Calculating laplacian and gradient

Question

consider the scalar field Φ(x,y,z) = 3*sin(-x)*sin(-y)

a) calculate the gradient of Φ

b) calculate the laplacian of Φ

c) is this field harmonic ? No/Yes

Answer 1

a. The gradient of a scalar field is a vector made up of the partial derivatives with respect to each relevant variable (x, y, and z here). Gradient = <∂/∂x(Φ), ∂/∂y(Φ), ∂/∂z(Φ)>. Doing the actual calculations:

∂/∂x(Φ) = 3sin(-y)*cos(-x)*(-1) = -3cos(-x)sin(-y). Using the fact that sin(-y) = -sin(-y) and cos(-x) = cos(x) because of the properties of odd and even functions, it simplifies to 3cos(x)sin(y).

∂/∂y(Φ) = 3sin(-x)*cos(-y), and by the same reasoning, it simplifies to 3sin(x)cos(y).

∂/∂z(Φ) = 0, because there are no z-terms in the original scalar-valued function.

So the gradient is <3cos(x)sin(y), 3sin(x)cos(y), 0>.

b. The Laplacian, however, is not a vector: it's a scalar operator defined by ∂²/∂x²(Φ) + ∂²/∂y²(Φ) + ∂²/∂z²(Φ). Basically, it's the divergence of the gradient. We can take partial derivatives of each component of the gradient and just add them:

-3sin(x)sin(y) + -3sin(x)sin(y) + 0 = -6sin(x)sin(y).

c. A field is harmonic if the Laplacian is zero, which obviously isn't the case here. So Φ is not harmonic.

Answer 2

a) Φ = 3sinxsiny

Φ_x = 3cosxsiny; Φ_y = 3sinxcosy; Φ_z = 0; gradient Φ:∇Φ = 3cosxsinyi + 3sinxcosyj;

b) Φ_xx = - 3sinxsiny; Φ_yy = - 3sinxsiny; Φ_zz = 0; laplacian Φ: ∇²Φ = -3sinxsinyi - 3sinxsinyj

c) Φ_xx + Φ_yy + Φ_zz = - 3sinxsiny - 3sinxsiny + 0 = - 6sinxsiny not equel identical 0.

So this field not harmonic