1 5CCM211A and 6CCM211B Outline Solutions 2 Part II Part II (a) Make the change of variable ξ = by. (b) We have from df dx - bf (x)= φ(x) that d (f 0 )(ξ ) - b b f (ξ )= b φ(ξ ), or, integrating by parts in the first term -iξ b f (ξ ) - b b f (ξ )= b φ(ξ ) using the fact that f (x) → 0 as |x|→ 0 —where? — important! On the other hand, we have b φ(ξ )= Z ∞ 0 e ixξ e -bx dx = 1 b - iξ . Hence b f (ξ )= -1 b 2 + ξ 2 . Therefore, using Fourier inversion, f (x)= 1 2π Z ∞ ∞ -e -ixξ b 2 + ξ 2 dξ = - 1 2b e -b|x| , where we use part (a) for the last equality. Check: show with this f that df dx - bf (x)= e - bx for positive x> 0, and is zero for negative x< 0. At x = 0 the function φ is discontinuous and ‘so’ f is only continuous but not differentiable at zero. (So f is not a ‘classical’ solution at x = 0, only a ‘distributional solution’ (see he web notes) but we will ignore this —se3 3rd and 4th year courses for details.)
