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: Find the unique solution to the boundary value problem for Laplace's equation in the square [0,1]x[0,1] with the given boundary conditions: \(u(0,y) = u(1,y) = u(x,1) = 0\) \(u(x,0) = x(1-x)\) Answer: The unique solution to the boundary value problem is given by: \(u(x,y) = \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n^3\pi^3}\sin(n\pi x)\cosh(n\pi y)\)

Step-by-step solution

Formulate Eigenvalue Problem using Separation of Variables

Assume that the solution \(u(x,y)\) can be separated into the product of two functions of independent variables, i.e. \(u(x,y) = X(x)Y(y)\). Plug this assumption into the Laplace equation \(u_{xx} + u_{yy} = 0\): $$ X''(x)Y(y) + X(x)Y''(y) = 0 $$ Now, divide by \(X(x)Y(y)\) to separate the variables: $$ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 $$ Since the LHS of this equation is a function of x and the RHS is a function of y, both are given as equal to a constant, say \(-k^2\): $$ \frac{X''(x)}{X(x)} = -k^2 \text{ and } \frac{Y''(y)}{Y(y)} = k^2 $$

Solve the Eigenvalue Problem for X(x)

For \(X(x)\), we now have the following ODE: $$ X''(x) + k^2 X(x) = 0 $$ With the given boundary conditions for X(x) \(u(0,y)=0\) and \(u(1,y)=0\): $$ X(0) = 0 \text{ and } X(1) = 0 $$ This is the Sturm-Liouville eigenvalue problem with boundary conditions. Solving this problem gives: $$ X_n(x) = \sin(n\pi x) $$ and $$ k_n = n \pi \quad\text{for } n=1, 2, 3, \dots $$

Solve the ODE for Y(y)

For \(Y(y)\), we have the following ODE: $$ Y''(y) - k_n^2 Y(y) = 0 $$ Solving this problem gives: $$ Y_n(y) = A_n \cosh\left(n \pi y\right) + B_n \sinh\left(n \pi y\right) $$

Solve for B_n

We already have the boundary condition \(u(x,1)=0\), which implies: $$ \sin(n\pi x)\left[A_n \cosh(n\pi) + B_n \sinh(n\pi)\right] = 0 $$ Since the sine term cannot be zero for all x, we conclude that \(B_n=0\) for all n.

Solve for A_n

Next, use the remaining boundary condition \(u(x,0)=x(1-x)\): $$ x(1-x) = \sum_{n=1}^{\infty}A_n\sin(n\pi x) $$ We find the coefficients \(A_n\) via a Fourier sine series: $$ A_n = \frac{2}{1-x} \int_0^1 x(1-x) \sin(n\pi x) dx $$ Compute this integral to obtain: $$ A_n = \frac{2}{n^3\pi^3}(-1)^{n+1} $$ Now, the unique solution to the boundary value problem is given by: $$ u(x,y) = \sum_{n=1}^{\infty}A_n X_n(x) Y_n(y) = \sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n^3\pi^3}\sin(n\pi x)\cosh(n\pi y) $$

Key concepts

Separation of Variables

Separation of Variables is a method used to solve partial differential equations (PDEs), where the solution is expressed as the product of functions that each depend on a single coordinate. This approach simplifies the problem by transforming a PDE into a set of ordinary differential equations (ODEs) that can be solved independently.

In the exercise, we apply this technique to solve Laplace's equation, a well-known PDE. By assuming the solution can be written as a product of two functions, such as \( u(x,y) = X(x)Y(y) \), we separate the PDE into two ODEs in terms of \( x \) and \( y \) respectively. Successfully separating the variables allows us to tackle each function individually, which is significantly more straightforward than handling them together in the PDE form.

Eigenvalue Problem

The Eigenvalue Problem arises when we are dealing with equations that have non-trivial solutions only for certain values of a parameter—these are the eigenvalues. In the context of our boundary value problem, these special values correspond to permissible frequencies of sine waves that meet the boundary conditions.

During separation of variables, we set each side of the resulting ODE equal to a constant which inevitably leads us to an eigenvalue problem. Specifically, for the function \( X(x) \), we find that when the eigenvalue \( k_n \) is set to \( n \pi \), where \( n \) is an integer, the solution satisfies the boundary conditions. The resulting eigenfunctions \( X_n(x) = \text{sin}(n\pi x) \) represent a valid separation and form part of the complete solution to the boundary value problem.

Fourier Sine Series

The Fourier Sine Series is an infinite series that represents a given function over a certain interval using sine functions. In our boundary value problem, after determining \( X(x) \) and \( Y(y) \), only the amplitudes (coefficients) \( A_n \) of the sine functions in the series are unknown.

To find these coefficients, we apply the condition at the boundary \( y = 0 \), which leads us to match the given boundary condition with a Fourier Sine Series. The integral formula provided in the exercise allows us to calculate the coefficients. The uniqueness of the Fourier coefficients thanks to properties of the sine functions guarantees that the series faithfully represents the boundary condition and hence contributes to the final solution of the PDE.

Laplace's Equation

Laplace's Equation is a second-order linear PDE given by \( u_{xx} + u_{yy} = 0 \) that appears in various areas of physics, including electromagnetism, astronomy, and fluid dynamics. In this context, it's typically associated with steady-state problems, where the solution represents a field, such as temperature, that does not change over time.

In this exercise, we solve a two-dimensional version of Laplace's Equation within a unit square domain. The boundary conditions describe fixed values of the solution along the edges of the square. Solving this PDE requires considerable mathematical techniques including separation of variables, understanding eigenvalues, and applying Fourier series to express non-trivial boundary conditions, culminating in an elegant expression for \( u(x, y) \).