# Answers to Some of the Problems

Chapter 3

1.

$g\left(t|\tau \right)={e}^{-k\left(t-\tau \right)}H\left(t-\tau \right)$

2.

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

3.

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

4.

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

5.

$g\left(t|\tau \right)=\left[{e}^{2\left(t-\tau \right)}-{e}^{t-\tau }\right]H\left(t-\tau \right)$

6.

$g\left(t|\tau \right)=\left(t-\tau \right){e}^{-2\left(t-\tau \right)}H\left(t-\tau \right)$

7.

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

8.

$g\left(t|\tau \right)=\mathrm{sin}\left(t-\tau \right)H\left(t-\tau \right)$

9.

$g\left(t|\tau \right)=\left[{e}^{t-\tau }-1\right]H\left(t-\tau \right)$

18.

$g\left(x|\xi \right)=\frac{\left(L-{x}_{>}\right)\left(\alpha +{x}_{<}\right)}{L+\alpha }$

and

$g\left(x|\xi \right)=2\sum _{n=1}^{\infty }\frac{1+{\alpha }^{2}{k}_{n}^{2}\mathrm{sin}\left[{k}_{n}\left(L-\xi \right)\right]\mathrm{sin}\left[{k}_{n}\left(L-x\right)\right]}{\left[\alpha +L\left(1+{\alpha }^{2}{k}_{n}^{2}\right)\right]{k}_{n}^{2}},$

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

19.

$g\left(x|\xi \right)=\left(1+{x}_{<}\right)\left(L-1-{x}_{>}\right)/L$

and

$g\left(x|\xi \right)=\frac{2{e}^{x+\xi }}{{e}^{2L}-1}+\frac{2{L}^{3}}{{\text{π}}^{\text{2}}}\sum _{n=1}^{\infty }\frac{{\phi }_{n}\left(\xi \right){\phi }_{n}\left(x\right)}{{n}^{2}\left({n}^{2}{\text{π}}^{\text{2}}+1\right)},$

where

20.

$g\left(x|\xi \right)=\frac{\left(1+{x}_{<}\right)\left(L+1-{x}_{>}\right)}{2+L}$

and

$g\left(x|\xi \right)=2\sum _{n=1}^{\infty }\frac{{\phi }_{n}\left(\xi \right){\phi }_{n}\left(x\right)}{\left(2+L+{k}_{n}^{2}L\right){k}_{n}^{2}},$

where ${\phi }_{n}\left(x\right)=\mathrm{sin}\left({k}_{n}x\right)+{k}_{n}\mathrm{cos}\left({k}_{n}x\right)$ and kn is the nth root of $\mathrm{tan}\left(kL\right)=2k/\left({k}^{2}-1\right).$

21.

$g\left(x|\xi \right)=\frac{a}{3}-{x}_{>}+\frac{{x}^{2}+{\xi }^{}}{}$

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