The Question:
if u=f(r) prove that
d2u/dx2+d2u/dy2=f"(r)+1/r f'(r)
where r=(x2+y2)½.
The Solution
f is a function of r, and r is a function of x and y.
use r_x and r_y to denote the (partial) differentials of r wrt x, and wrt y, etc.
use f' and f'' to denote the differentials of f wrt its single argument, r.
r = (x^2+y^2)^(1/2)
r_x = x.r^(-1) = x/r, r_xx = 1/r + x(-r^(-2))r_x = 1/r - x^2/r^3
r_y = y.r^(-1) = y/r, r_yy = 1/r + y(-r^(-2))r_y = 1/r - y^2/r^3
The differentials of r
r_x = x/r, r_xx = 1/r -x^2/r^3
r_y = y/r, r_yy = 1/r -y^2/r^3
du/dx = df/dx = (df/dr)(dr/dx) = f'*r_x
d2u/dx2 = d(f'*r_x)/dx = r_x*d(f')/dx + f'*d(r_x/dx)
d2u/dx2 = r_x*(d(f')/dr)(dr/dx) + f'*r_xx
d2u/dx2 = (r_x)^2*f'' + f'*r_xx
similary,
d2u/dy2 = (r_y)^2*f'' + f'*r_yy
d2u/dx2 + d2/udy2 = (r_x)^2*f'' + f'*r_xx + (r_y)^2*f'' + f'*r_yy
d2u/dx2 + d2u/dy2 = f''{(r_x)^2 + (r_y)^2} + f'{r_xx + r_yy}
substituting for the differentials of r, from above,
d2u/dx2 + d2u/dy2 = f''{(x/r)^2 + (y/r)^2} + f'{(1/r - x^2/r^3) + (1/r - y^2/r^3)}
d2u/dx2 + d2u/dy2 = f''{(x^2 + y^2)/r^2} + f'{(1/r + 1/r - (x^2 + y^2)/r^3)}
d2u/dx2 + d2u/dy2 = f''{r^2/r^2} + f'{(2/r - r^2/r^3)}
d2u/dx2 + d2u/dy2 = f'' + f'{(2/r - 1/r)}
d2u/dx2 + d2u/dy2 = f'' + (1/r)f'