Chapter 15

We show that \(\frac{1}{4} \bar{F}_{4} F_{4}=I:\) $$\frac{1}{4} \bar{F}_{4} F_{4}=\frac{1}{4}\left(\begin{array}{rrrr}1 & 1 & 1 & 1 \\\1 & -i & -1 & i \\ 1 & -1 & 1 & -1 \\\1 & i & -1 & -i\end{array}\right)\left(\begin{array}{rrrr}1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\\1 & -1 & 1 & -1 \\\1 & -i & -1 & i\end{array}\right)=\frac{1}{4}\left(\begin{array}{cccc} 4 & 0 & 0 & 0 \\\0 & 4 & 0 & 0 \\\0 & 0 & 4 & 0 \\\0 & 0 & 0 & 4\end{array}\right)=I$$ Thus \(F_{4}^{-1}=\frac{1}{4} \bar{F}_{4}\)

The boundary-value problem is $$a^{2} \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial t^{2}}, \quad 0 < x< L, \quad t > 0$$ $$u(0, t)=0, \quad u(L, t)=0, \quad t>0$$ $$u(x, 0)=A \sin \frac{\pi}{L} x,\left.\quad \frac{\partial u}{\partial t}\right|_{t=0}=0$$ Transforming the partial differential equation gives $$\frac{d^{2} U}{d x^{2}}-\left(\frac{s}{a}\right)^{2} U=-\frac{s}{a^{2}} A \sin \frac{\pi}{L} x$$ Using undetermined coefficients we obtain $$U(x, s)=c_{1} \cosh \frac{s}{a} x+c_{2} \sinh \frac{s}{a} x+\frac{A s}{s^{2}+a^{2} \pi^{2} / L^{2}} \sin \frac{\pi}{L} x$$ The transformed boundary conditions, \(U(0, s)=0, U(L, s)=0\) give in turn \(c_{1}=0\) and \(c_{2}=0 .\) Therefore $$U(x, s)=\frac{A s}{s^{2}+a^{2} \pi^{2} / L^{2}} \sin \frac{\pi}{L} x$$ and $$u(x, t)=A \mathscr{L}-1\left\\{\frac{s}{s^{2}+a^{2} \pi^{2} / L^{2}}\right\\} \sin \frac{\pi}{L} x=A \cos \frac{a \pi}{L} t \sin \frac{\pi}{L} x$$

(a) The result follows by letting \(\tau=u^{2}\) or \(u=\sqrt{\tau}\) in erf \((\sqrt{t})=\frac{2}{\sqrt{\pi}} \int_{0}^{\sqrt{t}} e^{-u^{2}} d u.\) (b) Using \(\mathscr{L}\left\\{t^{-1 / 2}\right\\}=\frac{\sqrt{\pi}}{s^{1 / 2}}\) and the first translation theorem, it follows from the convolution theorem that $$\begin{aligned} \mathscr{L}\\{\operatorname{erf}(\sqrt{t})\\} &=\frac{1}{\sqrt{\pi}} \mathscr{L}\left\\{\int_{0}^{t} \frac{e^{-\tau}}{\sqrt{\tau}} d \tau\right\\}=\frac{1}{\sqrt{\pi}} \mathscr{L}\\{1\\} \mathscr{L}\left\\{t^{-1 / 2} e^{-t}\right\\}=\left.\frac{1}{\sqrt{\pi}} \frac{1}{s} \mathscr{L}\left\\{t^{-1 / 2}\right\\}\right|_{s \rightarrow s+1} \\\&=\frac{1}{\sqrt{\pi}} \frac{1}{s} \frac{\sqrt{\pi}}{\sqrt{s+1}}=\frac{1}{s \sqrt{s+1}}\end{aligned}$$

since erfc \((\sqrt{t})=1-\operatorname{erf}(\sqrt{t})\) we have $$\mathscr{L}\\{\operatorname{erfc}(\sqrt{t})\\}=\mathscr{L}\\{1\\}-\mathscr{L}\\{\operatorname{erf}(\sqrt{t})\\}=\frac{1}{s}-\frac{1}{s \sqrt{s+1}}=\frac{1}{s}\left[1-\frac{1}{\sqrt{s+1}}\right].$$

The solution of $$a^{2} \frac{d^{2} U}{d x^{2}}-s^{2} U=0$$ is in this case $$U(x, s)=c_{1} e^{-(x / a) s}+c_{2} e^{(x / a) s}$$ since \(\lim _{x \rightarrow \infty} u(x, t)=0\) we have \(\lim _{x \rightarrow \infty} U(x, s)=0 .\) Thus \(c_{2}=0\) and $$U(x, s)=c_{1} e^{-(x / a) s}$$ If \(\mathscr{L}\\{u(0, t)\\}=\mathscr{L}\\{f(t)\\}=F(s)\) then \(U(0, s)=F(s) .\) From this we have \(c_{1}=F(s)\) and $$U(x, s)=F(s) e^{-(x / a) s}$$ Hence, by the second translation theorem, $$u(x, t)=f\left(t-\frac{x}{a}\right) \mathscr{U}\left(t-\frac{x}{a}\right)$$

From formulas (5) and (6) in the text $$\begin{aligned} A(\alpha) &=\int_{0}^{3} x \cos \alpha x d x=\left.\frac{x \sin \alpha x}{\alpha}\right|_{0} ^{3}-\frac{1}{\alpha} \int_{0}^{3} \sin \alpha x d x \\ &=\frac{3 \sin 3 \alpha}{\alpha}+\left.\frac{\cos \alpha x}{\alpha^{2}}\right|_{0} ^{3}=\frac{3 \alpha \sin 3 \alpha+\cos 3 \alpha-1}{\alpha^{2}} \end{aligned}$$ and $$\begin{aligned} B(\alpha) &=\int_{0}^{3} x \sin \alpha x d x=-\left.\frac{x \cos \alpha x}{\alpha}\right|_{0} ^{3}+\frac{1}{\alpha} \int_{0}^{3} \cos \alpha x d x \\ &=-\frac{3 \cos 3 \alpha}{\alpha}+\left.\frac{\sin \alpha x}{\alpha^{2}}\right|_{0} ^{3}=\frac{\sin 3 \alpha-3 \alpha \cos 3 \alpha}{\alpha^{2}} \end{aligned}$$ Hence $$\begin{aligned} f(x) &=\frac{1}{\pi} \int_{0}^{\infty} \frac{(3 \alpha \sin 3 \alpha+\cos 3 \alpha-1) \cos \alpha x+(\sin 3 \alpha-3 \alpha \cos 3 \alpha) \sin \alpha x}{\alpha^{2}} d \alpha \\ &=\frac{1}{\pi} \int_{0}^{\infty} \frac{3 \alpha(\sin 3 \alpha \cos \alpha x-\cos 3 \alpha \sin \alpha x)+\cos 3 \alpha \cos \alpha x+\sin 3 \alpha \sin \alpha x-\cos \alpha x}{\alpha^{2}} d \alpha \\ &=\frac{1}{\pi} \int_{0}^{\infty} \frac{3 \alpha \sin \alpha(3-x)+\cos \alpha(3-x)-\cos \alpha x}{\alpha^{2}} d \alpha \end{aligned}$$

By the sifting property, $$\mathscr{F}\\{\delta(x)\\}=\int_{-\infty}^{\infty} \delta(x) e^{i \alpha x} d x=e^{i \alpha 0}=1$$

The solution of Problem 3 can be written $$u(x, t)=\frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} d \alpha-\frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-k \alpha^{2} t} d \alpha.$$ Using $$\int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} d \alpha=\pi / 2$$ the last line becomes $$u(x, t)=u_{0}-\frac{2 u_{0}}{\pi} \int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-k \alpha^{2} t} d \alpha.$$

We already know that \(f * \delta=\delta * f .\) Then, by the sifting property, $$(f * \delta)(x)=\int_{\infty}^{\infty} f(\tau) \delta(x-\tau) d \tau=\int_{-\infty}^{\infty} f(\tau) \delta(\tau-x) d \tau=f(x)$$

Using the Fourier sine transform we find $$U(\alpha, t)=c e^{-k \alpha^{2} t}.$$ Now $$\mathscr{F}_{S}\\{u(x, 0)\\}=U(\alpha, 0)=\int_{0}^{1} \sin \alpha x d x=\frac{1-\cos \alpha}{\alpha}.$$ From this we find \(c=(1-\cos \alpha) / \alpha\) and so $$U(\alpha, t)=\frac{1-\cos \alpha}{\alpha} e^{-k \alpha^{2} t}$$ and $$u(x, t)=\frac{2}{\pi} \int_{0}^{\infty} \frac{1-\cos \alpha}{\alpha} e^{-k \alpha^{2} t} \sin \alpha x d \alpha.$$

From formula (5) in the text $$A(\alpha)=\int_{0}^{\infty} e^{-x} \cos \alpha x d x$$ Recall \(\mathscr{L}\\{\cos k t\\}=s /\left(s^{2}+k^{2}\right) .\) If we set \(s=1\) and \(k=\alpha\) we obtain $$A(\alpha)=\frac{1}{1+\alpha^{2}}$$ Now $$B(\alpha)=\int_{0}^{\infty} e^{-x} \sin \alpha x d x$$ Recall \(\mathscr{L}\\{\sin k t\\}=k /\left(s^{2}+k^{2}\right) .\) If we set \(s=1\) and \(k=\alpha\) we obtain $$B(\alpha)=\frac{\alpha}{1+\alpha^{2}}$$ Hence $$f(x)=\frac{1}{\pi} \int_{0}^{\infty} \frac{\cos \alpha x+\alpha \sin \alpha x}{1+\alpha^{2}} d \alpha$$

From formulas (5) and (6) in the text $$\begin{aligned} A(\alpha) &=\int_{-1}^{1} e^{x} \cos \alpha x d x \\ &=\frac{e(\cos \alpha+\alpha \sin \alpha)-e^{-1}(\cos \alpha-\alpha \sin \alpha)}{1+\alpha^{2}} \\ &=\frac{2(\sinh 1) \cos \alpha-2 \alpha(\cosh 1) \sin \alpha}{1+\alpha^{2}} \end{aligned}$$ and $$\begin{aligned} B(\alpha) &=\int_{-1}^{1} e^{x} \sin \alpha x d x \\ &=\frac{e(\sin \alpha-\alpha \cos \alpha)-e^{-1}(-\sin \alpha-\alpha \cos \alpha)}{1+\alpha^{2}} \\ &=\frac{2(\cosh 1) \sin \alpha-2 \alpha(\sinh 1) \cos \alpha}{1+\alpha^{2}} \end{aligned}$$ Hence $$f(x)=\frac{1}{\pi} \int_{0}^{\infty}[A(\alpha) \cos \alpha x+B(\alpha) \sin \alpha x] d \alpha$$

We first compute $$\frac{\sinh a \sqrt{s}}{s \sinh \sqrt{s}}=\frac{e^{a \sqrt{s}}-e^{-a \sqrt{s}}}{s\left(e^{\sqrt{s}}-e^{-\sqrt{s}}\right)}=\frac{e^{(a-1) \sqrt{s}}-e^{-(a+1) \sqrt{s}}}{s\left(1-e^{-2 \sqrt{s}}\right)}$$ $$\begin{array}{l}=\frac{e^{(a-1) \sqrt{s}}}{s}\left[1+e^{-2 \sqrt{s}}+e^{-4 \sqrt{s}}+\cdots\right]-\frac{e^{-(a+1) \sqrt{s}}}{s}\left[1+e^{-2 \sqrt{s}}+e^{-4 \sqrt{s}}+\cdots\right] \\\=\left[\frac{e^{-(1-a) \sqrt{s}}}{s}+\frac{e^{-(3-a) \sqrt{s}}}{s}+\frac{e^{-(5-a) \sqrt{s}}}{s}+\cdots\right] \\\\\quad-\left[\frac{e^{-(1+a) \sqrt{s}}}{s}+\frac{e^{-(3+a) \sqrt{s}}}{s}+\frac{e^{-(5+a) \sqrt{s}}}{s}+\cdots\right]\\\=\sum_{n=0}^{\infty}\left[\frac{e^{-(2 n+1-a) \sqrt{s}}}{s}-\frac{e^{-(2 n+1+a) \sqrt{s}}}{s}\right]\end{array}$$ Then $$\mathscr{L}^{-1}\left\\{\frac{\sinh a \sqrt{s}}{s \sinh \sqrt{s}}\right\\}=\sum_{n=0}^{\infty}\left[\mathscr{L}^{-1}\left\\{\frac{e^{-(2 n+1-a) \sqrt{s}}}{s}\right\\}-\mathscr{L}^{-1}\left\\{-\frac{e^{-(2 n+1+a) \sqrt{s}}}{s}\right\\}\right]$$ $$\begin{array}{l}=\sum_{n=0}^{\infty}\left[\operatorname{erfc}\left(\frac{2 n+1-a}{2 \sqrt{t}}\right)-\operatorname{erfc}\left(\frac{2 n+1+a}{2 \sqrt{t}}\right)\right] \\\=\sum_{n=0}^{\infty}\left(\left[1-\operatorname{erf}\left(\frac{2 n+1-a}{2 \sqrt{t}}\right)\right]-\left[1-\operatorname{erf}\left(\frac{2 n+1+a}{2 \sqrt{t}}\right)\right]\right) \\\=\sum_{n=0}^{\infty}\left[\operatorname{erf}\left(\frac{2 n+1+a}{2 \sqrt{t}}\right)-\operatorname{erf}\left(\frac{2 n+1-a}{2 \sqrt{t}}\right)\right].\end{array}$$

We use $$U(x, s)=c_{1} \cosh \frac{s}{a} x+c_{2} \sinh \frac{s}{a} x$$ Now \(U(0, s)=0\) implies \(c_{1}=0,\) so \(U(x, s)=c_{2} \sinh (s / a) x .\) The condition \(E d U /\left.d x\right|_{x=L}=F_{0}\) then yields \(c_{2}=F_{0} a / E s \cosh (s / a) L \) and so $$U(x, s)=\frac{a F_{0}}{E s} \frac{\sinh (s / a) x}{\cosh (s / a) L}=\frac{a F_{0}}{E s} \frac{e^{(s / a) x}-e^{-(s / a) x}}{e^{(s / a) L}+e^{-(s / a) L}}$$ $$=\frac{a F_{0}}{E s} \frac{e^{(s / a)(x-L)}-e^{-(s / a)(x+L)}}{1+e^{-2 s L / a}}$$ $$=\frac{a F_{0}}{E}\left[\frac{e^{-(s / a)(L-x)}}{s}-\frac{e^{-(s / a)(3 L-x)}}{s}+\frac{e^{-(s / a)(5 L-x)}}{s}-\cdots\right]$$ $$-\frac{a F_{0}}{E}\left[\frac{e^{-(s / a)(L+x)}}{s}-\frac{e^{-(s / a)(3 L+x)}}{s}+\frac{e^{-(s / a)(5 L+x)}}{s}-\cdots\right]$$ $$=\frac{a F_{0}}{E} \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{e^{-(s / a)(2 n L+L-x)}}{s}-\frac{e^{-(s / a)(2 n L+L+x)}}{s}\right]$$ and $$u(x, t)=\frac{a F_{0}}{E} \sum_{n=0}^{\infty}(-1)^{n}\left[\mathscr{L}-1\left\\{\frac{e^{-(s / a)(2 n L+L-x)}}{s}\right\\}-\mathscr{L}-1\left\\{\frac{e^{-(s / a)(2 n L+L+x)}}{s}\right\\}\right]$$ $$=\frac{a F_{0}}{E} \sum_{n=0}^{\infty}(-1)^{n}\left[\left(t-\frac{2 n L+L-x}{a}\right) \mathscr{U}\left(t-\frac{2 n L+L-x}{a}\right)\right.$$ $$\left.-\left(t-\frac{2 n L+L+x}{a}\right)^{\prime} U\left(t-\frac{2 n L+L+x}{a}\right)\right]$$

The 8 th roots of unity, \(\omega_{8}^{1}, \omega_{8}^{2}, \ldots, \omega_{8}^{8}\) are shown in the solution of Problem 7 above. The points in the complex plane are equally spaced on the perimeter of the unit circle.

$$\begin{aligned}\int_{a}^{b} e^{-u^{2}} d u &=\int_{a}^{0} e^{-u^{2}} d u+\int_{0}^{b} e^{-u^{2}} d u=\int_{0}^{b} e^{-u^{2}} d u-\int_{0}^{a} e^{-u^{2}} d u \\\&=\frac{\sqrt{\pi}}{2} \operatorname{erf}(b)-\frac{\sqrt{\pi}}{2} \operatorname{erf}(a)=\frac{\sqrt{\pi}}{2}[\operatorname{erf}(b)-\operatorname{erf}(a)] \end{aligned}$$

Using the Fourier sine transform we obtain $$U(\alpha, t)=c_{1} \cos \alpha a t+c_{2} \sin \alpha a t.$$ Now $$\mathscr{F}_{S}\\{u(x, 0)\\}=\mathscr{F}_{S}\left\\{x e^{-x}\right\\}=\int_{0}^{\infty} x e^{-x} \sin \alpha x d x=\frac{2 \alpha}{\left(1+\alpha^{2}\right)^{2}}=U(\alpha, 0).$$ Also, $$\mathscr{F}_{S}\left\\{u_{t}(x, 0)\right\\}=\left.\frac{d U}{d t}\right|_{t=0}=0.$$ This last condition gives \(c_{2}=0 .\) Then \(U(\alpha, 0)=2 \alpha /\left(1+\alpha^{2}\right)^{2}\) yields \(c_{1}=2 \alpha /\left(1+\alpha^{2}\right)^{2}\). Therefore $$U(\alpha, t)=\frac{2 \alpha}{\left(1+\alpha^{2}\right)^{2}} \cos \alpha a t$$ and $$u(x, t)=\frac{4}{\pi} \int_{0}^{\infty} \frac{\alpha \cos \alpha a t}{\left(1+\alpha^{2}\right)^{2}} \sin \alpha x d \alpha.$$

The function is odd. Thus from formula (11) in the text $$B(\alpha)=\int_{0}^{\infty} x e^{-x} \sin \alpha x d x$$ Now recall $$\mathscr{L}\\{t \sin k t\\}=-\frac{d}{d s} \mathscr{L}\\{\sin k t\\}=2 k s /\left(s^{2}+k^{2}\right)^{2}$$ If we set \(s=1\) and \(k=\alpha\) we obtain $$B(\alpha)=\frac{2 \alpha}{\left(1+\alpha^{2}\right)^{2}}$$ Hence from formula (10) in the text $$f(x)=\frac{4}{\pi} \int_{0}^{\infty} \frac{\alpha \sin \alpha x}{\left(1+\alpha^{2}\right)^{2}} d \alpha$$

We use $$U(x, s)=c_{1} e^{-\sqrt{s} x}+c_{2} e^{\sqrt{s} x}$$ The condition \(\lim _{x \rightarrow \infty} u(x, t)=0\) implies \(\lim _{x \rightarrow \infty} U(x, s)=0,\) so we define \(c_{2}=0 .\) Hence $$U(x, s)=c_{1} e^{-\sqrt{s} x}$$ The transform of \(u(0, t)=f(t)\) is \(U(0, s)=F(s) .\) Therefore $$U(x, s)=F(s) e^{-\sqrt{s} x}$$ and $$u(x, t)=\mathscr{L}^{-1}\left\\{F(s) e^{-x \sqrt{s}}\right\\}=\frac{x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{f(t-\tau) e^{-x^{2} / 4 \tau}}{\tau^{3 / 2}} d \tau$$

We use the Fourier sine transform with respect to \(x\) to obtain $$U(\alpha, y)=c_{1} \cosh \alpha y+c_{2} \sinh \alpha y.$$ The transforms of \(u(x, 0)=f(x)\) and \(u(x, 2)=0\) give, in turn, \(U(\alpha, 0)=F(\alpha)\) and \(U(\alpha, 2)=0 .\) The first condition gives \(c_{1}=F(\alpha)\) and the second condition then yields $$c_{2}=-\frac{F(\alpha) \cosh 2 \alpha}{\sinh 2 \alpha}.$$ Hence $$\begin{aligned} U(\alpha, y) &=F(\alpha) \cosh \alpha y-\frac{F(\alpha) \cosh 2 \alpha \sinh \alpha y}{\sinh 2 \alpha} \\ &=F(\alpha) \frac{\sinh 2 \alpha \cosh \alpha y-\cosh 2 \alpha \sinh \alpha y}{\sinh 2 \alpha} \\ &=F(\alpha) \frac{\sinh \alpha(2-y)}{\sinh 2 \alpha} \end{aligned}$$ and $$u(x, y)=\frac{2}{\pi} \int_{0}^{\infty} F(\alpha) \frac{\sinh \alpha(2-y)}{\sinh 2 \alpha} \sin \alpha x d \alpha.$$

We solve the three boundary-value problems: Using separation of variables we find the solution of the first problem is $$u_{1}(x, y)=\sum_{n=1}^{\infty} A_{n} e^{-n y} \sin n x \quad \text { where } \quad A_{n}=\frac{2}{\pi} \int_{0}^{\pi} f(x) \sin n x d x.$$ Using the Fourier sine transform with respect to \(y\) gives the solution of the second problem: $$u_{2}(x, y)=\frac{200}{\pi} \int_{0}^{\infty} \frac{(1-\cos \alpha) \sinh \alpha(\pi-x)}{\alpha \sinh \alpha \pi} \sin \alpha y d \alpha.$$ Also, the Fourier sine transform with respect to \(y\) gives the solution of the third problem: $$u_{3}(x, y)=\frac{2}{\pi} \int_{0}^{\infty} \frac{\alpha \sinh \alpha x}{\left(1+\alpha^{2}\right) \sinh \alpha \pi} \sin \alpha y d \alpha.$$ The solution of the original problem is $$u(x, y)=u_{1}(x, y)+u_{2}(x, y)+u_{3}(x, y).$$

(a) From formula (7) in the text with \(x=2,\) we have $$\frac{1}{2}=\frac{2}{\pi} \int_{0}^{\infty} \frac{\sin \alpha \cos \alpha}{\alpha} d \alpha=\frac{1}{\pi} \int_{0}^{\infty} \frac{\sin 2 \alpha}{\alpha} d \alpha$$ If we let \(\alpha=x\) we obtain $$\int_{0}^{\infty} \frac{\sin 2 x}{x} d x=\frac{\pi}{2}$$ (b) If we now let \(2 x=k t\) where \(k > 0\), then \(d x=(k / 2) d t\) and the integral in part (a) becomes

Transforming the partial differential equation gives $$k \frac{d^{2} U}{d x^{2}}-s U=-\frac{r}{s}$$ Using undetermined coefficients we obtain $$U(x, s)=c_{1} e^{-\sqrt{s / k} x}+c_{2} e^{\sqrt{s / k} x}+\frac{r}{s^{2}}$$ The condition \(\lim _{x \rightarrow \infty} \partial u / \partial x=0\) implies \(\lim _{x \rightarrow \infty} d U / d x=0,\) so we define \(c_{2}=0 .\) The transform of the remaining boundary condition gives \(U(0, s)=0 .\) This condition yields \(c_{1}=-r / s^{2} .\) Thus $$U(x, s)=r\left[\frac{1}{s^{2}}-\frac{e^{-\sqrt{s / k} x}}{s^{2}}\right]$$ Using the Table of Laplace transforms and the convolution theorem we obtain $$u(x, t)=r \mathscr{L}^{-1}\left\\{\frac{1}{s^{2}}-\frac{1}{s} \cdot \frac{e^{-\sqrt{s / k} x}}{s}\right\\}=r t-r \int_{0}^{t} \operatorname{erfc}\left(\frac{x}{2 \sqrt{k \tau}}\right) d \tau$$

The solution of $$\frac{d^{2} U}{d x^{2}}-s U=-u_{0}-u_{0} \sin \frac{\pi}{L} x$$ is $$U(x, s)=c_{1} \cosh (\sqrt{s} x)+c_{2} \sinh (\sqrt{s} x)+\frac{u_{0}}{s}+\frac{u_{0}}{s+\pi^{2} / L^{2}} \sin \frac{\pi}{L} x$$ The transformed boundary conditions \(U(0, s)=u_{0} / s\) and \(U(L, s)=u_{0} / s\) give, in turn, \(c_{1}=0\) and \(c_{2}=0\) Therefore $$U(x, s)=\frac{u_{0}}{s}+\frac{u_{0}}{s+\pi^{2} / L^{2}} \sin \frac{\pi}{L} x$$ and $$u(x, t)=u_{0} \mathscr{L}^{-1}\left\\{\frac{1}{s}\right\\}+u_{0} \mathscr{L}^{-1}\left\\{\frac{1}{s+\pi^{2} / L^{2}}\right\\} \sin \frac{\pi}{L} x=u_{0}+u_{0} e^{-\pi^{2} t / L^{2}} \sin \frac{\pi}{L} x$$

Using the Fourier transform with respect to \(x\) gives $$U(\alpha, y)=c_{1} \cosh \alpha y+c_{2} \sinh \alpha y.$$ The transform of the boundary condition \(\partial u /\left.\partial y\right|_{y=0}=0\) is \(d U /\left.d y\right|_{y=0}=0 .\) This condition gives \(c_{2}=0\) Hence $$U(\alpha, y)=c_{1} \cosh \alpha y.$$ Now by the given information the transform of the boundary condition \(u(x, 1)=e^{-x^{2}}\) is \(U(\alpha, 1)=\sqrt{\pi} e^{-\alpha^{2} / 4}.\) This condition then gives \(c_{1}=\sqrt{\pi} e^{-\alpha^{2} / 4} \cosh \alpha .\) Therefore $$U(\alpha, y)=\sqrt{\pi} \frac{e^{-\alpha^{2} / 4} \cosh \alpha y}{\cosh \alpha}$$ and $$\begin{aligned} U(x, y) &=\frac{1}{2 \sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-\alpha^{2} / 4} \cosh \alpha y}{\cosh \alpha} e^{-i \alpha x} d \alpha=\frac{1}{2 \sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-\alpha^{2} / 4} \cosh \alpha y}{\cosh \alpha} \cos \alpha x d \alpha \\ &=\frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-\alpha^{2} / 4} \cosh \alpha y}{\cosh \alpha} \cos \alpha x d \alpha. \end{aligned}$$

(a) From the identity $$\sin A \cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)]$$ we have $$\begin{aligned} \sin \alpha \cos \alpha x &=\frac{1}{2}[\sin (\alpha+\alpha x)+\sin (\alpha-\alpha x)] \\ &=\frac{1}{2}[\sin \alpha(1+x)+\sin \alpha(1-x)] \\ &=\frac{1}{2}[\sin \alpha(x+1)-\sin \alpha(x-1)] \end{aligned}$$ Then $$\frac{2}{\pi} \int_{0}^{\infty} \frac{\sin \alpha \cos \alpha x}{\alpha} d \alpha=\frac{1}{\pi} \int_{0}^{\infty} \frac{\sin \alpha(x+1)-\sin \alpha(x-1)}{\alpha} d \alpha$$ (b) Noting that $$\begin{aligned} F_{b} &=\frac{1}{\pi} \int_{0}^{b} \frac{\sin \alpha(x+1)-\sin \alpha(x-1)}{\alpha} d \alpha \\ &=\frac{1}{\pi}\left[\int_{0}^{b} \frac{\sin \alpha(x+1)}{\alpha} d \alpha-\int_{0}^{b} \frac{\sin \alpha(x-1)}{\alpha} d \alpha\right] \end{aligned}$$ and letting \(t=\alpha(x+1)\) so that \(d t=(x+1) d \alpha\) in the first integral and \(t=\alpha(x-1)\) so that \(d t=(x-1) d \alpha\) in the second integral we have $$F_{b}=\frac{1}{\pi}\left[\int_{0}^{b(x+1)} \frac{\sin t}{t} d t-\int_{0}^{b(x-1)} \frac{\sin t}{t} d t\right]$$ since \(\operatorname{Si}(x)=\int_{0}^{x}[(\sin t) / t] d t,\) this becomes $$F_{b}=\frac{1}{\pi}[\operatorname{Si}(b(x+1))-\operatorname{Si}(b(x-1))]$$ (c) In Mathematica we define \(\mathbf{f}\left[\mathbf{b}_{-}\right]:=(\mathbf{1} / \mathbf{P} \mathbf{i})(\sin \operatorname{Integral}[\mathbf{b}(\mathbf{x}+\mathbf{1})]-\sin \operatorname{Integral}[\mathbf{b}(\mathbf{x}-\mathbf{1})] .\) Graphs of \(F_{b}(x)\) for \(b=4,6,15,\) and 75 are shown below.

From the Table of Laplace transforms we have $$\int_{0}^{\infty} e^{-s t} \frac{\sin a t}{t} d t=\arctan \frac{a}{s}$$ and $$\int_{0}^{\infty} e^{-s t} \frac{\sin a t \cos b t}{t} d t=\frac{1}{2} \arctan \frac{a+b}{s}+\frac{1}{2} \arctan \frac{a-b}{s}.$$ Identifying \(\alpha=t, x=a,\) and \(y=s,\) the solution of Problem 14 is $$\begin{aligned} u(x, y) &=\frac{100}{\pi} \int_{0}^{\infty} \frac{1-\cos \alpha}{\alpha} e^{-\alpha y} \sin \alpha x d \alpha \\ &=\frac{100}{\pi}\left[\int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-\alpha y} d \alpha-\int_{0}^{\infty} \frac{\sin \alpha x \cos \alpha}{\alpha} e^{-\alpha y} d \alpha\right] \\ &=\frac{100}{\pi}\left[\arctan \frac{x}{y}-\frac{1}{2} \arctan \frac{x+1}{y}-\frac{1}{2} \arctan \frac{x-1}{y}\right]. \end{aligned}$$

We use $$U(x, s)=c_{1} \cosh \sqrt{\frac{s}{k}} x+c_{2} \sinh \sqrt{\frac{s}{k}} x+\frac{u_{0}}{s}$$ The transformed boundary conditions \(d U /\left.d x\right|_{x=0}=0\) and \(U(1, s)=0\) give, in turn, \(c_{2}=0\) and \(c_{1}=-u_{0} / s \cosh \sqrt{s / k} .\) Therefore $$U(x, s)=\frac{u_{0}}{s}-\frac{u_{0} \cosh \sqrt{s / k} x}{s \cosh \sqrt{s / k}}=\frac{u_{0}}{s}-u_{0} \frac{e^{\sqrt{s / k}} x+e^{-\sqrt{s / k} x}}{s\left(e^{\sqrt{s / k}}+e^{-\sqrt{s / k}}\right)}$$ $$=\frac{u_{0}}{s}-u_{0} \frac{e^{\sqrt{s / k}(x-1)}+e^{-\sqrt{s / k}(x+1)}}{s\left(1+e^{-2 \sqrt{s / k}}\right)}$$ $$=\frac{u_{0}}{s}-u_{0}\left[\frac{e^{-\sqrt{s / k}(1-x)}}{s}-\frac{e^{-\sqrt{s / k}(3-x)}}{s}+\frac{e^{-\sqrt{s / k}(5-x)}}{s}-\cdots\right]$$ $$-u_{0}\left[\frac{e^{-\sqrt{s / k}(1+x)}}{s}-\frac{e^{-\sqrt{s / k}(3+x)}}{s}+\frac{e^{-\sqrt{s / k}(5+x)}}{s}-\cdots\right]$$ $$=\frac{u_{0}}{s}-u_{0} \sum_{n=0}^{\infty}(-1)^{n}\left[\frac{e^{-(2 n+1-x) \sqrt{s} / \sqrt{k}}}{s}+\frac{e^{-(2 n+1+x) \sqrt{s} / \sqrt{k}}}{s}\right]$$ and $$\begin{aligned} u(x, t) &=u_{0} \mathscr{L}^{-1}\left\\{\frac{1}{s}\right\\}-u_{0} \sum_{n=0}^{\infty}(-1)^{n}\left[\mathscr{L}^{-1}\left\\{\frac{e^{-(2 n+1-x) \sqrt{s} / \sqrt{k}}}{s}\right\\}-\mathscr{L}^{-1}\left\\{\frac{e^{-(2 n+1+x) \sqrt{s} / \sqrt{k}}}{s}\right\\}\right] \\ &=u_{0}-u_{0} \sum_{n=0}^{\infty}(-1)^{n}\left[\operatorname{erfc}\left(\frac{2 n+1-x}{2 \sqrt{k t}}\right)-\operatorname{erfc}\left(\frac{2 n+1+x}{2 \sqrt{k t}}\right)\right] \end{aligned}$$

since erf \((0)=0\) and \(\lim _{x \rightarrow \infty}\) erf \((x)=1,\) we have $$\lim _{t \rightarrow \infty} u(x, t)=50[\text { erf }(0)-\text { erf }(0)]=0$$ and $$\lim _{x \rightarrow \infty} u(x, t)=50[\text { erf }(\infty)-\operatorname{erf}(\infty)]=50[1-1]=0.$$

(a) We use $$U(x, s)=c_{1} e^{-(s / a) x}+c_{2} e^{(s / a) x}+\frac{v_{0}^{2} F_{0}}{\left(a^{2}-v_{0}^{2}\right) s^{2}} e^{-\left(s / v_{0}\right) x}$$ The condition \(\lim _{x \rightarrow \infty} u(x, t)=0\) implies \(\lim _{x \rightarrow \infty} U(x, s)=0,\) so we must define \(c_{2}=0 .\) Consequently $$U(x, s)=c_{1} e^{-(s / a) x}+\frac{v_{0}^{2} F_{0}}{\left(a^{2}-v_{0}^{2}\right) s^{2}} e^{-\left(s / v_{0}\right) x}$$ The remaining boundary condition transforms into \(U(0, s)=0 .\) From this we find $$c_{1}=-v_{0}^{2} F_{0} /\left(a^{2}-v_{0}^{2}\right) s^{2}$$ Therefore, by the second translation theorem $$U(x, s)=-\frac{v_{0}^{2} F_{0}}{\left(a^{2}-v_{0}^{2}\right) s^{2}} e^{-(s / a) x}+\frac{v_{0}^{2} F_{0}}{\left(a^{2}-v_{0}^{2}\right) s^{2}} e^{-\left(s / v_{0}\right) x}$$ and $$\begin{aligned} u(x, t) &=\frac{v_{0}^{2} F_{0}}{a^{2}-v_{0}^{2}}\left[\mathscr{L}^{-1}\left\\{\frac{e^{-\left(x / v_{0}\right) s}}{s^{2}}\right\\}-\mathscr{L}^{-1}\left\\{\frac{e^{-(x / a) s}}{s^{2}}\right\\}\right] \\ &=\frac{v_{0}^{2} F_{0}}{a^{2}-v_{0}^{2}}\left[\left(t-\frac{x}{v_{0}}\right)^{2} U\left(t-\frac{x}{v_{0}}\right)-\left(t-\frac{x}{a}\right)^{2} U\left(t-\frac{x}{a}\right)\right] \end{aligned}$$ In the case when \(v_{0}=a\) the solution of the transformed equation is $$U(x, s)=c_{1} e^{-(s / a) x}+c_{2} e^{(s / a) x}-\frac{F_{0}}{2 a s} x e^{-(s / a) x}$$ The usual analysis then leads to \(c_{1}=0\) and \(c_{2}=0 .\) Therefore $$U(x, s)=-\frac{F_{0}}{2 a s} x e^{-(s / a) x}$$ and $$u(x, t)=-\frac{x F_{0}}{2 a} \mathscr{L}-1\left\\{\frac{e^{-(x / a) s}}{s}\right\\}=-\frac{x F_{0}}{2 a} \mathscr{U}\left(t-\frac{x}{a}\right)$$

We use $$U(x, s)=c_{1} e^{-\sqrt{s+h} x}+c_{2} e^{\sqrt{s+h} x}$$ The condition \(\lim _{x \rightarrow \infty} u(x, t)=0\) implies \(\lim _{x \rightarrow \infty} U(x, s)=0,\) so we take \(c_{2}=0 .\) Therefore $$U(x, s)=c_{1} e^{-\sqrt{s+h} x}$$ The Laplace transform of \(u(0, t)=u_{0}\) is \(U(0, s)=u_{0} / s\) and so $$U(x, s)=u_{0} \frac{e^{-\sqrt{s+h} x}}{s}$$ and $$u(x, t)=u_{0} \mathscr{L}^{-1}\left\\{\frac{e^{-\sqrt{s+h} x}}{s}\right\\}=u_{0} \mathscr{L}^{-1}\left\\{\frac{1}{s} e^{-\sqrt{s+h} x}\right\\}$$ From the first translation theorem, $$\mathscr{L}^{-1}\left\\{e^{-\sqrt{s+h} x}\right\\}=e^{-h t} \mathscr{L}^{-1}\left\\{e^{-x \sqrt{s}}\right\\}=e^{-h t} \frac{x}{2 \sqrt{\pi t^{3}}} e^{-x^{2} / 4 t}$$ Thus, from the convolution theorem we obtain $$u(x, s)=\frac{u_{0} x}{2 \sqrt{\pi}} \int_{0}^{t} \frac{e^{-h \tau-x^{2} / 4 \tau}}{\tau^{3 / 2}} d \tau$$

(a) Letting \(C(x, s)=\mathscr{L}\\{c(x, t)\\}\) we obtain $$\frac{d^{2} C}{d x^{2}}-\frac{s}{k} C=0 \quad \text { subject to }\left.\quad \frac{d C}{d x}\right|_{x=0}=-A$$ The solution of this initial-value problem is $$C(x, s)=A \sqrt{k} \frac{e^{-(x / \sqrt{k}) \sqrt{s}}}{\sqrt{s}}$$ so that $$c(x, t)=A \sqrt{\frac{k}{\pi t}} e^{-x^{2} / 4 k t}$$ (c) \(\int_{0}^{\infty} c(x, t) d x=\left.A k \operatorname{erf}\left(\frac{x}{2 \sqrt{k t}}\right)\right|_{0} ^{\infty}=A k(1-0)=A k\)