Answers to Some of the Problems

$g (t | τ) = e^{- k (t - τ)} H (t - τ)$

$g (t | τ) = \frac{1}{4} [e^{3 (t - τ)} - e^{- (t - τ)}] H (t - τ)$

$g (t | τ) = \frac{1}{2} [e^{- (t - τ)} - e^{- 3 (t - τ)}] H (t - τ)$

$g (t | τ) = \frac{1}{2} \sin [2 (t - τ)] e^{t - τ} H (t - τ)$

$g (t | τ) = [e^{2 (t - τ)} - e^{t - τ}] H (t - τ)$

$g (t | τ) = (t - τ) e^{- 2 (t - τ)} H (t - τ)$

$g (t | τ) = \frac{1}{6} [e^{3 (t - τ)} - e^{- 3 (t - τ)}] H (t - τ)$

$g (t | τ) = \sin (t - τ) H (t - τ)$

$g (t | τ) = [e^{t - τ} - 1] H (t - τ)$

18.

$g (x | ξ) = \frac{(L - x_{>}) (α + x_{<})}{L + α}$

and

$g (x | ξ) = 2 \sum_{n = 1}^{\infty} \frac{1 + α^{2} k_{n}^{2} \sin [k_{n} (L - ξ)] \sin [k_{n} (L - x)]}{[α + L (1 + α^{2} k_{n}^{2})] k_{n}^{2}},$

where kn is the nth root of tan(kL) = −αk.

19.

$g (x | ξ) = (1 + x_{<}) (L - 1 - x_{>}) / L$

and

$g (x | ξ) = \frac{2 e^{x + ξ}}{e^{2 L} - 1} + \frac{2 L^{3}}{π^{2}} \sum_{n = 1}^{\infty} \frac{φ_{n} (ξ) φ_{n} (x)}{n^{2} (n^{2} π^{2} + 1)},$

where $φ_{n} (x) = \sin (n π x / L) + n π \cos (n π x / L) / L .$

20.

$g (x | ξ) = \frac{(1 + x_{<}) (L + 1 - x_{>})}{2 + L}$

and

$g (x | ξ) = 2 \sum_{n = 1}^{\infty} \frac{φ_{n} (ξ) φ_{n} (x)}{(2 + L + k_{n}^{2} L) k_{n}^{2}},$

where $φ_{n} (x) = \sin (k_{n} x) + k_{n} \cos (k_{n} x)$ and kn is the nth root of $\tan (k L) = 2 k / (k^{2} - 1) .$

21.

$g (x | ξ) = \frac{a}{3} - x_{>} + x^{2} + ξ$

Get Green's Functions with Applications, 2nd Edition now with the O’Reilly learning platform.

O’Reilly members experience books, live events, courses curated by job role, and more from O’Reilly and nearly 200 top publishers.

Green's Functions with Applications, 2nd Edition by Dean G. Duffy