Problem 5
Question
Consider $$ \frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}} $$ with \(\frac{\partial u}{\partial x}(0, t)=0, \frac{\partial u}{\partial x}(L, t)=-h u(L, t)\), and \(u(x, 0)=f(x)\). (a) Solve if \(h>0\). (b) Solve if \(h<0\).
Step-by-Step Solution
Verified Answer
For \(h > 0\), the general solution is \(u(x, t)= A \cos(\sqrt{h}x) e^{-k \sqrt{h}t}\) and for \(h < 0\), it is \(u(x, t)= A \cos(\sqrt{-h'}x) e^{-k \sqrt{-h'}t}\) with \(h' > 0\). The specific solutions would depend on the form of \(f(x)\).
1Step 1: Set up the equation
Given the PDE \(\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}}\) with boundary conditions \(\frac{\partial u}{\partial x}(0, t)=0\), \(\frac{\partial u}{\partial x}(L,t)=-h u(L, t)\) and initial condition \(u(x, 0)=f(x)\). We are asked to solve the equation for two cases: (a) when \(h > 0\) and (b) when \(h < 0\).
2Step 2: Solve for \(h > 0\)
For \(h > 0\) and using the separation of variables, the equation can be written as \(\frac{dT}{T}=\frac{k d^2X}{X dx^2}\). By solving this, the general solution can be found as \(u(x, t)= X(x)T(t)\), where \(X(x) = A \cos(\sqrt{h}x)\) and \(T(t) = e^{-k \sqrt{h}t}\). Further, applying the given boundary and initial conditions may yield the particular solution depending on \(f(x)\).
3Step 3: Solve for \(h < 0\)
For \(h < 0\), we substitute \(h = -h'\) where \(h' > 0\). The equation thus assumes the same form and can be solved similarly to obtain \(u(x, t)= A \cos(\sqrt{-h'}x)e^{-k \sqrt{-h'}t}\). However, since the sign of \(h\) has changed, the second boundary condition becomes a complex number multiplying \(u(L, t)\). This requires a different approach to satisfy the boundary condition, which will also greatly depend on \(f(x)\).
Key Concepts
Boundary ConditionsSeparation of VariablesInitial Value ProblemHeat Equation
Boundary Conditions
In the study of partial differential equations (PDEs), like the heat equation given in the problem, 'boundary conditions' are crucial guidelines that describe the behavior of a function on the edges of its domain. These conditions are necessary for uniquely determining the solution to a PDE.
The boundary conditions given in the exercise are of two types. The first one, \( \frac{\partial u}{\partial x}(0, t)=0 \), is a Neumann condition, which prescribes the derivative along the spatial boundary at \( x=0 \). The second one, \( \frac{\partial u}{\partial x}(L, t)=-h u(L, t) \), is a Robin condition, a mixture of Neumann and Dirichlet conditions. It includes the function value \( u(L, t) \) directly, ensuring a relationship between the function and its derivative at \( x=L \).
To solve the heat equation with these boundary conditions, one must find a function \( u(x, t) \) that satisfies the heat equation while also adhering to these edge behaviors. The challenge lies in the complexity added by the Robin condition, which leads to specific solutions for positive and negative values of \( h \).
The boundary conditions given in the exercise are of two types. The first one, \( \frac{\partial u}{\partial x}(0, t)=0 \), is a Neumann condition, which prescribes the derivative along the spatial boundary at \( x=0 \). The second one, \( \frac{\partial u}{\partial x}(L, t)=-h u(L, t) \), is a Robin condition, a mixture of Neumann and Dirichlet conditions. It includes the function value \( u(L, t) \) directly, ensuring a relationship between the function and its derivative at \( x=L \).
To solve the heat equation with these boundary conditions, one must find a function \( u(x, t) \) that satisfies the heat equation while also adhering to these edge behaviors. The challenge lies in the complexity added by the Robin condition, which leads to specific solutions for positive and negative values of \( h \).
Separation of Variables
Separation of variables is a powerful mathematical technique used to solve partial differential equations, where the PDE is reduced to a set of simpler ordinary differential equations (ODEs). The key assumption of this method is that the solution can be expressed as the product of functions, each dependent on only one of the variables.
In our given heat equation, this technique assumes the solution takes the form \( u(x, t) = X(x)T(t) \), with \( X(x) \) only a function of \( x \) and \( T(t) \) only a function of \( t \). By substituting this product into the heat equation and simplifying, separate functions of space and time yield separate equations which can then be solved, using the appropriate boundary and initial conditions.
This manageable approach simplifies the problem considerably but does require careful matching of the boundary conditions to each separate function, \( X(x) \) and \( T(t) \), to ensure a coherent solution that addresses the complexity of the original PDE.
In our given heat equation, this technique assumes the solution takes the form \( u(x, t) = X(x)T(t) \), with \( X(x) \) only a function of \( x \) and \( T(t) \) only a function of \( t \). By substituting this product into the heat equation and simplifying, separate functions of space and time yield separate equations which can then be solved, using the appropriate boundary and initial conditions.
This manageable approach simplifies the problem considerably but does require careful matching of the boundary conditions to each separate function, \( X(x) \) and \( T(t) \), to ensure a coherent solution that addresses the complexity of the original PDE.
Initial Value Problem
An initial value problem in the context of PDEs involves finding a solution that satisfies the equation at the initial moment, which is typically at time \( t=0 \). This is represented by the initial condition \( u(x, 0)=f(x) \) in the given exercise.
This condition provides the initial state of the system described by the PDE and is integral to the uniqueness and existence of the solution. When employing separation of variables method, the initial condition influences the temporal part of the solution, which will be shaped as \( T(t) \) and must align with the initial state function \( f(x) \).
In the case of the heat equation, the initial value problem describes the temperature distribution along the rod at the initial time, which then evolves according to the heat equation. The challenge is to ensure that the evolving solution remains constant at the specified initial condition, to model a realistic physical scenario.
This condition provides the initial state of the system described by the PDE and is integral to the uniqueness and existence of the solution. When employing separation of variables method, the initial condition influences the temporal part of the solution, which will be shaped as \( T(t) \) and must align with the initial state function \( f(x) \).
In the case of the heat equation, the initial value problem describes the temperature distribution along the rod at the initial time, which then evolves according to the heat equation. The challenge is to ensure that the evolving solution remains constant at the specified initial condition, to model a realistic physical scenario.
Heat Equation
The heat equation, \( \frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}} \), epitomizes a fundamental PDE in thermal physics, representing the distribution of heat (or temperature) in a given domain over time. The function \( u(x, t) \) denotes the temperature at a point \( x \) and time \( t \), and \( k \) is the thermal diffusivity constant of the material.
In the context of this problem, solving the heat equation entails calculating the temperature distribution along a rod, considering both the initial temperature distribution and how it changes at the boundaries as dictated by boundary conditions. Whether \( h > 0 \) or \( h < 0 \), these will impact the nature of the solution, representing different physical situations, such as an insulated rod versus one in contact with a medium at a different temperature.
Understanding the heat equation and its implications in various scenarios allows for a deeper comprehension of heat transfer mechanisms and the mathematical models that describe them.
In the context of this problem, solving the heat equation entails calculating the temperature distribution along a rod, considering both the initial temperature distribution and how it changes at the boundaries as dictated by boundary conditions. Whether \( h > 0 \) or \( h < 0 \), these will impact the nature of the solution, representing different physical situations, such as an insulated rod versus one in contact with a medium at a different temperature.
Understanding the heat equation and its implications in various scenarios allows for a deeper comprehension of heat transfer mechanisms and the mathematical models that describe them.
Other exercises in this chapter
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