Chapter 2: Problem 2
Find solutions of the exponential type \(e^{\alpha x+\beta y}\) for the equations (a) \(u_{x x}-\frac{1}{c^{2}} u_{t t}=0\), (b) \(u_{x x}-\frac{1}{k} u_{t}=0\) (c) \(u_{x x}+u_{y y}=0\).
Short Answer
Expert verified
(a) \( u(x, t) = e^{\pm \frac{\beta}{c} x + \beta t} \), (b) \( u(x, t) = e^{\alpha x + k\alpha^2 t} \), (c) \( u(x, y) = e^{i\beta x + \beta y} \).
Step by step solution
01
Analyze part (a)
Consider the given equation \( u_{xx} - \frac{1}{c^2} u_{tt} = 0 \) and look for solutions of the form \( u(x, t) = e^{\alpha x + \beta t} \). Calculate the second derivatives: \( u_{xx} = \alpha^2 e^{\alpha x + \beta t} \) and \( u_{tt} = \beta^2 e^{\alpha x + \beta t} \). After substituting these into the original equation, we obtain \( \alpha^2 e^{\alpha x + \beta t} - \frac{1}{c^2} \beta^2 e^{\alpha x + \beta t} = 0 \). This simplifies to \( \alpha^2 - \frac{1}{c^2} \beta^2 = 0 \).
02
Solve for part (a) parameters
From \( \alpha^2 - \frac{1}{c^2} \beta^2 = 0 \), solve for \( \alpha \) and \( \beta \). This equation implies \( \alpha^2 = \frac{1}{c^2} \beta^2 \) or \( \alpha = \pm \frac{\beta}{c} \). Solutions for (a) are of the form \( u(x, t) = e^{\pm \frac{\beta}{c} x + \beta t} \).
03
Analyze part (b)
Consider the equation \( u_{xx} - \frac{1}{k} u_t = 0 \). Assume a solution of form \( u(x, t) = e^{\alpha x + \beta t} \). Calculating derivatives gives \( u_{xx} = \alpha^2 e^{\alpha x + \beta t} \) and \( u_t = \beta e^{\alpha x + \beta t} \). Substituting into the equation results in \( \alpha^2 e^{\alpha x + \beta t} - \frac{1}{k} \beta e^{\alpha x + \beta t} = 0 \). Simplifying gives \( \alpha^2 - \frac{1}{k} \beta = 0 \).
04
Solve for part (b) parameters
From \( \alpha^2 - \frac{1}{k} \beta = 0 \), solve for \( \alpha \) and \( \beta \). This implies \( \beta = k\alpha^2 \). Solutions for (b) are of the form \( u(x, t) = e^{\alpha x + k\alpha^2 t} \).
05
Analyze part (c)
Consider the equation \( u_{xx} + u_{yy} = 0 \) and assume a solution of form \( u(x, y) = e^{\alpha x + \beta y} \). Calculating second derivatives gives \( u_{xx} = \alpha^2 e^{\alpha x + \beta y} \) and \( u_{yy} = \beta^2 e^{\alpha x + \beta y} \). Substitute these into the equation to obtain \( \alpha^2 e^{\alpha x + \beta y} + \beta^2 e^{\alpha x + \beta y} = 0 \). This simplifies to \( \alpha^2 + \beta^2 = 0 \).
06
Solve for part (c) parameters
From \( \alpha^2 + \beta^2 = 0 \), solve for \( \alpha \) and \( \beta \). Since both squared terms must be negative, choose \( \alpha = i\beta \) or \( \beta = i\alpha \). Solutions for (c) are of the form \( u(x, y) = e^{i\beta x + \beta y} \) or \( u(x, y) = e^{\alpha x + i\alpha y} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wave Equation
The wave equation is a classic example of a hyperbolic partial differential equation (PDE). It models the distribution of waves, such as sound waves or electromagnetic waves, as they travel through a medium. The standard form of the wave equation is given by:
\[ u_{xx} - \frac{1}{c^2} u_{tt} = 0 \]
where \(u(x, t)\) represents the wave function,\( c \) is the speed of the wave, \( u_{xx} \) is the second spatial derivative, and \( u_{tt} \) is the second time derivative.
\[ u(x, t) = e^{\pm \frac{\beta}{c} x + \beta t} \]
These solutions describe waves moving in positive and negative directions along the x-axis.
\[ u_{xx} - \frac{1}{c^2} u_{tt} = 0 \]
where \(u(x, t)\) represents the wave function,\( c \) is the speed of the wave, \( u_{xx} \) is the second spatial derivative, and \( u_{tt} \) is the second time derivative.
- For solutions in the form of exponentials, substitute \(u(x, t) = e^{\alpha x + \beta t}\) into the equation.
- Calculate the respective second derivatives \(u_{xx} = \alpha^2 e^{\alpha x + \beta t}\) and \(u_{tt} = \beta^2 e^{\alpha x + \beta t}\).
- This substitution transforms the PDE into an algebraic equation: \(\alpha^2 - \frac{1}{c^2} \beta^2 = 0\).
\[ u(x, t) = e^{\pm \frac{\beta}{c} x + \beta t} \]
These solutions describe waves moving in positive and negative directions along the x-axis.
Heat Equation
The heat equation is a fundamental example of a parabolic PDE used to describe the distribution of heat (or variation in temperature) in a given region over time. The standard form of the heat equation is:
\[ u_{xx} - \frac{1}{k} u_t = 0 \]
where \(u(x, t)\) represents the temperature distribution,\( k \) denotes the thermal diffusivity, \( u_{xx} \) is the second spatial derivative, and \( u_t \) the first time derivative.
\[ u(x, t) = e^{\alpha x + k\alpha^2 t} \]
This result shows how temperature evolves with time, emphasizing the heat diffusion aspect of the problem.
\[ u_{xx} - \frac{1}{k} u_t = 0 \]
where \(u(x, t)\) represents the temperature distribution,\( k \) denotes the thermal diffusivity, \( u_{xx} \) is the second spatial derivative, and \( u_t \) the first time derivative.
- The work seeks out solutions of the form \(u(x, t) = e^{\alpha x + \beta t}\).
- By substituting into the heat equation, it reduces to: \(\alpha^2 e^{\alpha x + \beta t} - \frac{1}{k} \beta e^{\alpha x + \beta t} = 0\).
- Simplifying the expression leads to \(\alpha^2 = \frac{1}{k} \beta\).
\[ u(x, t) = e^{\alpha x + k\alpha^2 t} \]
This result shows how temperature evolves with time, emphasizing the heat diffusion aspect of the problem.
Laplace's Equation
Laplace's equation is a pivotal second-order elliptic PDE appearing in many areas such as electrostatics, fluid dynamics, and potential theory. It is given by:
\[ u_{xx} + u_{yy} = 0 \]
where \(u(x, y)\) is the unknown function, \( u_{xx} \) and \( u_{yy} \) are the second partial derivatives relative to \(x\) and \(y\), respectively.
\[ u(x, y) = e^{i\beta x + \beta y} \] or \[ u(x,y) = e^{\alpha x + i\alpha y} \]
These solutions illustrate harmonic functions—functions that exhibit no change over a given distance or angle.
\[ u_{xx} + u_{yy} = 0 \]
where \(u(x, y)\) is the unknown function, \( u_{xx} \) and \( u_{yy} \) are the second partial derivatives relative to \(x\) and \(y\), respectively.
- The target is to find solutions of the form \(u(x, y) = e^{\alpha x + \beta y}\).
- Upon computing second derivatives and substituting into the equation, one obtains: \(\alpha^2 e^{\alpha x + \beta y} + \beta^2 e^{\alpha x + \beta y} = 0\).
- This simplifies to \(\alpha^2 + \beta^2 = 0\).
\[ u(x, y) = e^{i\beta x + \beta y} \] or \[ u(x,y) = e^{\alpha x + i\alpha y} \]
These solutions illustrate harmonic functions—functions that exhibit no change over a given distance or angle.
Exponential Solutions
Exponential solutions are a powerful technique often used to solve differential equations, including PDEs. They assume solutions in forms similar to \(e^{\alpha x + \beta y}\), which simplifies solving the equations by transforming PDEs into algebraic equations. Let's explore its relevance in different contexts:
- For wave equations, exponentials portray constant-amplitude waves traveling through space and time.
- In the heat equation's context, exponential solutions represent changing temperatures diffusing through a medium.
- For Laplace's equation, these solutions often deal with boundary problems in codifying steady states or potential fields.