Chapter 9: Problem 26
Solve the one-dimensional diffusion equation subject to the following conditions: (a) \(c(0, t)=c(1, t)=0\), $$ c(x, 0)=\sin \pi x . $$ (b) \(c(0, t)=c(L, t)=0\), $$ c(x, 0)=\sin \left(\frac{2 \pi}{L} x\right)+\sin \left(\frac{3 \pi}{L} x\right) $$
Short Answer
Expert verified
For part (a), the solution is \(c(x, t) = \sin \pi x \ e^{ -\pi^2 D t} \) and for part (b), the solution is \(c(x, t) = \sin (\frac{2 \pi}{L}x ) \ e^{ - ( \frac{2 \pi}{L})^2 D t} + \sin (\frac{3 \pi}{L} x ) \ e^{- \frac{3^{2} \pi^{2}}{L^{2}} D t }\).
Step by step solution
01
Title - Understand the Diffusion Equation
The one-dimensional diffusion equation is given by \(\frac{\text{∂}c}{\text{∂}t} = D \frac{\text{∂}^2 c}{\text{∂}x^2}\), where \(D\) is the diffusion coefficient. Our goal is to find the function \(c(x, t)\) that satisfies this equation.
02
Title - Apply Separation of Variables
Assume a solution of the form \(c(x, t) = X(x)T(t)\). Substituting this into the diffusion equation, we get \(X(x)\frac{\text{d} T(t)}{\text{d} t} = D T(t) \frac{\text{d}^2 X(x)}{\text{d} x^2}\). Dividing both sides by \(X(x)T(t)\) gives \(\frac{1}{T(t)}\frac{\text{d} T(t)}{\text{d} t} = D \frac{1}{X(x)}\frac{\text{d}^2 X(x)}{\text{d} x^2} = -\beta\), where \(\beta\) is a separation constant.
03
Title - Solve the Temporal Part
The temporal part of the separation of variables yields \(\frac{\text{d} T(t)}{\text{d} t} = -\beta T(t)\). Solving this differential equation, we get \(T(t) = A e^{-\beta t}\), where \(A\) is a constant.
04
Title - Solve the Spatial Part
The spatial part of the separation of variables equation yields \(\frac{\text{d}^2 X(x)}{\text{d} x^2} = -\frac{\beta}{D} X(x)\). Solving this differential equation, we get \(X(x) = B \sin\bigg(\frac{\beta^{1/2}}{D^{1/2}}x\bigg) + C \cos\bigg(\frac{\beta^{1/2}}{D^{1/2}}x\bigg)\).
05
Title - Apply the Boundary Conditions for Part (a)
Given the boundary conditions for part (a), \(c(0, t) = 0\) and \(c(1, t) = 0\), we apply these to the spatial solution: \(X(0) = B \sin (0) + C \cos (0) = C = 0\) and \(X(1) = B\sin \bigg(\frac{\beta^{1/2}}{D^{1/2}}1\bigg) = 0\). For a non-trivial solution, \(\sin \bigg(\frac{\beta^{1/2}}{D^{1/2}}\bigg) = 0\), giving \(\beta = n^2 \pi^2 D\) where \(n\) is an integer.
06
Title - Form the General Solution for Part (a)
Considering the initial condition \(c(x, 0)\) for part (a), given \(c(x, 0) = \sin \pi x\), the general solution for part (a) becomes: \(c(x, t) = \sin \pi x \ e^{ -\pi^2 D t}\).
07
Title - Solve Boundary Conditions for Part (b)
Given the boundary conditions \(c(0, t) = 0\) and \(c(L, t) = 0\), using the same method as in previous steps, spatial solutions \(X(x)\) can be written in the form: \(k = n \frac{\pi}{L}\) where \(n\) must be integers.
08
Title - Form the General Solution for Part (b)
The initial condition in part (b) balances into these spatial solutions. So the solutions get value from given \(c(x, 0)=\sin (\frac{2 \pi}{L} x)+\sin (\frac{3 \pi}{L} x) \), giving: \(c(x, t) = \sin (\frac{2 \pi}{L}x) \ e^{ - (\frac{2 \pi}{L})^2 D t} + \sin (\frac{3 \pi}{L} x ) \ e^{- \frac{3^{2} \pi^{2}}{L^{2}} D t } \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
separation of variables
In solving partial differential equations like the one-dimensional diffusion equation, we can apply a method called 'separation of variables.' This technique simplifies complex problems by assuming that the solution can be separated into functions of each variable individually. For instance, we let the solution be in the form \( c(x,t) = X(x)T(t) \). Substituting this into our diffusion equation \(\frac{∂c}{∂t} = D \frac{∂^2 c}{∂x^2}\), we obtain an expression where each side of the equation depends on a different variable (either x or t). This allows us to separate the original equation into two simpler ordinary differential equations—one for time and one for space.
This method is generally useful in physics and engineering problems where boundary and initial conditions are well-defined, making the original partial differential equation easier to solve.
This method is generally useful in physics and engineering problems where boundary and initial conditions are well-defined, making the original partial differential equation easier to solve.
boundary conditions
Boundary conditions play a crucial role in solving the diffusion equation because they specify the behavior of the solution at the boundaries of the domain. For part (a) of our problem, we have boundary conditions: \(c(0, t) = 0\) and \(c(1, t) = 0\). These conditions mean that the concentration \(c\) must be zero at the ends of the interval at all times. When applying separation of variables, these boundary conditions help determine the form of the spatial solution \(X(x)\). For instance, the boundary condition \(X(0) = 0\) often results in removing the cosine term in the general solution. Similarly, for part (b), the boundary conditions are \(c(0, t) = 0\) and \(c(L, t) = 0\). These conditions ensure that the function \(X(x)\) satisfies zero concentration at both ends of the domain, L.
Thus, boundary conditions are essential in specifying unique solutions for differential equations and making sure they align with physical constraints.
Thus, boundary conditions are essential in specifying unique solutions for differential equations and making sure they align with physical constraints.
initial conditions
Initial conditions specify the state of the system at the beginning of the observation or experiment. These conditions are vital for determining the specific solution to our diffusion equation. For part (a) of the problem, the initial condition is \(c(x, 0) = \sin \pi x\). This tells us the starting concentration profile along the interval. When we solve for the temporal and spatial parts using separation of variables, our final solution must satisfy this initial condition. In part (b), the initial condition is \(c(x, 0) = \sin \left(\frac{2 \pi}{L} x\right) + \sin \left(\frac{3 \pi}{L} x\right)\). This condition offers a more complex start profile comprising multiple sine functions.
Both initial conditions are used to fully determine the constants and forms of the functions in our general solution. Without initial conditions, the solution would remain ambiguous or undefined within the context of the practical problem.
Both initial conditions are used to fully determine the constants and forms of the functions in our general solution. Without initial conditions, the solution would remain ambiguous or undefined within the context of the practical problem.
diffusion coefficient
The diffusion coefficient \(D\) is a parameter that quantifies the rate at which particles, energy, or other physical quantities spread. In the one-dimensional diffusion equation, \(\frac{∂c}{∂t} = D \frac{∂^2 c}{∂x^2}\), \(D\) controls how quickly the concentration changes over time. A higher diffusion coefficient indicates faster spreading. It units are \([ \text{𝑎𝑟𝑒𝑎} / \text{𝑡𝑖𝑚𝑒} ]\), reflecting how diffusion is taking place in the system.
When solving the diffusion equation with separation of variables, \(D\) appears in the spatial and temporal parts of the separated equations. Specifically, in \(\frac{∂^2 X(x)}{∂ x^2} = -\frac{\beta}{D} X(x)\) and \(\frac{∂ T(t)}{∂ t}= -\beta T(t)\), where \(\beta\) is a separation constant connected to \(D\).
The diffusion coefficient embodies the physical essence of diffusion in various fields like heat conduction, population dynamics, and chemical concentration.
When solving the diffusion equation with separation of variables, \(D\) appears in the spatial and temporal parts of the separated equations. Specifically, in \(\frac{∂^2 X(x)}{∂ x^2} = -\frac{\beta}{D} X(x)\) and \(\frac{∂ T(t)}{∂ t}= -\beta T(t)\), where \(\beta\) is a separation constant connected to \(D\).
The diffusion coefficient embodies the physical essence of diffusion in various fields like heat conduction, population dynamics, and chemical concentration.