Problem 2 - Arbeitsbuch Mathematik für Ingenieure

Question

Lösen Sie das Dirichlet-Problem $$ \begin{aligned} \Delta u &=0 & \text { für } x \in(0,1), y \in(0,1) \\ u(0, y) &=0, & u(1, y)=\sin (2 \pi y) & \text { für } y \in(0,1) \\ u(x, 0) &=0, \quad u(x, 1)=0 & \text { für } x \in(0,1) \end{aligned} $$ mit Hilfe eines Produktansatzes.

Step-by-Step Solution

Answer

The solution is: \(u(x, y) = \sin(2 \pi x) \sinh(2 \pi y)\).

Step 1: Understand the Problem

We are solving the Dirichlet problem for the Laplace equation \(abla^2 u = 0\) over the region \(0 < x < 1\), \(0 < y < 1\) with given boundary conditions.

Step 2: Assume Solution Form

Assume that the solution can be written as a product of two functions, one in terms of \(x\) and one in terms of \(y\): \(u(x, y) = X(x)Y(y)\).

Step 3: Substitute into Laplace Equation

Substitute \(u(x, y) = X(x)Y(y)\) into the Laplace equation \(abla^2 u = 0\).This gives:\( X''(x)Y(y) + X(x)Y''(y) = 0\).

Step 4: Separate Variables

Divide both sides by \(X(x)Y(y)\):\(\frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0\).This must equal a constant, \( -\beta\). Hence:\( X''(x) = -\beta X(x)\) and \(Y''(y) = \beta Y(y)\).

Step 5: Solve the Differential Equations

Solve the ordinary differential equations obtained earlier:\( X''(x) = -\beta X(x)\) yields \(X(x) = C_1 \cos(\beta x) + C_2 \sin(\beta x)\).\(Y''(y) = \beta Y(y)\) yields \(Y(y) = D_1 e^{\beta y} + D_2 e^{-\beta y}\).

Step 6: Apply Boundary Conditions at y = 0 and y = 1

Apply boundary conditions: \(u(x, 0) = 0\) and \(u(x, 1) = 0\).This imposes conditions on \(Y(y)\):1. \(Y(0) = 0\Rightarrow D_1 + D_2 = 0\)2. \(Y(1) = 0\Rightarrow D_1 e^{\beta} + D_2 e^{-\beta} = 0\).These form a system of equations with nontrivial solutions when \(\beta = n \pi\) for \(n = 1, 2, 3, \ldots\).

Step 7: Solve for X(x) using Boundary Conditions at x = 0 and x = 1

Apply boundary conditions: \(u(0, y) = 0\) and \(u(1, y) = \sin(2 \pi y)\).For non-trivial solutions, use \(\beta = 2 \pi\). Then \(X(x) = C_1 \cos(2 \pi x) + C_2 \sin(2 \pi x)\).Since \(u(0, y) = 0\), we get:\(C_1 = 0\). Using \(u(1, y) = \sin(2 \pi y)\), \(C_2 = 1\).

Step 8: Construct the Complete Solution

With \(C_1 = 0\) and \(C_2 = 1\), \(X(x) = \sin(2 \pi x)\).Since we have \(Y(y) = e^{2 \pi y} - e^{-2 \pi y} = \sinh(2 \pi y)\), the complete solution is: \(u(x, y) = \sin(2 \pi x) \sinh(2 \pi y)\).

Key Concepts

Laplace equation

The Laplace equation, represented as \(abla^2 u = 0\), is a fundamental PDE (partial differential equation) in mathematical physics. It expresses that the divergence of the gradient of a function is zero. This equation arises in contexts like electrostatics, fluid dynamics, and potential theory. For this particular problem, the Laplace equation describes the behavior of the function \(u(x, y)\) within a given region. Solving this equation helps us understand how the function \(u\) varies based on the boundary conditions imposed on \(u\).
Every term in the Laplace equation needs to be zero when summed, reflecting a steady-state scenario without any internal sources or sinks. This foundational concept is crucial for solving a variety of physical problems.
The process of solving the Laplace equation typically involves finding a function \(u\) that meets certain boundary conditions (values of the function at the edges of the region of interest). These boundary conditions are crucial to determine the exact form of the solution.

boundary conditions

Boundary conditions specify the behavior of a solution \(u\) on the boundaries of the domain. In our problem, we have:

• \(u(0, y) = 0\)
• \(u(1, y) = \sin(2 \pi y)\)
• \(u(x, 0) = 0\)
• \(u(x, 1) = 0\)

These conditions ensure that the solution is unique and correspond to fixed values on the edges of our rectangular region \(0 < x < 1, \ 0 < y < 1\). By applying these constraints, we can narrow down the potential forms of the solution so that it fits perfectly within the specified boundaries.
For example, the boundary condition \(u(1, y) = \sin(2 \pi y)\) means that at \(x = 1\), the solution must match the \(sin(2 \pi y)\) function across the \y\ axis. This kind of information is critical for constructing an accurate solution using methods like separation of variables.

separation of variables

Separation of variables is a powerful method used to solve partial differential equations like the Laplace equation. This technique assumes that the solution can be written as the product of functions, each depending only on one of the independent variables. Here, we assumed \(u(x, y) = X(x)Y(y)\).
By substituting this form into the Laplace equation, the PDE is transformed into two ordinary differential equations (ODEs), each depending on one variable, making the problem simpler to solve.
The separation process works as follows:

• Substitute the product form into the PDE: \(X''(x)Y(y) + X(x)Y''(y) = 0\)
• Divide both sides by \(X(x)Y(y)\): \(\frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0\)

This must be equal to a constant, generally denoted as \(-\beta\). Thus we get two ODEs, one for each variable. This method simplifies complex PDEs by reducing them into more manageable ODE forms.

ordinary differential equations

Ordinary differential equations (ODEs) arise when we separate variables in the context of solving PDEs. For the Laplace equation, we obtain two ODEs:

• \(X''(x) = -\beta X(x)\)
• \(Y''(y) = \beta Y(y)\)

Solving these ODEs provides us with general solutions:

• For \(X(x)\): \(C_1 \cos(\beta x) + C_2 \sin(\beta x)\)
• For \(Y(y)\): \(D_1 e^{\beta y} + D_2 e^{-\beta y}\)

The constants \(C_1, C_2, D_1, \text{and} D_2\) are found using boundary conditions, ensuring that the solution fits the original problem's constraints perfectly. In our case, we determined that non-trivial solutions exist when \(\beta = 2 \pi\). The solution for \(X(x)\) and \(Y(y)\) then leads us to the complete solution: \(u(x, y) = \sin(2 \pi x) \sinh(2 \pi y)\).