Dirichlet in the Upper-Half Plane Jenna Nelson and Gloria Mayorga Occidental College April 22, 2016 Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 1 / 17
Overview 1 Evaluating real improper integrals 2 Using results to solve Dirichlet problem in upper-half plane Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 2 / 17
a. Real Improper Integrals Use the techniques presented in Section 6.6 to establish the integral formulas cos(s) πe a s 2 ds = + a2 a sin(s) s 2 + a 2 ds = 0 To minimize confusion, we will make the substitution s = x cos(s) s 2 + a 2 ds sin(s) s 2 + a 2 ds cos(x) x 2 + a 2 dx sin(x) x 2 + a 2 dx Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 3 / 17
a. Real Improper Integrals Referencing back to Euler s Formula, we see that: f (x)e iαx dx = f (x) = f (x) cos(αx)dx + i 1 x 2 + a 2, α = 1 1 1 x 2 + a 2 eix dx = x 2 cos xdx + i + a2 f (x) sin(αx)dx 1 x 2 sin xdx + a2 Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 4 / 17
a. Real Improper Integrals We can evaluate this real integral by considering the complex contour integral f (z)e iz 1 dz, f (z) = z 2 + a 2 C C The contour C consists of [, ] on the x-axis and the semicircle C R in the upper half plane 1 1 1 z 2 + a 2 eiz dz = C R z 2 + a 2 eiz dz + x 2 + a 2 eix dx Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 5 / 17
a. Real Improper Integrals From Cauchy s Residue Theorem, we know that f (z)dz = 2πi Res(f (z), z o ) C For the contour integral we re considering: 1 z 2 + a 2 eiz dz = 2πi Res(f (z)e iz, z o ) C The roots are z o = ± ia, both of order 1, but we only need to consider z o = ia: Res(f (z)e iz, ia) = g(ia) h (ia) = eiz 2z = e a z=ia 2ia C 1 z 2 + a 2 eiz dz = 2πi( e a 2ia ) = πe a a Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 6 / 17
a. Real Improper Integrals Using our result from the previous slide, we get C R 1 z 2 + a 2 eiz dz + 1 x 2 + a 2 eix dx = πe a a From Theorem 6.6.2, the integral over the semicircle becomes zero as R, which leaves us with only the real portion 1 x 2 + a 2 eix dx = πe a a Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 7 / 17
a. Real Improper Integrals 1 1 πe a x 2 cos xdx + i + a2 x 2 sin xdx = + i0 + a2 a Substituting back in s for x, we find the results to the initial integrals: cos s πe a s 2 ds = + a2 a and sin s s 2 + a 2 ds = 0 Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 8 / 17
Solve the Dirichlet Problem in the upper half-plane y > 0 subject to the boundary condition φ(x, 0) = f (x) = cos(x), < x <. [Hint: Make the substitution s = t x and use the formulas in part (a).] Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 9 / 17
Theorem (Poisson Integral Formula for the Half-Plane) If f (x) is a piece-wise continuous and bounded function on < x <, then the solution to the Dirichlet problem in the upper half-plane y > 0 with boundary condition φ(x, y) = f (x) at all points of continuity of f is given by φ(x, y) = y f (t) π (x t) 2 + y 2 dt Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 10 / 17
f (t) represents the boundary condition, which is given by φ(x, 0) = f (x) = cos(x) Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 11 / 17
f (t) represents the boundary condition, which is given by The integral becomes φ(x, 0) = f (x) = cos(x) φ(x, y) = y π cos(t) (x t) 2 + y 2 dt Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 11 / 17
f (t) represents the boundary condition, which is given by The integral becomes φ(x, 0) = f (x) = cos(x) φ(x, y) = y π cos(t) (x t) 2 + y 2 dt Note: (x t) 2 = (t x) 2. Our integral can now be written as φ(x, y) = y π cos(t) (t x) 2 + y 2 dt Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 11 / 17
To help us solve for φ(x, y), we make the substitution s = t x, which also gives us ds = dt Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 12 / 17
To help us solve for φ(x, y), we make the substitution s = t x, which also gives us ds = dt. φ(x, y) = y π cos(t) (t x) 2 + y 2 dt Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 12 / 17
To help us solve for φ(x, y), we make the substitution s = t x, which also gives us ds = dt. φ(x, y) = y π φ(x, y) = y π cos(t) (t x) 2 + y 2 dt cos(s + x) s 2 + y 2 ds Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 12 / 17
Using the trig identity cos(s + x) = cos(s) cos(x) sin(s) sin(x) Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 13 / 17
Using the trig identity cos(s + x) = cos(s) cos(x) sin(s) sin(x) We can rewrite our integral as φ(x, y) = y π cos(s) cos(x) sin(s) sin(x) s 2 + y 2 ds Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 13 / 17
Using the trig identity cos(s + x) = cos(s) cos(x) sin(s) sin(x) We can rewrite our integral as φ(x, y) = y π φ(x, y) = y π cos(s) cos(x) sin(s) sin(x) s 2 + y 2 ds cos(s) cos(x) s 2 + y 2 ds y π sin(s) sin(x) s 2 + y 2 ds Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 13 / 17
Using the trig identity cos(s + x) = cos(s) cos(x) sin(s) sin(x) We can rewrite our integral as φ(x, y) = y π φ(x, y) = y π φ(x, y) = y π cos(x) cos(s) cos(x) sin(s) sin(x) s 2 + y 2 ds cos(s) cos(x) s 2 + y 2 ds y π cos(s) s 2 + y 2 ds y π sin(x) sin(s) sin(x) s 2 + y 2 ds sin(s) s 2 + y 2 ds Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 13 / 17
We showed in (a) that cos(s) πe a s 2 ds = + a2 a and sin(s) s 2 + a 2 ds = 0 Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 14 / 17
We showed in (a) that cos(s) πe a s 2 ds = + a2 a and sin(s) s 2 + a 2 ds = 0 From these results and by setting a = y φ(x, y) = y π cos(x) cos(s) s 2 + y 2 ds y π sin(x) sin(s) s 2 + y 2 ds Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 14 / 17
We showed in (a) that cos(s) πe a s 2 ds = + a2 a and sin(s) s 2 + a 2 ds = 0 From these results and by setting a = y φ(x, y) = y π cos(x) cos(s) s 2 + y 2 ds y π sin(x) sin(s) s 2 + y 2 ds φ(x, y) = y π cos(x)πe y y 0 Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 14 / 17
Final solution to the given Dirichlet problem: φ(x, y) = cos(x)e y Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 15 / 17
Final solution to the given Dirichlet problem: φ(x, y) = cos(x)e y To confirm solution, check that Laplace s Equation is satisfied 2 φ x 2 + 2 φ y 2 = 0 cos(x)e y + cos(x)e y = 0 Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 15 / 17
Final solution to the given Dirichlet problem: φ(x, y) = cos(x)e y To confirm solution, check that Laplace s Equation is satisfied 2 φ x 2 + 2 φ y 2 = 0 cos(x)e y + cos(x)e y = 0 and that the boundary condition is satisfied, φ(x, 0) = cos(x). Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 15 / 17
References Dennis Zill and Patrick Shanahan (2015) Complex Analysis: A First Course with Applications 3rd Ed. Some Consequences of the Residue Theorem Section 6.6. Poisson Integral Formula Section 7.4. Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 16 / 17
Thank You Jenna Nelson and Gloria Mayorga (Oxy) Dirichlet April 22, 2016 17 / 17