Problem 17
Question
Solve the given boundary-value problem by an appropriate integral transform. Make assumptions about boundedness where necessary. $$ \begin{aligned} &\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial u}{\partial t}, \quad 0 < x <\pi, \quad t > 0 \\ &u(0, t)=1, u(\pi, t)=1, \quad t > 0 \\ &u(x, 0)=1+\sin 2 x, \quad 0 < x <\pi \end{aligned} $$
Step-by-Step Solution
Verified Answer
The solution is \( u(x, t) = 1 + e^{-4t} \sin 2x \).
1Step 1: Analyze the Problem
The given problem is a heat equation, expressed as \( \frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial u}{\partial t} \). We need to find the function \( u(x, t) \) that satisfies this partial differential equation (PDE), along with the boundary conditions \( u(0, t)=1 \) and \( u(\pi, t)=1 \), and the initial condition \( u(x, 0)=1+\sin 2x \).
2Step 2: Apply Separation of Variables
We assume a solution of the form \( u(x, t) = X(x)T(t) \). This implies that \( X''(x)T(t) = X(x)T'(t) \). Dividing through by \( X(x)T(t) \), we obtain \( \frac{X''(x)}{X(x)} = \frac{T'(t)}{T(t)} = -\lambda \), where \( \lambda \) is a separation constant.
3Step 3: Solve for Spatial Part \( X(x) \)
The equation \( X''(x) + \lambda X(x) = 0 \) must be solved with the boundary conditions \( X(0)=1 \) and \( X(\pi)=1 \). Assuming \( X(x) \) is in the form of a sinusoidal function, we consider \( \lambda = n^2 \). The solution can be expressed as \( X(x) = a_n \cos(nx) + b_n \sin(nx) \). By applying the boundary conditions, we discern that these must be particular combinations to maintain \( X(0) = 1 \) and \( X(\pi) = 1 \).
4Step 4: Solve for Temporal Part \( T(t) \)
The equation \( T'(t) = -n^2T(t) \) can be solved by simple integration, yielding \( T(t) = c_n e^{-n^2t} \).
5Step 5: Combine Results to Form General Solution
The product solution is \( u(x, t) = \sum_{n=0}^{\infty} [a_ne^{-n^2t} \cos(nx) + b_ne^{-n^2t} \sin(nx)] \). Given that \( u(0, t) = 1 \) and \( u(\pi, t) = 1 \), these conditions help determine the coefficients necessary to satisfy the boundary conditions.
6Step 6: Apply Initial Condition
To find the coefficients \( a_n \) and \( b_n \), apply the initial condition \( u(x, 0) = 1 + \sin 2x \). You equate \( \sum_{n=0}^{\infty} (a_n \cos(nx) + b_n \sin(nx)) \) to \( 1 + \sin 2x \), identifying \( a_0 = 1, a_n = 0 \) for \( n eq 0 \), \( b_2 = 1 \), and \( b_n = 0 \) for \( n eq 2 \).
7Step 7: Express the Final Solution
The final solution satisfying all the conditions is given by \( u(x, t) = 1 + e^{-4t} \sin 2x \).
Key Concepts
Boundary-Value ProblemHeat EquationSeparation of VariablesPartial Differential Equation
Boundary-Value Problem
A boundary-value problem is a mathematical scenario involving a differential equation along with a set of additional constraints, known as boundary conditions. These conditions specify values that a solution to the differential equation must satisfy at the boundaries of the domain.
In the problem we're solving, the differential equation is defined in the region between 0 and \( \pi \). The boundary conditions provided are crucial for obtaining a specific solution, which contrasts with initial value problems that only have conditions at a starting point.
In the problem we're solving, the differential equation is defined in the region between 0 and \( \pi \). The boundary conditions provided are crucial for obtaining a specific solution, which contrasts with initial value problems that only have conditions at a starting point.
- Boundary conditions here are given by \( u(0, t) = 1 \) and \( u(\pi, t) = 1 \), which must hold true for all values of \( t > 0 \).
- The initial condition \( u(x, 0) = 1 + \sin 2x \) specifies the solution's behavior at the beginning of the time domain.
Heat Equation
The heat equation is a fundamental partial differential equation (PDE) in mathematical physics. It describes the distribution of heat (or variation in temperature) in a given region over time.
In our exercise, the heat equation is expressed as:
This second derivative essentially describes how "curved" the temperature distribution is; greater curvature suggests a stronger tendency for temperature to change. The heat equation is used to model many physical systems beyond just temperature, such as diffusion and fluid flow.
In our exercise, the heat equation is expressed as:
- \( \frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial u}{\partial t} \)
This second derivative essentially describes how "curved" the temperature distribution is; greater curvature suggests a stronger tendency for temperature to change. The heat equation is used to model many physical systems beyond just temperature, such as diffusion and fluid flow.
Separation of Variables
Separation of variables is a widely used mathematical method for solving partial differential equations (PDEs), like the heat equation. The main idea is to assume that the solution can be expressed as the product of functions, each depending on a single coordinate.
For the heat equation, we assumed a solution of the form \( u(x, t) = X(x)T(t) \), where \( X(x) \) is a function of space and \( T(t) \) is a function of time. This assumption splits the equation into two ordinary differential equations:
For the heat equation, we assumed a solution of the form \( u(x, t) = X(x)T(t) \), where \( X(x) \) is a function of space and \( T(t) \) is a function of time. This assumption splits the equation into two ordinary differential equations:
- Spatial part: \( \frac{X''(x)}{X(x)} = -\lambda \), yielding a sinusoidal solution.
- Temporal part: \( \frac{T'(t)}{T(t)} = -\lambda \), leading to exponential decay over time.
Partial Differential Equation
Partial differential equations (PDEs) involve functions of multiple variables and their partial derivatives. They are essential in describing systems with spatial and temporal dynamics.
In our context, the heat equation is a second-order linear PDE that models the flow of heat in one-dimensional space over time.
In our context, the heat equation is a second-order linear PDE that models the flow of heat in one-dimensional space over time.
- Unlike ordinary differential equations (ODEs), which deal with functions of a single variable, PDEs use multiple variables. This complexity requires more advanced solution techniques, like separation of variables or integral transforms.
- The heat equation in this problem describes how the temperature at any point in space changes over time. Its solution allows us to understand how heat is distributed across a medium like a metal rod or a thin slab.
Other exercises in this chapter
Problem 17
Solve the given integral equation for the function \(f\). $$ \int_{0}^{\infty} f(x) \cos \alpha x d x=e^{-\alpha} $$
View solution Problem 17
Use the Laplace transform to solve the heat equation \(u_{x x}=u_{t}, x>0, t>0\) subject to the given conditions. $$ \begin{aligned} &u(0, t)=60+40 q(t-2), \qua
View solution Problem 18
Solve the given integral equation for the function \(f\). $$ \int_{0}^{\infty} f(x) \sin \alpha x d x=\left\\{\begin{array}{lr} 1, & 01 \end{array}\right. $$
View solution Problem 18
Use the Laplace transform to solve the heat equation \(u_{x x}=u_{t}, x>0, t>0\) subject to the given conditions. $$ \begin{aligned} &u(0, t)=\left\\{\begin{arr
View solution