Example
Example
In[6]:=
Clear[p1,p2,p3,p4]
p1=Plot[t+(Pi/2), {t,-Pi,0}, DisplayFunction->Identity,
Ticks->{{-Pi,Pi},Automatic}];
p2=Plot[(Pi/2)-t, {t,0,Pi}, DisplayFunction->Identity,
Ticks->{{-Pi,Pi},Automatic}];
p3=Plot[t-(3Pi/2), {t,Pi,2Pi}, DisplayFunction->Identity,
Ticks->{{Pi,2Pi},Automatic}];
p4=Plot[(5Pi/2)-t, {t,2Pi,3Pi}, DisplayFunction->Identity,
Ticks->{{2Pi,3Pi},Automatic}];
inputF=Show[p1,p2,p3,p4,DisplayFunction->$DisplayFunction,
Ticks->{{-Pi,Pi,2Pi,3Pi},Automatic}];
Since F(t) is an even function, the bk terms will all be zero. Let's find the ak terms.
In[7]:=
a0=(1/2Pi)(Integrate[t+(Pi/2),{t,-Pi,0}]+Integrate[(Pi/2)-t,{t,0,Pi}])
Out[7]=
0
In[8]:=
a[n_]:=a[n]=(1/Pi)(Integrate[(t+(Pi/2)) u[t,Pi,n],{t,-Pi,0}]+
Integrate[((Pi/2)-t) u[t,Pi,n],{t,0,Pi}])
Table[a[k],{k,1,10}]
Out[8]=
4 4 4 4 4
{--, 0, ----, 0, -----, 0, -----, 0, -----, 0}
Pi 9 Pi 25 Pi 49 Pi 81 Pi
Can you see the pattern?
Let's put undetermined coefficients to work here. We need to solve
u''+0.02u'+25u=4/(k^2 Pi) cos(kt) (for k odd).
Note that with L=Pi, u[t,Pi,k]=cos(kt)
We assume a solution of the form ck cos(kt) + dk sin(kt)
In[9]:=
testsol[t_]:=c Cos[k t]+d Sin[k t]
In[10]:=
D[testsol[t],{t,2}]+.02 D[testsol[t],t]+25 testsol[t]
Out[10]=
2 2
-(c k Cos[k t]) - d k Sin[k t] +
25 (c Cos[k t] + d Sin[k t]) +
0.02 (d k Cos[k t] - c k Sin[k t])
We want to set this equal to 4/(k^2 Pi)cos(kt) (k odd) and solve for the constants c and d.
In[11]:=
Simplify[%]
Out[11]=
2
25 c Cos[k t] + 0.02 d k Cos[k t] - c k Cos[k t] +
2
25 d Sin[k t] - 0.02 c k Sin[k t] - d k Sin[k t]
Equating cos(kt) terms we get 25c +.02d k -c k^2=4/(k^2 Pi)
Equating sin(kt) terms we get 25d-.02c k -d k^2=0
In[12]:=
Clear[c,d]
Solve[{25 c +.02 d k - c k^2==4/(k^2 Pi),
25 d-.02 c k - d k^2==0},{c,d}];
cc=Evaluate[c/.Flatten[%][[1]]]
dd=Evaluate[d/.Flatten[%%][[2]]]
Out[12]=
2
-4. (25. - 1. k )
------------------------------------
2 4 6
-1963.5 k + 157.078 k - 3.14159 k
Out[13]=
-0.08 k
------------------------------------
2 4 6
-1963.5 k + 157.078 k - 3.14159 k
Note: Here % means the last calculation,
%% means two calculations back.
This is just a trick to get the output into a variable.
Remember, we only want ODD cos(k t) terms. That is what the
{n,1,10,2} is for.
In[14]:=
uk=Table[(cc/.k->n)Cos[n t]+(dd/.k->n)Sin[n t],{n,1,10,2}]
Out[14]=
{0.0530516 Cos[t] + 0.0000442097 Sin[t],
0.00884182 Cos[3 t] + 0.0000331568 Sin[3 t],
0. Cos[5 t] + 0.509296 Sin[5 t],
-6
-0.00108265 Cos[7 t] + 6.31546 10 Sin[7 t],
-7
-0.000280694 Cos[9 t] + 9.0223 10 Sin[9 t]}
In[15]:=
uparticular=Sum[uk[[j]],{j,1,5}]
Out[15]=
0.0530516 Cos[t] + 0.00884182 Cos[3 t] + 0. Cos[5 t] -
0.00108265 Cos[7 t] - 0.000280694 Cos[9 t] +
0.0000442097 Sin[t] + 0.0000331568 Sin[3 t] +
-6
0.509296 Sin[5 t] + 6.31546 10 Sin[7 t] +
-7
9.0223 10 Sin[9 t]
What about the homogeneous part? From above we have M=1, C=.02, K=25, so
C^2<4MK and u(t) (homogeneous) is Exp[-.01 t](ACos(5 t) + BSin(5 t). As
t->infinity, the homogeneous part contributes nothing to the steady state
solution (do you see why?).
In[16]:=
sol=Plot[uparticular,{t,-Pi,3Pi}];
In[17]:=
Show[inputF,sol];
The steady state solution is almost harmonic (amplitudes are all close) with
frequency five times that of F(t).
Notice that the sin(5 t) portion of the solution dominates: the other
coefficients are quite small.
Up to Forced Oscillations