Question

Solve the following boundary-value problem: $$ \begin{cases}u_{t}=4 u_{x x}, & -\infty2 .\end{cases} $$

Short answer

The solution is \( u(x,t) = \frac{1}{2} \left[ \text{erf} \left( \frac{x+2}{2\sqrt{2t}} \right) - \text{erf} \left( \frac{x-2}{2\sqrt{2t}} \right) \right] \).

Step-by-step solution

Identify the Problem Type

This problem is an initial boundary value problem for the heat equation, given by \( u_t = 4u_{xx} \). The domain is infinite in the \( x \)-direction. The initial condition is specified as a piecewise function \( f(x) \).

Use the Heat Equation Solution Formulation

For the heat equation \( u_t = k u_{xx} \) with \( k=4 \), the solution, given an initial condition \( u(x,0) = f(x) \), can be expressed as the convolution with the heat kernel: \[ u(x,t) = \frac{1}{\sqrt{8\pi t}} \int_{-\infty}^{\infty} e^{-\frac{(x-y)^2}{8t}} f(y) \, dy \]

Define the Heat Kernel

The heat kernel for this problem is \[ G(x,t) = \frac{1}{\sqrt{8\pi t}} e^{-\frac{x^2}{8t}} \]This defines how heat propagates through the infinite medium over time.

Set Up the Integral with the Initial Condition

Plug the initial condition \( f(x) \) into the integral: \[ u(x,t) = \frac{1}{\sqrt{8\pi t}} \left( \int_{-2}^{2} e^{-\frac{(x-y)^2}{8t}} \, dy \right) \] Here, the integration limits change according to where \( f(x) = 1 \), which is when \(-2 \leq y \leq 2\).

Solve the Integral

To evaluate \[ \int_{-2}^{2} e^{-\frac{(x-y)^2}{8t}} \, dy \]note that it's a definite integral of a Gaussian function over a finite interval. A substitution \( z = x-y \) can simplify it to \[ \int_{x-2}^{x+2} e^{-\frac{z^2}{8t}} \, dz \].

Apply Gaussian Integral Formulations

The integral \[ \int e^{-az^2} \, dz \] is solved using known properties of the error function, typically given as \[ \text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} \, dt \].Substitution gives values based on making the equation match this structure.

Express the Solution Using the Error Function

Thus, the solution is expressed in terms of the error function, evaluating it at \((x \/ - 2)\/\sqrt{8t}\) and \((x \/ + 2)\/\sqrt{8t}\). Hence,\[ u(x,t) = \frac{1}{2} \left[ \text{erf} \left( \frac{x+2}{2\sqrt{2t}} \right) - \text{erf} \left( \frac{x-2}{2\sqrt{2t}} \right) \right] \]

Key concepts

Heat Equation

The heat equation is a fundamental partial differential equation (PDE) that describes how heat distributes itself in a given region over time. It is central to the study of heat conduction and diffusion processes.
The general form of the heat equation is \( u_t = k u_{xx} \), where:

• \( u(x,t) \) represents the temperature at point \( x \) and time \( t \).
• \( u_t \) is the partial derivative of \( u \) with respect to time, representing the rate of change of temperature.
• \( u_{xx} \) is the second partial derivative with respect to space, indicating how temperature changes across the spatial dimension.
• \( k \) is the thermal diffusion coefficient that plays a key role in determining the rate of heat distribution.

In our specific problem, \( k = 4 \), which signifies the speed at which heat diffuses through the medium.

Initial Condition

Initial conditions are crucial in solving PDEs like the heat equation, as they define the state of the system at the beginning of the observation period (i.e., at \( t = 0 \)).
For our boundary value problem, the initial condition is given by the piecewise function \( f(x) \):

• \( f(x) = 1 \) for values of \( x \) such that \( |x| \leq 2 \).
• \( f(x) = 0 \) when \( |x| > 2 \).

This means that at time \( t = 0 \), the temperature is 1 unit within the interval \([-2, 2]\) and 0 outside of it. The role of the initial condition here is to serve as the starting point for how heat will spread over time, deeply influencing the system's evolution.

Heat Kernel

The heat kernel is instrumental in solving the heat equation, especially in infinite or large domains. It represents the fundamental solution reflecting how an initial amount of heat disperses over time.
In our context, the heat kernel is defined as:\[ G(x,t) = \frac{1}{\sqrt{8\pi t}} e^{-\frac{x^2}{8t}} \]This equation expresses:

• How an initial point source of heat diffuses through the medium over time \( t \).
• The Gaussian nature of the function, characterized by the exponential term, dictates the symmetric and rapidly decaying nature of heat distribution from the source point.
• The prefactor \( \frac{1}{\sqrt{8\pi t}} \) ensures the proper normalization, reflecting the conservation of total heat.

By convolving the initial condition \( f(x) \) with the heat kernel, one obtains the temperature distribution \( u(x,t) \) for any future time \( t \).

Error Function

The error function, abbreviated as \( \text{erf} \), is a mathematical function closely associated with the normal distribution and is pivotal in solving integrals involving Gaussian functions.
In our boundary value problem, we use the error function to manage the definite integral encountered in the convolution of the initial condition with the heat kernel. Here, the error function is defined as:\[ \text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} \, dt \]This function arises naturally when integrating the Gaussian function since it effectively handles the area under the curve from 0 to \( z \).
In our heat equation problem, solutions involve expressions like:\[ u(x,t) = \frac{1}{2} \left[ \text{erf} \left( \frac{x+2}{2\sqrt{2t}} \right) - \text{erf} \left( \frac{x-2}{2\sqrt{2t}} \right) \right] \]This captures the cumulative distribution aspect of heat as it spreads from a localized region, adjusting for boundaries defined by our piecewise initial conditions.