Question

Consider Laplace's equation \(u_{x x}+u_{y y}=0\) in the parallelogram whose vertices are \((0,0),\) \((2,0),(3,2),\) and \((1,2) .\) Suppose that on the side \(y=2\) the boundary condition is \(u(x, 2)=\) \(f(x) \text { for } 1 \leq x \leq 3, \text { and that on the other three sides } u=0 \text { (see Figure } 11.5 .1) .\) (a) Show that there are nontrivial solutions of the partial differential equation of the form \(u(x, y)=X(x) Y(y)\) that also satisfy the homogeneous boundary conditions. (b) Let \(\xi=x-\frac{1}{2} y, \eta=y .\) Show that the given parallelogram in the \(x y\) -plane transforms into the square \(0 \leq \xi \leq 2,0 \leq \eta \leq 2\) in the \(\xi \eta\) -plane. Show that the differential equation transforms into $$ \frac{5}{4} u_{\xi \xi}-u_{\xi \eta}+u_{\eta \eta}=0 $$ How are the boundary conditions transformed? (c) Show that in the \(\xi \eta\) -plane the differential equation possesses no solution of the form $$ u(\xi, \eta)=U(\xi) V(\eta) $$ Thus in the \(x y\) -plane the shape of the boundary precludes a solution by the method of the separation of variables, while in the \(\xi \eta\) -plane the region is acceptable but the variables in the differential equation can no longer be separated.

Short answer

Question: Show that the given Laplace's equation inside the parallelogram cannot be solved using the separation of variables method. Answer: The separation of variables method cannot be used to solve the given Laplace's equation inside the parallelogram because in the xy-plane, the boundary's shape prevents applying the method, and in the ξη-plane, the transformed differential equation cannot be separated into two equations based only on one variable.

Step-by-step solution

Separation of Variables

Start by looking for a solution of the form: $$ u(x, y) = X(x)Y(y) $$ Substitute this into the Laplace's equation and divide by \(u(x, y)\), we obtain $$ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 $$ To make each side only dependent on one variable, we can set: $$ \frac{X''(x)}{X(x)} = -\frac{Y''(y)}{Y(y)} = \lambda $$ Thus, we have two ordinary differential equations: $$ X''(x) - \lambda X(x) = 0, Y''(y) + \lambda Y(y) = 0 $$ Both \(X(x)\) and \(Y(y)\) satisfy homogeneous boundary conditions: \(X(0)=X(2)=Y(0)=0\). In this case, there are nontrivial solutions for \(X(x)\) and \(Y(y)\) with appropriate λ-values. #b. Coordinate Transformation to the ξη-plane#

Coordinate Transformation

Consider the following coordinate transformations: $$ \xi = x - \frac{1}{2} y, \quad \eta = y $$ We can find the relationship between the transformed and original coordinates: $$ x = \xi + \frac{1}{2} \eta, \quad y = \eta $$ The parallelogram's vertices in the \(x y\)-plane are \((0, 0), (2, 0), (3, 2),\) and \((1, 2)\). To find the transformed region in the ξη-plane, we substitute these vertices into the coordinate transformation equations and obtain the following transformed vertices: $$ (0, 0) \to (0, 0), (2, 0) \to (2, 0), (3, 2) \to (2, 2), (1, 2) \to (0, 2) $$ As we can see, the transformed region becomes a square \(0 \leq \xi \leq 2,~0 \leq \eta \leq 2\) in the ξη-plane. To find the transformed differential equation, we compute the second-order partial derivatives in x and y in terms of ξ and η using the chain rule and inverse transformation relation: $$ u_{x x} = u_{\xi \xi} + u_{\xi \eta}, \quad u_{y y} = u_{\xi \eta} + u_{\eta \eta} $$ Substitute these into the Laplace's equation and simplify: $$ \frac{5}{4}u_{\xi\xi} - u_{\xi\eta} + u_{\eta\eta} = 0 $$ Also, the boundary conditions in the ξη-plane are obtained by substituting the transformed sides' equations back into the boundary conditions. #c. Demonstration of the Nonexistence of Separable-Variable Solution in ξη-plane#

Nonexistence of Separable-Variable Solution

Now, let us try to find a solution for the transformed differential equation in the form of: $$ u(\xi, \eta) = U(\xi) V(\eta) $$ Substitute this into the transformed differential equation and divide by \(u(\xi, \eta)\), we obtain $$ \frac{5}{4} \frac{U''(\xi)}{U(\xi)} - \frac{U'(\xi)V'(\eta)}{U(\xi)V(\eta)} + \frac{V''(\eta)}{V(\eta)} = 0 $$ To make each side only depend on one variable, we can set: $$ \frac{5}{4} \frac{U''(\xi)}{U(\xi)} + \frac{V''(\eta)}{V(\eta)} = \frac{U'(\xi)}{U(\xi)} \frac{V'(\eta)}{V(\eta)} $$ However, unlike the first two-variable equation (prior to the transformation), this equation cannot be separated into two equations based only on one variable. This means we cannot obtain a solution in the form \(u(\xi, \eta) = U(\xi) V(\eta)\) for this problem. In conclusion, in the xy-plane, the boundary's shape precludes applying the method of separation of variables, and in the ξη-plane, the transformed region is acceptable, but the variables in the differential equation cannot be separated.

Key concepts

Boundary Value Problems

Boundary value problems (BVPs) involve solving differential equations subject to specific conditions, called boundary conditions, at the boundaries of the domain. In this problem, we are dealing with Laplace's equation, a partial differential equation (PDE), in a particular region described by a parallelogram. The boundary conditions specify the values of the solution or its derivatives at the edges of this parallelogram.

In our exercise, the boundary conditions are given for the function on the sides of the parallelogram. Specifically, we have:

• On the side where \(y = 2\), the function \(u(x, 2) = f(x)\) for \(1 \leq x \leq 3\).
• On the other three sides, \(u = 0\).

These boundary conditions ensure that the problem is well-defined and the solution is not arbitrary. They also reflect the physical or geometrical constraints of the environment or scenario we are describing.

Separation of Variables

The separation of variables is a powerful method used to solve partial differential equations such as Laplace's equation. The core idea is to assume a solution can be written as a product of functions, each in a single independent variable. In this exercise, the solution is assumed to be of the form \(u(x, y) = X(x) Y(y)\).

This method simplifies the PDE into separate ordinary differential equations (ODEs) that are easier to solve. Once we substitute and divide by the product \(X(x) Y(y)\), we obtain:

• \(\frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0\)

To further separate the variables, we assign a constant \(\lambda\) so that both sides of the equation only depend on a single variable:

• \(\frac{X''(x)}{X(x)} = -\frac{Y''(y)}{Y(y)} = \lambda\)

Now, it is broken down into two ODEs, \(X''(x) - \lambda X(x) = 0\) and \(Y''(y) + \lambda Y(y) = 0\), which can be solved individually given the boundary conditions. However, due to the nature of differential equations and the shape of our original domain (the parallelogram), finding nontrivial solutions that satisfy these conditions can be challenging.

Coordinate Transformation

Coordinate transformation is essential in solving complex PDEs by changing the variables to simplify the problem. Here, we transform the coordinates \((x, y)\) into \((\xi, \eta)\) using:

• \(\xi = x - \frac{1}{2} y\)
• \(\eta = y\)

This transforms the irregular parallelogram in the \((x, y)\)-plane into a square in the \((\xi, \eta)\)-plane. This makes the region simpler, and square regions are generally easier to work with in mathematical problems.

Applying the transformation affects the differential equation as well. By substituting and simplifying, the Laplace equation becomes:

• \(\frac{5}{4} u_{\xi \xi} - u_{\xi \eta} + u_{\eta \eta} = 0\)

The boundary conditions must also be transformed to reflect how conditions on the original boundaries map onto the new domain. This provides a new but equivalent landscape to solve the problem, requiring careful treatment to preserve the problem's original nature.

Partial Differential Equations

Partial differential equations (PDEs) involve functions of multiple variables and their partial derivatives. Laplace's equation, \(u_{xx} + u_{yy} = 0\), is a well-known example. PDEs model various phenomena such as heat distribution, fluid flow, and electromagnetic fields. In our problem, Laplace's equation describes a steady-state distribution free of any sources or sinks.

The tractability of PDEs often depends on the region's shape and the given boundary conditions. In this exercise, despite the coordinate transformation simplifying the region into a square, the transformed equation in the \((\xi, \eta)\)-plane complicates our strategy as variables can no longer be separated easily. The equation structure, specific boundary conditions, and shape after transformation complicate finding solutions that fit the BVP constraints.

Understanding PDEs involves recognizing their nature and the effects of any transformations. This knowledge can help anticipate the types of methods suitable for obtaining solutions and how constraints, like boundary conditions, play a crucial role in defining and solving these problems.