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Magazine Chapter 11: Problem 6
L枚sen Sie das folgende Anfangs-Randwertproblem: $$ \begin{aligned} \frac{\partial u}{\partial t} &=2 \cdot \frac{\partial^{2} u}{\partial x^{2}} & & \text { f眉r } x \in(0,1), t>0 \\ u(x, 0) &=\sin 3 \pi x & \text { f眉r } x \in(0,1) \\ u(0, t) &=u(1, t)=0 & \text { f眉r } t>0 \end{aligned} $$.
Short Answer
Expert verified
The solution is \( u(x, t) = \sin 3 \pi x e^{-18 \pi^2 t} \).
Step by step solution
01
Identify the type of problem
This is a partial differential equation (PDE) with initial and boundary conditions. Specifically, it's a heat equation: \( \frac{\frac{\partial u}{\partial t}} = 2 \frac{\frac{\partial^{2} u}{\partial x^{2}}} \) with given initial and boundary conditions.
02
Apply separation of variables
Assume a solution of the form \( u(x,t) = X(x)T(t) \). Substitute this form into the PDE and separate variables to get two ordinary differential equations (ODEs).
03
Substitute and separate variables
Substitute \( u(x,t) = X(x)T(t) \) into the PDE:\[ X(x) \frac{dT(t)}{dt} = 2T(t) \frac{d^2X(x)}{dx^2} \].Divide both sides by \( X(x)T(t) \) to separate variables:\[ \frac{1}{T(t)} \frac{dT(t)}{dt} = 2 \frac{1}{X(x)} \frac{d^2X(x)}{dx^2} = -\rho \] where \( \rho \) is a separation constant.
04
Solve the ODEs
Solve the resulting ODEs: For \( T(t) \):\[ \frac{dT}{dt} + \rho T = 0 \] which has solution \( T(t) = A e^{-\rho t} \).For \( X(x) \):\[ \frac{d^2 X}{dx^2} + \frac{\rho}{2} X = 0 \] which has solution \( X(x) = B \sin \frac{\sqrt{2\rho}}{2} x + C \cos \frac{\sqrt{2\rho}}{2} x \).
05
Apply boundary conditions to determine constants
Use boundary conditions \( u(0, t) = u(1, t) = 0 \) to determine constants:When \( x = 0, X(0) = 0 \) gives \( C = 0 \).When \( x = 1, X(1) = 0 \) gives \( B \sin \frac{\sqrt{2\rho}}{2} = 0 \). This implies \( \frac{\sqrt{2\rho}}{2} = n\pi \) for integer \( n \). Therefore, \( \rho = 2 n^2\pi^2 \). The solution for \( X(x) \) is reduced to \( X(x) = B \sin n \pi x \).
06
Form the general solution
Combine the solutions for \( X(x) \) and \( T(t) \) to form the general solution:\[ u(x,t) = \sum_{n=1}^{\infty} B_n \sin n \pi x e^{-2 n^2 \pi^2 t} \].
07
Apply initial condition
Use the initial condition \( u(x,0) = \sin 3 \pi x \) to determine the coefficients:Comparing to the initial condition gives \( B_n = \delta_{n,3} \), indicating it is only the third term that survives. Therefore, the solution is:\[ u(x,t) = \sin 3 \pi x e^{-18 \pi^2 t}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heat Equation
The heat equation is a fundamental partial differential equation (PDE) used to describe the distribution of heat (or temperature variations) in a given region over time. It is mathematically represented as:
\ \[ \frac{\frac{\text{鈭倉u}{\text{鈭倉t}} = k \frac{\frac{\text{鈭倉^2u}{\text{鈭倉x^2}} \]
Where:
* \(\frac{\text{鈭倉u}{\text{鈭倉t}\) represents the time rate of change of temperature.
* \(\frac{\frac{\text{鈭倉^2u}{\text{鈭倉x^2}}\) represents the spatial rate of change of temperature along the x-axis.
* \(k\) is the thermal diffusivity constant of the material.
This equation assumes the heat conducts through the medium uniformly.
\ \[ \frac{\frac{\text{鈭倉u}{\text{鈭倉t}} = k \frac{\frac{\text{鈭倉^2u}{\text{鈭倉x^2}} \]
Where:
* \(\frac{\text{鈭倉u}{\text{鈭倉t}\) represents the time rate of change of temperature.
* \(\frac{\frac{\text{鈭倉^2u}{\text{鈭倉x^2}}\) represents the spatial rate of change of temperature along the x-axis.
* \(k\) is the thermal diffusivity constant of the material.
This equation assumes the heat conducts through the medium uniformly.
Separation of Variables
Separation of variables is a mathematical method used to simplify partial differential equations (PDEs) into simpler, separate ordinary differential equations (ODEs). This approach relies on assuming that the solution can be written as a product of functions, each dependent on a single variable.
For instance, assume \(u(x, t) = X(x)T(t)\), and substitute this form into the PDE.
This yields:
\ \[ X(x) \frac{dT(t)}{dt} = k T(t) \frac{d^2X(x)}{dx^2} \]
Dividing both sides by \(X(x)T(t)\) separates the equation into:
\ \[ \frac{\frac{1}{T(t)} \frac{dT(t)}{dt}} = k \frac{\frac{1}{X(x)} \frac{d^2X(x)}{dx^2}} = -\rho \]
Here, \(\rho\) is a separation constant. By separating the variables, we obtain two simpler ODEs: one for time \(T(t)\) and one for space \(X(x)\).
For instance, assume \(u(x, t) = X(x)T(t)\), and substitute this form into the PDE.
This yields:
\ \[ X(x) \frac{dT(t)}{dt} = k T(t) \frac{d^2X(x)}{dx^2} \]
Dividing both sides by \(X(x)T(t)\) separates the equation into:
\ \[ \frac{\frac{1}{T(t)} \frac{dT(t)}{dt}} = k \frac{\frac{1}{X(x)} \frac{d^2X(x)}{dx^2}} = -\rho \]
Here, \(\rho\) is a separation constant. By separating the variables, we obtain two simpler ODEs: one for time \(T(t)\) and one for space \(X(x)\).
Boundary Conditions
Boundary conditions specify the behavior of the solution on the boundaries of the domain. For the heat equation, we often consider:
1. **Dirichlet Boundary Condition:** Specifies the value of the function at the boundary. Example: \(u(0, t) = 0\) or \(u(1, t) = 0\).
2. **Neumann Boundary Condition:** Specifies the value of the derivative of the function at the boundary. Example: \(\frac{\text{鈭倉u}{\text{鈭倉x}(0, t) = 0\).
3. **Mixed Boundary Condition:** Combines both Dirichlet and Neumann conditions.
In our example, the conditions are:
\[ \begin{aligned} \ u(0, t) &= 0 \ u(1, t) &= 0 \ \text{for} \ t > 0 \ \ \end{aligned} \]
These imply that the temperature is fixed at 0 at the boundaries of the domain \(x = 0\) and \(x = 1\).
1. **Dirichlet Boundary Condition:** Specifies the value of the function at the boundary. Example: \(u(0, t) = 0\) or \(u(1, t) = 0\).
2. **Neumann Boundary Condition:** Specifies the value of the derivative of the function at the boundary. Example: \(\frac{\text{鈭倉u}{\text{鈭倉x}(0, t) = 0\).
3. **Mixed Boundary Condition:** Combines both Dirichlet and Neumann conditions.
In our example, the conditions are:
\[ \begin{aligned} \ u(0, t) &= 0 \ u(1, t) &= 0 \ \text{for} \ t > 0 \ \ \end{aligned} \]
These imply that the temperature is fixed at 0 at the boundaries of the domain \(x = 0\) and \(x = 1\).
Initial Conditions
Initial conditions specify the state of the system at the beginning of the observation (initial time, typically \(t = 0\)). For the heat equation, the initial condition specifies the initial temperature distribution.
In our case, we have:
\[ u(x, 0) = \sin 3 \pi x \]
This tells us that at time \(t = 0\), the temperature distribution along the rod is given by the sine function \( \sin 3 \pi x \). The initial condition is essential in determining the coefficients for the general solution of the PDE. It ensures that the solution accurately reflects the physical initial state of the system.
By applying the initial condition, we derive specific terms and constants that satisfy both the initial and boundary conditions, leading to the unique solution to our problem.
In our case, we have:
\[ u(x, 0) = \sin 3 \pi x \]
This tells us that at time \(t = 0\), the temperature distribution along the rod is given by the sine function \( \sin 3 \pi x \). The initial condition is essential in determining the coefficients for the general solution of the PDE. It ensures that the solution accurately reflects the physical initial state of the system.
By applying the initial condition, we derive specific terms and constants that satisfy both the initial and boundary conditions, leading to the unique solution to our problem.
