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

$I=\dfrac{\pi}{n\sin(2\pi/n)}$ .