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}{ll} u_{x x}+u_{y y}=0, & 0

Short answer

Question: Using the Separation of Variables and the Fourier series expansion methods, find the solution to the Laplace equation, \(u_{xx}+u_{yy}=0\), on the rectangular domain \(0\leq x\leq 1\), \(0\leq y\leq \pi\), with the following boundary conditions: 1. \(u_x(0,y)=0\) 2. \(u_x(1,y)=\pi^2 - y^2\) 3. \(u_y(x,0)=0\) 4. \(u(x,\pi)=0\) Summarize the steps involved in finding the solution.

Step-by-step solution

Separation of variables and defining X(x) and Y(y)

Let's assume that u(x,y) can be separated as a product of two functions, one depending on x alone and the other depending on y alone: \(u(x, y) = X(x)Y(y)\). We can now substitute this into the Laplace equation, \(u_{xx} + u_{yy} = 0\): $$ X''(x)Y(y) + X(x)Y''(y) = 0. $$

Separate X(x) and Y(y) equations

Now, divide both sides of the equation by \(X(x)Y(y)\) to separate the variables: $$ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0. $$ Since the left-side of the equation depends only on x and the right-side depends only on y, both sides must be equal to some constant. Let's denote the constant as \(-\lambda\): $$ \frac{X''(x)}{X(x)} = \lambda, \quad \frac{Y''(y)}{Y(y)} = -\lambda. $$

Solving X(x) equation

Now, let's solve the X(x) equation: $$ X''(x) = \lambda X(x). $$ We are given that \(u_x(0, y) = 0\) and \(u_x(1, y) = \pi^2 - y^2\). Therefore, \(X'(0)Y(y) = 0\) and \(X'(1)Y(y) = \pi^2 - y^2\). Since \(Y(y) \neq 0\), we must have \(X'(0) = 0\) and \(X'(1)=(\pi^2 - y^2)/Y(y)\), which must be constant (let's denote it as \(A\)): \(X'(0) = 0\) and \(X'(1) = A\). Since \(X''(x) = \lambda X(x)\), the solution to the differential equation is a linear combination of \(\sin(\sqrt{\lambda}x)\) and \(\cos(\sqrt{\lambda}x)\). Taking the derivative of the general solution, and using the boundary conditions \(X'(0) = 0\) and \(X'(1) = A\), gives the value of \(\lambda\) and a unique form of X(x).

Solving Y(y) equation

Now, let's solve the Y(y) equation: $$ Y''(y) = -\lambda Y(y). $$ Using the boundary conditions \(u_y(x, 0) = 0\) (which means \(Y'(0) = 0\)) and \(u(x, \pi) = 0\) (which means \(Y(\pi) = 0\)), we can find the value of the constant \(-\lambda\) and a unique form of Y(y).

Combine X(x) and Y(y)

Now, using the X(x) and Y(y) solutions found in Step 3 and Step 4, we can find the general solution u(x, y) as a product of X(x) and Y(y).

Fourier series expansion

Finally, since \(u_x(1, y) = \pi^2 - y^2\), we can use Fourier series expansion to represent this function as an infinite sum of sine and cosine functions. By integrating this Fourier series term by term, we can find the corresponding X(x) and Y(y) solutions. Then find the final solution \(u(x,y)\) as the product of those obtained X(x) and Y(y) solutions with the corresponding Fourier coefficients.

Graph the surface u(x,y)

The final step is to graph the surface \(u=u(x,y)\) with \(0 \leq x \leq 1\) and \(0 \leq y \leq \pi\), showing the domain within which our solution is defined. Using software like Matlab or Mathematica, we can visualize the surface by inputting the function and the specified domain.

Key concepts

Separation of Variables

In the context of solving partial differential equations, the method of separation of variables is an essential technique. It assumes that the solution can be written as a product of functions, each depending on a different variable. For the boundary value problem given, we start by positing that the solution \( u(x, y) \) can be expressed as \( X(x)Y(y) \). This decomposition allows us to substitute into the Laplace equation, \[ u_{xx} + u_{yy} = 0, \] leading to an equation where each side depends exclusively on one variable. After separating, you'll have two ordinary differential equations (ODEs): one for \( X(x) \) and another for \( Y(y) \). This is a powerful approach because it reduces a potentially difficult partial differential equation (PDE) into more manageable ODEs. Ultimately, solving these can provide insights or even an exact solution to our original problem.

Laplace Equation

The Laplace equation is a second-order partial differential equation given by \[ abla^2 u = u_{xx} + u_{yy} = 0. \] In physics, it often describes phenomena such as electric potential in a charge-free region or steady-state heat distribution. In our boundary value problem, the equation applies to a rectangular domain, with specified boundary conditions at the edges.Boundary conditions are crucial here. They ensure that the unique solution to this PDE satisfies the physical constraints or geometric limits of the problem. For example, in our problem, conditions like \( u_y(x, 0) = 0 \) and \( u(x, \pi) = 0 \) are integral in determining the specific solution. Solving the Laplace equation here involves meeting these conditions while satisfying the broader equation structure.

Fourier Series

Fourier series expansions come into play when needing to express functions as an infinite sum of sines and cosines. This technique is particularly useful when you have complicated boundary conditions, such as \( u_x(1, y) = \pi^2 - y^2 \), which might be challenging to incorporate directly into the solution.By decomposing functions using Fourier series, you convert them into a form that can be more straightforwardly matched with solutions derived via separation of variables. The Fourier coefficients are determined by integrating over the domain, making it possible to tailor a solution that meshes accurately with boundary conditions.In solving our boundary value problem, Fourier series not only provides a way to meet the condition at \( x = 1 \), but it also contributes to constructing the full solution \( u(x, y) = X(x)Y(y) \).

Differential Equation Solution

The aim in solving a differential equation, specifically the boundary value problem cited here, is to find the function \( u(x, y) \) that satisfies both the equation itself and the accompanying boundary conditions. After employing separation of variables, we are left with ordinary differential equations that are generally easier to solve.For instance, when analyzing the equation for \( X(x) \), its solution involves constants that are determined by the boundary conditions \( X'(0) = 0 \) and \( X'(1) = A \), where A is related to another boundary condition. Similarly, solving the equation for \( Y(y) \) requires taking boundary conditions into account, like \( Y'(0) = 0 \) and \( Y(\pi) = 0 \).Ultimately, combining these solutions with the conditions helps us obtain the desired solution for the entire problem. The methodical capture of boundary conditions and clear formulation of \( X(x) \) and \( Y(y) \) ensures that the final product \( u(x, y) \) meets all given requirements.