Supplementary Materials
Prob 1. $u_{x}+(\sin x+y)u_{y}=0,$ $u(0,y)=y.$
Solution:
$\dfrac{dx}{1}=\dfrac{dy}{\sin x+y},$ or $\dfrac{dy}{dx}=y+\sin x.$
The characteristic curves are given by $y=ce^{x}+\dfrac{1}{D-1}\sin x=ce^{x}+\dfrac{D+1}{D^{2}-1}\sin x,$
or $(y+\dfrac{1}{2}\cos x+\dfrac{1}{2}\sin x)e^{-x}=c.$
Therefore, $u(x,y)=f(ye^{-x}+\dfrac{e^{-x}}{2}\cos x+\dfrac{e^{-x}}{2}\sin x).$
From the initial condition, $f(y+1/2)=y,$ or $f(\rho )=\rho -1/2.$ We get
$u(x,y)=ye^{-x}+\dfrac{e^{-x}}{2}\cos x+\dfrac{e^{-x}}{2}\sin x-1/2.$
Prob 2.Find $u(x,t)$ which satisfies the partial differential equation
$(\dfrac{\partial ^{2}}{\partial x^{2}}-m^{2})u(x,t)=\dfrac{\partial u(x,t)}{\partial t},$ $t>0,$ $-\infty$ < $x$ < $\infty ,$
as well as the initial condition $u(x,0)=\delta (x)$, the Dirac delta function. The boundary conditions are $u(\pm \infty ,t)=0.$
Solution:
Let $u(x,t)=\int_{-\infty }^{\infty }\dfrac{dk}{2\pi }e^{ikx}U(k,t),$ then
$\dfrac{dU}{dt}=-(k^{2}+m^{2})U\Longrightarrow U(k,t)=c(k)e^{-(k^{2}+m^{2})t}.$
Thus $u(x,t)=\int_{-\infty }^{\infty }\dfrac{dk}{2\pi }e^{ikx}c(k)e^{-(k^{2}+m^{2})t}.$
From the initial condition, we get $c(k)=\int_{-\infty }^{\infty }dx $ $e^{-ikx}\delta (x)=1,$ or
$u(x,t)=\int_{-\infty }^{\infty }\dfrac{dk}{2\pi }e^{ikx}e^{-(k^{2}+m^{2})t}, $
a Gaussian integral. Since $k^{2}t-ikx=t(k-ix/2t)^{2}+x^{2}/(4t),$ we get
$u(x,t)=\dfrac{e^{-m^{2}t}e^{-x^{2}/4t}}{\sqrt{4\pi t}}.$
Prob 3. Find $u(\overrightarrow{x},t)$ which satisfies the wave equation
$\nabla^{2}u=\dfrac{\partial^{2}u}{\partial t^{2}}$, $ t>0,$
as well as the initial conditions
$u(\overrightarrow{x},0)=a(\overrightarrow{x})$ , $u_{t}(\overrightarrow{x},0)=b(\overrightarrow{x}).$
It is assumed that the equation holds for all $\overrightarrow{x},$ which is the position vector in an $n$ dimensional space. We also assume that $u(\overrightarrow{x},t)$ vanishes at spatial infinity.
Solution:
We put
$u(\overrightarrow{x},t)=\int\dfrac{d^{n}k}{(2\pi)^{n}}e^{i\overrightarrow {k}\cdot\overrightarrow{x}}U(\overrightarrow{k},t),$
where $n$ is the number of spatial dimensions. The wave equation requires
$-\omega^{2}U(\overrightarrow{k},t)=\dfrac{\partial U}{\partial t},$ $\omega=\sqrt{\overrightarrow{k}^{2}}.$
The wave equation requires
$(\dfrac{d^{2}}{dt^{2}}+\omega^{2})U(\overrightarrow{k},t)=0.$
Thus
$U(\overrightarrow{k},t)=A(\overrightarrow{k})\cos\omega t+\dfrac {B(\overrightarrow{k})}{\omega}\sin\omega t.$
Therefore,
$u(\overrightarrow{x},t)=\int\dfrac{d^{n}\overrightarrow{k}}{(2\pi)^{n}}e^{i\overrightarrow{k}\cdot\overrightarrow{x}}[A(\overrightarrow{k})\cos\omega t+\dfrac{B(\overrightarrow{k})}{\omega}\sin\omega t].$
The initial conditions give
$A(\overrightarrow{k})=\int d^{n}x^{\prime}$ $e^{-i\overrightarrow{k}\cdot\overrightarrow{x^{\prime}}}a(\overrightarrow{x}),$ $B(\overrightarrow{k})=\int d^{n}x^{\prime}$ $e^{-i\overrightarrow {k}\cdot\overrightarrow{x^{\prime}}}b(\overrightarrow{x}).$
The quantity above can be expressed directly with $a(\overrightarrow{x})$ and $b(\overrightarrow{x}):$
$u(\overrightarrow{x},t)=\int d^{n}\overrightarrow{x^{\prime}}[G_{1}(\overrightarrow{x}-\overrightarrow{x^{\prime}},t)a(\overrightarrow{x^{\prime}})+G_{2}(\overrightarrow{x}-\overrightarrow{x^{\prime}},t)b(\overrightarrow{x^{\prime}})],$ (1)
where
$G_{1}(\overrightarrow{x},t)=\int\dfrac{d^{n}k}{(2\pi)^{n}}e^{i\overrightarrow{k}\cdot\overrightarrow{x}}\cos\omega t,$ $(2a)$
and
$G_{2}(\overrightarrow{x},t)=\int\dfrac{d^{n}k}{(2\pi)^{n}}\dfrac {e^{i\overrightarrow{k}\cdot\overrightarrow{x}}\sin\omega t}{\omega}.$ $(2b)$
Note that
$\dfrac{\partial G_{2}}{\partial t}=G_{1}.$
Aside: The integrals in (2a) and (2b) can be evaluated out explicitly: We write
$\overrightarrow{k}=\omega\overrightarrow{n},$
where $\overrightarrow{n}$ is a unit vector. Then
$\overrightarrow{k}\cdot\overrightarrow{x}=\omega r\cos\theta,$ $d^{n}k=\omega^{n-1}d\omega d\Omega,$
Where $\Omega$ denotes the solid angle. Thus we have
$G_{1}(\overrightarrow{x},t)=\int d\Omega\int_{0}^{\infty}\dfrac{d\omega}{(2\pi)^{n}}\omega^{n-1}e^{i\omega r\cos\theta}\cos\omega t,$ (2c)
and
$G_{2}(\overrightarrow{x},t)=\int d\Omega\int_{0}^{\infty}\dfrac{d\omega}{(2\pi)^{n}}\omega^{n-1}e^{i\omega r\cos\theta}\dfrac{\sin\omega t}{\omega}.$ (2d)
1. One spacial dimension ($n=1$)
We have from (2a) that
$G_{1}(\overrightarrow{x},t)=\int_{-\infty}^{\infty}\dfrac{dk}{2\pi}e^{ikx}\cos\omega t$
Since $\cos\omega t=\cos kt,$ regardless of the sign of $k,$
We have,
$G_{1}(x,t)=\dfrac{1}{2}\int_{-\infty}^{\infty}\dfrac{dk}{2\pi}[e^{ik(x+t)}+e^{ik(x-t)}]=\dfrac{\delta(x+t)+\delta(x-t)}{2}.$
From (2b), we have
$G_{2}(\overrightarrow{x},t)=\int\dfrac{dk}{2\pi}\dfrac{e^{ikx}\sin\omega t}{\omega}.$
Since
$\dfrac{\sin\omega t}{\omega}=\dfrac{\sin kt}{k},$
whether $k$ is positive or negative, we have
$G_{2}(x,t)=\dfrac{1}{2i}\int_{-\infty}^{\infty}\dfrac{dk}{2\pi}\dfrac{e^{ik(x+t)}-e^{ik(x-t)}}{k}=\dfrac{1}{2}\int_{x-t}^{x+t}d\tau \int_{-\infty}^{\infty}\dfrac{dk}{2\pi}e^{ik\tau}=\dfrac{1}{2}\int_{x-t}^{x+t}d\tau\delta(\tau),$
which means
$G_{2}(x,t)=\dfrac{1}{2},$ if $-t$ < $x$ < $t$,
$ =0,$ otherwise.
It is then straightforward to find that
$u(x,t)=\dfrac{a(x-t)+a(x+t)}{2}+\dfrac{\int_{x-t}^{x+t}dx^{\prime}b(x^{\prime})}{2},$
which is (4.29) in the textbook.
2. Two-spatial dimension ($n=2$)
We have from (1) and (2) that
$G_{2}(\overrightarrow{x},t)=\int\dfrac{d^{2}\overrightarrow{k}}{(2\pi)^{2}}e^{i\overrightarrow{k}\cdot\overrightarrow{x}}\dfrac{\sin\omega t}{\omega }=\int_{0}^{2\pi}d\theta\int_{0}^{\infty}\dfrac{d\omega}{4\pi^{2}}e^{i\omega r\cos\theta}\dfrac{e^{i\omega t-\epsilon\omega}-e^{-i\omega t-\epsilon\omega}}{2i},$
where we have introduced a cutoff factor $e^{-\epsilon\omega},$ with $\epsilon$ infinitesimal. Carrying out the integration over $\omega$, we get
$G_{2}(\overrightarrow{x},t)=\dfrac{1}{8\pi^{2}}\int_{0}^{2\pi}d\theta (\dfrac{1}{r\cos\theta+t+i\epsilon}-\dfrac{1}{r\cos\theta-t+i\epsilon})$
$=\dfrac{1}{8\pi^{2}}\int_{0}^{2\pi}d\theta(\dfrac{1}{t+i\epsilon+r\cos\theta }+\dfrac{1}{t-i\epsilon+r\cos\theta}),$
where we have replaced $\cos\theta$ by $-\cos\theta$ in the second denominator or, equivalently, $\theta\rightarrow\pi+\theta$. Thus
$G_{2}(\overrightarrow{x},t)=\dfrac{1}{4\pi^{2}}\operatorname{Re}\int _{0}^{2\pi}d\theta\dfrac{1}{t+i\epsilon+r\cos\theta}.$
If $t>r,$ the denominators in the integrand above never vanish, and we may drop $\epsilon$ in the expression. Using the result
$\int_{0}^{2\pi}d\theta\dfrac{1}{a+b\cos\theta}=\dfrac{2\pi}{\sqrt{a^{2}-b^{2}}},$ $a>b,$
we get
$G_{2}(\overrightarrow{x},t)=\dfrac{1}{2\pi}\dfrac{1}{\sqrt{t^{2}-r^{2}}},$ $t>r.$
If $t$ < $r$, the integral for $G_{2}$ is of the form
$I\equiv\int_{0}^{2\pi}d\theta\dfrac{1}{a+i\epsilon+b\cos\theta},$ $a$ and $b$ both positive, $a$ < $b.$
Let
$z=e^{i\theta},$
then
$I=\dfrac{2}{ib}\oint_{c} dz\dfrac{1}{z^{2}+2z(a/b+i\epsilon)+1},$
where $c$ is the unit circle with center at the origin. The poles of the integrand above are
$z=-a/b-i\epsilon\pm i\sqrt{1-a^{2}/b^{2}}.$
The pole inside the unit circle is
$z=-a/b-i\epsilon+i\sqrt{1-a^{2}/b^{2}}.$
Thus
$I=\dfrac{2\pi}{i\sqrt{b^{2}-a^{2}}},$
which is purely imaginary. Thus
$G_{2}(\overrightarrow{x},t)=0,$ $t$ < $r.$
To summarize, we have
$G_{2}(\overrightarrow{x},t)=\dfrac{1}{2\pi}\dfrac{1}{\sqrt{t^{2}-r^{2}}},$ $t>r,$
$ =0,$ $ t$ < $r.$
3. Three spatial dimension ($n=3$)
We have
$G_{2}(\overrightarrow{x},t)=\int\dfrac{d^{3}k}{(2\pi)^{3}}\dfrac {e^{i\overrightarrow{k}\cdot\overrightarrow{x}}\sin\omega t}{\omega}=\int d(\cos\theta)\int_{0}^{\infty}\dfrac{\omega d\omega}{(2\pi)^{2}}e^{i\omega r\cos\theta}\sin\omega t.$
Carrying out the integration over $\theta,$ we get,
$G_{2}(\overrightarrow{x},t)=\int_{0}^{\infty}\dfrac{d\omega}{4\pi^{2}}\dfrac{e^{i\omega r}-e^{-i\omega r}}{ir}\sin\omega t.$
Since the integrand above is an even function of $\omega,$ we have
$G_{2}(\overrightarrow{x},t)=\int_{-\infty}^{\infty}\dfrac{d\omega}{4\pi^{2} }\dfrac{e^{i\omega r}-e^{-i\omega r}}{2ir}\sin\omega t=\int_{-\infty}^{\infty }\dfrac{d\omega}{4\pi^{2}}\dfrac{e^{i\omega r}}{ir}\sin\omega t=\int_{-\infty }^{\infty}\dfrac{d\omega}{4\pi^{2}}\dfrac{e^{i\omega r}}{ir}\dfrac{e^{i\omega t}-e^{-i\omega t}}{2i}$
$=\dfrac{1}{4\pi r}[\delta(r-t)-\delta(r+t)]=\dfrac{1}{4\pi r}\delta(r-t), $
as $r$ and $t$ are both positive.
Aside: From the results above, we see that
$G_{1}(\overrightarrow{x},t)=G_{2}(\overrightarrow{x},t)=0$ if $t$ < $r,$
where
$r\equiv\left\vert \overrightarrow{x}\right\vert .$
This means the wave cannot propagate faster that the speed of unity.
Problem 4. Find the Green function for the wave equation satisfying
$\dfrac{\partial^{2}G(x-x^{\prime})}{\partial t^{2}}-\nabla^{2}G(x-x^{\prime})=\delta^{4}(x-x^{\prime})$ , $ t$ > $0.$
This Green function is assumed to vanish at spatial infinity and for $t$ < $t^{\prime}.$ In the above, $x$ is a four-vector, i.e. the spatial component of $x$ is the 3-vector $\overset{\rightarrow}{x}$ and the time component of $x$ is $t.$
Solution:
Expressing the delta function in the equation above by its Fourier integral, we have
$\dfrac{\partial^{2}G(x-x^{\prime})}{\partial t^{2}}-\nabla^{2}G(x-x^{\prime})=\int\dfrac{d^{4}k}{(2\pi)^{4}}e^{-ik\cdot(x-x^{\prime})},$ where $d^{4}k\equiv dk_{0}d^{3}k$ and $k\cdot x\equiv k_{0}t-\overset {\rightarrow}{k}\cdot\overset{\rightarrow}{x}.$
Thus we have
$G(x-x^{\prime})=\dfrac{1}{\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}}\int\dfrac{d^{4}k}{(2\pi)^{4}}e^{-ik\cdot(x-x^{\prime})}=-\int\dfrac{d^{4}k}{(2\pi)^{4}}\dfrac{e^{-ik\cdot(x-x^{\prime})}}{k_{0}^{2}-\omega^{2}}=-\int\dfrac{d^{3}k\text{ }e^{i\overset{\rightarrow}{k}\cdot(\overset{\rightarrow}{x}-\overset{\rightarrow}{x}^{\prime})}}{(2\pi)^{3}}\int_{-\infty}^{\infty}\dfrac{dk_{0}}{2\pi}\dfrac{e^{-ik_{0}(t-t^{\prime})}}{k_{0}^{2}-\omega^{2}}$
The integrand has poles at $k_{0}=\pm\omega.$ Thus the integral is divergent unless we specify how the contour passes by the points $k_{0}=\pm\omega$ in the complex $k_{0}$-plane. For example, we may define the integral to be the principle value of this divergent integral, but then the integral will not satisfy the requirement of vanishing for $t$ < $t^{\prime}$. We note that when $t$ < $t^{\prime},$ the integrand vanishes exponentially in the upper-half $k_{0}% $-plane. Thus we define the path of integration to be infinitesimally above the real axis of the $k_{0}$-plane. With such a definition, we get, by closing the contour upstairs.
$G(x-x^{\prime})=0,$ $t$ < $t^{\prime},$
satisfying the initial condition.
For $t>t^{\prime},$ we close the contour downstairs, picking up the residue of the poles at $k_{0}=\pm\omega,$ we get,
$G(x-x^{\prime})=-\int\dfrac{d^{3}k\text{ }e^{i\overset{\rightarrow}{k}\cdot(\overset{\rightarrow}{x}-\overset{\rightarrow}{x}^{\prime})}}{(2\pi)^{3}}(-2\pi i)\dfrac{1}{2\pi}\dfrac{e^{i\omega(t-t^{\prime})}-e^{-i\omega(t-t^{\prime})}}{-2\omega}=\int\dfrac{d^{3}k\text{ }e^{i\overset{\rightarrow}{k}\cdot(\overset{\rightarrow}{x}-\overset{\rightarrow}{x}^{\prime})}}{(2\pi)^{3}}\dfrac{\sin\omega(t-t^{\prime})}{\omega},$
which is $G_{2}(x-x^{\prime})$ as given by (2b).
Problem 5. Find the solution $u(x^{\prime})$\ which satisfies the wave equation with a source:
$\dfrac{\partial^{2}u(x^{\prime})}{\partial t^{\prime2}}-\nabla^{\prime 2}u(x^{\prime})=\rho(x^{\prime})$, $ t^{\prime}>0,$ (3)
as well as the initial conditions
$u(\overrightarrow{x^{\prime}},0)=a(\overrightarrow{x^{\prime}})$ , $u_{t}(\overrightarrow{x^{\prime}},0)=b(\overrightarrow{x^{\prime}}).$
In the above, $\rho(x)$ is a given function; $x^{\prime}$ is the four vector $(t^{\prime},\overrightarrow{x^{\prime}})$. It is assumed that the equation holds for all $\overrightarrow{x^{\prime}}$ and all $t^{\prime}>0. $ We assume that the solution vanishes at spatial infinity.
Solution:
The PDE satisfied by the Green function is
$[\dfrac{\partial^{2}}{\partial t^{2}}-\nabla^{2}]G(x-x^{\prime})=\delta^{4}(x-x^{\prime}).$ (4)
Multiplying (3) by $G(x-x^{\prime})$ and integrate over all $x^{\prime}$ and all $t^{\prime}>0,$ we get
$\int d^{3}x^{\prime}\int_{0}^{\infty}dt^{\prime}[G(x-x^{\prime })(\dfrac{\partial^{2}}{\partial t^{\prime2}}-\nabla^{\prime2})u(x^{\prime })]=\int d^{3}x^{\prime}\int_{0}^{\infty}dt^{\prime}G(x-x^{\prime}% )\rho(x^{\prime}).$ (5)
Doing integraton by parts twice, we get
$\int_{0}^{\infty}dt^{\prime}[G(x-x^{\prime})(\dfrac{\partial^{2}}{\partial t^{\prime2}}u(x^{\prime})]$
$=[G(x-x^{\prime})\dfrac{\partial u(x^{\prime})}{\partial t^{\prime}}-\dfrac{\partial G(x-x^{\prime})}{\partial t^{\prime}}u(x^{\prime})]_{t^{\prime}=0}^{t^{\prime}=\infty}+\int_{0}^{\infty}dt^{\prime}u(x^{\prime})\dfrac{\partial^{2}}{\partial t^{\prime2}}G(x-x^{\prime}).$
Note:
(a) $G(x-x^{\prime})$ vanishes at $t^{\prime}=\infty$ $\ $(because $t-t^{\prime}$ is then negative)$;$
(b) $\dfrac{\partial G(x-x^{\prime})}{\partial t^{\prime}}=-\dfrac{\partial G(x-x^{\prime})}{\partial t}$ (because $G$ is a function of $(t-t^{\prime})$ )
Utilizing (a) and (b), we find
$\int_{0}^{\infty}dt^{\prime}[G(x-x^{\prime})(\dfrac{\partial^{2}}{\partial t^{\prime2}}u(x^{\prime})]$
$=-G(\overrightarrow{x}-\overrightarrow{x}^{\prime},t)b(\overrightarrow {x}^{\prime})-$ $\dfrac{\partial G(\overrightarrow{x}-\overrightarrow {x}^{\prime},t)}{\partial t}a(\overrightarrow{x}^{\prime})+\int_{0}^{\infty }dt^{\prime}u(x^{\prime})\dfrac{\partial^{2}}{\partial t^{2}}G(x-x^{\prime}).$
Similarly, by doing integration by parts twice, we get
$\int d^{3}x^{\prime}G(x-x^{\prime})(-\nabla^{\prime2})u(x^{\prime})=\int d^{3}x^{\prime}$ $u(x^{\prime})(-\nabla^{2})G(x-x^{\prime}).$
The above is true because $G(x-x^{\prime})$ vanishes at the spatial infinity. Therefore, the surface terms from integration by parts vanish. Thus (5) becomes,
$\int d^{3}x^{\prime}\int_{0}^{\infty}dt^{\prime}$ $u(x^{\prime })[(\dfrac{\partial^{2}}{\partial t^{2}}-\nabla^{2})G(x-x^{\prime})]$
$=\int d^{3}x$ $[G(\overrightarrow{x}-\overrightarrow{x}^{\prime },t)b(\overrightarrow{x}^{\prime})+\dfrac{\partial G(\overrightarrow {x}-\overrightarrow{x}^{\prime},t)}{\partial t}a(\overrightarrow{x}^{\prime })]+\int d^{3}x^{\prime}\int_{0}^{\infty}dt^{\prime}G(x-x^{\prime})\rho(x^{\prime}).$
By (4), the left side of the equation above is simply $u(x).$ Thus we have
$u(x)=\int d^{3}x$ $[G(\overrightarrow{x}-\overrightarrow{x}^{\prime },t)b(\overrightarrow{x}^{\prime})+\dfrac{\partial G(\overrightarrow {x}-\overrightarrow{x}^{\prime},t)}{\partial t}a(\overrightarrow{x}^{\prime })]+\int d^{3}x^{\prime}\int_{0}^{\infty}dt^{\prime}G(x-x^{\prime})\rho(x^{\prime}).$ (6)
Notice that if $\rho(x)=0,$ the equation above is just (1) in problem 3.
Problem 6. Use the Green function method to find the solution $u(x^{\prime})$\ which satisfies the heat equation with a heat source:
$\dfrac{\partial u(x^{\prime})}{\partial t^{\prime}}-\nabla^{\prime 2}u(x^{\prime})=\rho(x^{\prime})$, $ t^{\prime}>0,$ (7)
as well as the initial conditions
$u(\overrightarrow{x^{\prime}},0)=a(\overrightarrow{x^{\prime}})$ .
Solution:
Let
$(\dfrac{\partial}{\partial t}-\nabla^{2})G(x-x^{\prime})=\delta ^{(4)}(x-x^{\prime}),$ (8a)
or
$(\dfrac{\partial}{\partial t}-\nabla^{2})G(x-x^{\prime})=\int\dfrac{d^{4}% k}{(2\pi)^{4}}e^{-ik\cdot(x-x^{\prime})}.$ (8b)
The initial condition for the Green function is that
$G(x-x^{\prime})=0,$ $t$ < $t^{\prime}.$
The boundary condition is that it vanishes at spatial infinity. The solution of (8b) satisfying these conditions is
$G(x-x^{\prime})=\int\dfrac{d^{4}k}{(2\pi)^{4}}\dfrac{e^{-ik\cdot(x-x^{\prime})}}{-ik_{0}+\omega^{2}}.$ (9)
We get, closing the contour of integration over $k_{0}$ in the lower-half plane,
$G(x-x^{\prime})=\int\dfrac{d^{3}k}{(2\pi)^{3}}e^{-\omega^{2}(t-t^{\prime}% )}e^{i\overset{\rightarrow}{k}\cdot(\overset{\rightarrow}{x}\cdot \overset{\rightarrow}{x}^{\prime})},$ $t>t^{\prime}.$ (10)
We multiply (7) by $G(x-x^{\prime})$ and integrate over all $\overset{\rightarrow}{x}^{\prime}$ and $t^{\prime}>0:$
$\int d^{3}x\int_{0}^{\infty}dt^{\prime}[G(x-x^{\prime})(\dfrac{\partial }{\partial t^{\prime}}-\nabla^{\prime2})u(x^{\prime})]=\int d^{3}x^{\prime }\int_{0}^{\infty}dt^{\prime}G(x-x^{\prime})\rho(x^{\prime}).$ (11)
Carrying out an integration by parts, we get
$\int_{0}^{\infty}dt^{\prime}G(x-x^{\prime})\dfrac{\partial}{\partial t^{\prime}}u(x^{\prime})=-G(\overset{\rightarrow}{x}-\overset{\rightarrow}% {x}^{\prime},t)a(x^{\prime})+\int_{0}^{\infty}dt^{\prime}u(x^{\prime}% )\dfrac{\partial G(x-x^{\prime})}{\partial t}.$ (12)
Similarly,
$\int_{0}^{\infty}dt^{\prime}G(x-x^{\prime})\nabla^{\prime2}u(x^{\prime})=\int_{0}^{\infty}dt^{\prime}u(x^{\prime})\nabla^{2}G(x-x^{\prime}).$ (13)
Using (12) and (13), (11) becomes
$\int d^{3}x\int_{0}^{\infty}dt^{\prime}[u(x^{\prime})(\dfrac{\partial }{\partial t^{\prime}}-\nabla^{\prime2})G(x-x^{\prime})]=\int d^{3}xG(\overset{\rightarrow}{x}-\overset{\rightarrow}{x}^{\prime},t)a(x^{\prime })+\int d^{3}x^{\prime}\int_{0}^{\infty}dt^{\prime}G(x-x^{\prime})\rho(x^{\prime}).$
By (8a), the equation above is
$u(x)=\int d^{3}xG(\overset{\rightarrow}{x}-\overset{\rightarrow}{x}^{\prime },t)a(x^{\prime})+\int d^{3}x^{\prime}\int_{0}^{\infty}dt^{\prime }G(x-x^{\prime})\rho(x^{\prime}).$ (14)
If $\rho(x)=0,$(14) is (4.41)\ in the textbook with $dx$ replaced by $d^{3}x$.
Chapter 1
1.Find the general solution of $\dfrac{d^{2}y}{dx^{2}}-y=2+x^{2}+e^{x}.$
Solution:
$m^{2}=1\Rightarrow m=\pm1.$ Thus the complementary solution is
$Y=Ae^{x}+Be^{-x}.$
We also have
$\dfrac{1}{D^{2}-1}2=-2$ , $\dfrac{1}{D^{2}-1}x^{2}=-(1+D^{2}+\cdot\cdot\cdot)x^{2}=-x^{2}-2.$
$\dfrac{1}{D^{2}-1}e^{x}=e^{x}\dfrac{1}{(D+1)^{2}-1}1=e^{x}\dfrac{1}{D^{2}+2D}1$
$=e^{x}\dfrac{1}{D(D+2)}1=e^{x}\dfrac{1}{D(2)}1=\dfrac{xe^{x}}{2}.$
Therefore, the general solution of the differential equation is $y=Ae^{x}+Be^{-x}-4-x^{2}+\dfrac{xe^{x}}{2}.$
2. Find the general solution of the ordinary differential equation $\dfrac{d^{4}y}{dx^{4}}+4y=\sin2x+2x+e^{x}.$
Solution:
$m^{4}=-4\Longrightarrow m=1+i,-1+i,-1-i,1-i.$
Thus the complementary solution is
$Y(x)=e^{x}(a\cos x+b\sin x)+e^{-x}(c\cos x+d\sin x).$
A particular solution is
$y_{p}(x)=\dfrac{1}{D^{4}+4}[\sin2x+2x+e^{x}]=\dfrac{\sin2x}{20}+\dfrac{x}{2}+\dfrac{e^{x}}{5}.$
The general solution is
$y(x)=e^{x}(a\cos x+b\sin x)+e^{-x}(c\cos x+d\sin x)+\dfrac{\sin2x}{20}+\dfrac{x}{2}+\dfrac{e^{x}}{5}.$
3. Find the general solution of the ordinary differential equation $\dfrac{d^{4}y}{dx^{4}}-y=3e^{2x}+e^{x}+2x.$
Solution:
$m^{4}-1=0\Rightarrow m=\pm1,\pm i.$
Thus
$Y=ae^{x}+be^{-x}+c\cos x+d\sin x.$
We also have
$y_{P}=\dfrac{1}{D^{4}-1}(3e^{2x}+e^{x}+2x)$
$ =\dfrac{e^{2x}}{5}+\dfrac{1}{(D-1)(D+1)(D^{2}+1)}e^{x}-(1+D^{4}+\cdot\cdot\cdot)2x$
$ =\dfrac{e^{2x}}{5}+\dfrac{1}{(D-1)}\dfrac{1}{4}e^{x}-2x$
$ =\dfrac{e^{2x}}{5}+e^{x}\dfrac{1}{D}\dfrac{1}{4}-2x=\dfrac{e^{2x}}{5}+\dfrac{e^{x}x}{4}-2x.$
Therefore,
$y=ae^{x}+be^{-x}+c\cos x+d\sin x+\dfrac{e^{2x}}{5}+\dfrac{xe^{x}}{4}-2x.$
4. Find the general solution for $\dfrac{d^{3}y}{dx^{3}}-y=2+\cos x.$
Solution:
The complementary solution satisfies $\dfrac{d^{3}Y}{dx^{3}}-Y=0.$ Thus
$Y=e^{mx},m^{3}=1\Rightarrow m_{n}=e^{2in\pi/3}.$
The three roots are: $m_{0}=1,$ $m_{1}=(-1+i\sqrt{3})/2,$ $m_{2}=(-1+i\sqrt {3})/2.$
We get
$Y=Ae^{x}+e^{-x/2}[B\cos(x\sqrt{3}/2)+C\sin(x\sqrt{3}/2)].$
The particular solution is
$y_{p}=\dfrac{1}{D^{3}-1}(2+\cos x)=-2-\dfrac{1}{D+1}\cos x$
$=-2-\dfrac{D-1}{D^{2}-1}\cos x=-2-\dfrac{-\sin x-\cos x}{-1-1}.$
We have
$y=Ae^{x}+e^{-x}[B\cos(x\sqrt{3}/2)+C\sin(x\sqrt{3}/2)]-2-\dfrac{\sin x+\cos x}{2}.$
5. Consider the equation of motion
$\overset{\cdot\cdot}{x}+a\overset{\cdot}{x}+x=b\cos t,$ $\overset{\cdot}{x}\equiv\dfrac{dx}{dt},$ $a>0.$
which describes a harmonic oscillator under a linear damping force as well as an external harmonic force of the same frequency. Find the general solution of this equation.
Solution:
$m^{2}$ $+am+1=0\Rightarrow$ $\ m_{1}=-a/2+\sqrt{a^{2}/4-1},$ $m_{2}=-a/2-\sqrt{a^{2}/4-1}.$
If $a$ < $2,$ the two roots above are $-a/2\pm i\omega,$ where $\omega =\sqrt{1-a^{2}/4}$, and the complementary solution $X$ is given by
$X=e^{-at/2}(c\cos\omega t+d\sin\omega t),$ where $c$ and $d$ are constants.
We note the motion is a damped oscillation with the angular fequency $\omega$, which is smaller than unity--the angular frequency when $a=0$.
If $a>2$, the two roots are $-a/2\pm\sqrt{a^{2}/4-1}.$ The complementry solution is
$X=e^{-at/2}(ce^{\sqrt{a^{2}/4-1}t}+de^{-\sqrt{a^{2}/4-1}t}).$
Both solutions above describe motions vanishing as $t\rightarrow\infty$ without oscillations.
If $a=2,$
$X=e^{-t}(c+dt).$
Both solutions vanish as $t\rightarrow\infty$ without oscillations.
The particular solution is
$x_{p}=\dfrac{1}{D^{2}+aD+1}b\cos t=\dfrac{1}{aD}b\cos t=\dfrac{b\sin t}{a}.$
The general solution is
$x=X+\dfrac{b\sin t}{a}.$
6. Show that the particular solution of a linear differential equation with constant coefficients can always be expressed by an integral.
Solution:
Let the equation be
$\dfrac{d^{n}y}{dx^{n}}+a_{n-1}\dfrac{d^{n-1}y}{dx^{n-1}}+\cdot\cdot\cdot+a_{1}\dfrac{dy}{dx}+a_{0}y=f(x),$
where $a_{m}$, $m=0,1\cdot\cdot\cdot$ $n,$ are constants.
We have
$Y=c_{1}e^{m_{1}x}+c_{2}e^{m_{2}x}+\cdot\cdot\cdot+c_{n}e^{m_{n}x},$
where $m_{j}$, $j=1,2\cdot\cdot\cdot n$ are the roots of the algebric equation
$m^{n}+a_{n-1}m^{n-1}+\cdot\cdot\cdot+a_{1}m+a_{0}=0.$
We assume here that they are distinct. If the algebraic equation has multiple roots, only slight modifications are needed.
The particular solution is given by
$y_{p}=\dfrac{1}{D^{n}+a_{n-1}D^{n-1}+\cdot\cdot\cdot+a_{1}D+a_{0}}f(x).$
We shall write
$\dfrac{1}{D^{n}+a_{n-1}D^{n-1}+\cdot\cdot\cdot+a_{1}D+a_{0}}=\dfrac {1}{(D-m_{1})(D-m_{2})\cdot\cdot\cdot(D-m_{n})}.$
By partial fraction, we get
$\dfrac{1}{D^{n}+a_{n-1}D^{n-1}+\cdot\cdot\cdot+a_{1}D+a_{0}}$
$=\dfrac{1}{(m_{1}-m_{2})\cdot\cdot\cdot(m_{1}-m_{n})}\dfrac{1}{D-m_{1}}+\dfrac{1}{(m_{2}-m_{1})\cdot\cdot\cdot(m_{2}-m_{n})}\dfrac{1}{D-m_{2}}$
$+\cdot\cdot\cdot+\dfrac{1}{(m_{n}-m_{1})\cdot\cdot\cdot(m_{n}-m_{n-1})}\dfrac{1}{D-m_{n}}.$
Since
$\dfrac{1}{D-m}f(x)=e^{mx}\dfrac{1}{D}[e^{-mx}f(x)]=$ $e^{mx}\int^{x}e^{-mx^{\prime}}f(x^{\prime})dx^{\prime},$
we have
$y_{p}(x)=\int^{x}[\dfrac{e^{m_{1}(x-x^{\prime})}}{(m_{1}-m_{2})(m_{1}-m_{3})\cdot\cdot\cdot(m_{1}-m_{n})}+\dfrac{e^{m_{2}(x-x^{\prime})}}{(m_{2}-m_{1})(m_{2}-m_{3})\cdot\cdot\cdot(m_{2}-m_{n})}$ $+\cdot$
$\cdot\cdot+\dfrac{e^{m_{n}(x-x^{\prime})}}{(m_{n}-m_{1})(m_{n}-m_{2})\cdot\cdot\cdot(m_{n}-m_{n-1})}]$ $f(x^{\prime})dx^{\prime}.$
Chapter 2
1.As we know, the sum of two complex numbers can be represented as that of two vectors in a two dimensional space. How about the product of two complex numbers?
Solution:
Let $ \alpha\equiv a_{1}+ia_{2}$, $\beta\equiv b_{1}+ib_{2}.$
These two complex numbers can be reprsented by the vectors
$\overset{\rightarrow}{v}\equiv a_{1}\overset{\rightarrow}{e_{1}}+a_{2}\overset{\rightarrow}{e_{2}}$, $\overset{\rightarrow}{u}\equiv b_{1}\overset{\rightarrow}{e_{1}}+b_{2}\overset{\rightarrow}{e_{2}}.$
where $\overset{\rightarrow}{e_{1}}$ and $\overset{\rightarrow}{e_{2}}$ are unit vectors in the direction of the $x$-axis and the $y$-axis, respectively.
We have
$\alpha^{\ast}\beta=(a_{1}-ia_{2})(b_{1}+ib_{2})=(a_{1}b_{1}+a_{2}b_{2})+i(a_{1}b_{2}-a_{2}b_{1}).$
The real part of $\alpha^{\ast}\beta$ is the scalar product of $\overset{\rightarrow}{v}$ and $\overset{\rightarrow}{u};$ the imaginary part of $\alpha^{\ast}\beta$ is the cross product of $\overset{\rightarrow}{v}$ and $\overset{\rightarrow}{u}$.
2. Find the real part and the imaginary part of $e^{iz}$ as well as those of $\cos z$. For what values of $z$ is the real part of $e^{iz}$ equal to $\cos z$?
Solution:
$e^{iz}=e^{ix-y}=e^{-y}(\cos x+i\sin x)=u+iv,$
$e^{-iz}=e^{-ix+y}=e^{y}(\cos x-i\sin x).$
Thus
cos$z=\dfrac{e^{iz}+e^{-iz}}{2}=\cos x\cosh y-i\sin x\sinh y.$
If $u$ is equal to $\cos z$, then
$e^{-y}\cos x=\cos x\cosh y-i\sin x\sinh y.$
The real part and the imaginary part of the equation above are:
$(e^{-y}-\cosh y)\cos x=0$\ \ \ and \ \ $\sin x\sinh y=0.$
If $y\neq0$, these two equalities give $\sin x=\cos x=0$, which is not possible.
Both equalities are satisfied if
$y=0.$
Thus the real part of $e^{iz}$ is equal to $\cos z$ only if $z$ is real.
3. For what constant value (or values ) of $a$ is the function $u=x+ax^{3}$ the real part of an analytic function? Find the imaginary part of this analytic function. Express this analytic function as a function of $z$.
Solution:
$u_{xx}=6ax$ , $u_{yy}=0$ .
Thus $\nabla^{2}u=0$ only if $a=0,$ in which case
$u=x$.
The conjugate of $x$, denoted as $v$, satisfies
$v_{y}=u_{x}=1$ and $v_{x}=-u_{y}=0.$
We get
$v=y+c,$
and
$u+iv=z+ic,$
where $c$ is a constant.
4. Find the roots of $\ (z+1)^{8}=(z^{2}-1)^{8}$ .
Solution:
The equation above gives
$(z^{2}-1)=e^{i2\pi n/8}(z+1),$ $n=0,1,2\cdot\cdot\cdot,7$ ,
or
$(z+1)(z-1-e^{i2\pi n/8})=0,$ $n=0,1,2\cdot\cdot\cdot,7.$
Thus the roots are
$z=-1$
and
$z=1+e^{i2\pi n/8},$ $n=0,1,2\cdot\cdot\cdot,7$ .
5. Find the Laurent series expanded around $z=0$ for $f(z)=\dfrac{1}{z(z-2)}$.
Specify the region in which each of these series is valid.
Solution:
In the region $|z|$ < $2$ , we have
$f(z)=-\dfrac{1}{2z(1-z/2)}=-\dfrac{1}{2z}(1+\dfrac{z}{2}+\cdot\cdot \cdot+\dfrac{z^{n}}{2^{n}}+\cdot\cdot\cdot)$ .
And in the region $|z|>2$ , we have
$f(z)=\dfrac{1}{z^{2}}\dfrac{1}{1-2/z}=\dfrac{1}{z^{2}}(1+\dfrac{2}{z}+\cdot\cdot\cdot+\dfrac{2^{n}}{z^{n}}+\cdot\cdot\cdot).$
6.Evaluate the following integrals:
a.$\int_{-\infty}^{\infty}\dfrac{(1+\cos\pi x)dx}{(1-x^{2})},$
b.$\int_{-\infty}^{\infty}\dfrac{\sin^{2}x}{1+x^{2}}dx,$
c. $\int_{-\infty}^{\infty}\dfrac{1+3x^{2}}{(1+x^{2})(4+x^{2})}dx$.
Solution:
a. We get
$J\equiv\int_{-\infty}^{\infty}\dfrac{(1+e^{i\pi x})dx}{(1-x^{2})}=0$ by closing contour upstairs. ( Note that $J$ is a convergent integral.) Thus
$I=$Re $J=0.$
b. $I=\int_{-\infty}^{\infty}\dfrac{\sin^{2}x}{1+x^{2}}dx=-\dfrac{1}{4}\int_{-\infty}^{\infty}\dfrac{e^{2ix}-2+e^{-2ix}}{1+x^{2}}dx.$
Let
$J\equiv\int_{-\infty}^{\infty}\dfrac{e^{2ix}-1}{1+x^{2}}dx.$
By closing contour upstairs, we get
$J=2\pi i\dfrac{e^{-2}-1}{2i}=-\pi(1-e^{-2}).$
Thus
$I=-$Re $J/2=\pi(1-e^{-2})/2.$
c. By closing contour upstairs, we get
$\int_{-\infty}^{\infty}\dfrac{1+3x^{2}}{(1+x^{2})(4+x^{2})}dx=2\pi i[\dfrac{-2}{(2i)(3)}+\dfrac{-11}{(-3)(4i)}]=7\pi/6$.
7. Evaluate the following integrals:
a. $\int_{0}^{2\pi}\dfrac{1}{3-2\sin\theta}d\theta.$
b. $\int_{-\infty}^{\infty}(\dfrac{1}{x^{2}+1})^{3}dx.$
c. $\int_{-\infty}^{\infty}\dfrac{\cos(\pi x/2)}{x^{2}-1}dx.$
Solution:
a. Let $z=e^{i\theta}.$ We have
$I=- \oint \nolimits_{c}\dfrac{dz}{z^{2}-3iz-1},$
where $c$ is a unit circle with its center at the origin. The integrand above has two simple poles:
$z_{1}=i(3+\sqrt{5})/2,$ $z_{2}=i(3-\sqrt{5})/2.$
Since $z_{2}$ is the pole inside $c$,
$I=2\pi i$ Residue$(z_{2})=-\dfrac{2\pi}{(3-\sqrt{5})-3}=\dfrac{2\pi}{\sqrt {5}}.$
b. Closing the contour upstairs, we get
$I=2\pi i$Res$(i).$
Since the integrand has a third-order pole at $i$, we get
Res$(i)=\dfrac{1}{2}\underset{z\rightarrow i}{\lim}\dfrac{d^{2}}{dz^{2}}\dfrac{1}{(z+i)^{3}}=\dfrac{6}{(2i)^{5}}=\dfrac{3}{16i}.$
We have $I=\dfrac{3\pi}{8}.$
c. Since both the numerator and the denominator vanish at $x=\pm1,$ the integrand is entire. By deforming the contour, we get
$I=\dfrac{1}{2}\int_{c}\dfrac{e^{iz\pi/2}+e^{-iz\pi/2}}{z^{2}-1}dz,$
where $c$ passes around the points $z=\pm1$ from above. By closing the contour upstairs, we find the first term in the integrand equal to zero. We get
$I=\dfrac{1}{2}\int_{c}\dfrac{e^{-iz\pi/2}}{z^{2}-1}dz=-\pi i(-\dfrac{i}{2}+\dfrac{i}{-2})=-\pi.$
8. Evaluate the following integrals with the use of the Cauchy residue theorem.
a.$\int_{0}^{\infty}\dfrac{dx}{(9+x^{2})(4+x^{2})},$
b.$\int_{-\infty}^{\infty}\dfrac{\sin^{2}x}{x^{2}}dx,$
Solution:
a. $I=\dfrac{1}{2}\int_{-\infty}^{\infty}\dfrac{dx}{(9+x^{2})(4+x^{2})}.$
Closing the contour upstairs, we get
$I=\pi i[\dfrac{1}{(6i)(-5)}+\dfrac{1}{(5)(4i)}]=\pi/60.$
b. Since the integrand is an entire function, we deform the contour and get
$I=-\dfrac{1}{4}\int_{c}\dfrac{e^{2iz}-2+e^{-2iz}}{z^{2}}dz,$
where $c$ is a contour in the upper half-plane joining $-\infty$ and $\infty$. The first two terms in the integrand give zero. We get
$I=\dfrac{2\pi i}{4}\underset{z\rightarrow0}{\lim}\dfrac{de^{-2iz}}{dz}=\pi.$
9. Evaluate the following integrals:
a. $\int_{-\infty}^{\infty}\dfrac{\sin cx}{(x+a)^{2}+b^{2}}dx,$
where $a$ is a real number while $b$ and $c$ are both real and positive.
b. $\int_{-\infty}^{\infty}\dfrac{x^{2}}{x^{4}+1}dx.$
c. $\int_{C}\dfrac{e^{1/z}}{z}dz$, where $c$ is a circle around the origin in the counterclockwise direction.
Solution:
a. $I=$Im$\int_{-\infty}^{\infty}\dfrac{e^{icx}}{(x+a)^{2}+b^{2}}dx=$Im$[2\pi i$Res$(-a+ib)]$
$ =$Im$[2\pi i\dfrac{e^{-iac}e^{-bc}}{2ib}]=-\dfrac{\pi e^{-bc}\sin ac}{b}.$
b. Closing the contour upstairs, we get $I=(2\pi i)[$Res$(e^{i\pi/4})+$Res$(-e^{-i\pi/4})].$
Since
Res($e^{i\pi/4}$)=$\dfrac{z^{2}}{4z^{3}}]_{z=\exp(i\pi/4)}$ ,
Res($-e^{-i\pi/4}$)=$\dfrac{z^{2}}{4z^{3}}]_{z=-\exp(-i\pi/4)\text{ ,}}$
$I=(2\pi i)[\dfrac{1}{4(e^{i\pi/4})}+\dfrac{1}{4(-e^{-i\pi/4})}]=(2\pi i)\dfrac{-2i\sin(\pi/4)}{4}=\pi\sqrt{2}/2.$
c. We have
$I=\int_{C}\dfrac{e^{1/z}}{z}dz=\int_{C}\dfrac{1}{z}(1+z+z^{2}/2+\cdot\cdot\cdot)dz.$
By the Cauchy residue theorem, we get
$I=2\pi i,$
10. Evaluate the integral $\ I\equiv\int_{0}^{\infty}\dfrac{x}{1+x^{n}}dx$.
Solution:
Let $\ J\equiv\int_{C}\dfrac{z}{1+z^{n}}dz,$
where $C$ is the contour of a pie with two of its boundaries the infinite ray of $\theta=0$ and that of $\theta=2\pi/n.$ Then
$J=(1-e^{4i\pi/n})I$ .
By the Cauchy residue theorem, we have
$J=\dfrac{2\pi i}{ne^{i(n-2)\pi/n}}$.
Thus
$I=\dfrac{\pi}{n\sin(2\pi/n)}$ .