The problem is to solve the partial differential equation: $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ This is Laplace's equation in two dimensions.

1. Problem Description

The problem is to solve the partial differential equation:

$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$

This is Laplace's equation in two dimensions.

2. Solution Steps

Laplace's equation can be solved using several techniques, including separation of variables, Fourier transforms, or complex analysis. Let's use separation of variables. Assume a solution of the form $u(x, y) = X(x)Y(y)$. Substituting this into the equation yields:

$X''(x)Y(y) + X(x)Y''(y) = 0$

Divide both sides by $X(x)Y(y)$:

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

Now, rearrange the equation:

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

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 $-k^2$:

$\frac{X''(x)}{X(x)} = -k^2$

$\frac{Y''(y)}{Y(y)} = k^2$

This leads to two ordinary differential equations:

$X''(x) + k^2 X(x) = 0$

$Y''(y) - k^2 Y(y) = 0$

The general solutions are:

$X(x) = A \cos(kx) + B \sin(kx)$

$Y(y) = C \cosh(ky) + D \sinh(ky)$

Thus, a general solution to Laplace's equation is of the form:

$u(x, y) = (A \cos(kx) + B \sin(kx))(C \cosh(ky) + D \sinh(ky))$

Other solutions can be found by choosing different separation constants (e.g. $k^2$ instead of $-k^2$, or even 0). The actual solution will depend on the specific boundary conditions.

3. Final Answer

The general solution to the Laplace equation $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ can be expressed as

$u(x, y) = (A \cos(kx) + B \sin(kx))(C \cosh(ky) + D \sinh(ky))$. The specific solution depends on the boundary conditions.