Introduction
Laplace's equation in two dimensions, ∇²u = ∂²u/∂x² + ∂²u/∂y² = 0, describes steady-state physical phenomena such as electrostatic potential and steady heat distribution. When defined over a rectangular domain with prescribed Dirichlet boundary conditions, the boundary value problem can be solved analytically using the separation of variables technique.
1. Separation of Variables
Assume a separable solution of the form:
$$u(x, y) = X(x)Y(y)$$
Differentiating and substituting into the governing PDE ∂²u/∂x² + ∂²u/∂y² = 0 yields:
$$X''(x)Y(y) + X(x)Y''(y) = 0 \implies \frac{X''(x)}{X(x)} = -\frac{Y''(y)}{Y(y)}$$
Since the left-hand side depends solely on $x$ and the right-hand side solely on $y$, both must equal a common constant. To satisfy the homogeneous Dirichlet conditions $u(0, y) = u(l, y) = 0$, choose a negative separation constant $-p^2$ (where $p > 0$):
$$\frac{X''}{X} = -p^2 \quad \implies \quad X''(x) + p^2 X(x) = 0$$
$$\frac{Y''}{Y} = p^2 \quad \implies \quad Y''(y) - p^2 Y(y) = 0$$
2. Solving the Spatial ODEs and Applying Boundary Conditions
The general solutions to these ordinary differential equations are:
- Solution for X(x): $X(x) = C_1 \cos(px) + C_2 \sin(px)$
- Solution for Y(y): $Y(y) = C_3 \cosh(py) + C_4 \sinh(py)$
Apply the homogeneous boundary conditions sequentially:
- Condition 1: $u(0, y) = 0 \implies X(0)Y(y) = 0 \implies X(0) = 0$.
Substituting $x = 0$: $C_1 \cos(0) + C_2 \sin(0) = 0 \implies C_1 = 0$. - Condition 2: $u(l, y) = 0 \implies X(l) = 0$.
With $C_1 = 0$, $C_2 \sin(pl) = 0$. For non-trivial solutions ($C_2 \neq 0$), $\sin(pl) = 0 \implies pl = k\pi$, leading to the eigenvalues:
$$p_k = \frac{k\pi}{l}, \quad k = 1, 2, 3, \dots$$ - Condition 3: $u(x, 0) = 0 \implies Y(0) = 0$.
Substituting $y = 0$: $C_3 \cosh(0) + C_4 \sinh(0) = 0 \implies C_3 = 0$.
3. Superposition and Non-Homogeneous Condition
Combining the valid solutions for each mode $k$, the general solution by linear superposition is:
$$u(x, y) = \sum_{k=1}^{\infty} B_k \sin\left(\frac{k\pi x}{l}\right) \sinh\left(\frac{k\pi y}{l}\right)$$
Applying the fourth boundary condition at $y = a$:
$$u(x, a) = \sum_{k=1}^{\infty} B_k \sinh\left(\frac{k\pi a}{l}\right) \sin\left(\frac{k\pi x}{l}\right) = \sin\left(\frac{n\pi x}{l}\right)$$
4. Evaluation of the Mode Coefficient
Comparing both sides by the orthogonality of sinusoidal functions on the interval $[0, l]$:
- For $k \neq n$, $B_k = 0$.
- For $k = n$, $B_n \sinh\left(\frac{n\pi a}{l}\right) = 1 \implies B_n = \frac{1}{\sinh\left(\frac{n\pi a}{l}\right)}$.
Thus, the infinite series collapses into a single unique closed-form harmonic solution:
$$u(x, y) = \frac{\sin\left(\frac{n\pi x}{l}\right) \sinh\left(\frac{n\pi y}{l}\right)}{\sinh\left(\frac{n\pi a}{l}\right)}$$
Conclusion
The resulting closed-form solution satisfies both Laplace's equation and all four prescribed boundary conditions. Because the non-homogeneous boundary condition comprises a single pure spatial mode, the superposition collapses without requiring a full Fourier series expansion.