Problem 15
Question
Solve the given boundary-value problem by an appropriate integral transform. Make assumptions about boundedness where necessary. Show that a solution of the BVP \(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0,-\infty < x < \infty, 0 < y < 1\) \(\left.\frac{\partial u}{\partial y}\right|_{y=0}=0, u(x, 1)=f(x),-\infty < x <\infty\) is \(u(x, y)=\frac{1}{\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty} f(t) \frac{\cosh \alpha y \cos \alpha(t-x)}{\cosh \alpha} d t d \alpha .\)
Step-by-Step Solution
Verified Answer
The solution is found using a Fourier transform, yielding the integral representation given.
1Step 1: Define the Problem
We are tasked with finding a solution to the boundary-value problem for the Laplace equation, \( \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 \), in the domain \( -\infty < x < \infty, 0 < y < 1 \), with boundary conditions \( \left.\frac{\partial u}{\partial y}\right|_{y=0}=0 \) and \( u(x, 1)=f(x) \) for \( -\infty < x < \infty \).
2Step 2: Recognize the Appropriate Transform
Observe that the boundary condition at \( y=0 \) suggests the use of a Fourier transform in the \( x \)-variable, which will help in simplifying the partial differential equation to an ordinary differential equation in the \( y \)-variable.
3Step 3: Apply the Fourier Transform
Take the Fourier transform of \( u(x, y) \) with respect to \( x \): \[ U(\alpha, y) = \int_{-\infty}^{\infty} u(x, y) e^{-i \alpha x} \, dx. \] Use this transform on the differential equation to get:\[ -\alpha^2 U(\alpha, y) + \frac{\partial^2 U}{\partial y^2}(\alpha, y) = 0. \]
4Step 4: Solve the Ordinary Differential Equation
The resulting transformed equation is: \[ \frac{\partial^2 U}{\partial y^2} - \alpha^2 U = 0. \]Solve this ordinary differential equation for \( U(\alpha, y) \), knowing the general solution is:\[ U(\alpha, y) = A(\alpha) \cosh(\alpha y), \]as the hyperbolic sine term doesn't satisfy the boundedness condition as \( y \to 0 \).
5Step 5: Apply Boundary Conditions in Transform Space
Apply the given boundary conditions in the transform space. For \( y=1 \), we have: \[ U(\alpha, 1) = \int_{-\infty}^{\infty} f(x) e^{-i \alpha x} \, dx = \hat{f}(\alpha). \]Thus, \[ A(\alpha) \cosh \alpha = \hat{f}(\alpha). \]
6Step 6: Substitute Back and Invert the Transform
Substitute the boundary condition solution into \( U(\alpha, y) \):\[ U(\alpha, y) = \frac{\hat{f}(\alpha)}{\cosh(\alpha)} \cosh(\alpha y). \]Now, invert the Fourier transform to find \( u(x, y) \):\[ u(x, y) = \int_{-\infty}^{\infty} \frac{\hat{f}(\alpha) \cosh(\alpha y)}{\cosh(\alpha)} e^{i \alpha x} \, d\alpha. \]
7Step 7: Express the Solution in Terms of the Given Function
Recognize that the inverse Fourier transform of \( \hat{f}(\alpha) \) is just the original function \( f(x) \), which leads us to the integral representation: \[ u(x, y) = \frac{1}{\pi} \int_{0}^{\infty} \int_{-\infty}^{\infty} f(t) \frac{\cosh(\alpha y) \cos(\alpha [t-x])}{\cosh(\alpha)} \, dt \, d\alpha. \]
Key Concepts
Boundary-Value ProblemFourier TransformOrdinary Differential EquationIntegral Transform
Boundary-Value Problem
Boundary-value problems (BVPs) are vital in the study of differential equations. They occur when a differential equation is supplemented by constraints, known as boundary conditions. For the Laplace equation, which is a second-order partial differential equation, BVPs are essential for ensuring solutions are not only determined but also applicable to real-world scenarios.
In BVPs, the solution must satisfy the differential equation within a domain and meet boundary conditions set on the domain's boundaries. In our exercise, for instance, the solution is demanded to hold the Laplace equation while satisfying specific boundary rules: \( \frac{\partial u}{\partial y} |_{y=0}=0 \) and \( u(x, 1)=f(x) \) for \( -\infty < x < \infty \).
This constraint ensures the solution is unique and satisfies the physical or theoretical context of the problem. BVPs frequently arise in heat conduction, fluid dynamics, and other physics domains. They are solved using various mathematical techniques, one prominent method being integral transforms.
In BVPs, the solution must satisfy the differential equation within a domain and meet boundary conditions set on the domain's boundaries. In our exercise, for instance, the solution is demanded to hold the Laplace equation while satisfying specific boundary rules: \( \frac{\partial u}{\partial y} |_{y=0}=0 \) and \( u(x, 1)=f(x) \) for \( -\infty < x < \infty \).
This constraint ensures the solution is unique and satisfies the physical or theoretical context of the problem. BVPs frequently arise in heat conduction, fluid dynamics, and other physics domains. They are solved using various mathematical techniques, one prominent method being integral transforms.
Fourier Transform
The Fourier transform is a powerful mathematical tool used to convert functions in the time or spatial domain into the frequency domain. It decomposes a function into an infinite sum of sines and cosines, which helps simplify the analysis and solution of differential equations.
This transform is particularly useful when dealing with periodic problems, as it translates a complex equation into one that might be simpler to address, often turning partial differential equations into ordinary differential equations. In the problem at hand, this tool is used on the function \( u(x, y) \) via the \( x \)-variable. The Fourier transform looks like this:
\[ U(\alpha, y) = \int_{-\infty}^{\infty} u(x, y) e^{-i \alpha x} \, dx. \]
Post-transform, our task is to solve a new, ordinary differential equation. Using the Fourier transform provides us with a clean mechanism to tackle BVPs efficiently, especially for equations defined over infinite or semi-infinite domains.
This transform is particularly useful when dealing with periodic problems, as it translates a complex equation into one that might be simpler to address, often turning partial differential equations into ordinary differential equations. In the problem at hand, this tool is used on the function \( u(x, y) \) via the \( x \)-variable. The Fourier transform looks like this:
\[ U(\alpha, y) = \int_{-\infty}^{\infty} u(x, y) e^{-i \alpha x} \, dx. \]
Post-transform, our task is to solve a new, ordinary differential equation. Using the Fourier transform provides us with a clean mechanism to tackle BVPs efficiently, especially for equations defined over infinite or semi-infinite domains.
Ordinary Differential Equation
Ordinary differential equations (ODEs) are equations involving functions of only one independent variable. The process of applying the Fourier transform in our problem simplifies our Laplace equation, a partial differential equation, into an ODE with respect to the \( y \)-variable.
Once transformed, the original problem reduces to solving the classical second-order linear ODE:
\[ \frac{\partial^2 U}{\partial y^2} - \alpha^2 U = 0. \]
The solution to this ODE relies on recognizing the features of hyperbolic trigonometric functions, such as the hyperbolic cosine \( \cosh(\alpha y) \). Since it satisfies the boundary conditions in terms of boundedness, it forms part of the ODE solution. Solving ODEs effectively brings us closer to unveiling the behavior of the original function \( u(x, y) \) and eventually to the solution for our entire boundary-value problem.
Once transformed, the original problem reduces to solving the classical second-order linear ODE:
\[ \frac{\partial^2 U}{\partial y^2} - \alpha^2 U = 0. \]
The solution to this ODE relies on recognizing the features of hyperbolic trigonometric functions, such as the hyperbolic cosine \( \cosh(\alpha y) \). Since it satisfies the boundary conditions in terms of boundedness, it forms part of the ODE solution. Solving ODEs effectively brings us closer to unveiling the behavior of the original function \( u(x, y) \) and eventually to the solution for our entire boundary-value problem.
Integral Transform
Integral transforms, like the Fourier transform, are techniques that convert differential equations into a form that is often simpler to solve. They work by integrating the product of the known function and a kernel function over a specific range.
The key advantage of integral transforms lies in their capacity to change the focus of the problem. Instead of dealing with a differential equation directly, the problem is shifted to solving a related but potentially simpler equation. For instance, using the Fourier transform in the boundary-value problem converts the Laplace equation into an ordinary differential equation, which is generally more straightforward when addressing classic boundary problems.
Integral transforms also aid in inversion processes, allowing us to revert back to the original function once a solution is found in the transform domain. This utility is particularly evident in our final solution expression for \( u(x, y) \) in our exercise. The dual nature of these transforms in both analysis and solution construction underlines their critical role in mathematical physics and engineering applications.
The key advantage of integral transforms lies in their capacity to change the focus of the problem. Instead of dealing with a differential equation directly, the problem is shifted to solving a related but potentially simpler equation. For instance, using the Fourier transform in the boundary-value problem converts the Laplace equation into an ordinary differential equation, which is generally more straightforward when addressing classic boundary problems.
Integral transforms also aid in inversion processes, allowing us to revert back to the original function once a solution is found in the transform domain. This utility is particularly evident in our final solution expression for \( u(x, y) \) in our exercise. The dual nature of these transforms in both analysis and solution construction underlines their critical role in mathematical physics and engineering applications.
Other exercises in this chapter
Problem 15
Find the cosine and sine integral representations of the given function. $$ f(x)=x e^{-2 x}, x>0 $$
View solution Problem 15
Use the Laplace transform to solve the heat equation \(u_{x x}=u_{t}, x>0, t>0\) subject to the given conditions. $$ \begin{aligned} &u(0, t)=f(t), \quad \lim _
View solution Problem 16
Use an appropriate Fourier integral transform to solve the given boundary- value problem. Make assumptions about boundedness where necessary. $$ \begin{aligned}
View solution Problem 16
Use the Laplace transform to solve the heat equation \(u_{x x}=u_{t}, x>0, t>0\) subject to the given conditions. $$ \left.\frac{\partial u}{\partial x}\right|_
View solution