Laplace Equation and Harmonic Functions

The given equation is the two-dimensional Laplace equation:

\begin{equation*} u_{xx} + u_{yy} = 0 \end{equation*}

A function $u(x, y)$ is a solution to the Laplace equation in a domain if its second partial derivatives exist and satisfy the equation at every point in the domain. Such functions are called harmonic functions.

We need to check which of the given functions satisfy this equation in the unit disk $D = \{(x, y) \mid x^2 + y^2 < 1\}$. To do this, we calculate the second partial derivatives with respect to $x$ ($u_{xx}$) and $y$ ($u_{yy}$) for each function and check if their sum is zero.

Checking Each Function for Laplace Equation

Option 1: $u(x, y) = x^5 + 2x^2y^3 - y^5$

First, find the first partial derivatives:

• $u_x = \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x^5 + 2x^2y^3 - y^5) = 5x^4 + 4xy^3$
• $u_y = \frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(x^5 + 2x^2y^3 - y^5) = 6x^2y^2 - 5y^4$

Next, find the second partial derivatives:

• $u_{xx} = \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(5x^4 + 4xy^3) = 20x^3 + 4y^3$
• $u_{yy} = \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}(6x^2y^2 - 5y^4) = 12x^2y - 20y^3$

Now, sum the second partial derivatives:

$u_{xx} + u_{yy} = (20x^3 + 4y^3) + (12x^2y - 20y^3) = 20x^3 + 12x^2y - 16y^3$

Since $20x^3 + 12x^2y - 16y^3$ is not identically zero for all $(x, y)$ in the unit disk, this function is not a solution to the Laplace equation.

Option 2: $u(x, y) = x^2 + 2xy - y^2$

First, find the first partial derivatives:

• $u_x = \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x^2 + 2xy - y^2) = 2x + 2y$
• $u_y = \frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(x^2 + 2xy - y^2) = 2x - 2y$

Next, find the second partial derivatives:

• $u_{xx} = \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(2x + 2y) = 2$
• $u_{yy} = \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}(2x - 2y) = -2$

Now, sum the second partial derivatives:

$u_{xx} + u_{yy} = 2 + (-2) = 0$

Since $u_{xx} + u_{yy} = 0$, this function is a solution to the Laplace equation in the unit disk.

Option 3: $u(x, y) = \cos(y)e^x + \sin(x)e^y$

First, find the first partial derivatives:

• $u_x = \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(\cos(y)e^x + \sin(x)e^y) = \cos(y)e^x + \cos(x)e^y$
• $u_y = \frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(\cos(y)e^x + \sin(x)e^y) = -\sin(y)e^x + \sin(x)e^y$

Next, find the second partial derivatives:

• $u_{xx} = \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(\cos(y)e^x + \cos(x)e^y) = \cos(y)e^x - \sin(x)e^y$
• $u_{yy} = \frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}(-\sin(y)e^x + \sin(x)e^y) = -\cos(y)e^x + \sin(x)e^y$

Now, sum the second partial derivatives:

$u_{xx} + u_{yy} = (\cos(y)e^x - \sin(x)e^y) + (-\cos(y)e^x + \sin(x)e^y) = 0$

Since $u_{xx} + u_{yy} = 0$, this function is a solution to the Laplace equation in the unit disk.

Option 4: $u(x, y) = \frac{1+x}{1+2 x+x^2+y^2} = \frac{1+x}{(1+x)^2+y^2}$

This function can be recognized as the real part of a complex analytic function. Consider the function $f(z) = \frac{1}{z+1}$ where $z = x + iy$.

$f(z) = \frac{1}{(x+iy)+1} = \frac{1}{(x+1)+iy} = \frac{(x+1)-iy}{((x+1)+iy)((x+1)-iy)} = \frac{x+1 - iy}{(x+1)^2 + y^2}$

The real part of $f(z)$ is $\text{Re}(f(z)) = \frac{x+1}{(x+1)^2 + y^2}$, which matches the given function $u(x,y)$.

A function $f(z)$ is analytic in a domain if its derivative exists at every point in that domain. The function $f(z) = \frac{1}{z+1}$ is analytic everywhere except where the denominator is zero, i.e., $z+1 = 0$, which means $z = -1$. The point $z = -1$ corresponds to $(x, y) = (-1, 0)$.

The domain in question is the unit disk $D = \{(x, y) \mid x^2 + y^2 < 1\}$. The point $(-1, 0)$ has $x^2+y^2 = (-1)^2 + 0^2 = 1$. This point lies on the boundary of the unit disk, not inside the open disk $D$. Therefore, the function $f(z) = \frac{1}{z+1}$ is analytic throughout the unit disk $D$.

A property of analytic functions is that their real and imaginary parts are harmonic functions (i.e., they satisfy the Laplace equation) in the domain where the function is analytic. Since $u(x, y)$ is the real part of an analytic function $f(z)$ in the unit disk $D$, $u(x, y)$ must be a harmonic function in $D$.

Thus, $u_{xx} + u_{yy} = 0$ for this function in the unit disk. So, Option 4 is a solution to the Laplace equation.

Summary of Solutions

Based on the analysis, the functions that satisfy the Laplace equation $u_{xx} + u_{yy} = 0$ in the unit disk are those from Options 2, 3, and 4.

• Option 2: $x^2 + 2xy - y^2$
• Option 3: $\cos(y)e^x + \sin(x)e^y$
• Option 4: $\frac{1+x}{1+2 x+x^2+y^2}$