1. Using Einstein Summation notation, prove the identity:

BAABABBABA )()()()()( ∇⋅−∇⋅+⋅∇−⋅∇=××∇ [10 points] We begin by writing the cross product of A and B as:

C = kjijki BAC ε=⇒×BA So then our identity, CBA ×∇=××∇ )( ,

We consider )()()( kjn

ijkimnkjn

ijkmnikjijkn

mnim BAx

BAx

BAx ∂

∂=

∂∂

=∂∂

=×∇ εεεεεεC

where the last step follows as a cyclic permutation of the indices mni. We apply the epsilon-delta relationship to find:

)()( kjn

kmjnknjm BAx

C∂∂

−=×∇ δδδδ

We expand the derivative via the product rule to obtain:

][][)()( kn

jjn

kkmjnkn

jjn

kknjmkjn

kmjnknjm Bx

AAx

BBx

AAx

BBAx ∂

∂+

∂∂

−∂∂

+∂∂

=∂∂

− δδδδδδδδ

In the first bracketed expression, we must have (from the product of Kronecker deltas) that j = m and also that k = n; in the second bracketed expression we must have that j = n and also that k = m, making these subsitutions we get:

mn

nnn

mnn

mmn

n Bx

AAx

BBx

AAx

B∂∂

−∂∂

−∂∂

+∂∂ (1)

We recall that in Einstein summation notation, repeated indices are summed, so that

∇⋅=∂∂

⋅∇=∂∂

⋅= AABAi

iii

ii xAA

xBA ;;

This means that the first term in (1) is AB )( ∇⋅ ; the second term is ; the third term is and the last term is -

)( BA ⋅∇)( AB ⋅∇− BA )( ∇⋅ . Summing these four terms proves the

requested identity.

2. Spherical coordinates represent an example of an orthogonal coordinate system. The transformation from Cartesian to spherical coordinates is given by:

θφθφθ cos,sinsin,cossin rzryrx ===

a) Find the scale factors for the spherical coordinate system. In calculating these scale factors, you may take as a given that the system is orthogonal and you may make explicit use of that fact in your calculations. If you do so, explain how you are exploiting the orthogonal nature of spherical coordinates. [10 points] The key time saving piece of information is that you could use explicitly the fact that the spherical coordinate system is orthogonal. This means you know that all cross terms, i.e.,

φθφθ dddrdddr ;; all sum to zero. We start by taking derivatives of the transformation equations:

φφθθφθφθ drdrdrdx sinsincoscoscossin −+=

φφθθφθφθ drdrdrdy cossinsincossinsin ++=

θθθ drdrdz sincos −=

We sum the squares of each term to find:

222222 )sinsin()coscos()cos(sin)()()( φφθθφθφθ drdrdrdzdydx −++=++ + +

222 )cossin()sincos()sin(sin φφθθφθφθ drdrdr ++22 )sin()(cos θθθ drdr −+

Squaring all terms, using the Pythagorean theorem when applicable, and grouping by differentials, we find:

222222222 )(sin)()()()()( θθθ drdrdrdzdydx ++=++ From this, we can read off that h1 =hr = 1; h2 = hθ = r; h3 = hφ= r sin θ b) If r is the position vector show that 3=⋅∇ r for spherical coordinates. [10 pts] We use the expression for divergence in spherical coordinates (given on the table of results):

⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

=⋅∇ )()()(1213

3312

2321

1321

hhFq

hhFq

hhFqhhh

F

In this case, the vector is the position vector r. In spherical coordinates, there is only an r component of this vector, so our expression above becomes:

3)sin(sin1)0()0()sin(

sin1 23

2222

22 =⎥⎦⎤

⎢⎣⎡∂∂

=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

+∂∂

=⋅∇ θθφθ

θθ

rrr

rrrr

r

3. Consider the function:

kjiv )()tanln()( 223323 yxzzxyzex y ++++++=

Calculate:

∫∫ ⋅S

dSnv

where S is the surface of a sphere of radius 2 centered on the origin. [10 pts] The key point here is to recognize this as an example of the divergence theorem, so that:

∫ ∫ ⋅=⋅∇V S

dSd nvv τ

So that all we have to do is find the divergence of v and integrate that function over the volume of the sphere. While the given function v is quite ugly, it has a very simple divergence, namely:

Div v = 3(x2 + y2 + z2) = 3r2

Our integral then becomes:

∫ ∫∫∫ ∫∫∫==⋅∇V

dddrrddrdrrd φθθφθθτ sin3)sin(3422v

The angular part of this integral is simply 4π, leaving a simple radial integral evaluated from r = 0, 2, so our final result is 12π (25/5). 4. For any odd function f(x), find the value of:

∫−

3

1

)()( dxxxf δ

where δ(x) is the Dirac Delta function. Show your work clearly and/or explain the reasoning behind your answer. [10 pts]

We realize that because of the sifting property of the Dirac Delta function, the value of this integral is equal to f(0). We are told that the function is odd, and we can use a number of ways of showing that f(0) for an odd function is 0, so that the value of this integral is zero. [For instance, we can argue that any odd function can be expanded in a Taylor series around x0 =0; this series will have the form a1x + a3x3+a5x5 +…for an odd function. Clearly, the value of f(0) = 0]

5. Determine the Fourier series for f(x) = x + 1 on the interval (-π, π). Write the first 3 non-zero terms of each summation. You may use symmetry arguments to ease calculations; if you do so, state your reasoning explicitly. [10] This problem was worked out in its entirety in the solution set to homework set #8, question no. 6. Please refer to: http://www.luc.edu/faculty/dslavsk/homework/phys301hw8s.pdf

6. An eager student of mathematical physics observes that:

∫∞

− =0

1dxen nx

From this, our student argues that any function can be expanded as a series of the form:

∑ −=n

nxnecxf )( (2)

where the cn are suitably chosen constants.

Is our student correct? Can any function be meaningfully expanded in an infinite series of the form of eq. (2)? Explain your reasoning. (Your grade will be based on your reasoning; an answer of ‘yes’ or ‘no’ unaccompanied by appropriate justification will earn no credit.) [10]

Our eager student is not correct. In order to expand f(x) in terms of a basis set of functions, that basis set of functions must represent a complete, orthogonal set of functions on the interval in question. Fourier series are based on the orthogonality of trig functions; expansion in terms of Legendre polynomials is based on the orthogonality of Legendre polynomials on (-1, 1).

The proposed basis set above, e-nx is not an orthogonal set of functions. Orthogonal functions have the property:

∫ = mnmn dxxfxf δ)()(

where δmn is the Kronecker delta. We can show easily that mnmxnx

nmdxee δ≠

+=−

∞−∫

1

0

so that these are not orthogonal functions and cannot serve as the basis set for expansion.

7. A radioactive nucleus is created by the neutron bombardment of another nucleus and is destroyed via its own radioactive decay. If the rate of production is a constant given by Q, and the decay constant of the nucleus is given by λ, the following differential equation describes the time rate of change of the number of radioactive nuclei:

)()( tNQdt

tdN λ−=

Where N(t) is the number of nuclei at a particular time t.

a) Find the expression for N(t) if N(0) = 0. [10]

This is a very straightforward example of a first order differential equation that can be solved by separating variables. We group similar variables on each side of the equation to find:

∫ ∫ +=−−

⇒=−

⇒=−

CttNQdttNQ

tdNdttNQ

tdN )(ln1)(

)()(

)( λλλλ

Multiplying through and exponentiating both sides yields:

λλ /)()( tCeQtN −−=

We can evaluate the constant C by using the initial condition that N(0)=0, showing that C=Q, and we write our solution as:

λ

λ )1()(teQtN

−−=

b) What is the limiting value of N(t) as t ∞→ ? Explain why this is what you would expect from the original differential equation. [5 pts]

As t grows very large, the exponent in the answer above goes toward zero, and the steady-state population of N tends toward N(t) = Q/λ . This is expected from the original

differential equation since as t grows large, we expect that dN(t)/dt goes to zero in the steady-state (that is the meaning of steady-state). Under these conditions we can solve the original equation algebraically and find that when dN(t)/dt= 0, N(t) = Q/λ as we find.

8. Using series solution techniques, solve the differential equation:

0=−′′ xyy

Derive the recursion relation, and use this to write the first three non-zero terms of each solution to the differential equation. [10 pts]

This problem was solved in its entirety as problem #1 in homework #12. Please refer to:

http://www.luc.edu/faculty/dslavsk/homework/phys301hw12s.pdf

9. Bessel’s equation is written as:

0)( 222 =−+′+′′ ypxyxyx

where p is any number (positive, negative, an integer or a fraction).

a) Derive the indicial equation for Bessel’s equation and solve for the roots of that equation. [5 pts]

This differential equation satisfies the Fuchsian conditions, so we can use the method of Frobenius with a trial solution of ∑

=0nn xay += rn

We substitute this into the original differential equation and find:

∑ ∑ ∑ ∑ =−+++−++ +++++

0 0 0

22 0)()1)(( rnn

rnn

rnn

rnn xapxaxarnxarnrn

0 (1)

The indicial equation is determined by setting n = 0 in the summations containing the lowest power of x. These are the first, second and last summations. Setting n =0 in those terms yields:

prprprrr ±=⇒=⇒=−+− 222 0)1(

b) Derive the recursion relation; derive the expressions for an for n even and separately for n odd. [5 pts]

To find the recursion relation, go back to eq. (1) above. We now must re-index the third sum so that all powers of x are equal. We set k=n+2, and obtain:

∑ ∑ ∑ ∑ =−+++−++ ++−

++

0 0 2

22 0)()1)(( rn

nrn

nrn

nrn

n xapxaxarnxarnrn0

Note that the third sum now has x to the same power as the other sums, but now starts at the lower limit of n=2. We can gain useful information by “stripping out” the n=1 term from the other sums. First, recall that we have already “stripped out” the n=0 term when we derived the indicial equation.

If we set n=1 and “strip out” that term, we find:

[ ] 00)1()1( 112 =⇒=−+++ aaprrr

For values of n greater than or equal to 2:

[ ])2(

0)()1)(( 22

2

pnna

aaaprnrnrn nnnn +

−=⇒=+−++−++ −

− (2)

In the last step, we use the fact that r=p for the larger root of the indicial equation. We can look at this recursion relation and realize that it will generate two branches; one for even n and one for odd n. However, we already determined that a1=0; this coupled with the recursion relation tells us that all aodd=0; and the recursion relation (2) holds for all even n.

c) In class we solved this equation for p = ½; find the solution for p = -1/2 and show explicitly how this is related to a well known trig function. [5 pts]

Use the recursion relation in (2) with p = -1/2 to find:

!656;

!434;

2:

)1(04

602

40

22 aaa

aaaa

anna

a nn

−=

⋅−

==⋅

−=

−=

−−

= −

Using these coefficients with r= - ½ , we can substitute back into our trial solution of and get the series solution: ∑

=

=0n

n xay +rn

xxaxxxxay cos...)!6!4!2

1( 2/10

6422/1

0−− =+−+−=

(A number of students found the coefficients correctly, but omitted the x-1/2 term from the final result.)

10. Consider Laplace’s Equation in Cartesian coordinates:

02 =∇ V

Assume a trial solution of the form )()(),( yYxXyxV = . Substitute this solution into Laplace’s equation and derive the general form of the solution to this equation in Cartesian coordinates. Without any boundary conditions you can proceed no further than writing a general solution. [10 pts]

We begin by writing Laplace’s equation in Cartesian coordinates (in 2 dimensions):

02

2

2

22 =

∂∂

+∂∂

=∇yV

xVV (1)

We substitute the trial solution )()(),( yYxXyxV = into (1), obtaining:

0=′′+′′ YXYX (2)

We divide (2) by XY to obtain:

0=′′

+′′

YY

XX (3)

As we have discussed in class and read in the text, the first term is a function of X only and the second term is a function of Y only. Because of this, we realize that each term must equal a constant, and we call this constant the separation constant. We can then rewrite (3) as two separate ordinary differential equations:

22 kYYandk

XX

=′′

−=′′

(4)

These equations can be rewritten as:

00 22 =−′′=+′′ YkYandXkX

These equations have very well known solutions:

kyky DeCeYkxBkxAX

−+=

+= )sin()cos(

Since our trial solution was )()(),( yYxXyxV = , we can multiply these two solutions to find our general solution form:

))(sincos()()(),( kyky DeCekxBkxAyYxXyxV −++==

With boundary conditions we could determine the constants A, B, C, D and k and find the series solutions that solves Laplace’s equation; however, without those boundary conditions this is the extent of our analysis.