Problem 4
Question
The solution of Problem 3 can be written $$u(x, t)=\frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} d \alpha-\frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-k \alpha^{2} t} d \alpha.$$ Using $$\int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} d \alpha=\pi / 2$$ the last line becomes $$u(x, t)=u_{0}-\frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-k \alpha^{2} t} d \alpha.$$
Step-by-Step Solution
Verified Answer
Apply the given integral to simplify: final solution is \( u(x, t) = u_{0} - \frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-k \alpha^{2} t} \, d \alpha \).
1Step 1: Understand the Problem
We have the solution for a problem in the form of an integral to evaluate, specifically focusing on functions depending on two parameters: space \(x\) and time \(t\). We aim to simplify the expression using a known integral.
2Step 2: Apply Given Integral
Use the given result \( \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} \, d \alpha = \frac{\pi}{2} \) to simplify the first integral in the solution expression for \(u(x, t)\).
3Step 3: Simplify Original Expression
The original expression for \(u(x, t)\) is: \[u(x, t)=\frac{2 u_{0}}{\pi} \left( \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} \, d \alpha \right) - \frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-k \alpha^{2} t} \, d \alpha.\] Applying the given result, the expression becomes: \[u(x, t)=\frac{2 u_{0}}{\pi} \cdot \frac{\pi}{2} - \frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-k \alpha^{2} t} \, d \alpha.\]
4Step 4: Simplify Coefficients
Evaluate the constant expression \( \frac{2 u_{0}}{\pi} \cdot \frac{\pi}{2} \) which simplifies as follows:\[ \frac{2 u_{0}}{\pi} \cdot \frac{\pi}{2} = u_{0}. \]Thus, the expression for \(u(x, t)\) becomes: \[ u(x, t) = u_{0} - \frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-k \alpha^{2} t} \, d \alpha. \]
5Step 5: Validate the Final Expression
The final expression for \(u(x, t)\) matches the given form. We have simplified the original integral correctly using the known result, and then re-evaluated the constant to obtain the final form.
Key Concepts
Integral CalculusPartial Differential EquationsBoundary Value Problems
Integral Calculus
Integral calculus is an area of mathematics that helps us to calculate areas under curves, total accumulations, and solve problems involving continuous change. In this exercise, we deal with integrals that involve trigonometric functions like \( \sin(\alpha x) \), making our integral a bit more complex.
Using integral calculus, we evaluate two key expressions, each integrating from zero to infinity with respect to parameter \( \alpha \). The first important integral involves the formula \( \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} \ d\alpha = \frac{\pi}{2} \). This tells us the area under the curve of the function \( \frac{\sin \alpha x}{\alpha} \) across a certain range. This result is crucial in simplifying the original expression for \( u(x, t) \).
By applying the result of this known integral, we effectively reduce the complexity of the integral expression in the solution process of the exercise.
Using integral calculus, we evaluate two key expressions, each integrating from zero to infinity with respect to parameter \( \alpha \). The first important integral involves the formula \( \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} \ d\alpha = \frac{\pi}{2} \). This tells us the area under the curve of the function \( \frac{\sin \alpha x}{\alpha} \) across a certain range. This result is crucial in simplifying the original expression for \( u(x, t) \).
By applying the result of this known integral, we effectively reduce the complexity of the integral expression in the solution process of the exercise.
- The formula allows for the calculation of indefinite integrals where the antiderivative is hard to find.
- It's essential for solving various integral problems in physics and engineering.
Partial Differential Equations
Partial Differential Equations (PDEs) are equations that involve multiple variables and their partial derivatives. They are fundamental in describing the behavior of physical systems across multiple dimensions.
In our exercise, the function \( u(x, t) \) represents a solution influenced by both spatial \( x \) and temporal \( t \) variables, typical of PDEs. Understanding PDEs requires recognizing how changes in space and time affect the behavior of a system—whether it's temperature distribution, wave propagation, or other phenomena.
In context of the Fourier Transform, PDEs often simplify or transform into easier problems that allow us to analyze or solve systems efficiently. Specifically, the term \( e^{-k\alpha^2 t} \) hints at diffusion processes, showcasing the rate at which influences dissipate over time. Here, the partial derivative concerning \( t \) points towards heat or wave equation solutions.
In our exercise, the function \( u(x, t) \) represents a solution influenced by both spatial \( x \) and temporal \( t \) variables, typical of PDEs. Understanding PDEs requires recognizing how changes in space and time affect the behavior of a system—whether it's temperature distribution, wave propagation, or other phenomena.
In context of the Fourier Transform, PDEs often simplify or transform into easier problems that allow us to analyze or solve systems efficiently. Specifically, the term \( e^{-k\alpha^2 t} \) hints at diffusion processes, showcasing the rate at which influences dissipate over time. Here, the partial derivative concerning \( t \) points towards heat or wave equation solutions.
- PDEs can describe various natural phenomena.
- They provide a rich framework for modeling continuous systems in engineering and physics.
Boundary Value Problems
Boundary Value Problems (BVPs) involve finding a solution to differential equations with specific conditions set at the boundaries. In practice, these 'boundaries' translate to conditions like fixed temperatures or restrictions on waves at certain points.
Our exercise suggests \( u(x, t) \) satisfies some conditions at the limits of \( x \) or \( t \). Boundary conditions can be essential for determining specific solutions to differential equations, guiding how physical systems must behave at the edges.
For instance, transforming an integral using known boundary conditions simplifies computation and aids in obtaining exact solutions in real-world scenarios.
Our exercise suggests \( u(x, t) \) satisfies some conditions at the limits of \( x \) or \( t \). Boundary conditions can be essential for determining specific solutions to differential equations, guiding how physical systems must behave at the edges.
For instance, transforming an integral using known boundary conditions simplifies computation and aids in obtaining exact solutions in real-world scenarios.
- Boundary conditions personalize a general solution to meet specific physical situations.
- They help ensure that mathematical solutions are both practical and realistic.
Other exercises in this chapter
Problem 3
From formulas (5) and (6) in the text $$\begin{aligned} A(\alpha) &=\int_{0}^{3} x \cos \alpha x d x=\left.\frac{x \sin \alpha x}{\alpha}\right|_{0} ^{3}-\frac{
View solution Problem 3
By the sifting property, $$\mathscr{F}\\{\delta(x)\\}=\int_{-\infty}^{\infty} \delta(x) e^{i \alpha x} d x=e^{i \alpha 0}=1$$
View solution Problem 4
We already know that \(f * \delta=\delta * f .\) Then, by the sifting property, $$(f * \delta)(x)=\int_{\infty}^{\infty} f(\tau) \delta(x-\tau) d \tau=\int_{-\i
View solution Problem 5
Using the Fourier sine transform we find $$U(\alpha, t)=c e^{-k \alpha^{2} t}.$$ Now $$\mathscr{F}_{S}\\{u(x, 0)\\}=U(\alpha, 0)=\int_{0}^{1} \sin \alpha x d x=
View solution