Chapter 12: Problem 53
Solve the nonhomogeneous initial-boundary value problem.
$$
\begin{array}{l}
u_{t}=u_{x x}-6 x, \quad 0
Short Answer
Expert verified
The final solution of the given initial-boundary value problem for the one-dimensional heat equation is:
$$
u(x, t) = (3 + 2x - x^2) + \sum_{n=1}^{\infty} (A_n\cos{n\pi x} + B_n\sin{n\pi x})e^{-n^2\pi^2 t}
$$
where \(A_n\) and \(B_n\) are Fourier series coefficients obtained by solving the corresponding integrals. This solution represents the temperature distribution of the one-dimensional space in terms of position \(x\) and time \(t\).
Step by step solution
01
Identify the given PDE and boundary conditions
We are given the following PDE and boundary conditions:
$$
\begin{cases}
u_t = u_{xx} - 6x, & 0 < x < L, t > 0 \\
u(0, t) = 3, & t > 0 \\
u_x(1, t) = 2, & t > 0 \\
u(x, 0) = x^3 - x^2 + x + 3, & 0 \le x \le 1
\end{cases}
$$
Here, \(u(x,t)\) represents the temperature distribution of the one-dimensional space at position \(x\) and time \(t\). Our goal is to find the function \(u(x,t)\) that satisfies this PDE and boundary conditions.
02
Separate variables
Assume that the solution \(u(x,t)\) can be separated into the product of two functions: \(X(x)\) and \(T(t)\). So we can write \(u(x,t) = X(x)T(t)\).
Now, substitute this separated form into the PDE, and divide both sides by \(X(x)T(t)\):
$$
\frac{T'(t)}{T(t)} = \frac{X''(x) - 6xX(x)}{X(x)}
$$
03
Find the particular solution
Notice that the left side of the above equation is a function of time, whereas the right side is a function of position. Thus, both sides must be equal to a constant, which we will call \(-\lambda\):
$$
\begin{cases}
\frac{T'(t)}{T(t)} = -\lambda \\
\frac{X''(x)-6xX(x)}{X(x)} = -\lambda
\end{cases}
$$
To find the particular solution of \(u(x,t)\) we need to first solve the following boundary value problem:
$$
\begin{cases}
X''(x) - 6xX(x) = 0 \\
X(0) = 3 \\
X'(1) = 2
\end{cases}
$$
The general solution of this ODE for \(u_p(x,t)\) is given by \(u_p(x,t) = X(x) = Ax^2 + Bx + C\), with \(A, B, C\) being constants. By substituting the given boundary values we get:
$$
u_p(0) = 3 = C \\
u_p'(1) = 2 = 2A + B
$$
Let \(u_p(x,t) = X_p(x) = 3 + 2x - x^2\).
04
Subtract the particular solution
Now subtract the particular solution from the PDE and boundary conditions:
$$
\begin{cases}
(v_t = u_t - X_p'(x)) = u_{xx} \\
v(0, t) = (u(0, t) - 3) = 0 \\
v_x(1, t) = (u_x(1, t) - 2) = 0 \\
v(x, 0) = (u(x, 0) - X_p(x)) = x^3 - 2x^2 + 2x
\end{cases}
$$
05
Find the Fourier series coefficients
We can find the Fourier coefficients by solving the integrated equations for \(A_n\) and \(B_n\):
$$
A_n = 2\int_{0}^{1} (x^3 - 2x^2 + 2x)\cos{n\pi x}\, dx \\
B_n = 2\int_{0}^{1} (x^3 - 2x^2 + 2x)\sin{n\pi x}\, dx
$$
After solving the integrals, we obtain the coefficients \(A_n\) and \(B_n\).
06
Express v(x, t) as a Fourier series
Now express the function \(v(x,t)\) as a sum of Fourier series:
$$
v(x, t) = \sum_{n=1}^{\infty} (A_n\cos{n\pi x} + B_n\sin{n\pi x})e^{-n^2\pi^2 t}
$$
07
Add the particular solution
To find the final solution \(u(x, t)\), we add the particular solution we found in Step 3:
$$
u(x, t) = u_p(x, t) + v(x, t) = (3 + 2x - x^2) + \sum_{n=1}^{\infty} (A_n\cos{n\pi x} + B_n\sin{n\pi x})e^{-n^2\pi^2 t}
$$
This is the solution to the given initial-boundary value problem, representing the temperature distribution of the one-dimensional space as a function of position and time.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Differential Equations
Partial Differential Equations (PDEs) are mathematical expressions involving unknown multivariable functions and their partial derivatives. These equations are fundamental to describing various physical phenomena such as heat transfer, wave propagation, fluid dynamics, and more. The equation presented in our example, (u_t = u_{xx} - 6x), characterizes the temperature distribution within a one-dimensional rod over time, considering both the spatial second derivative (indicative of heat diffusion) and a non-homogeneous term (-6x), which could represent an external heat source.
PDEs can be classified based on the degree of the highest partial derivative involved as either linear or nonlinear, and as homogeneous or nonhomogeneous. The equation in our exercise is a second-order linear nonhomogeneous PDE. To solve such PDEs, one typically resorts to analytical methods, such as separation of variables or numerical techniques for more complex scenarios.
PDEs can be classified based on the degree of the highest partial derivative involved as either linear or nonlinear, and as homogeneous or nonhomogeneous. The equation in our exercise is a second-order linear nonhomogeneous PDE. To solve such PDEs, one typically resorts to analytical methods, such as separation of variables or numerical techniques for more complex scenarios.
Fourier Series
The Fourier series is a powerful mathematical tool used to represent periodic functions as an infinite sum of sine and cosine functions. Each term in the series is harmonically related to the original function. In the context of PDE solutions, Fourier series help in solving problems with periodic boundary conditions by decomposing the solution into simpler, sine and cosine terms.
Key to this approach is calculating the series coefficients, which are obtained by integrating the original function multiplied by the respective sine and cosine terms over one period. As shown in the solution steps, finding these coefficients involves solving integrals that represent the projection of the original function onto the basis functions of sines and cosines. This process allows the function to be reconstructed as closely as possible within the given boundary conditions.
Key to this approach is calculating the series coefficients, which are obtained by integrating the original function multiplied by the respective sine and cosine terms over one period. As shown in the solution steps, finding these coefficients involves solving integrals that represent the projection of the original function onto the basis functions of sines and cosines. This process allows the function to be reconstructed as closely as possible within the given boundary conditions.
Boundary Conditions
Boundary conditions are constraints necessary for the unique solution of PDEs. They describe the behavior of the solution at the edges of the domain, which in our case, is the physical spatial limits of the rod. There are several types of boundary conditions such as Dirichlet, Neumann, and Robin, each specifying different physical constraints.
In our exercise, the Dirichlet condition u(0, t) = 3 specifies the value of the solution at one end of the rod, while the Neumann condition u_x(1, t) = 2 prescribes the temperature gradient at the other end. Together, these conditions determine the uniqueness of the solution by anchoring the temperature distribution at the domain boundaries. Properly applying boundary conditions is essential for physical relevance and accuracy of the PDE solution.
In our exercise, the Dirichlet condition u(0, t) = 3 specifies the value of the solution at one end of the rod, while the Neumann condition u_x(1, t) = 2 prescribes the temperature gradient at the other end. Together, these conditions determine the uniqueness of the solution by anchoring the temperature distribution at the domain boundaries. Properly applying boundary conditions is essential for physical relevance and accuracy of the PDE solution.
Separation of Variables
Separation of variables is a technique used to solve PDEs by reducing them to a set of simpler, ordinary differential equations (ODEs). The method assumes that the PDE solution can be expressed as a product of functions, each depending on a single independent variable. In our problem, we assume u(x,t) = X(x)T(t), and then separate the PDE into two ODEs with respect to space and time.
After separating, one can solve for each function individually, with each ODE often requiring its own set of boundary conditions derived from the original PDE's boundary conditions. The end goal is to construct a solution to the original PDE by combining the solutions to the individual ODEs. This technique taps into the superposition principle and is especially effective for linear PDEs with separable variables and specific boundary conditions.
After separating, one can solve for each function individually, with each ODE often requiring its own set of boundary conditions derived from the original PDE's boundary conditions. The end goal is to construct a solution to the original PDE by combining the solutions to the individual ODEs. This technique taps into the superposition principle and is especially effective for linear PDEs with separable variables and specific boundary conditions.