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, \quad 0

Short answer

(a) \(u(x, y) = \sum_{n=0}^{\infty} A_n \sin\left(\frac{(2n+1)\pi x}{6}\right) \sin\left(\frac{(2n+1)\pi y}{6}\right)\) (b) \(u(x, y) = \sum_{n=0}^{\infty} A_n \cos\left(\frac{(2n+1)\pi x}{6}\right) \sin\left(\frac{(2n+1)\pi y}{6}\right)\) (c) \(u(x, y) = \sum_{n=0}^{\infty} A_n \cos\left(\frac{(2n+1)\pi x}{6}\right) \cos\left(\frac{(2n+1)\pi y}{6}\right)\) (d) \(u(x, y) = \sum_{n=0}^{\infty} A_n \sin\left(\frac{(2n+1)\pi x}{6}\right) \cos\left(\frac{(2n+1)\pi y}{6}\right)\) Answer: (a) \(u(x, y) = \sum_{n=0}^{\infty} A_n \sin\left(\frac{(2n+1)\pi x}{6}\right) \sin\left(\frac{(2n+1)\pi y}{6}\right)\)

Step-by-step solution

Separate the Variables

Assume that the solution \(u(x, y)\) can be written as a product of two functions: \(u(x, y) = X(x)Y(y)\). Substitute this into the PDE: $$ X''(x)Y(y) + X(x)Y''(y) = 0 $$

Obtain Two Ordinary Differential Equations (ODEs)

Divide the equation by \(X(x)Y(y)\) and rearrange terms: $$ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 $$ Since the left side depends only on \(x\) and the right side depends only on \(y\), both sides must be equal to a constant, say \(-\lambda^2\). We obtain two ODEs: $$ X''(x) = -\lambda^2 X(x) \quad \text{and} \quad Y''(y) = \lambda^2 Y(y) $$

Solve the ODEs

Let's solve the first ODE for \(X(x)\): $$ X''(x) = -\lambda^2 X(x) $$ The general solution is given by: $$ X(x) = A\cos(\lambda x) + B\sin(\lambda x) $$ For the second ODE, \(Y''(y) = \lambda^2 Y(y)\), the general solution is, $$ Y(y) = C\cosh(\lambda y) + D\sinh(\lambda y) $$

Apply the Boundary Conditions

Using the boundary conditions, we can find restrictions on the coefficients and the values of \(\lambda\): 1) \(u(x, 0) = X(x)Y(0) = 0\). Since \(X(x)\) cannot be zero for a nontrivial solution, we must have \(Y(0) = 0\). This implies that \(C = 0\), and the expression for \(Y(y)\) simplifies to: $$ Y(y) = D\sinh(\lambda y) $$ 2) \(u_y(x, 1) = X(x)Y'(1) = 0\). Since \(X(x)\) cannot be zero, we must have \(Y'(1) = 0\). Differentiating \(Y(y)\) with respect to \(y\) and setting \(y=1\), we obtain: $$ \lambda D\cosh(\lambda) = 0 $$ Since \(D\neq 0\), we must have \(\cosh(\lambda) = 0\) which is impossible for real \(\lambda\). Thus, we must use complex values for \(\lambda\), denoting the complex values as \(\lambda = ik\), where \(k\) is real. The ODEs and their general solutions are now: $$ X''(x) = -k^2 X(x) \quad \text{and} \quad X(x) = A\cos(kx) + B\sin(kx) $$ $$ Y''(y) = -(ik)^2 Y(y) \quad \text{and} \quad Y(y) = C\cosh(i k y) + D\sinh(i k y) $$ Using Euler's formula, we can express the last solution with sines and cosines: $$ Y(y) = E\sin(ky) + F\cos(ky) $$

Apply Remaining Boundary Conditions

Using the boundary conditions \(u(0, y) = y(2y^2 - 9y + 12)\) and \(u_x(3, y) = 0\), we can determine the coefficients and values of \(k\) for the general solution: For \(u(0, y) = y(2y^2 - 9y + 12)\), we need \(X(0)Y(y) = y(2y^2 - 9y + 12)\), and since \(X(0) = A\), we must have \(Y(y) = y(2y^2 - 9y + 12)\) when \(x = 0\). Thus, \(F = 0\) and the solution for \(Y(y)\) becomes: $$ Y(y) = E\sin(ky) $$ For \(u_x(3, y) = 0\), we need \(X'(3)Y(y) = 0\). Since \(Y(y)\) cannot be zero, we must have \(X'(3) = 0\), which implies that \(Bk\cos(3k) - A\sin(3k) = 0\). Since \(B\) cannot be zero, we must have \(\cos(3k) = 0\), which implies that \(3k\) is an odd multiple of \(\frac{\pi}{2}\). Thus, \(k = \frac{(2n + 1)\pi}{6}\) for integers \(n\).

Obtain the Solution in Terms of Sums

The general form of the solution is now: $$ u(x, y) = \sum_{n=0}^{\infty} A_n \sin\left(\frac{(2n+1)\pi x}{6}\right) \sin\left(\frac{(2n+1)\pi y}{6}\right) $$

Find Coefficients Through Orthogonality

To determine the coefficients \(A_n\), we can use the orthogonality of the sine functions with respect to both \(x\) and \(y\): $$ A_n = \frac{2}{3}\frac{2}{1}\int_0^1 \int_0^3 y\left(2y^2 - 9y + 12\right) \sin\left(\frac{(2n+1)\pi x}{6}\right)\sin\left(\frac{(2n+1)\pi y}{6}\right) dx dy $$ To find the integral, we have to rewrite the integral in terms of double angle sine functions and simplify the expression. It is important to integrate twice with respect to both variables \(x\) and \(y\). The result will be the expression for the coefficients \(A_n\).

Obtain the Final Solution

Once the coefficients \(A_n\) are found, plug them back into the general solution: $$ u(x, y) = \sum_{n=0}^{\infty} A_n \sin\left(\frac{(2n+1)\pi x}{6}\right) \sin\left(\frac{(2n+1)\pi y}{6}\right) $$ This is the unique solution of the given boundary value problem.

Key concepts

Partial Differential Equations

Partial differential equations (PDEs) are equations that involve rates of change with respect to more than one variable. Unlike ordinary differential equations, which depend on a single variable, PDEs are crucial in modeling phenomena involving space and time. For example, in physics, they describe waves, heat, and quantum mechanics. In the given exercise, the PDE is \(u_{xx} + u_{yy} = 0\), which is a Laplace's equation. PDEs are vital tools for physicists, engineers, and mathematicians as they seek to describe complex systems across multiple dimensions.

• Partial derivatives: Rates of change with respect to one of several variables.
• Grid of points: Solutions may involve analyzing behavior across a continuous region.
• Jacobi and Gauss-Seidel methods: Techniques in numerical solutions for PDEs.

Separation of Variables

Separation of variables is a classic technique used to solve PDEs. The main idea is to assume that the solution can be written as a product of functions, each depending on a single variable. For the given problem, we assume \(u(x, y) = X(x)Y(y)\). This transforms a complex PDE into simpler ordinary differential equations (ODEs) that can be solved individually. Once the ODEs are solved, their solutions are combined to form the complete solution to the original PDE.

• Assumption: The solution is the product of single-variable functions.
• Transformation: Converts PDE into ODEs.
• Application: Useful in problems with symmetrical or constant boundary conditions.

Laplace's Equation

Laplace's equation, \(u_{xx} + u_{yy} = 0\), is a second-order PDE often used in fields like electromagnetics, fluid dynamics, and potential theory. This equation describes situations where a quantity is at equilibrium, such as temperature distribution in a metal plate over time. Solving Laplace's equation involves finding a function \(u(x, y)\) that satisfies the equation across the specified domain and adheres to the given boundary conditions. In this context, boundary value problems often provide a snapshot of stationary state conditions.

• Equilibrium: Solutions represent steady states where no net change occurs.
• Harmonic functions: Solutions are typically smooth and continuous.
• Applications: Widely applicable in modeling physical phenomena.

Orthogonality of Sine Functions

The orthogonality of sine functions is a key property that simplifies solving boundary value problems using separation of variables. Two sine functions are orthogonal if their integral over a specific interval equals zero, unless the functions are identical. In this exercise, orthogonality is used to compute the coefficients in the solution expansion. By integrating the product of sine functions over the domain, we can find expressions for these coefficients due to their orthogonal nature. This method ensures that each term in the solution meets both the PDE and boundary conditions effectively.

• Integral property: Orthogonal functions integrate to zero over their interval.
• Coefficient determination: Compute coefficients by projecting onto orthogonal functions.
• Efficiency: Greatly simplifies the task of finding expansion terms for solutions.