14A (Complex Methods)

L [y] (s) = \frac{1}{π} \int_{0}^{\infty} t^{- 1/2} exp (- s t - \frac{a ^{2}}{4 t}) d t .

Using the supplied Gaussian integral after

t = x^{2}

gives

L [y] (s) = \frac{e ^{- ∣ a ∣ s}}{s} .

b

t

gives

U_{xx} - s U = - f (x) .

The bounded resolvent solution supplied in the hint can be written

U (x, s) = \int_{- \infty}^{\infty} \frac{e ^{- s ∣ x - ξ ∣}}{2 s} f (ξ) d ξ .

Part (a) inverts the kernel, yielding the heat kernel

K (∣ x - ξ ∣, t) = \frac{1}{4 π t} exp (- \frac{( x - ξ ) ^{2}}{4 t}) .

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