Chapter 9: Problem 2
In each exercise, use representation (8) to obtain the solution of the two-
dimensional heat equation \(u_{t}(x, y, t)=\kappa\left[u_{x x}(x, y, t)+u_{y
y}(x, y, t)\right], 0
Short Answer
Expert verified
The final solution for u(x, y, t) is:
$$u(x, y, t) = 8 \sin \frac{\pi x}{a}\sin\frac{\pi y}{b4} e^{-\kappa (\pi^2/a^2 + \pi^2/b^2)t} - \sin\frac{2\pi x}{a}\sin\frac{3\pi y}{b} e^{-\kappa (4\pi^2/a^2 + 9\pi^2/b^2)t}$$
Here, u(x, y, t) is a combination of the eigenfunctions and the coefficients A_nm, which were determined from the given initial condition.
Step by step solution
01
Separate the Variables
To separate the variables, we assume the solution, \(u(x, y, t)\), has the form of \(X(x)Y(y)T(t)\). Plugging this into the heat equation gives us:
$$X(x)Y(y)T'(t) = \kappa [X''(x)Y(y)T(t)+ X(x)Y''(y)T(t)]$$
02
Obtain Ordinary Differential Equations (ODEs)
Divide both sides of the equation above by \(\kappa X(x)Y(y)T(t)\) to simplify the equation:
$$\frac{T'(t)}{\kappa T(t)} = \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)}$$
Notice that the left-hand side now only depends on \(t\), and the right-hand side only depends on \(x\) and \(y\). Thus, we can treat them separately with a constant:
$$\frac{T'(t)}{\kappa T(t)} = -\lambda^2 = \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)}$$
Here, we've let the separation constant be \(-\lambda^2\), where \(\lambda^2 > 0\) to obtain harmonic solutions.
03
Solve the ODEs for X and Y
Now, we have three ODEs to solve:
1. \(T'(t) = -\kappa \lambda^2 T(t)\)
2. \(X''(x) = -\lambda^2 X(x)\)
3. \(Y''(y) = \mu^2 Y(y)\)
Let's introduce a new constant, \(\mu^2\), such that \(\all{}-\lambda^2 = -\nu^2 - \mu^2\), where \(\nu^2 > 0\).
We can then solve each of the ODEs with the proper boundary conditions, which are \(X(0) = Y(0) = X(a) = Y(b) = 0\).
04
Find the Eigenvalues and Eigenfunctions
Solving the ODEs with their boundary conditions gives us the eigenvalues and eigenfunctions for \(X(x)\) and \(Y(y)\):
$$X_n(x) = \sin\frac{n\pi x}{a}, \quad\text{where}\quad n = 1, 2, \cdots$$
$$Y_m(y) = \sin\frac{m\pi y}{b}, \quad\text{where}\quad m = 1, 2, \cdots$$
The final solution for \(u(x, y, t)\) can then be written as a sum of products of the eigenfunctions. From the given initial condition, we can determine the coefficients required to solve the heat equation. In the current problem, the initial condition is given by:
$$f(x,y) = 8\sin\frac{\pi x}{a}\sin\frac{\pi y}{b} - \sin\frac{2\pi x}{a}\sin\frac{3\pi y}{b}$$
05
Expanding in Terms of Eigenfunctions
Now the task is to represent the initial condition as an expansion of product eigenfunctions functions:
$$f(x,y) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} A_{nm} \sin\frac{n\pi x}{a} \sin\frac{m\pi y}{b}$$
Here, \(A_{nm}\) can be found by taking the inner product of both sides with respect to both \(x\) and \(y\).
06
Calculate the Coefficients \(A_{nm}\)
The coefficients \(A_{nm}\) can be calculated using the following inner product:
$$A_{nm} = \frac{4}{ab}\int_0^a\int_0^b f(x, y) \sin\frac{n\pi x}{a} \sin\frac{m\pi y}{b} dxdy$$
By comparing the given initial condition with our expansion, we can easily find that:
$$A_{11}=8, \quad A_{23}=-1, \quad A_{nm}=0 \ \ \text{for other values of \(n\) and \(m\).}$$
07
Write the Final Solution using the Initial Condition
Combine everything back to get the expression of \(u(x, y, t)\) in terms of the eigenfunctions and the coefficients \(A_{nm}\):
$$u(x, y, t) = 8 \sin \frac{\pi x}{a}\sin\frac{\pi y}{b4} e^{-\kappa (\pi^2/a^2 + \pi^2/b^2)t} - \sin\frac{2\pi x}{a}\sin\frac{3\pi y}{b} e^{-\kappa (4\pi^2/a^2 + 9\pi^2/b^2)t}$$
The provided seven-step process yields the solution to the given two-dimensional heat equation with the given initial condition using the separation of variables method. It is important to note thet this solution is for the particular case of initial condition \(f(x, y)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separation of Variables
In the study of partial differential equations like the two-dimensional heat equation, the method of separation of variables is an essential tool. This method assumes that the solution can be written as a product of functions, each depending on only one of the variables involved. For our exercise, we assume that the solution is in the form of
This method works best in situations where the domain and boundary conditions are nice and regular, like rectangles or circles.
By dividing the transformed equation by the function of time, space, and constants, each side of the equation depends on only one variable and can therefore be separated.
In our example, once the variables are separated, each ODE can be individually analyzed and solved.
- \( u(x, y, t) = X(x)Y(y)T(t) \)
This method works best in situations where the domain and boundary conditions are nice and regular, like rectangles or circles.
By dividing the transformed equation by the function of time, space, and constants, each side of the equation depends on only one variable and can therefore be separated.
In our example, once the variables are separated, each ODE can be individually analyzed and solved.
Eigenvalues and Eigenfunctions
Once we separate variables in our equation, we arrive at several ordinary differential equations (ODEs). The solutions to these ODEs involve eigenvalues and eigenfunctions. These mathematical concepts arise naturally in problems dealing with linear operators, like our heat equation.
- Solving the spatial ODEs involves finding the eigenvalues, which are specific scalar values where a non-zero solution exists that satisfies the boundary conditions.
- The corresponding eigenfunctions are the solutions to the equation for these eigenvalues and often have simple forms, like sine or cosine functions.
- \(X''(x) = -\lambda^2 X(x)\)
- \(Y''(y) = \mu^2 Y(y)\)
- \(X_n(x) = \sin\frac{n\pi x}{a}\)
- \(Y_m(y) = \sin\frac{m\pi y}{b}\)
Boundary Conditions
Boundary conditions define how a function behaves at the boundary of its domain, significantly shaping the solution of differential equations.
In the given problem, the boundary conditions determine that the solution \(u(x,y,t)\) must vanish along the edges of the rectangle:
By ensuring the solution satisfies these constraints, the series expansion of the solution is uniquely defined. It essentially forces the eigenfunctions (e.g., sine functions in this rectangular domain) to meet these zero values at the boundaries.
Without the correct application of boundary conditions, the solution would not be accurate or applicable in the practical context of heat distribution within a fixed boundary.
In the given problem, the boundary conditions determine that the solution \(u(x,y,t)\) must vanish along the edges of the rectangle:
- \(x = 0, x = a\)
- \(y = 0, y = b\)
By ensuring the solution satisfies these constraints, the series expansion of the solution is uniquely defined. It essentially forces the eigenfunctions (e.g., sine functions in this rectangular domain) to meet these zero values at the boundaries.
Without the correct application of boundary conditions, the solution would not be accurate or applicable in the practical context of heat distribution within a fixed boundary.
Initial Condition Modeling
Initial condition modeling is just as important as boundary conditions since it involves specifying the state of the system at the starting time, \(t = 0\).
In our problem, the initial condition provides how heat is distributed over the domain initially:
Using the eigenfunctions obtained from the separation of variables, we express this initial condition as an expansion to find the coefficients required in our final solution.
Evaluating the initial conditions via eigenfunction expansions allows us to assemble the complete solution to the heat equation, fully considering both the temporally and spatially distributed nature of the heat across the given area.
In our problem, the initial condition provides how heat is distributed over the domain initially:
- \(u(x, y, 0) = f(x, y)\).
Using the eigenfunctions obtained from the separation of variables, we express this initial condition as an expansion to find the coefficients required in our final solution.
Evaluating the initial conditions via eigenfunction expansions allows us to assemble the complete solution to the heat equation, fully considering both the temporally and spatially distributed nature of the heat across the given area.