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Chapter 10: Problem 13

Find a formal solution to the initial-boundary value problem $$\frac{\partial u}{\partial t}=2 \frac{\partial^{2} u}{\partial x^{2}}+4 x, 0<\mathcal{x}<\pi,t>0,$$ $$u(0, t)=u(\pi, t)=0, \quad t>0,$$ $$u(x, 0)=\sin x,0<\mathcal{x}<\pi$$

Short Answer

Expert verified
Unfortunately, this problem doesn't have a general solution due to difficulties in solving the second ordinary differential equation. But we're able to illustrate how to analyze a given partial differential equation and how to apply the method of separation of variables and the boundary conditions.

Step by step solution

01

Identify the standard form

First, identify the standard form of the given equation. It is a heat equation, which is specified as a parabolic partial differential equation that describes the distribution of heat in a certain region over time. In this case, the equation is given by \(\frac{\partial u}{\partial t} = 2\frac{\partial^2 u}{\partial x^2} + 4x\).
02

Apply the separation of variables

Apply the separation of variables by letting \(u(x,t)=X(x)T(t)\). Substituting into the given partial differential equation results in the equation \(X''(x)T(t) = 0.5T'(t)X(x) + 2xT(t)X(x)\). This equation can be simplified by dividing both sides by \(X(x)T(t)\), which gives us \(\frac{T'(t)}{0.5T(t)} = \frac{X''(x) + 4xX(x)}{X(x)}\). Since the left-hand side is a function of \(t\) and the right-hand side is a function of \(x\), both sides must equal to a constant. We denote this constant as -\(\lambda^2\). Thus, we have two ordinary differential equations: \(\frac{T'(t)}{0.5T(t)} = -\lambda^2\) and \(\frac{X''(x) + 4xX(x)}{X(x)} = -\lambda^2\).
03

Solve the ordinary differential equations

From the first ordinary differential equation, we get \(T'(t) + \lambda^2 0.5T(t) = 0\), which is a first order linear differential equation. Its solution is given by \(T(t) = Ce^{-\lambda^2 0.5t}\), where \(C\) is a constant. The second equation can be rewritten as \(X''(x) + 4xX(x) + \lambda^2 X(x)= 0\). Unfortunately, this equation does not have a known method of solution except for special cases of \(\lambda^2\). This means, we are unable to derive a general solution.
04

Apply the boundary conditions

Next, apply the boundary conditions \(u(0,t) = u(\pi,t) = 0\) and \(u(x,0) = \sin(x)\) which gives us \(X(0)T(t) = X(\pi)T(t) = 0\) and \(X(x)T(0) = \sin(x)\). The first two conditions imply that \(X(0) = X(\pi) = 0\), and the last condition implies \(X(x) = \sin(x)\). However, due to the complicated nature of the second ordinary differential equation, we can't find a general solution.

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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 equations used to describe functions with multiple variables and their derivatives. They show how a certain condition described by a function changes when two or more variables change. In our exercise, we're looking at how heat distribution changes over time and position.
  • Function with Derivatives: PDEs involve variables and their partial derivatives, which means each derivative considers one variable at a time.
  • Describes Physical Phenomena: PDEs model real-world processes like wave movements, heat flow, and more.
  • Complexity Levels: Depending on how variables and derivatives are arranged, PDEs can vary in difficulty.
The heat equation given in the exercise is a type of partial differential equation. It helps determine how heat distributes along a rod over time.
Heat Equation
The heat equation is a popular PDE that describes how heat evolves over time within a region. Given its importance in physical sciences and engineering, understanding this equation is crucial for those fields.
  • Form: Typically expressed as \(\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}+ f(x)\). Here, \(u\) represents temperature, \(\alpha\) is the thermal diffusivity, and \(f(x)\) incorporates external heat sources.
  • Application: It's used to model heat distribution in various objects, like metals or fluids, ensuring efficient thermal management.
  • Importance: The equation helps predict how temperature changes, which is essential for designing engines, electronics, and many other technologies.
In the exercise, the equation reflects not just the heat conduction (the term \(2 \frac{\partial^2 u}{\partial x^2}\)) but also a constant heat source \(4x\), impacting how we solve for temperature distribution.
Separation of Variables
Separation of variables is a technique used to simplify solving PDEs. It lets us break down a complex problem into simpler components by assuming solutions can be represented as the product of functions, each in a single variable.
  • Method: Assume \(u(x,t) = X(x)T(t)\), separating effects over time and space.
  • Process: Substitute this assumption into the PDE to create two ordinary differential equations (ODEs), one for each function \(X\) and \(T\).
  • Requirement: The right side depends only on \(x\), and the left side only on \(t\), both must equal a constant to maintain equality.
This method was applied but couldn't fully solve the problem due to the complexity of the resulting ODEs.
Boundary Conditions
Boundary conditions are essential for determining the specific solutions to PDEs. They give us additional information that allows us to uniquely determine the solution to a boundary value problem.
  • Purpose: Ensure the solution meets the problem's physical constraints or characteristics at the boundaries (start or end points).
  • Types: Dirichlet (specifies function values at boundaries) and Neumann (specifies derivative values at boundaries).
  • In Our Exercise: Conditions like \(u(0, t) = u(\pi, t) = 0\) and \(u(x, 0) = \sin(x)\) help define initial and boundary behaviors of heat.
The conditions direct how to approach solving the equations, but complex situations can limit solution feasibility, as seen in this example.

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Most popular questions from this chapter

$$ \frac{\partial^{2} u}{\partial t^{2}} $$ \(=\frac{\partial^{2} u}{\partial x^{2}}+x \sin t\), \(0\)<\(x\)<\(\pi\), \(t>0\) $$ u(0, t) $$ \(=u(\pi, t)=0\) \(t>0\), $$ u(x, 0)=0 $$, \(0\)<\(x\)<\(\pi\), $$ \frac{\partial u}{\partial t}(x, 0) $$ \(=0\), \(0\)<\(x\)<\(\pi\)

In Section \(8.8,\) it was shown that the Legendre polynomials \(P_{n}(x)\) are orthogonal on the interval \([-1,1]\) with respect to the weight function \(w(x) \equiv 1 .\) Using the fact that the first three Legendre polynomials are $$P_{0}(x) \equiv 1, P_{1}(x)=x, P_{2}(x)=(3 / 2) x^{2}-(1 / 2)$$ find the first three coeffcients in the expansion $$f(x)=c_{0} P_{0}(x)+c_{1} P_{1}(x)+c_{2} P_{2}(x)+\cdots$$ where $$f(x)$$ is the function $$f(x) : \left\\{\begin{array}{cc}{-1,} & {-1< x < 0} \\ {1,} & {0 < x < 1}\end{array}\right.$$

Gibbs Phenomenon. Josiah Willard Gibbs, who was awarded the first American doctorate in engineering (Yale, 1863 , observed that near points of discontinuity of \(f\) , the partial sums of the Fourier series for \(f\) may over-shoot by approximately 9\(\%\) of the jump, regardless of for the function $$f(x)=\left\\{\begin{aligned}-1, &-\pi < x < 0 \\ 1, & 0 < x < \pi \end{aligned}\right.$$ whose Fourier series has the partial sums $$\begin{array}{r}{f_{2 n-1}(x)=\frac{4}{\pi}\left[\sin x+\frac{1}{3} \sin 3 x\right.} \\ {+\cdots+\frac{\sin (2 n-1) x}{(2 n-1)}}\end{array}$$ To verify this for \(f(x),\) proceed as follows: (a) Show that $$\begin{aligned} \pi(\sin x) f_{2 n-1}^{\prime}(x) &=4 \sin x[\cos x+\cos 3 x\\\\+\cdots+\cos (2 n-1) x ] \\ &=2 \sin 2 n x \end{aligned}$$ (b) Infer from part (a) and the figure that the maximum occurs at \(x=\pi /(2 n)\) and has the value $$f_{2 n-1}\left(\frac{\pi}{2 n}\right)=\frac{4}{\pi}\left[\sin \frac{\pi}{2 n}+\frac{1}{3} \sin \frac{3 \pi}{2 n}\right.$$ $$+\cdots+\frac{1}{2 n-1} \sin \frac{(2 n-1) \pi}{2 n} ]$$ (c) Show that if one approximates $$\int_{0}^{\pi} \frac{\sin x}{x} d x$$ each interval as the place to evaluate the integrand, then $$\begin{aligned} \int_{0}^{\pi} \frac{\sin x}{x} d x \approx \frac{\sin (\pi / 2 n)}{\pi / 2 n} & \frac{\pi}{n}+\cdots \\ &+\frac{\sin [(2 n-1) \pi / 2 n]}{(2 n-1) \pi / 2 n} \frac{\pi}{n} \\ &=\frac{\pi}{2} f_{2 n-1}\left(\frac{\pi}{2 n}\right) \end{aligned}$$ (d) Use the result of part (c) to show that the overshoot satisfies $$\lim _{n \rightarrow \infty} f_{2 n-1}\left(\frac{\pi}{2 n}\right)=\frac{2}{\pi} \int_{0}^{\pi} \frac{\sin x}{x} d x$$ (e) Using the result of part (d) and a numerical integration algorithm (such as Simpson's rule Appendix C) for the sine integral function $$S i(z) :=\int_{0}^{z} \frac{\sin x}{x} d x$$ show that \(\lim _{n \rightarrow \infty} f_{2 n-1}(\pi /(2 n)) \approx 1.18 .\) Thus, the approximations overshoot the true value of \(f\left(0^{+}\right)=1\) by \(0.18,\) or 9\(\%\) of the jump from \(f\left(0^{-}\right)\) to \(f\left(0^{+}\right) .\)

\(f(\theta)=\cos ^{2} \theta, \quad-\pi \leq \theta \leq \pi\)

$$\begin{array} { l l } { y ^ { \prime \prime } + \lambda y = 0 ; } & { 0 < x < \pi } \\ { y ^ { \prime } ( 0 ) = 0 , } & { y ( \pi ) = 0 } \end{array}$$
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