A Special S-L Equation

Consider the Sturm-Liouville equation
y''+ky=0
y(-L)=y(L) y'(-L)=y'(L)
on [-L,L].
Using Sturm-Liouville notation, r(x)=1, p(x)=1, and q(x)=0.
The eigenvalue k=0 has corresponding eigenfunction 1 (in fact any constant function). The eigenvalues (nPi/L)^2 (n a positive integer) have linearly independent eigenfunctions un(x)=Cos[nPix/L] and vn(x)=Sin[nPix/L].

In[2]:=

  Clear[u,v,L]

In[3]:=

  u[x_,L_,n_]:=Cos[n Pi x/L]

In[4]:=

  ColumnForm[Table[u[x,L,k],{k,0,10}]]

Out[4]=

  1
      Pi x
  Cos[----]
       L
      2 Pi x
  Cos[------]
        L
      3 Pi x
  Cos[------]
        L
      4 Pi x
  Cos[------]
        L
      5 Pi x
  Cos[------]
        L
      6 Pi x
  Cos[------]
        L
      7 Pi x
  Cos[------]
        L
      8 Pi x
  Cos[------]
        L
      9 Pi x
  Cos[------]
        L
      10 Pi x
  Cos[-------]
         L

Note that u[x,L,0]=1 and this will be used.

In[5]:=

  v[x_,L_,n_]:=Sin[n Pi x/L]

In[6]:=

  ColumnForm[Table[v[x,L,k],{k,0,10}]]

Out[6]=

  0
      Pi x
  Sin[----]
       L
      2 Pi x
  Sin[------]
        L
      3 Pi x
  Sin[------]
        L
      4 Pi x
  Sin[------]
        L
      5 Pi x
  Sin[------]
        L
      6 Pi x
  Sin[------]
        L
      7 Pi x
  Sin[------]
        L
      8 Pi x
  Sin[------]
        L
      9 Pi x
  Sin[------]
        L
      10 Pi x
  Sin[-------]
         L

Note that v[x,L,0]=0 and will not be used.

Verify these by substituting into the boundary value problem.

Check the first seven:

In[7]:=

  Table[D[u[x,L,k],{x,2}]+(k Pi/L)^2 u[x,L,k],{k,0,6}]

Out[7]=

  {0, 0, 0, 0, 0, 0, 0}

In[8]:=

  Table[D[v[x,L,k],{x,2}]+(k Pi/L)^2 v[x,L,k],{k,1,6}]

Out[8]=

  {0, 0, 0, 0, 0, 0}

By the Sturm-Liouville Theorem, we know that
un and um and vn and vm are orthogonal with respect to the weight function p(x)=1.

Check many at a time:

In[9]:=

  Table[Integrate[u[x,L,k]u[x,L,m],{x,-L,L}],{k,0,4},{m,0,4}];
  MatrixForm[%]

Out[9]=

  2 L   0     0     0     0
  
  0     L     0     0     0
  
  0     0     L     0     0
  
  0     0     0     L     0
  
  0     0     0     0     L

In[10]:=

  Table[Integrate[v[x,L,k]v[x,L,m],{x,-L,L}],{k,1,4},{m,1,4}];
  MatrixForm[%]

Out[10]=

  L   0   0   0
  
  0   L   0   0
  
  0   0   L   0
  
  0   0   0   L

In addition, since the un and vn are linearly independent, we have that un and vn are othogonal to each other.

In[11]:=

  Table[Integrate[u[x,L,k]v[x,L,m],{x,-L,L}],{k,0,4},{m,1,4}];
  MatrixForm[%]

Out[11]=

  0   0   0   0
  
  0   0   0   0
  
  0   0   0   0
  
  0   0   0   0
  
  0   0   0   0

We will also require the following. These facts will make our lives easier when finding Fourier coefficients of the eigenfunction expansions.

In[12]:=

  Table[Integrate[u[x,L,k]u[x,L,k],{x,-L,L}],{k,0,4}]

Out[12]=

  {2 L, L, L, L, L}

In[13]:=

  Table[Integrate[v[x,L,k]v[x,L,k],{x,-L,L}],{k,1,4}]

Out[13]=

  {L, L, L, L}

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