Question

Consider the heat equation \(u_{t}(x, t)=\kappa u_{x x}(x, t)\), where \(\kappa\) is a positive constant. (a) Suppose we make the change of time variable \(\tau=\kappa t\). Show that the heat equation transforms into \(\partial u / \partial \tau=\partial^{2} u / \partial x^{2}\). Therefore, if we know a solution of \(\partial u / \partial t=\partial^{2} u / \partial x^{2}\), we can obtain a solution of \(u_{t}=\kappa u_{x x}\) by replacing \(t\) with \(\kappa t\). (b) Show that \(u(x, t)=e^{-\pi^{2} t} \sin \pi x\) is a solution of the initial-boundary value problem $$ \begin{array}{lc} u_{t}(x, t)=u_{x x}(x, t), & 0

Short answer

Question: Verify that the function \(u(x, t) = 5 e^{-3\pi^2 t} \sin \pi x\) satisfies the initial-boundary value problem: $$ \begin{cases} u_t(x, t) = 3 u_{xx}(x, t), & 0<x<1, \quad 0<t<\infty \\ u(0, t)=u(1, t)=0, & 0 \leq t<\infty \\ u(x, 0)=5\sin \pi x, & 0 \leq x \leq 1 \end{cases} $$

Step-by-step solution

Transform the Heat Equation with Time Variable Change

Let's make the change of time variable \(\tau=\kappa t\). We want to find \(\partial u / \partial \tau\) and rewrite the heat equation in terms of \(\tau\) and \(x\). To find \(\partial u / \partial \tau\), apply the chain rule: $$\frac{\partial u}{\partial \tau} = \frac{\partial u}{\partial t} \frac{\partial t}{\partial \tau}$$ We know \(\frac{\partial \tau}{\partial t} = \kappa\), and therefore \(\frac{\partial t}{\partial \tau} = \frac{1}{\kappa}\). Plugging this into the equation above we get: $$\frac{\partial u}{\partial \tau} = u_{t}(x, t) \frac{1}{\kappa}$$ Replacing \(u_t(x, t) = \kappa u_{xx}(x, t)\), we get: $$\frac{\partial u}{\partial \tau} = u_{xx}(x, t)$$

Verify that \(u(x, t)=e^{-\pi^{2} t} \sin \pi x\) is a Solution of the Initial-boundary Value Problem

Now we need to check that \(u(x, t)=e^{-\pi^{2} t} \sin \pi x\) satisfies the initial-boundary value problem: $$ \begin{cases} u_t(x, t) = u_{xx}(x, t), & 0<x<1, \quad 0<t<\infty \\ u(0, t)=u(1, t)=0, & 0 \leq t<\infty \\ u(x, 0)=\sin \pi x, \quad 0 \leq x \leq 1 \end{cases} $$ First, find the partial derivatives with respect to \(t\) and \(x\): 1) \(u_t(x, t) = -\pi^2 e^{-\pi^2 t} \sin \pi x\) 2) \(u_{xx}(x, t) = e^{-\pi^2 t} \pi^2 \sin \pi x\) Since \(u_t(x, t) = u_{xx}(x, t)\), the first condition is satisfied. Now, check the boundary conditions: 1) \(u(0, t) = e^{-\pi^2 t} \sin (0) = 0\) 2) \(u(1, t) = e^{-\pi^2 t} \sin \pi = 0\) The boundary conditions are satisfied. Finally, check the initial condition: $$u(x, 0)=e^{-\pi^2 (0)} \sin \pi x = \sin \pi x $$ All conditions are satisfied. Therefore, \(u(x, t) = e^{-\pi^{2} t} \sin \pi x\) is a solution of the initial-boundary value problem.

Solve the New Initial-boundary Value Problem

We'll now solve the new problem: $$ \begin{cases} u_t(x, t) = 3 u_{xx}(x, t), & 0<x<1, \quad 0<t<\infty \\ u(0, t)=u(1, t)=0, & 0 \leq t<\infty \\ u(x, 0)=5\sin \pi x, & 0 \leq x \leq 1 \end{cases} $$ From step 1, we know that the heat equation can be transformed into \(\partial u / \partial \tau = \partial^{2} u / \partial x^{2}\). So, if we have a solution for \(\partial u / \partial t = \partial^{2} u / \partial x^{2}\), we can replace \(t\) with \(\kappa t\) to obtain a solution for \(u_t(x, t) = 3 u_{xx}(x, t)\). Using \(\kappa = 3\), we have: $$u(x, t) = e^{-\pi^{2} (3t)} \sin \pi x$$ Now, we need a solution with the initial condition \(u(x, 0) = 5 \sin \pi x\). The solution above has the form \(u(x, 0) = \sin \pi x\). Multiply the solution by a factor of \(5\) to satisfy the initial condition: $$u(x, t) = 5 e^{-3\pi^{2} t} \sin \pi x$$ This is the solution to the new initial-boundary value problem.

Key concepts

Partial Differential Equations

Partial differential equations (PDEs) are mathematical expressions that involve the rates of change with respect to multiple variables. In the context of the heat equation, a common example of a PDE is \( u_t(x, t) = \kappa u_{xx}(x, t) \). Here, \( u \) represents the temperature distribution in a given medium, \( t \) denotes time, and \( x \) is the spatial coordinate.

In real-world terms, the heat equation models the diffusion of heat in an object over time. \( \kappa \) is a constant related to the thermal diffusivity of the material. Understanding PDEs is crucial in multiple domains such as physics, engineering, and finance as they describe the behavior of various physical phenomena.

When solving PDEs like the heat equation, we apply methods such as separation of variables, Fourier series, and transforms, aiming to find a function that satisfies the equation over the given domain. The robustness of these methods allows us to tackle complex boundary and initial conditions, providing a streamlined path to solving otherwise complex problems.

Initial-boundary Value Problem

The initial-boundary value problem is a type of problem that requires finding a solution to a PDE within a specified range (boundary) and establishes the state of the system at the starting point in time (initial condition).

The exercise uses \( u(x, 0) = \sin \pi x \) for the initial condition and specifies zero temperature at both ends of the spatial domain through the boundary conditions \( u(0, t) = u(1, t) = 0 \) over time. These conditions model a physical scenario where the ends of a rod are maintained at constant temperature (zero), while the initial heat distribution is given by \( \sin \pi x \).

It is imperative that the solution to PDEs respects both the initial condition and the boundary conditions to accurately reflect the physical scenario. The example provides an analytical solution, a method which aims to solve the mathematical problem exactly, offering a direct insight into how the system behaves over time under the constraints imposed by the boundary and initial conditions.

Solution of Differential Equations

Solving differential equations involves finding a function (or set of functions) that meets the criteria set out by the equation itself, as well as any boundary and initial conditions. In the case of our heat equation exercise, the solution follows a process of verifying that a proposed function is indeed the solution and then adapting it to new conditions.

Initially, we find that \( u(x, t) = e^{-\pi^2 t} \sin \pi x \) is a solution to the given initial-boundary value problem. To adapt this solution to a modified scenario with different constants, the concept of similarity solutions comes into play. By transforming the time variable and scaling the solution, we create a new function \( u(x, t) = 5e^{-3\pi^2 t} \sin \pi x \) that meets the altered initial condition of \( u(x, 0) = 5 \sin \pi x \).

This method shows how initial conditions affect the form of the solution but do not change the fundamental nature of the solution's dependency on space and time. Educational practice in differential equations often involves these transformations and adaptations to reinforce the student's understanding of how solutions can be modified while remaining valid within the mathematical framework of the problem.