text{erfc}\) comes from its ability to model "spreading-out" behavior mathematically, providing insights into how a process behaves beyond just solving the PDE. It's an excellent example of benefiting from known transforms to streamline the calculation and enhance understanding.