Question

In Exercises \(1-16\) apply the definition developed in Example 1 to solve the boundary value problem. (Use Theorem 11.3 .5 where it applies.) Where indicated by \(\mathrm{C}\), graph the surface \(u=u(x, y), 0 \leq x \leq a\), \(0 \leq y \leq b\) $$ \begin{array}{l} u_{x x}+u_{y y}=0, \quad 0

Short answer

Question: Determine the final solution to the given boundary value problem of the Laplace Equation \(u_{xx} + u_{yy} = 0\) with boundary conditions \(u(x, 0) = x^{2}(2-x), u(x, 3) = 0, u(0, y) = 0, \text{ and } u(2, y) = 0\). Answer: The final solution to the given boundary value problem is: $$ u(x,y) = \sum_{n=1}^\infty B_n\sin n\pi x(C_n\cosh n\pi y + D_n\sinh n\pi y) $$ where the coefficients \(B_n\), \(C_n\), and \(D_n\) are determined by applying the boundary condition \(u(x, 3) = 0\).

Step-by-step solution

1. Substitute and separate the PDE

First, rewrite the given equation with the variables \(u(x,y)=X(x)Y(y)\): $$ (X''(x)Y(y) + X(x)Y''(y))=0 $$ Now, divide both sides by \(X(x)Y(y)\) to get: $$ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 $$

2. Solve the separated ODEs

Notice that the left side depends only on x and the right side depends only on y. So, we can set each side to a constant, say \(-k^2\): $$ \frac{X''(x)}{X(x)} = -k^2, \quad \frac{Y''(y)}{Y(y)} = k^2 $$ Now, we have two ordinary differential equations (ODEs): $$ X''(x) + k^2X(x) = 0 $$ $$ Y''(y) - k^2Y(y) = 0 $$

3. Determine the general solution of \(u(x,y)\)

Solve the first ODE for \(X(x)\): $$ X(x) = A\cos kx + B\sin kx $$ Solve the second ODE for \(Y(y)\): $$ Y(y) = C\cosh ky + D\sinh ky $$ Now the general solution of \(u(x,y)\) is: $$ u(x,y) = (A\cos kx + B\sin kx)(C\cosh ky + D\sinh ky) $$

4. Apply the boundary conditions

We are given the following boundary conditions: $$ u(x, 0) = x^{2}(2-x), \quad u(x, 3) = 0, \quad u(0, y) = 0, \quad u(2, y) = 0 $$ First, use the condition \(u(x,0) = x^2 (2-x)\): $$ u(x,0) = (A\cos kx + B\sin kx)(C) $$ Next, use the condition \(u(0,y) = 0\): $$ u(0,y) = (A)(C\cosh ky) = 0$$ Since \(C\) can't be zero (otherwise the solution would be zero), we have \(A=0\). Now, the general solution becomes: $$ u(x,y) = B\sin kx(C\cosh ky + D\sinh ky) $$ Next, use the condition \(u(2, y) = 0\): $$ u(2,y) = B\sin (2k)(C\cosh ky + D\sinh ky) = 0 $$ Since \(u(x,y)\) cannot always be zero, we have \(\sin (2k)=0\) which means \(k=n\pi\) for \(n\in\mathbb{Z}\). Now, we rewrite the general solution as: $$ u_n(x,y) = B_n\sin n\pi x(C_n\cosh n\pi y + D_n\sinh n\pi y) $$ The infinite sum of these solutions is: $$ u(x,y) = \sum_{n=1}^\infty u_n(x,y) = \sum_{n=1}^\infty B_n\sin n\pi x(C_n\cosh n\pi y + D_n\sinh n\pi y) $$ Finally, use the condition \(u(x, 3) = 0\) to determine the coefficients \(B_n\), \(C_n\), and \(D_n\), and obtain the final solution: $$ u(x,y) = \sum_{n=1}^\infty B_n\sin n\pi x(C_n\cosh n\pi y + D_n\sinh n\pi y) $$

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are a type of mathematical equation that involves multiple independent variables. These equations are critical in physics and engineering because they describe a wide range of physical phenomena, from heat transfer to wave propagation.

In the exercise provided, a specific type of PDE known as Laplace's equation, \( u_{xx} + u_{yy} = 0 \), is given. This equation arises in contexts where the function \( u(x, y) \) represents a steady-state situation, such as the distribution of temperature in a solid.

• PDEs incorporate derivatives with respect to more than one variable.
• They play a crucial role in mathematical modeling of systems with spatial and temporal dynamics.
• Simplifying PDEs often involves separation of variables or substitution methods to reduce them to simpler Ordinary Differential Equations (ODEs).

Understanding PDEs is essential for solving complex problems that require modeling and simulation of real-world scenarios.

Separation of Variables

Separation of Variables is a powerful analytical method used to solve PDEs like the one in our exercise. The basic idea is to assume that the solution \( u(x, y) \) can be written as a product of two functions, each depending on a single variable, such that \( u(x, y) = X(x)Y(y) \).

This method simplifies the problem by reducing the PDE to a pair of ODEs. In the step-by-step solution, the equation \( \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 \) is obtained after division by \( X(x)Y(y) \). Each side is then set equal to a constant, allowing us to solve them separately.

• This technique is beneficial when boundary conditions are also separable.
• Separation of Variables is especially useful for linear, homogeneous PDEs with constant coefficients.
• Once separated, the solutions to each ODE are combined to form the complete solution to the original PDE.

By utilizing this method, complex PDEs can be reduced to more manageable ODEs, making the problem-solving process more straightforward.

Boundary Conditions

In boundary value problems, boundary conditions are essential as they specify the solution behavior at the domain's boundaries. They ensure the uniqueness of the solution and can guide the integration process of differential equations.

The exercise outlines various boundary conditions: \( u(x, 0) = x^2(2-x) \), \( u(x, 3) = 0 \), \( u(0, y) = 0 \), and \( u(2, y) = 0 \). These conditions dictate how \( u(x, y) \) behaves on the edges of the defined region.

• They can be categorized as Dirichlet (fixed value) or Neumann (fixed derivative) conditions.
• Proper application of boundary conditions leads to a system of equations that determines arbitrary constants in the solution.
• In the problem, boundary conditions are applied sequentially to simplify and solve for the constants in the separated ODE solutions.

Accurately implementing these conditions is vital for deriving meaningful and applicable solutions to PDEs.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) contain functions of a single variable and their derivatives. They are simpler compared to PDEs but form an integral part of the separation of variables process.

After separating the PDE, two ODEs arise: \( X''(x) + k^2X(x) = 0 \) and \( Y''(y) - k^2Y(y) = 0 \). Solving these equations involves finding functions \(X(x)\) and \(Y(y)\) that satisfy each differential equation.

• The solutions often involve trigonometric functions like sine and cosine, or exponential and hyperbolic functions, depending on the nature of the roots of the characteristic equation.
• These solutions are subjected to boundary conditions to find specific solutions for the problem at hand.
• Upon solving the ODEs and applying boundary conditions, the general solution forms a series that includes all possible solutions combined.

Understanding ODEs is essential, as they are foundational to solving more complex PDEs through the method of separation of variables.