Harmonic functions from analytic ones

Mathematics · Complex Analysis · ages 20-21

Name ______________________ Date ____________

At a point where u is harmonic, u_xx = 7. What is u_yy at that point?
Answer: ______________
For a twice continuously differentiable function u(x,y), which expression is its planar Laplacian Δu?
- u_x + u_y
- u_xx + u_yy
- u_xx - u_yy
- u_xy + u_yx
For f(z) = z² with z = x + iy, what is its real part u(x,y)?
- 2xy + y²
- x² - y²
- x² + y²
If f = u + iv is analytic on an open domain, then u_xx + u_yy = 0 throughout that domain.
Circle one: True False
An analytic function has the form f(z) = cz, where c is a complex constant. Its real part is u(x,y) = 3x + 2y. What is the imaginary part of c?
Answer: ______________
For u = x² - y² and a proposed conjugate v = 2xy + x, calculate u_y + v_x at (2,1). A nonzero result rules out analyticity.
Answer: ______________
For u = x² - y² + 6xy, integrating v_y = u_x gives v = 2xy + 3y² + h(x). Which condition on h makes v_x = -u_y hold?
- h'(x) = -6x
- h'(x) = 6x
- h'(x) = -6y
- h'(x) = 0
Let u be a planar electrostatic potential and v its harmonic conjugate. At a point with ∇u ≠ 0, the electric field E = -∇u crosses v's level curve at a right angle.
Circle one: True False
For u = x³ - 3xy² on the plane, which is its harmonic conjugate, up to a real constant?
- x³ + y³
- 3x²y + y³
- 3x²y - y³
For u = log r on the punctured plane, a local conjugate is θ. One loop around 0 increases θ by 2π. Which conclusion follows?
- The real part is not harmonic, so a local conjugate cannot exist anywhere on this domain.
- A global conjugate exists because harmonicity guarantees a global partner on every open domain.
- Adding one fixed constant removes the angle change and makes the conjugate single-valued.
- Local conjugates exist, but no global single-valued conjugate exists on this domain.

Answer key

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Harmonic functions from analytic ones W1-mt_jqajP39XI4-s1

-7 · Harmonicity gives 7 + u_yy = 0, so u_yy = -7.
u_xx + u_yy · The planar Laplacian adds the second derivative in x to the second derivative in y.
x² - y² · Using i² = -1 gives (x + iy)² = x² - y² + i(2xy). The real part is x² - y².
True · Analyticity gives smooth components and the Cauchy-Riemann equations. Differentiating those equations gives cancelling mixed partials.
-2 · Writing c = A + iB gives Re(cz) = Ax - By. Thus A = 3 and -B = 2, so Im(c) = -2.
1 · Here u_y = -2y and v_x = 2y + 1. Their sum is 1, so the second Cauchy-Riemann equation fails.
h'(x) = -6x · Here -u_y = 2y - 6x and v_x = 2y + h'(x). Equating them gives h'(x) = -6x.
False · CR gives ∇u·∇v = 0. Since ∇v is normal to v's level curve, E = -∇u is tangent to that curve, not perpendicular to it.
3x²y - y³ · Integrating v_y = 3x² - 3y² gives v = 3x²y - y³ + h(x). The second equation gives h'(x) = 0.
Local conjugates exist, but no global single-valued conjugate exists on this domain. · A single-valued conjugate must return to its starting value after a loop. The change of 2π persists after adding any constant.