Lecture 5

In the Lecture we discuss the use of complex numbers and complex functions to solve linear second

order (in)homogeneous ordinary differential equations by using the Ansatz of complex exponential time

dependence. Here we will discuss the tools available in Mathematica to study such equations. In

particular, recall that we used the Solve command for algebraic equations. Now we want to employ the

DSolve command for differential equations.

First consider the homogeneous (undriven) equation for a mass on a spring with viscous damping.

Note that DSolve expects derivatives to be indicated by primes, 's, and that the desire function or

depedent variable (second argument) and independent variable (third argument) be clearly indicated.

We have

In[89]:= DSolve@m x''@tD + b x'@tD + k x@tD 0, x@tD, tD

Out[89]= ::x@tD ® ã

-b- b2-4 k m t

2 m C@1D + ã

-b+ b2-4 k m t

2 m C@2D>>

If we don't tell Mathematica that x ia function t, it tells us

In[90]:= DSolve@m x''@tD + b x'@tD + k x@tD 0, x, tD

Out[90]= ::x ® FunctionB8t<, ã

-b- b2-4 k m t

2 m C@1D + ã

-b+ b2-4 k m t

2 m C@2DF>>

We get just what we expect from our standard (real) exponential Ansatz, x[t] ~

eΑt

® Α = -b

2 m±

b2

4 m2

-k

m. Note also that Mathematica includes the 2 undetermined constants

(labeled in the form C[1], C[2]) which will be determined by initial conditions. This confirms the analysis

in Eq. (5.29) and following in the Lecture. Mathematica will also solve for these constants directly if the

inital conditions are provided in extra equations, as in

In[91]:= sol = DSolve@8m x''@tD + b x'@tD + k x@tD 0, x@0D == 1, x'@0D 0<, x@tD, tD

Out[91]= ::x@tD ®1

2 b2 - 4 k m

-b ã

-b- b2-4 k m t

2 m + b ã

-b+ b2-4 k m t

2 m + ã

-b- b2-4 k m t

2 m b2

- 4 k m + ã

-b+ b2-4 k m t

2 m b2

- 4 k m >>

To pull on the solution for manipulation we define

In[92]:= sol1 = x@tD . sol@@1DD

Out[92]=

1

2 b2 - 4 k m

-b ã

-b- b2-4 k m t

2 m + b ã

-b+ b2-4 k m t

2 m + ã

-b- b2-4 k m t

2 m b2

- 4 k m + ã

-b+ b2-4 k m t

2 m b2

- 4 k m

And test the initial conditions

In[93]:= sol1 . t ® 0

Out[93]= 1

In[94]:= sol1' . t ® 0

Out[94]= 0 &

Let's try it with constants defined via

In[95]:= k = 1; b = 1; m = 1;

In[96]:= sol2 = DSolve@8m x''@tD + b x'@tD + k x@tD 0, x@0D == 1, x'@0D 0<, x@tD, tD

Out[96]= ::x@tD ®1

3

ã-t2

3 CosB3 t

2

F + 3 SinB3 t

2

F >>

The expected damped oscillation that looks like (note the initial conditions are explicit in the plot).

In[97]:= Plot@x@tD . sol2@@1DD, 8t, 0, 10<, PlotRange ® 8-.2, 1<, AxesLabel ® 8"t", "x@tD"<D

Out[97]=

2 4 6 8 10

t

-0.2

0.2

0.4

0.6

0.8

1.0

x@tD

Now consider the inhomogeneous (harmonically driven) form of the equation (see Eqns. (5.23) to (5.25)

(note that Mathematica still rememebers our values for m, b and k)

In[98]:= sol3 = DSolve@m x''@tD + b x'@tD + k x@tD F 1 Cos@Ω0 t + j1D, x@tD, tD

2 Lec5_227_08.nb

Out[98]= ::x@tD ® ã-t2

C@2D CosB3 t

2

F +

ã-t2

C@1D SinB3 t

2

F + F1 - 3 CosB3 t

2

F CosBj1 -1

2

t I 3 - 2 Ω0MF -

3 CosB3 t

2

F CosB3 t

2

+ j1 + t Ω0F - CosBj1 -1

2

t I 3 - 2 Ω0MF SinB3 t

2

F -

CosB3 t

2

+ j1 + t Ω0F SinB3 t

2

F + 3 SinB3 t

2

F SinBj1 -1

2

t I 3 - 2 Ω0MF +

CosB3 t

2

F SinB3 t

2

- j1 - t Ω0F + CosB3 t

2

F SinB3 t

2

+ j1 + t Ω0F -

3 SinB3 t

2

F SinB3 t

2

+ j1 + t Ω0F - CosB3 t

2

F CosBj1 -1

2

t I 3 - 2 Ω0MF Ω0 +

CosB3 t

2

F CosB3 t

2

+ j1 + t Ω0F Ω0 - 3 CosBj1 -1

2

t I 3 - 2 Ω0MF

SinB3 t

2

F Ω0 + 3 CosB3 t

2

+ j1 + t Ω0F SinB3 t

2

F Ω0 + SinB3 t

2

F

SinBj1 -1

2

t I 3 - 2 Ω0MF Ω0 + 3 CosB3 t

2

F SinB3 t

2

- j1 - t Ω0F Ω0 -

3 CosB3 t

2

F SinB3 t

2

+ j1 + t Ω0F Ω0 + SinB3 t

2

F SinB3 t

2

+ j1 + t Ω0F Ω0 +

3 CosB3 t

2

F CosBj1 -1

2

t I 3 - 2 Ω0MF Ω0

2+

3 CosB3 t

2

F CosB3 t

2

+ j1 + t Ω0F Ω0

2-

CosBj1 -1

2

t I 3 - 2 Ω0MF SinB3 t

2

F Ω0

2- CosB

3 t

2

+ j1 + t Ω0F SinB3 t

2

F Ω0

2-

3 SinB3 t

2

F SinBj1 -1

2

t I 3 - 2 Ω0MF Ω0

2+

CosB3 t

2

F SinB3 t

2

- j1 - t Ω0F Ω0

2+ CosB

3 t

2

F SinB3 t

2

+ j1 + t Ω0F Ω0

2+

3 SinB3 t

2

F SinB3 t

2

+ j1 + t Ω0F Ω0

2+ 2 CosB

3 t

2

F CosBj1 -1

2

t I 3 - 2 Ω0MF

Ω0

3- 2 CosB

3 t

2

F CosB3 t

2

+ j1 + t Ω0F Ω0

3- 2 SinB

3 t

2

F

SinBj1 -1

2

t I 3 - 2 Ω0MF Ω0

3- 2 SinB

3 t

2

F SinB3 t

2

+ j1 + t Ω0F Ω0

3

I2 3 I-1 + 3 Ω0 - Ω0

2M I1 + 3 Ω0 + Ω0

2MM>>

Yikes - lots of apparent complexity so simplify

Lec5_227_08.nb 3

In[99]:= Simplify@x@tD . sol3@@1DDD

Out[99]=

1

1 - Ω0

2 + Ω0

4

ã-t2

-ãt2

F1 I-Cos@j1 + t Ω0D - Sin@j1 + t Ω0D Ω0 + Cos@j1 + t Ω0D Ω0

2M +

C@2D CosB3 t

2

F + C@1D SinB3 t

2

F I1 - Ω0

2+ Ω

0

4M

More muscle

In[100]:= FullSimplify@x@tD . sol3@@1DDD

Out[100]= ã-t2

C@2D CosB3 t

2

F + C@1D SinB3 t

2

F +

F1 HCos@j1 + t Ω0D + Ω0 HSin@j1 + t Ω0D - Cos@j1 + t Ω0D Ω0LL1 - Ω

0

2 + Ω0

4

Now we recognize the first (damped) term as the complementary solution, i.e., the solution to the

homogeneous equation (which knows about the damping and the natural frequency), while the second

term is the particular solution for the specific driving function (and driving frequency). To see this is the

same as Eq. (5.25) we would have to expand the single cosine function (with the 2 phases) in terms of

the phase due to the spring system.

Now with inital conditions and a specific diving force we have

In[101]:= F1 = 2; Ω0 = 1; j1 = Π 3;

In[102]:= sol4 =

DSolve@8m x''@tD + b x'@tD + k x@tD F 1 Cos@Ω0 t + j1D, x@0D == 1, x'@0D 0<, x@tD, tD

Out[102]= ::x@tD ®1

3

ã-t2

3 CosB3 t

2

F - 3 3 CosB3 t

2

F +

3 3 ãt2

Cos@tD CosB3 t

2

F2

+ 3 ãt2

CosB3 t

2

F2

Sin@tD - 3 SinB3 t

2

F -

3 SinB3 t

2

F + 3 3 ãt2

Cos@tD SinB3 t

2

F2

+ 3 ãt2

Sin@tD SinB3 t

2

F2

>>

In[103]:= FullSimplify@x@tD . sol4@@1DDD

Out[103]= 3 Cos@tD + Sin@tD +1

3

ã-t2

-3 I-1 + 3 M CosB3 t

2

F - I3 + 3 M SinB3 t

2

F

4 Lec5_227_08.nb

In[104]:= Plot@x@tD . sol4@@1DD, 8t, 0, 15<, PlotRange ® 8-2, 3<, AxesLabel ® 8"t", "x@tD"<D

Out[104]=

2 4 6 8 10 12 14

t

-2

-1

1

2

3

x@tD

Note there is a brief period (as in the previous plot) where the complementary solution damps out and

after which the initial conditions are "forgotten". Then for large times we settle into the behavior given

by the particular solution corresponding to the specific driving function. Note that the coefficients in the

complementary solution (C[1]) & C[2]) are chosen so that the full solution (complementary plus particu-

lar) obeys the given inital conditions, as illustrated in the plot. This is not the case for the individual bits

alone.

Complementary

In[105]:= PlotB1

3

ã-t2

-3 J-1 + 3 N CosB3 t

2

F - J3 + 3 N SinB3 t

2

F ,

8t, 0, 15<, PlotRange ® 8-2, 3<, AxesLabel ® 8"t", "x_comp@tD"<F

Out[105]=

2 4 6 8 10 12 14

t

-2

-1

1

2

3

x_comp@tD

Particular

Lec5_227_08.nb 5

In[106]:= PlotB 3 Cos@tD + Sin@tD, 8t, 0, 15<,

PlotRange ® 8-2, 3<, AxesLabel ® 8"t", "x_part@tD"<F

Out[106]=

2 4 6 8 10 12 14

t

-2

-1

1

2

3

x_part@tD

But the sum of the 2 (linear superposition) satisfies the inhomogeneous equation AND the initial

condiitons.

6 Lec5_227_08.nb