1. Chapter 3 Differential Equations
3.1 Introduction
Almost all the elementary and numerous advanced parts of theoretical physics are formulated in terms of differential equations (DE).
Newton’s Laws
Maxwell equations
Schrodinger and Dirac equations etc.
Since the dynamics of many physical systems involve just two derivatives, DE of second order occur most frequently in physics. e.g.,
acceleration in classical mechanics
the kinetic energy operator in quantum mechanics
Ordinary differential equation (ODE) Partial differential equation (PDE)

2. 1. Laplace's eq.: Examples of PDEs
This very common and important eq. occurs in studies of
a. electromagnetic phenomena, b. hydrodynamics,
c. heat flow, d. gravitation.
2. Poisson's eq.,
In contrast to the homogeneous Laplace eq., Poisson's eq. is non-homogeneous with a source term
3. The wave (Helmholtz) and time-independent diffusion eqs.,
These eqs. appear in such diverse phenomena as
a. elastic waves in solids, b. sound or acoustics,
c. electromagnetic waves, d. nuclear reactors.
4. The time-dependent diffusion eq.

3. 5. The time-dependent wave eq.,
where is a four-dimensional analog of the Laplacian.
6.The scalar potential eq.,
7.The Klein-Gordon eq., , and the corresponding vector eqs. in which  is replaced by a vector function.
8.The Schrodinger wave eq.
and
for the time-independent case.

4. Some general techniques for solving second-order PDEs
1.Separation of variables, where the PDE is split into ODEs that are related
by common constants which appear as eigenvalues of linear operators,
LY = lY, usually in variable.
2. Conversion of a PDE into an integral eq. using Green's functions
applies to inhomogeneous PDEs.
3. Other analytical methods such as the use of integral transforms.
4. Numerical calculations
Nonlinear PDEs
Notice that the above mentioned PDEs are linear (in ). Nonlinear ODEs an PDEs are a rapidly growing and important field.
The simplest nonlinear wave eq
Perhaps the best known nonlinear eq. of second is the Korteweg-deVries (KdV) eq.

5. 3.2 First-order Differential Equations
We consider here the general form of first-order DE:
(3.1)
The eq. is clearly a first-order ODE. It may or may not be linear, although we shall
treat the linear case explicitly later.
Separable variables
Frequently, the above eq. will have the special form

6. or
P(x)dx + Q(y) dy = 0
Integrating from (x0, y0) to (x, y) yields,
Since the lower limits and contribute constants, we may ignore the lower
limits of integration and simply add a constant of integration.
Example Boyle's Law
In differential form Boyle's gas law is
V-volume, P - Pressure. or
ln V + ln P = C.
If we set C = ln k, PV = k.

7. Since the RHS is zero, we look for an unknown function 
Exact Differential Equations
Consider
P(x,y) dx + Q(x,y) dy = 0 .
This eq. is said to be exact if we match the LHS of it to a differential dj,
Since the RHS is zero, we look for an unknown function 
j(x,y) = const. and dj= 0.
We have
and
The necessary and sufficient for our eq. to be exact is that the second, mixed
partial derivatives of j (assumed continuous) are independent of the order of differential:
If such j(x,y) exists then the solution is j(x,y)=C.

8. There always exists at least one integrating facts, a(x,y), such that
Linear First-order ODE
If f (x,y) has the form -p(x)y + q(x), then
(3.2)
It is the most general linear first-order ODE. If q(x) = 0, Eq.(3.2) is homogeneous (in y). A nonzero q(x) may represent a source or deriving term. The equation is linear; each term is linear in y or dy/dx. There are no higher powers; that is, y2, and no products, ydy/dx. This eq. may be solved exactly.

9. Let us look for an integrating factor a(x) so that
(3.3)
may be rewritten as
(3.4)
The purpose of this is to make the left –hand side of Eq.(3.2) a derivative so that it can be integrated—by inspection.
Expanding Eq. (3.4), we obtain
Comparison with Eq.(3.3) shows that we must require

10. Here is a differential equation for a(x) , with the variables a and x
separable. We separate variables, integrate, and obtain
(3.5)
as our integrating factor.
With a(x) known we proceed to integrate Eq.(3.4). This, of course, was the
point of introducing a(x) in the first place. We have
Now integrating by inspection, we have
The constants from a constant lower limit of integration are lumped into the
constant C. Dividing
by α(x) , we obtain

11. Finally, substituting in Eq.(3.5) for α yields
(3.6)
Equation (3.6) is the complete general solution of the linear,
first-order differential equation, Eq.(3.2). The portion
corresponds to the case q(x) =0 and is a general solution of the Homogeneous
differential equation. The other term in Eq.(3.6),
is a particular solution corresponding to the specific source term q(x) .

12. V
Example RL Circuit
For a resistance-inductance circuit Kirchhoff’s law leads to
for the current I(t) , where L is the inductance and R the resistance, both
constant. V(t) is the time-dependent impressed voltage.
From Eq.(3.5) our integrating factor α(t) is

13. Then by Eq.(3.6)
with the constant C to be determined by an initial condition (a boundary condition).
For the special case V(t)=V0 , a constant,
If the initial condition is I(0) =0, then
and

14. 3.3 SEPARATION OF VARIABLES
A very important PDE in physics
Electromagnetic field, wave transition etc.
Cartesian coordinates
Cylindrical coordinates
Spherical coordinates
m: mass of an electron; hbar: Plank constant

15. Using for the Laplacian. For the present let be a constant.
Certain partial differential equations can be solved by separation of variables. The method splits the partial differential equation of n variables into ordinary differential equations. Each separation introduces an arbitrary constant of separation . If we have n variables, we have to introduce n-1 constants, determined by the conditions imposed in the problem being solved.
Using for the Laplacian. For the present let be a constant.
Cartesian Coordinates
In Cartesian coordinates the Helmholtz equation becomes
(3.7)

16. where a second separation constant has been introduced.
Let
(3.7a)
Dividing by
and rearranging terms, we obtain
(3.8)
The left-hand side is a function of x alone, whereas the right-hand side depends only on y and z, but x , y , and z are all independent coordinates. The only possibility is setting each side equal to a constant, a constant of separation. We choose
(3.10)
(3.9)
Rearrange Eq.3.10
where a second separation constant has been introduced.

17. Similarly
(3.11)
(3.12)
introducing a constant by to produce a symmetric set of equations. Now we have three ordinary differential equations ((3.9),(3.11), and (3.12)) to replace Eq.(3.7). Our solution should be labeled according to the choice of our constants l, m ,and n ,that is ,
(3.13)
Subject to the conditions of the problem being solved.
We may develop the most general solution of Eq.(3.7)by taking a linear combination of solutions ,
(3.14)
Where the constant coefficients are determined by the boundary conditions

18. Example: Laplace equation in rectangular coordinatesz
Consider a rectangle box with dimensions (a,b,c) in the (x,y,z) directions. All surfaces of the box are kept at zero potential, except the surface z=c, which is at a potential V(x,y).
It is required to find the potential everywhere inside the box.
z=c
y=c y
x=a x

19. Example: Laplace equation in rectangular coordinates
Where
Then the solutions of the three ordinary differential equations are

20. Here the features of Fourier series have been used.
To have F=0 at x=a and y=b, we must have aa=np, and bb=mp. Then
Since F=V(x,y) at z=c:
We have the coefficients
Here the features of Fourier series have been used.

21. Circular Cylindrical Coordinates
With our unknown function ψ dependent on ρ, φ , and z , the Helmholtz equation becomes
(3.15) or
(3.16)
As before, we assume a factored form for ψ ,
(3.17)

22. Substituting into Eq.(3.16), we have
(3.18)
All the partial derivatives have become ordinary derivatives. Dividing by PΦZ
and moving the z derivative to the right-hand side yields
(3.19)
Then
(3.20)
And
(3.21)

23. Setting , multiplying by , and rearranging terms, we obtain
(3.22)
We may set the right-hand side to m2 and
(3.23)
Finally, for the ρ dependence we have
(3.24)
This is Bessel’s differential equation . The solution and their properties are presented in Chapter 6. -
The original Helmholtz equation has been replaced by three ordinary differential equations. A solution of the Helmholtz equation is
A general Sol.
(3.26)

24. Spherical Polar Coordinates
Let us try to separate the Helmholtz equation in spherical polar coordinates:
(3.27)
(3.29)
(3.30)

25. Multiplying Eq.(3.32) by and rearranging terms, we obtain
we use as the separation constant. Any constant will do, but this one will make life a little easier. Then
(3.31)
and
(3.32)
Multiplying Eq.(3.32) by and rearranging terms, we obtain
(3.33)

26. Again, the variables are separated
Again, the variables are separated. We equate each side to a constant Q and
finally obtain
Once more we have replaced a partial differential equation of three variables by three ordinary differential equations. Eq.(3.34) is identified as the associated Legendre equation in which the constant Q becomes l(l+1); l is an integer. If is a (positive) constant, Eq. (3.35) becomes the spherical Bessel equation.
Again, our most general solution may be written
(3.34)
(3.35)
(3.36)

27. The restriction that k^2 be a constant is necessarily
The restriction that k^2 be a constant is necessarily. The separation process will
Still be possible for k^2 as general as
In the hydrogen atom problem, one of the most important examples of the Schrodinger
Wave equation with a closed form solution is k^2=f(r)
Finally, as an illustration of how the constant m in Eq.(3.31) is restricted, we note that φ in cylindrical and spherical polar coordinates is an azimuth angle. If this is a classical problem, we shall certainly require that the azimuthal solution Φ(φ) be singled valued, that is,

28. This is equivalent to requiring the azimuthal solution to have a period of 2π or some integral multiple of it. Therefore m must be an integer. Which integer it is depends on the details of the problem. Whenever a coordinate corresponds to an axis of translation or to an azimuth angle the separated equation always has the form

29. Let us consider a general second order homogeneous DE (in y) as
3.4 Singular Points
Let us consider a general second order homogeneous DE (in y) as
y'' + P(x) y' + Q(x) y = 0 (3.40)
where y' = dy/dx. Now, if P(x) and Q(x) remain finite at x = , point x =
is an ordinary point. However, if either P(x) or Q(x) ( or both) diverges as
x approaches to , is a singular point.
Using Eq.(3.40), we are able to distinguish between two kinds of singular points

30. These definitions hold for all finite values of x0
These definitions hold for all finite values of x0. The analysis of is similar to the treatment of functions of a complex variable. We set x = 1/z, substitute into the DE, and then let By changing variables in the derivative, we have
(3.41)
(3.42)
Using these results, we transform Eq.(3.40) into
(3.43)
The behavior at x =  (z = 0) then depends on the behavior of the new
coefficients

31. and
as z 0. If these two expressions remain finite, point x =  is an ordinary
point. If they diverge no more rapidly than that 1/z and 1/ , respectively, x =
is a regular singular point, otherwise an irregular singular point.
Example
Bessel's eq. is
Comparing it with Eq. (3.40) we have
P(x) = 1/x, Q(x) = ,
which shows that point x = 0 is a regular singularity. As x   (z  0),
from Eq. (3.43), we have the coefficients
and
Since the Q(z) diverges as , point x =  is an irregular or essential singularity.

32. Irregular singularity
We list , in Table 3.4, several typical ODEs and their singular points.
Table 3.4
Regular singularity
Equation
Irregular singularity
Hypergeometric
___
0,1,
2. Legendre
___
-1,1,
3. Chebyshev
___
-1,1,
4. Confluent hypergeometric
5. Bessel
6. Laguerre
7. Simple harmonic oscillator
___
___
8. Hermite

33. A linear second-order homogeneous ODE
3.5 Series Solutions
(A)
(B)
The Bessel’s function
A linear second-order homogeneous ODE
y'' + P(x) y' + Q(x) y = 0.

34. Linear second-order ODE
In this section, we develop a method of a series expansion for obtaining one solution of the linear, second-order, homogeneous DE.
A linear, second-order, homogeneous ODE may be written in the form
y'' + P(x) y' + Q(x) y = 0.
the most general solution may be written as
Our physical problem may lead to a nonhomogeneous, linear, second-order DE
y'' + P(x) y' + Q(x) y = F(x).
Specific solution of this eq., yp, could be obtained by some special techniques. Obviously, we may add to yp any solution of the corresponding homogeneous eq.
Hence,
The constants c1 and c2 will eventually be fixed by boundary conditions

35. To seek a solution with the form
Fuchs’s Theorem:
We can always get at least one power-series solution, provided we are expanding about a point that is an ordinary point or at worst a regular singular point.

36. .
To illustrate the series solution, we apply the method to two important DEs.
First, the linear oscillator eq.
, (3.44)
with known solutions y = sin wx, cos wx.
We try
with k and al still undetermined. Note that k need not be an integer. By differentiating twice, we obtain

37. By substituting into Eq. (3.44), we have
(3.45)
From the uniqueness of power series the coefficients of each power of x on the LHS must vanish individually.
The lowest power is , for l = 0 in the first summation. The requirement that the coefficient vanishes yields
k(k-1) = 0.
Since, by definition, a0  0, we have
k(k-1) = 0
This eq., coming from the coefficient of the lowest power of x, is called the indicial equation. The indicial eq. and its roots are of critical importance to our analysis. k=0 or k=1
If k = 1, the coefficient (k+1)k of must vanish so that
= 0.
We set l = j+2 in the first summation and l' = j in the second. This results in

38. This is a two-term recurrence relation.
+2 (k+j+2)(k+j+1) = 0 or
This is a two-term recurrence relation.
We first try the solution k = 0. The recurrence relation becomes
which leads to

39. ……
So our solution is
If we choose the indicial eq. root k = 1, the recurrence relation becomes
Again, we have
For this choice, k = 1, we obtain

40. This series substitution, known as Frobenius' method, has given us two series solution of the linear oscillation eq. However, two points must be strongly emphasized:
(1) The series solution should be substituted back into the DE, to see if it works.
(2) The acceptability of a series solution depends on its convergence (including asymptotic convergence).
Expansion above
It is perfectly possible to write
Indeed, for the Legendre eq the choice
x0 = 1 has some advantages. The point x0 should not be chosen at an essential
singularity -or the method will probably fail.

41. Limitations of Series Approach
This attack on the linear oscillator eq. was perhaps a bit too easy.
To get some idea of what can happen we try to solve Bessel's eq.
(3.46)
Again, assuming a solution of the form

42. [(n+j)(n+j-1)+(n+j)- ] + =0.
we differentiate and substitute into Eq. (3.46). The result is
By setting l= 0, we get the coefficient of ,
a0[k(k-1) + k- ] = 0.
The indicial equation
with solution k = +n or -n. For the coefficients of x^(k+1), we obtain
For k = +n or -n (k is not equal 1/2), [ ] does not vanish and we must require = 0.
Proceeding to the coefficient of for k = n, we set l = j in the 1st, 2nd, and 4th terms and l = j-2 in the 3rd term. By requiring the resultant coefficient of to vanish, we obtain
[(n+j)(n+j-1)+(n+j)- ] =0.

43. When j j+2, this can be written for j≥0 as
(3.47)
which is the desired recurrence relation. Repeated application of this recurrence
relation leads to
, and so on, and in general
Inserting these coefficients in our assumed series solution, we have

44. The final summation is identified as the Bessel function Jn(x).
In summation form
(3.48)
The final summation is identified as the Bessel function Jn(x).
When k = -n and n is not integer, we may generate a second distinct series to be labeled J-n(x). However, when -n is a negative integer, trouble develops.
The second solution simply reproduces the first. We have failed to construct a second independent solutions for Bessel's eq. by this series technique when n is an integer.

45. Will this method always work? The answer is no!
SUMMARY
If we are expanding about an ordinary point or at worst about a regular singularity, the series substitution approach will yield at least one solution (Fuchs’s theorem).
Whether we get one or two distinct solutions depends on the roots of the indicial equation.
1. If the two roots of the indicial equation are equal, we can obtain only one solution by this series substitution method.
2. If the two roots differ by a non-integer number, two independent solutions may be obtained.
3. If the two roots differ by an integer, the larger of the two will yield a solution.
The smaller may or may not give a solution, depending on the behavior of the coefficients. In the linear oscillator equation we obtain two solutions; for Bessel’s equation, only one solution.

46. Regular and Irregular Singularities
The success of the series substitution method depends on the roots of the indicial eq. and the degree of singularity. To have clear understanding on this point, consider four simple eqs.
(3.49a)
(3.49b)
(3.49c)
(3.49d)

47. For the 1st eq., the indicial eq. is
k(k-1) - 6 =0,
giving k = 3, -2. Since the eq. is homogeneous in x ( counting as ), here is no recurrence relation. However, we are left with two perfectly good solution, and
For the 2nd eq., we have = 0, with no solution at all, for we have agreed that  0. The series substitution broke down at Eq. (3.49b) which has an irregular singular point at the origin.
Continuing with the Eq. (3.49c), we have added a term y'/x. The indicial eq. is , but again, there is no recurrence relation. The solutions are
y = , both perfectly acceptable one term series.
For Eq. (3.49d), (y'/x y'/x2), the indicial eq. becomes k = 0. There is a recurrence relation

48. Unless a is selected to make the series terminate we have
Hence our series solution diverges for all x ≠0.

49. 3.6 A Second Solution
In this section we develop two methods of obtaining a second independent solution: an integral method and a power series containing a logarithmic term. First, however we consider the question of independence of a set of function.
Linear Independence of Solutions
Given a set of functions, , the criterion for linear dependence is the existence of a relation of the form
(3.50)
in which not all the coefficients k are zero. On the other hand, if the only solution of Eq. (3.50) is k=0 for all , the set of functions is said to be linearly independent.
Let us assume that the functions  are differentiable. Then, differentiating Eq. (3.50) repeatedly, we generate a set of eq.

50. …
This gives us a set of homogeneous linear eqs. in which k are the unknown
quantities. There is a solution kl 0 only if the determinant of the coefficients
of the k's vanishes, .

51. This determinant is called the Wronskian.
1. If the wronskian is not equal to zero, then Eq.(3.50) has no solution other than k = 0. The set of functions is therefore independent.
2. If the Wronskian vanishes at isolated values of the argument, this does not necessarily prove linear dependence (unless the set of functions has only two functions). However, if the Wronskian is zero over the entire range of the variable. the functions  are linearly dependent over this range.
Example: Linear Independence
The solution of the linear oscillator eq. are The Wronskian becomes
and f1 and f2 are therefore linearly independent. For just two functions this means that one is not a multiple of the other, which is obviously true in this case.

52. You know that
but this is not a linear relation.
Example Linear Dependence
Consider the solutions of the one-dimensional diffusion eq.:
We have f1 = ex and f2 = e-x, and we add 3 = cosh x, also a solution.
Then
because the first and third rows are identical. Here, they are linearly
dependent, and indeed, we have
- 2 coshx = with kl  0.

53. A Second Solution
Returning to our 2nd order ODE
y'' + P(x) y' + Q(x) y = 0
Let y1 and y2 be two independent solutions. Then Wroskian is W=
By differentiating the Wronskian, we obtain
In the special case that P(x) = 0, i.e.
y'' + Q(x) y = 0, (3.52)
W = = constant.

54. Since our original eq. is homogeneous, we may multiply solutions and by whatever constants we wish and arrange to have W =1 ( or -1). This case P(x) = 0, appears more frequently than might be expected. ( in Cartesian coordinates, the radical dependence of (r) in spherical polar coordinates lack a first derivative). Finally, every linear 2nd-order ODE can be transformed into an eq. of the form of Eq.(3.52).
For the general case, let us now assume that we have one solution by a series substitution ( or by guessing). We now proceed to develop a 2nd, independent solution for which W  0.
We integrate from x1 =a to x1 = x to obtain
(3.53)

55. (3.54)
By combining Eqs. (3.53) and (3.54), we have
(3.55)
Finally, by integrating Eq. (3.55) from x2=b to x2=x we get
Here a and b are arbitrary constants and a term y1(x)y2(b)/y1(b) has been
dropped, for it leads to nothing new. As mentioned before, we can set
W(a) =1 and write

56. (3.56)
If we have the important special case of P(x) = 0. The above eq. reduces to
Now, we can take one known solution and by integrating can generate a
second independent solution.
Example
A Second Solution for the Linear Oscillator eq.
From y = 0 with P(x) = 0, let one solution be y1 = sin x.

57. Chapter 4. Orthogonal Functions* (Optional Reading)
4.1 Hermitian Operators (HO)
HO in quantum mechanics (QM)
As we know, is an HO operator. As is customary in QM, we simply assume that the wave functions satisfy appropriate boundary conditions: vanishing sufficiently strongly at infinity or having periodic behavior. The operator L is called Hermitian if
The adjoint A+ of an operator A is defined by
Clearly if A+ = A (self-adjoint) and satisfies the above mentioned boundary conditions, then A is Hermitian. The expectation value of an operator L is defined as