Cauchy Problem and Wave Equations
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Table of Contents
D’Alembert solution of the Cauchy problem for the one-dimensional wave equation
Welcome to our comprehensive guide on the D’Alembert Wave Equation—an essential method for solving the one-dimensional wave equation via the Cauchy problem. In this article, we present a clear, step-by-step derivation that transforms complex mathematical concepts into accessible insights. Whether you are a student or a seasoned mathematician, our detailed explanation and illustrative derivations will enrich your understanding and problem-solving skills. Let’s embark on this mathematical journey together!
Step by Step solution for the one-dimensional wave equation
The one-dimensional wave equation is a fundamental partial differential equation (PDE) that models phenomena such as vibrations, sound propagation, and light waves. The Cauchy problem involves finding a solution that satisfies given initial conditions. In our case, the PDE is given by:
Problem:
\begin{align}
& u_{tt}=c^{2}u_{xx}
\end{align}
with the initial conditions:
\begin{align}
& u (x, 0) = f (x), x \in \mathbb{R} \\
& u_{t} (x, 0) = g (x), x \in \mathbb{R}
\end{align}
Step 1: Deriving the Characteristic Equations
The characteric equation is \begin{align} & dx^{2} − c^{2}dt^{2} = 0 \nonumber\\ \implies & (dx + c~dt)(dx − c~dt) = 0 \nonumber\\ \implies & dx + c~dt= 0 \text{ and } dx − c~dt = 0 \nonumber\\ \implies & x + ct= A \text{ and } x – ct = B \end{align} where \( A \) and \( B \) are arbitrary constants.
Step 2: Introducing New Variables
Let \begin{align} \xi = x + ct \text{ and } \eta = x – ct \end{align} By differentiating these variables with respect to \( x \) and \( t \), we obtain: \begin{align} \xi_{x} =1,~\xi_{t} =c \text{ and } \eta_{x} =1,~\eta_{t} =-c\nonumber\\ \\ \end{align}
Step 3: Transforming the Derivatives
Using the chain rule, the partial derivative \( u_{x} \) is expressed as: \begin{align} & u_{x} = u_{\xi}\xi_{x}+u_{\eta}\eta_{x} \nonumber\\ \implies & u_{x} = u_{\xi}+u_{\eta} \\ \implies & \frac{\partial}{\partial x} = \frac{\partial}{\partial \xi}+\frac{\partial}{\partial \eta} \end{align} Similarly, the partial derivative \( u_{t} \) becomes: \begin{align} & u_{t} = u_{\xi}\xi_{t}+u_{\eta}\eta_{t} \nonumber\\ \implies & u_{t} = c~u_{\xi}-c~u_{\eta} \\ \implies & \frac{\partial}{\partial t} = c~\frac{\partial}{\partial \xi}-c~\frac{\partial}{\partial \eta} \end{align} Proceeding further, we determine the second-order derivatives: \begin{align} u_{xx} & = \frac{\partial}{\partial x}\Big(u_{x}\Big) \nonumber\\ & = \frac{\partial}{\partial x}\Big(u_{\xi}+u_{\eta}\Big)~\text{using (7)} \nonumber\\ & = \Big( \frac{\partial}{\partial \xi}+\frac{\partial}{\partial \eta} \Big)\Big(u_{\xi}+u_{\eta}\Big)~\text{using (8)} \nonumber\\ & = u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta} \end{align} and \begin{align} u_{tt} & = \frac{\partial}{\partial t}\Big(u_{t}\Big) \nonumber\\ & = \frac{\partial}{\partial t}\Big(c~u_{\xi}-c~u_{\eta}\Big)~\text{using (9)} \nonumber\\ & = \Big( c~\frac{\partial}{\partial \xi}-c~\frac{\partial}{\partial \eta} \Big)\Big( c~u_{\xi}-c~u_{\eta} \Big)~\text{using (10)} \nonumber\\ & = c^{2}u_{\xi\xi}-2c^{2}u_{\xi\eta}+c^{2}u_{\eta\eta} \end{align}
Step 4: Simplifying the Wave Equation
Substitute the expressions for \( u_{xx} \) and \( u_{tt} \) into the original PDE (1): \begin{align} & c^{2}u_{\xi\xi}-2c^{2}u_{\xi\eta}+c^{2}u_{\eta\eta}=c^{2}u_{\xi\xi}+2c^{2}u_{\xi\eta}+c^{2}u_{\eta\eta} \nonumber\\ \implies & c^{2}u_{\xi\eta}=0 \nonumber\\ \implies & u_{\xi\eta}=0 ~\because c\ne 0\nonumber\\ \implies & u_{\xi}=\omega(\xi) ~\text{where } \omega=\omega(\xi) \nonumber\\ \implies & u=\int \omega(\xi) d\xi+\psi(\eta) ~\text{where } \psi=\psi(\eta) \nonumber\\ \implies & u(\xi,\eta)= \phi(\xi) +\psi(\eta) ~\text{where } \phi(\xi)=\int \omega(\xi) d\xi \nonumber\\ \implies & u(x,t)= \phi(x+ct) +\psi(x-ct) ~\text{using (5) }\nonumber\\ \\ \implies & u_{t}(x,t)= c~\phi^{\prime}(x+ct) -c~\psi^{\prime}(x-ct) \nonumber\\ \\ \end{align}
Step 5: Incorporating the Initial Conditions
To determine the arbitrary functions \( \phi \) and \( \psi \), we apply the initial conditions (2) and (3). Setting \( t=0 \) in the general solution yields: \begin{align} & u (x, 0) = f (x)\nonumber\\ \implies & \phi(x) +\psi(x)= f (x)~\text{using (13) }\nonumber\\ \\ \end{align} and \begin{align} & u_{t} (x, 0) = g (x)\nonumber\\ \implies & c~\phi^{\prime}(x) -c~\psi^{\prime}(x) = g (x)~\text{using (14) }\nonumber\\ \implies & c~\phi(x) -c~\psi(x) = \int^{x}_{x_{0}}g (x)\,dx+\text{Constant} \nonumber\\ \implies & \phi(x) -\psi(x) =\frac{1}{c} \int^{x}_{x_{0}}g (\tau)\,d\tau+D \nonumber\\ \\ \end{align} Adding (15) and (16), we get \begin{align} \phi(x)=\frac{1}{2}f (x)+\frac{1}{2c} \int^{x}_{x_{0}}g (\tau)\,d\tau+\frac{D}{2} \nonumber\\ \\ \end{align} Substracting (16) from (15), we get \begin{align} \psi(x)=\frac{1}{2}f (x)-\frac{1}{2c} \int^{x}_{x_{0}}g (\tau)\,d\tau-\frac{D}{2} \nonumber\\ \\ \end{align} where \( D \) is an integration constant.
Step 6: Assembling the Final Solution
Substituting the expressions for \( \phi \) and \( \psi \) back into our general solution (13) yields : \begin{align} & u(x,t)= \phi(x+ct) +\psi(x-ct) \nonumber\\ \implies & u(x,t)= \frac{1}{2}f (x+ct)+\frac{1}{2c} \int^{x+ct}_{x_{0}}g (\tau)\,d\tau+\frac{D}{2} +\frac{1}{2}f (x-ct)-\frac{1}{2c} \int^{x-ct}_{x_{0}}g (\tau)\,d\tau-\frac{D}{2}\nonumber\\ \implies & u(x,t)= \frac{1}{2}\Big[f (x+ct)+f (x-ct)\Big]+\frac{1}{2c}\left[ \int^{x+ct}_{x_{0}}g (\tau)\,d\tau-\int^{x-ct}_{x_{0}}g (\tau)\,d\tau\right] \nonumber\\ \implies & u(x,t)= \frac{1}{2}\Big[f (x+ct)+f (x-ct)\Big]+\frac{1}{2c} \int^{x+ct}_{x-ct}g (\tau)\,d\tau \nonumber\\ \end{align} This is the soulution of the d’Alembert solution of the Cauchy problem for the one-dimensional wave equation.
Conclusion
In conclusion, we have meticulously derived the D’Alembert solution for the one-dimensional wave equation. By introducing characteristic variables and systematically transforming the PDE, we reduced the problem to determining two arbitrary functions defined by the initial conditions. This derivation not only illuminates the theoretical underpinnings of wave phenomena but also serves as a practical tool for solving a broad range of physical problems.
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FAQs
Partial Differential Equations
- What is a partial differential equation (PDE)?
A PDE is an equation that involves unknown multivariable functions and their partial derivatives. It describes how the function changes with respect to multiple independent variables.
- How do PDEs differ from ordinary differential equations (ODEs)?
Unlike ODEs, which involve derivatives with respect to a single variable, PDEs involve partial derivatives with respect to two or more independent variables.
- What are the common types of PDEs?
PDEs are generally classified into three types based on their characteristics:
- Elliptic: e.g., Laplace’s equation
- Parabolic: e.g., the heat equation
- Hyperbolic: e.g., the wave equation
- What role do boundary and initial conditions play?
- Boundary conditions specify the behavior of the solution along the edges of the domain.
- Initial conditions are used in time-dependent problems to define the state of the system at the start.
- What methods are commonly used to solve PDEs?
There are several techniques, including:
- Analytical methods like separation of variables, Fourier and Laplace transforms, and the method of characteristics
- Numerical methods such as finite difference, finite element, and spectral methods
- What is the method of separation of variables?
This method assumes that the solution can be written as a product of functions, each depending on only one of the independent variables. This assumption reduces the PDE to a set of simpler ODEs that can be solved individually.
- In which fields are PDEs applied?
PDEs model a wide range of phenomena across various fields including physics (heat transfer, fluid dynamics), engineering (stress analysis, electromagnetics), finance (option pricing models), and more.
- What distinguishes linear from nonlinear PDEs?
- Linear PDEs have terms that are linear with respect to the unknown function and its derivatives, making them more tractable analytically.
- Nonlinear PDEs include terms that are nonlinear, often leading to complex behaviors and requiring specialized methods for solution.
- How do you determine the order of a PDE?
The order of a PDE is defined by the highest derivative (partial derivative) present in the equation. For example, if the highest derivative is a second derivative, the PDE is second order.
- What are some common challenges in solving PDEs?
Challenges include finding closed-form analytical solutions, handling complex geometries and boundary conditions, and the significant computational effort required for accurate numerical solutions.
