Lesson 33 (9.7-2): Laplace's Equation (part 2)

The Governing Equation and Domain

We are solving **Laplace's Equation** in a rectangular domain \(0 < x < a\) and \(0 < y < b\):

The standard procedure is the **separation of variables**, assuming a solution of the form $u(x,y) = X(x)Y(y)$.

Substituting into Laplace's equation gives:

This separates the problem into two ordinary differential equations (ODEs):

Problem 1: Homogeneous Conditions on Three Sides

Boundary Conditions (BCs)

The boundary conditions are defined on the four sides of the rectangle:

• Bottom: $u(x,0) = 0$ (held at zero)
• Top: $u(x,b) = f(x)$ (condition we decide)
• Left (Insulated): $u_x(0,y) = 0$
• Right (Insulated): $u_x(a,y) = 0$

The homogeneous BCs are for the $X$ function:

• $u_x(0,y) = 0 \quad \rightarrow \quad X'(0) = 0$
• $u_x(a,y) = 0 \quad \rightarrow \quad X'(a) = 0$

And for the $Y$ function:

• $u(x,0) = 0 \quad \rightarrow \quad Y(0) = 0$

Solving for $X(x)$ (Eigenvalue Problem)

We solve $X'' - cX = 0$ with $X'(0) = 0$ and $X'(a) = 0$.

Case 1: $c = 0$

The ODE is $X'' = 0$, giving $X(x) = C_1x + C_2$. The derivative is $X'(x) = C_1$.

• $X'(0) = C_1 = 0$
• $X'(a) = C_1 = 0$

This gives a non-trivial solution $X(x) = C_2$. We choose $X_0 = 1$.

This corresponds to the **eigenvalue** $\lambda_0 = c = 0$ and **eigenfunction** $X_0 = 1$.

Case 2: $c < 0$ (Let $c = -\lambda^2$, where $\lambda > 0$)

The ODE is $X'' + \lambda^2 X = 0$, giving $X(x) = C_1 \cos(\lambda x) + C_2 \sin(\lambda x)$.

The derivative is $X'(x) = -\lambda C_1 \sin(\lambda x) + \lambda C_2 \cos(\lambda x)$.

• $X'(0) = \lambda C_2 \cos(0) = \lambda C_2 = 0 \quad \rightarrow \quad C_2 = 0$
• $X'(a) = -\lambda C_1 \sin(\lambda a) = 0$

For a non-trivial solution ($C_1 \neq 0$), we must have $\sin(\lambda a) = 0$.

Therefore, $\lambda a = n\pi$, where $n = 1, 2, 3, \ldots$.

The **eigenvalues** are $c_n = -\lambda_n^2$, where $\lambda_n = \frac{n\pi}{a}$.

The **eigenfunctions** are $X_n(x) = \cos\left(\frac{n\pi x}{a}\right)$ for $n = 1, 2, 3, \ldots$.

Combining both cases, the eigenvalues and eigenfunctions for $n = 0, 1, 2, 3, \ldots$ are:

Solving for $Y(y)$

We solve $Y'' + cY = 0$ with $Y(0) = 0$. Since $c = \lambda_n^2$, the equation is $Y'' + \lambda_n^2 Y = 0$ for $n=0$ and $Y'' - \lambda_n^2 Y = 0$ for $n \ge 1$.

Case 1: $n = 0$ (where $c_0 = 0$)

The ODE is $Y'' = 0$, giving $Y(y) = C_1y + C_2$.

• $Y(0) = C_2 = 0$.

Thus, $Y_0(y) = y$.

Case 2: $n = 1, 2, 3, \ldots$ (where $c_n = (n\pi/a)^2$)

The ODE is $Y'' - \left(\frac{n\pi}{a}\right)^2 Y = 0$. The general solution uses hyperbolic functions:

• $Y(0) = C_1 \cosh(0) + C_2 \sinh(0) = C_1(1) + C_2(0) = C_1 = 0$.

Thus, $Y_n(y) = \sinh\left(\frac{n\pi y}{a}\right)$.

General Solution and Final Boundary Condition

The product solutions $u_n(x,y) = X_n(x)Y_n(y)$ are:

• $n=0$: $u_0(x,y) = (1)(y) = y$
• $n \ge 1$: $u_n(x,y) = \cos\left(\frac{n\pi x}{a}\right)\sinh\left(\frac{n\pi y}{a}\right)$

The general solution is a linear combination of all $u_n$:

Applying the last BC, $u(x,b) = f(x)$:

This is a **Fourier Cosine Series** for $f(x)$ over $[0, a]$. The coefficients are found using the Euler-Fourier formulas. Let the full coefficients be $A_n$.

Determining Coefficients

For $n=0$:

For $n \ge 1$:

Problem 2: Semi-Infinite Domain

This problem models a long strip of plate where the domain is $0 < x < a$ and $y > 0$ (i.e., $y \to \infty$).

Boundary Conditions (BCs)

• Left (Insulated): $u_x(0,y) = 0$
• Right (Insulated): $u_x(a,y) = 0$
• Bottom: $u(x,0) = f(x)$
• Condition at $y \to \infty$: The temperature $u(x,y)$ must be bounded.

Solution for $X(x)$

The BCs for $X(x)$ are $X'(0) = 0$ and $X'(a) = 0$, which are the same as in Problem 1. Thus, the eigenvalues and eigenfunctions are the same:

Solution for $Y(y)$

We solve $Y'' - \lambda_n^2 Y = 0$.

Case 1: $n = 0$ (where $\lambda_0 = 0$)

The ODE is $Y'' = 0$, giving $Y_0(y) = C_1y + C_2$. For $u(x,y)$ to be bounded as $y \to \infty$, $C_1$ must be 0. Thus, $Y_0(y) = 1$ (by setting $C_2=1$).

Case 2: $n = 1, 2, 3, \ldots$ (where $\lambda_n > 0$)

The general solution in exponential form is:

The term $e^{\frac{n\pi}{a}y}$ goes to $\infty$ as $y \to \infty$. Since the temperature must be **bounded** at $y=\infty$, we set $C_1 = 0$.

Thus, $Y_n(y) = e^{-\frac{n\pi y}{a}}$.

General Solution and Final Boundary Condition

The product solutions $u_n(x,y) = X_n(x)Y_n(y)$ are:

• $n=0$: $u_0(x,y) = (1)(1) = 1$
• $n \ge 1$: $u_n(x,y) = e^{-\frac{n\pi y}{a}} \cos\left(\frac{n\pi x}{a}\right)$

The general solution is:

Applying the last BC, $u(x,0) = f(x)$:

This is again a **Fourier Cosine Series** for $f(x)$ over $[0, a]$. The coefficients $c_n$ are the standard Fourier cosine coefficients:

Determining Coefficients

For $n=0$:

For $n \ge 1$: