Question

2. (i) Verify that
u(x,y;λ) = e−λy cosλx, −∞ < λ < ∞
is a one parameter family of solutions of Laplaces’s equation in R2.
(ii) Find v(x,y;λ) = ∂ ∂λu(x,y;λ) and verify that v(x,y;λ), − ∞ < λ < ∞, is also a one parameter family of solutions of Laplace’s equation in R2.
(iii) For (x,y) in the upper half plane y > 0, the improper integral
v(x,y) = ∞ u(x,y;λ)dλ 0
is convergent. Evaluate this integral and show by direct computation that v(x,y) is a solution of Laplace’s equation in the upper half plane.

Answer 1

(i) We compute the derivatives and find that

$u_{xx}=-\lambda^2e^{-\lambda y}cos(\lambda x),\quad u_{yy}=\lambda^2e^{-\lambda y}cos(\lambda x)$

From the latter it follows that the Laplace equation is satisfied. I.e., $u_{xx}+u_{yy}=0$ .

(ii)

$u_{\lambda}=-\lambda e^{-\lambda y}cos(\lambda x)-e^{-\lambda y}sin(\lambda x),$

$u_{\lambda\lambda}=\lambda^2e^{-\lambda y}cos(\lambda x)+\lambda^2e^{-\lambda y}sin(\lambda x)+\lambda e^{-\lambda y}sin(\lambda x)-\lambda e^{-\lambda y}cos(\lambda x)=$

$=\lambda^2u-\lambda u+\lambda^2 v+\lambda v,$

where $v=e^{-\lambda y}sin(\lambda x)$ . We compute the second order derivatives and get $v_{xx}=-\lambda^2e^{-\lambda y}sin(\lambda x), v_{yy}=\lambda^2e^{-\lambda y}sin(\lambda x)$ .

Thus, $u_{\lambda\lambda}$ also satisfies the Laplace equation.

(iii) At first, we calculate the indefinite integral $\int e^{-\lambda y}cos(\lambda x)d\lambda=-\frac{1}{y}e^{-\lambda y}cos(\lambda x)+$

$-\frac{x}{y}\int e^{-\lambda y}sin(\lambda x)d\lambda=-\frac{1}{y}e^{-\lambda y}cos(\lambda x)+\frac{x}{y^2}e^{-\lambda y}sin(\lambda x)-\frac{x^2}{y^2}\int e^{-\lambda y}cos(\lambda x)d\lambda$

We denote $I=\int e^{-\lambda y}cos(\lambda x)d\lambda$ . Then $I=-\frac{1}{y}e^{-\lambda y}cos(\lambda x)+\frac{x}{y^2}e^{-\lambda y}sin(\lambda x)-\frac{x^2}{y^2}I$ .

From the latter we get: $I=-\frac{y}{x^2+y2}e^{-\lambda y}cos(\lambda x)+\frac{x}{x^2+y^2}e^{-\lambda y}sin(\lambda x)$

Using the latter, we get $v(x,y)=\int_0^{+\infty} e^{-\lambda y}cos(\lambda x)d\lambda=\frac{y}{x^2+y^2}$

The latter has the following derivatives:

$v_{xx}=\frac{2y(3x^2-y^2)}{(y^2+x^2)^3};\qquad v_{yy}=-\frac{2y(3x^2-y^2)}{(y^2+x^2)^3}$

From the latter it is clear that $v_{xx}+v_{yy}=0$ .