Show that any function given by omega = (ax)(by)e^-sqrt(a2+b2)z satisfies Laplace's equation in three dimensions. (d2 omega)/(d x2) + (d2 omega)/(d y2) + (d2 omega)/(d z2) = 0

Let $C_y = \cos(by)$ and $E_z = e^{-\sqrt{a^2+b^2}z}$. Then $\omega = \sin(ax) C_y E_z$.

$$\frac{\partial \omega}{\partial x} = a\cos(ax)C_y E_z$$ $$\frac{\partial^2 \omega}{\partial x^2} = -a^2\sin(ax)C_y E_z = -a^2\omega$$

Let $S_x = \sin(ax)$ and $E_z = e^{-\sqrt{a^2+b^2}z}$. Then $\omega = S_x \cos(by) E_z$.

$$\frac{\partial \omega}{\partial y} = S_x (-b\sin(by)) E_z = -b\sin(ax)\sin(by)E_z$$ $$\frac{\partial^2 \omega}{\partial y^2} = S_x (-b^2\cos(by)) E_z = -b^2\sin(ax)\cos(by)E_z = -b^2\omega$$

Let $S_x = \sin(ax)$ and $C_y = \cos(by)$. Let $k = \sqrt{a^2+b^2}$. Then $\omega = S_x C_y e^{-kz}$.

$$\frac{\partial \omega}{\partial z} = S_x C_y (-k e^{-kz}) = -k\sin(ax)\cos(by)e^{-kz}$$ $$\frac{\partial^2 \omega}{\partial z^2} = S_x C_y (-k)(-k e^{-kz}) = k^2\sin(ax)\cos(by)e^{-kz} = k^2\omega$$

Substitute $k = \sqrt{a^2+b^2}$:

$$\frac{\partial^2 \omega}{\partial z^2} = (\sqrt{a^2+b^2})^2\omega = (a^2+b^2)\omega$$

$$\frac{\partial^2 \omega}{\partial x^2} + \frac{\partial^2 \omega}{\partial y^2} + \frac{\partial^2 \omega}{\partial z^2} = -a^2\omega - b^2\omega + (a^2+b^2)\omega$$ $$= (-a^2 - b^2 + a^2 + b^2)\omega = 0\omega = 0$$

Since the sum is 0, the function $\omega$ satisfies Laplace's equation.