Question

For which of the following functions does the Laplacian vanish?

• A. \(xe^{y} - ye^{x}\)
• B. \(x\cos(y) - y\cos(x)\)
• C. \(e^{x+iy}\)
• D. \(yx^{2} - \dfrac{y^{3}}{3} - xy\)

Hint:

Hint:
Compute \(f_{xx}+f_{yy}\) for each option, or recognize which ones are the real/imaginary part of an analytic function of \(z=x+iy\).

Answer 1

The Correct Option is C, D

Step 1: Understanding the Concept:
The Laplacian of a function \(f(x,y)\) is \(\nabla^2 f = \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2}\). A function with \(\nabla^2 f = 0\) everywhere is called harmonic. We check each option by taking the two second partial derivatives directly and adding them.

Step 2: Check option (A).
For \(f = xe^{y} - ye^{x}\): \(f_x = e^{y} - ye^{x}\), so \(f_{xx} = -ye^{x}\). Also \(f_y = xe^{y} - e^{x}\), so \(f_{yy} = xe^{y}\).
\[ \nabla^2 f = xe^{y} - ye^{x} \]
This is not zero for general \(x,y\), so (A) is not harmonic.

Step 3: Check option (B).
For \(f = x\cos(y) - y\cos(x)\): \(f_x = \cos(y) + y\sin(x)\), so \(f_{xx} = y\cos(x)\). Also \(f_y = -x\sin(y) - \cos(x)\), so \(f_{yy} = -x\cos(y)\).
\[ \nabla^2 f = y\cos(x) - x\cos(y) \]
This is also not zero in general, so (B) is not harmonic.

Step 4: Check option (C).
For \(f = e^{x+iy}\), differentiating twice in \(x\) brings down a factor of \(1^2\), and differentiating twice in \(y\) brings down a factor of \(i^2=-1\):
\(f_{xx} = e^{x+iy}\) and \(f_{yy} = i^2 e^{x+iy} = -e^{x+iy}\).
\[ \nabla^2 f = e^{x+iy} - e^{x+iy} = 0 \]
So (C) is harmonic.

Step 5: Check option (D).
For \(f = yx^{2} - \dfrac{y^{3}}{3} - xy\): \(f_x = 2xy - y\), so \(f_{xx} = 2y\). Also \(f_y = x^{2} - y^{2} - x\), so \(f_{yy} = -2y\).
\[ \nabla^2 f = 2y - 2y = 0 \]
So (D) is also harmonic.

Final Answer:
Options (A) and (B) give a non-zero Laplacian, while (C) and (D) give zero. \[ \boxed{e^{x+iy} \text{ and } yx^{2} - \dfrac{y^{3}}{3} - xy} \]