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Analysis of the eigenvalues and eigenfunctions for   ′′
y   (x )  +  λy   (x )  =   0  for all possible homogeneous boundary conditions

Nasser M. Abbasi

February 5, 2018 compiled on — Monday February 05, 2018 at 12:48 AM

Contents

1 Summary of result
 1.1 case 1: boundary conditions y (0) = 0,y(L) = 0
 1.2 case 2: boundary conditions y (0) = 0,y′(L ) = 0
 1.3 case 3: boundary conditions y (0 ) = 0, y(L) + y′(L ) = 0
 1.4 case 4: boundary conditions   ′
y (0) = 0,y(L ) = 0
 1.5 case 5: boundary conditions   ′        ′
y (0) = 0,y (L) = 0
 1.6 case 6: boundary conditions y′(0) = 0,y(L ) + y′(L) = 0
 1.7 case 7: boundary conditions y (0 )+  y′(0) = 0,y(L ) = 0
 1.8 case 8: boundary conditions y(0)+  y′(0) = 0,y′(L) = 0
 1.9 case 9: boundary conditions y(0)+   ′                ′
y (0) = 0,y(L) + y (L ) = 0
2 Derivations
 2.1 case 1: boundary conditions y (0) = 0,y(L) = 0
 2.2 case 2: boundary conditions y (0) = 0,y′(L ) = 0
 2.3 case 3: boundary conditions y (0 ) = 0, y(L) + y′(L ) = 0
 2.4 case 4: boundary conditions   ′
y (0) = 0,y(L ) = 0
 2.5 case 5: boundary conditions   ′        ′
y (0) = 0,y (L) = 0
 2.6 case 6: boundary conditions y′(0) = 0,y(L ) + y′(L) = 0
 2.7 case 7: boundary conditions y (0 ) + y′(0) = 0,y(L) = 0
 2.8 case 8: boundary conditions y (0 ) + y′(0) = 0,y′(L ) = 0
 2.9 case 9: boundary conditions y(0) + y′(0) = 0,y(L ) + y ′(L ) = 0

The eigenvalues and eigenfunctions for y′′ + λy = 0  over 0 < x < L  for all possible combinations of homogeneous boundary conditions are derived analytically. For each boundary condition case, a plot of the first few normalized eigenfunctions are given as well as the numerical values of the first few eigenvalues for the special case when L =  π  .

1 Summary of result

This section is a summary of the results. It shows for each boundary conditions the eigenvalues found and the corresponding eigenfunctions, and the full solution. A partial list of the numerical values of the eigenvalues for L = π  is given and a plot of the first few normalized eigenfunctions.

1.1 case 1: boundary conditions y(0) = 0,y(L) = 0




eigenvalues
eigenfunctions



λ < 0  None None
λ = 0 None None
λ > 0        (  )
λn =   nπ 2     n = 1,2, 3,⋅⋅⋅
       L               (  --- )
Φn (x ) = cnsin √ λnx



Normalized eigenfunctions: For L  = 1,

               (      )
         √ --    ∘ ---
Φn (x ) =   2sin    λnx

For L =  π,

         ∘ --    (      )
            2-    ∘ ---
Φn (x) =    π sin    λnx

List of eigenvalues

{                       }
  π2  4π2  9π2 16 π2
  L2, L2-, L2-,-L2--,⋅⋅⋅

List of numerical eigenvalues when L = π

{1,4,8,16, 25,⋅⋅⋅}

This is a plot showing how the eigenvalues change in value


pict

This is a plot showing the corresponding normalized eigenfunctions for the first 4  eigenvalues. We see that the number of zeros for Φn (x)  is n − 1  inside the interval 0 < x <  π  . (not counting the end points). Hence Φ1(x)  which correspond to λ1 = 1  in this case, will have no zeros inside the interval. While Φ2 (x)  which correspond to λ2 =  4  in this case, will have one zero and so on.


pict

1.2 case 2: boundary conditions           ′
y(0) = 0,y(L ) = 0




eigenvalues
eigenfunctions



λ < 0  None None
λ = 0 None None
λ > 0        (  )2
λn =   n2πL       n = 1,3, 5,⋅⋅⋅               (√ --- )
Φn (x ) = cnsin   λnx



Normalized eigenfunctions: For L  = 1,

         √ --  ( ∘ ---)
Φn (x ) =   2sin    λnx

For L =  π,

         ∘ --    (∘ --- )
Φn (x) =    2-sin    λnx
            π

List of eigenvalues

{                         }
  π2-- 9π2- 25π2- 49π2-
  4L2, 4L2, 4L2 , 4L2 ,⋅⋅⋅

List of numerical eigenvalues when L = π

{0.25,2.25, 6.25, 12.25,20.25,⋅⋅⋅}

This is a plot showing how the eigenvalues change in value


pict

This is a plot showing the corresponding normalized eigenfunctions for the first 4  eigenvalues.


pict

1.3 case 3: boundary conditions                  ′
y(0) = 0,y(L) + y (L ) = 0




eigenvalues
eigenfunctions



λ < 0  None None
λ = 0 None None
λ > 0  roots of     (√ -- )   √ --
tan    λL  +    λ = 0                ( √--- )
Φn (x) = cnsin   λnx



Normalized eigenfunctions: For L  = π,

pict

The normalization constant in this case depends on the eigenvalue.

List of numerical eigenvalues when L  = π  (since there is no analytical solution)

{0.620,2.794,6.845,12.865,20.879, ⋅⋅⋅}

This is a plot showing how the eigenvalues change in value


pict

This is a plot showing the corresponding normalized eigenfunctions for the first 4  eigenvalues.


pict

1.4 case 4: boundary conditions  ′
y(0) = 0,y(L ) = 0




eigenvalues
eigenfunctions



λ < 0  None None
λ = 0 None None
λ > 0        (  )
λn =   n2πL 2     n = 1,3, 5,⋅⋅⋅               (√ --- )
Φn (x ) = cncos   λnx



Normalized eigenfunctions for L =  1

             (  --- )
˜Φ  =  √2-cos  ∘ λ  x      n =  1,3,5,⋅⋅⋅
  n               n

When L =  π

     ∘  --
˜       2-   ( ∘ ---)
Φn =    π cos    λnx       n = 1,3,5,⋅⋅⋅

List of eigenvalues

{ π2   9π2  25π2  49π2    }
  ---, ---, ----, ----,⋅⋅⋅
  4L2  4L2  4L2   4L2

List of numerical eigenvalues when L = π

{0.25,2.25, 6.25, 12.25,20.25,⋅⋅⋅}

This is a plot showing how the eigenvalues change in value


pict

This is a plot showing the corresponding normalized eigenfunctions for the first 4  eigenvalues.


pict

1.5 case 5: boundary conditions y′(0) = 0,y′(L) = 0




eigenvalues
eigenfunctions



λ < 0  None None
λ = 0 Yes constant say 1
λ > 0  λ  =  (nπ)2     n = 1,2, 3,⋅⋅⋅
  n    L Φ  (x ) = c cos(√ λ--x)
 n        n        n



Normalized eigenfunction when L = 1

      √ --   (∘ --- )
˜Φn =    2cos    λnx       n =  1,2,3,⋅⋅⋅

When L =  π

     ∘  --
        2    ( ∘ ---)
˜Φn =    --cos    λnx       n = 1,2,3,⋅⋅⋅
        π

For ˜Φ0   , When L = 1

˜Φ0 = 1

When L =  π

      ∘ --
˜       1-
Φ0 =    π

List of eigenvalues

{                         }
    π2- 4π2- 9π2-16-π2
  0,L2 ,L2 , L2 , L2  ,⋅⋅⋅

List of numerical eigenvalues when L = π

{0, 1,4,9,16,⋅⋅⋅}

This is a plot showing how the eigenvalues change in value


pict

This is a plot showing the corresponding normalized eigenfunctions for the first 4  eigenvalues.


pict

1.6 case 6: boundary conditions y′(0) = 0,y(L ) + y ′(L ) = 0




eigenvalues
eigenfunctions



λ < 0  None None
λ = 0 None None
λ > 0  Roots of √ --    (√ -- )
  λ tan    λL   = 1                 (√ ---)
Φn (x) = cncos    λnx



Normalized eigenfunctions for L =  π  are

pict

List of numerical eigenvalues when L  = π  (There is no analytical solution for the roots)

{0.147033, 1.48528, 4.576, 9.606, 16.622,⋅⋅⋅}

This is a plot showing how the eigenvalues change in value


pict

This is a plot showing the corresponding normalized eigenfunctions for the first 4  eigenvalues.


pict

1.7 case 7: boundary conditions y(0)+  y′(0) = 0,y (L) = 0




eigenvalues
eigenfunctions



λ < 0  Root of      (√ ----)   √ ----
tanh    − λL  =   − λ  (one root)             (√ ---- )  √ ----    (√ ---- )
Φ (x) = sinh    − λx  −   − λ cosh   − λx
λ = 0  None None
λ > 0  Roots of    (     )
tan  √ λL  =  √ λ-              (     )          (     )
Φ  (x) = sin √ λx   − √ λ-cos √ λx
  n



List of numerical eigenvalues when L =  π  (There is no analytical solution for the roots)

{− 0.992, 1.664, 5.631, 11.623,⋅⋅⋅}

This is a plot showing how the eigenvalues change in value


pict

This is a plot showing the corresponding eigenfunctions for the first 4  eigenvalues.


pict

1.8 case 8: boundary conditions y(0)+  y′(0) = 0,y ′(L ) = 0




eigenvalues
eigenfunctions



λ < 0  Root of      (√ ----)
tanh    − λL  =  √1−λ-  (one root)               ( √---- )   √ ----    (√ ----)
Φ −1(x) = sinh   − λx  −    − λ cosh   − λx
λ = 0  None None
λ > 0  Roots of    ( √ --)
tan    λL  =  −√1-
               λ              ( √-- )   √ --   (√ -- )
Φn (x) = sin   λx   −   λ cos   λx



List of numerical eigenvalues when L =  π  (There is no analytical solution for the roots)

{− 1.007,0.480,3.392, 8.376, 24,368,⋅⋅⋅}

This is a plot showing how the eigenvalues change in value


pict

This is a plot showing the corresponding eigenfunctions for the first 4  eigenvalues.


pict

1.9 case 9: boundary conditions y(0)+  y′(0) = 0,y (L) + y′(L ) = 0




eigenvalues
eigenfunctions



λ < 0  − 1  Φ− 1(x) = sinh (x) − cosh(x )
λ = 0  None None
λ > 0        (nπ)2
λn =   L        n = 1,2, 3,⋅⋅⋅             (√ ---)   √ ---   (√ --- )
Φn (x ) = sin    λnx  −   λn cos   λnx



List of eigenvalues

{     π2  4π2 9 π2 16 π2    }
  − 1,--, ---,----,-----,⋅⋅⋅
      L2  L2   L2   L2

List of numerical eigenvalues when L = π

{ − 1,1, 4,9,16,⋅⋅⋅}

This is a plot showing how the eigenvalues change in value


pict

This is a plot showing the corresponding eigenfunctions for the first 4  eigenvalues.  


pict

2 Derivations

2.1 case 1: boundary conditions y(0) = 0,y(L) = 0

Let the solution be        rx
y = Ae  . This leads to the characteristic equation

pict

Let λ < 0

In this case − λ  is positive and hence √ ----
  − λ  is also positive. Let √ ----
  − λ = μ  where μ >  0  . Hence the roots are ± μ  . This gives the solution

y = c1 cosh(μx ) + c2 sinh (μx )

First B.C. y(0) = 0  gives

0 = c
     1

The solution becomes

y(x) = c sinh (μx)
        2

The second B.C. y (L ) = 0  results in

0 = c2 sinh (μL )

But sinh (μL ) ⁄= 0  since μL ⁄= 0  , hence c2 = 0,  Leading to trivial solution. Therefore λ <  0  is not eigenvalue.

Let λ = 0  , The solution is

y (x) = c1 + c2x

First B.C. y(0) = 0  gives

0 = c
     1

The solution becomes

y (x) = c x
         2

Applying the second B.C. y (L ) = 0  gives

0 = c2L

Therefore c2 = 0,  leading to trivial solution. Therefore λ =  0  is not eigenvalue.

Let λ > 0  , The solution is

             ( √-- )         (√ --)
y (x ) = c1cos   λx   + c2sin    λx

First B.C. y(0) = 0  gives

0 = c1

The solution becomes

             (√ -- )
y(x) = c2 sin    λx

Second B.C. y(L ) = 0  gives

          (     )
           √ --
0 =  c2 sin    λL

Non-trivial solution implies    (√ -- )
sin   λL   =  0  or √ --
  λL =  nπ  for n = 1,2,3, ⋅⋅⋅ . Therefore

pict

The corresponding eigenfunctions are

           (∘  ---)
Φn = cn sin     λnx       n = 1,2,3,⋅⋅⋅

The normalized ˜Φn  eigenfunctions are now found. In this problem the weight function is r (x) = 1  , therefore solving for cn  from

pict

Hence

     ∘ ----------√------------
        --------4--λn---------
cn =    2√ λ-L − sin(2 √ λ-L)
            n             n

For example, when L =  1  the normalization constant becomes (since now √ ---
  λn =  nLπ = nπ  )

pict

For L = π  , the normalization constant becomes (since now √ λ--=  nπ = n
    n   π  )

pict

The normalization c
 n  value depends on the length. When L =  1

             (  --- )
Φ˜  = √2-sin  ∘ λ  x      n = 1, 2,3,⋅⋅⋅
  n              n

When L =  π

     ∘  --
˜       2-   (∘  ---)
Φn =    π sin     λnx       n = 1,2,3,⋅⋅⋅

2.2 case 2: boundary conditions y(0) = 0,y′(L ) = 0

Let the solution be y = Aerx  . This leads to the characteristic equation

pict

Let λ < 0

In this case − λ  is positive and hence √ ----
  − λ  is also positive. Let √ ----
  − λ = μ  where μ >  0  . Hence the roots are ± μ  . This gives the solution

y = c1 cosh(μx ) + c2 sinh (μx )

First B.C. gives

0 = c1

Hence solution becomes

y(x) = c sinh (μx)
        2

Second B.C. gives

pict

But cosh (μL )  can not be zero, hence only other choice is c2 = 0  , leading to trivial solution. Therefore λ <  0  is not eigenvalue.

Let λ = 0  , The solution is

y (x) = c1 + c2x

First B.C. gives

0 = c1

Hence solution becomes

y (x) = c2x

Second B.C. gives

pict

Leading to trivial solution. Therefore λ = 0  is not eigenvalue.

Let λ > 0  , the solution is

             ( √-- )         (√ --)
y (x ) = c1cos   λx   + c2sin    λx

First B.C. gives

0 = c1

Hence solution becomes

             (√ -- )
y(x) = c2 sin    λx

Second B.C. gives

pict

Non-trivial solution implies    (√ -- )
cos   λL   = 0  or √--
 λL  =  n2π   for n =  1,3,5,⋅⋅⋅ . Therefore

pict

The eigenvalues are

     ( nπ )2
λn =   ---       n = 1,3,5,⋅⋅⋅
       2L

The corresponding eigenfunctions are

           (∘  ---)
Φn = cn sin     λnx       n = 1,3,5,⋅⋅⋅

The normalized ˜Φ
 n  eigenfunctions are now found. Since the weight function is r (x) = 1  , therefore solving for cn  from

pict

As was done earlier, the above results in

     ∘ ----------√------------
        -√------4--λn(-√----)-
cn =    2  λnL − sin  2  λnL      n =  1,3,5,⋅⋅⋅

For L = 1  the normalization constant becomes (since now √---   nπ    nπ-
 λn =  2L =  2   )

pict

For L = π  , the normalization constant becomes (since now √ ---
  λn =  n2ππ = n2   )

pict

Therefore, for L = 1

      √ --   (∘ --- )
Φ˜n  =   2sin    λnx       n = 1, 3,5,⋅⋅⋅

For L = π

        --
     ∘  2    (∘  ---)
˜Φn =    --sin     λnx       n = 1,3,5,⋅⋅⋅
        π

2.3 case 3: boundary conditions y(0) = 0,y(L) + y′(L ) = 0

Let the solution be y = Aerx  . This leads to the characteristic equation

pict

Let λ < 0

In this case − λ  is positive and hence √ −-λ-  is also positive. Let √ − λ-= μ  where μ >  0  . Hence the roots are ± μ  . This gives the solution

y = c1 cosh(μx ) + c2 sinh (μx )

First B.C. y(0) = 0  gives

0 = c1

Hence solution becomes

y(x) = c2sinh (μx)

Second B.C. y(L ) + y ′(L ) = 0  gives

0 = c2(sinh (μL ) + μcosh (μx))

But sinh (μL ) ⁄= 0  since μL ⁄= 0  and cosh (μx )  can not be zero, hence c2 = 0,  Leading to trivial solution. Therefore λ < 0  is not eigenvalue.

Let λ = 0  , The solution is

y (x) = c + c x
         1   2

First B.C. y(0) = 0  gives

0 = c
     1

The solution becomes

y (x) = c x
         2

Second B.C.          ′
y(L ) + y (L ) = 0  gives

pict

Therefore c2 = 0,  leading to trivial solution. Therefore λ = 0  is not eigenvalue.

Let λ > 0  , The solution is

             (     )         (    )
               √--            √ --
y (x ) = c1cos   λx   + c2sin    λx

First B.C. y(0) = 0  gives

0 = c1

The solution becomes

             (√ -- )
y(x) = c2 sin    λx

Second B.C.          ′
y(L ) + y (L ) = 0  gives

       (   ( √ --)    √--   ( √ -- ))
0 =  c2 sin    λL  +   λ cos    λL

For non-trivial solution, we want    (√ -- )   √ --   ( √-- )
sin    λL   +   λ cos   λL   =  0  or     (√ -- )   √ --
tan    λL   +   λ = 0  Therefore the eigenvalues are given by the solution to

    (√ -- )   √ --
tan    λL   +   λ = 0

And the corresponding eigenfunction is

           (      )
            ∘  ---
Φn = cn sin     λnx       n = 1,2,3,⋅⋅⋅

The normalized ˜Φn  eigenfunctions are now found. Since the weight function is r (x) = 1  , therefore solving for c
 n  from

pict

As was done earlier, the above results in

     ∘ ----------√------------
        --------4--λn---------
cn =    2√ λ-L − sin (2√ λ-L)     n =  1,2,3,⋅⋅⋅
            n             n

Since there is no closed form solution to λn  as it is a root of nonlinear equation    ( √ --)   √ --
tan    λL  +   λ =  0  , the normalized constant is found numerically. For L = π  , the first few roots are

λn =  {0.620, 2.794,6.845,12.865,20.879,⋅⋅⋅}

In this case, the normalization constants depends on n  and are not the same as in earlier cases. The following small program was written to find the first 10  normalization constants and to verify that each will make ∫L  2   2(√ --- )
 0 cnsin    λnx  dx =  1

The normalized constants are found to be (for L = π  )

cn = {0.729448, 0.766385, 0.782173,0.788879, 0.792141,0.79393, 0.795006, 0.7957,0.796171, 0.796506}


pict

The above implies that the first normalized eigenfunction is

                   (  ------ )
Φ  = (0.729448 )sin  √ 0.620x
 1

And the second one is

                   (√ ------ )
Φ2 = (0.766385 )sin    2.794x

And so on.

2.4 case 4: boundary conditions y′(0) = 0,y(L ) = 0

Let the solution be y = Aerx  . This leads to the characteristic equation

pict

Let λ < 0

In this case − λ  is positive and hence √ −-λ-  is also positive. Let √ − λ-= μ  where μ >  0  . Hence the roots are ± μ  . This gives the solution

pict

First B.C. y′(0) = 0  gives

pict

Hence solution becomes

y (x) = c1cosh (μx)

Second B.C. y(L ) = 0  gives

0 = c1cosh (μL )

But cosh(μL )  can not be zero, hence c1 = 0,  Leading to trivial solution. Therefore λ <  0  is not eigenvalue.

Let λ = 0  , The solution is

y (x) = c1 + c2x

First B.C.  ′
y (0) = 0  gives

0 = c2

The solution becomes

y (x) = c1

Second B.C. y(L ) = 0  gives

0 = c1

Therefore c1 = 0,  leading to trivial solution. Therefore λ =  0  is not eigenvalue.

Let λ > 0  , The solution is

pict

First B.C.  ′
y (0) = 0  gives

pict

The solution becomes

             (     )
              √ --
y(x) = c1cos    λx

Second B.C. y(L ) = 0  gives

          (√ -- )
0 = c1 cos   λL

For non-trivial solution, we want    (      )
     √ --
cos    λL   = 0  or √ --    nπ
  λL =  2   for odd n = 1, 3,5,⋅⋅⋅ Therefore

     (    )2
λn =   nπ-       n = 1,3,5,⋅⋅⋅
       2L

The corresponding eigenfunctions are

           ( ∘ ---)
Φn = cn cos    λnx       n = 1,3,5,⋅⋅⋅

The normalized ˜
Φn  eigenfunctions are now found. In this problem the weight function is r (x) = 1  , therefore solving for cn  from

pict

Hence

     ∘ ----------√------------
                4  λn
cn =    -√----------(--√----)-
        2  λnL + sin 2  λnL

For example, when L =  1  the normalization constant becomes (since now √ ---
  λn =  nπ = nπ
        2L    2   )

pict

Which is the same when the eigenfunction was    (    )
sin  n2πL x . For L = π  , the normalization constant becomes (since now   ---
√ λn =  nπ = n
        2L   2   )

pict

The normalization cn  value depends on the length. When L =  1

      √ --   (∘ --- )
˜Φ  =    2cos    λ  x      n =  1,3,5,⋅⋅⋅
  n               n

When L =  π

     ∘  --
˜       2-   ( ∘ ---)
Φn =    π cos    λnx       n = 1,3,5,⋅⋅⋅

2.5 case 5: boundary conditions  ′        ′
y(0) = 0,y (L) = 0

Let the solution be        rx
y = Ae  . This leads to the characteristic equation

pict

Let λ < 0

In this case − λ  is positive and hence √ ----
  − λ  is also positive. Let √ ----
  − λ = μ  where μ >  0  . Hence the roots are ± μ  . This gives the solution

pict

First B.C. y′(0) = 0  gives

pict

Hence solution becomes

y (x) = c1cosh (μx)

Second B.C. y′(L) = 0  gives

0 = c μsinh (μL )
     1

But sinh (μL )  can not be zero since μL  ⁄= 0  , hence c  = 0,
 1  Leading to trivial solution. Therefore λ <  0  is not eigenvalue.

Let λ = 0  , The solution is

y (x) = c1 + c2x

First B.C. y′(0) = 0  gives

0 = c2

The solution becomes

y (x) = c1

Second B.C. y′(L) = 0  gives

0 = 0

Therefore c1   can be any value. Therefore λ = 0  is an eigenvalue and the corresponding eigenfunction is any constant, say 1  .

Let λ > 0  , The solution is

pict

First B.C. y′(0) = 0  gives

pict

The solution becomes

             (√ -- )
y(x) = c1cos    λx

Second B.C.  ′
y (L) = 0  gives

              (     )
        √ --    √ --
0 = − c1  λsin    λL

For non-trivial solution, we want    ( √ --)
sin    λL  =  0  or √ --
  λL =  nπ  for n = 1,2, 3,⋅⋅⋅ Therefore

     ( nπ-)2
λn =   L         n = 1,2,3,⋅⋅⋅

And the corresponding eigenfunctions are

               (√ --)
Φn (x) = cncos    λx       n = 1,2,3,⋅⋅⋅

The normalized ˜Φn  eigenfunctions are now found. In this problem the weight function is r (x) = 1  , therefore solving for cn  from

pict

As before, the above simplifies to

     ∘ -----------------------
                4√ λ--
cn =    -√----------n(--√----)-
        2  λnL + sin 2  λnL

For example, when L =  1  the normalization constant becomes (since now √ ---   nπ
  λn =  L  = nπ  )

pict

For L = π  , the normalization constant becomes (since now √ ---
  λn =  nLπ = n  )

pict

The normalization cn  value depends on the length. When L =  1

˜     √ --   (∘ --- )
Φn =    2cos    λnx       n =  1,2,3,⋅⋅⋅

When L =  π

     ∘  --
        2    ( ∘ ---)
˜Φn =    --cos    λnx       n = 1,2,3,⋅⋅⋅
        π

For n = 0  , corresponding to the λ0   eigenvalue, since the eigenfunction is taken as the constant 1  , then

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Therefore, When L = 1

˜
Φ0 = 1

When L =  π

      ∘ --
˜Φ  =    1-
  0     π

2.6 case 6: boundary conditions  ′                ′
y(0) = 0,y(L ) + y (L ) = 0

Let the solution be        rx
y = Ae  . This leads to the characteristic equation

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Let λ < 0

In this case − λ  is positive and hence √ ----
  − λ  is also positive. Let √ ----
  − λ = μ  where μ >  0  . Hence the roots are ± μ  . This gives the solution

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First B.C. y′(0) = 0  gives

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Hence solution becomes

y (x) = c1cosh (μx)

Second B.C. y(L ) + y ′(L ) = 0  gives

0 =  c (cosh(μL ) + μsinh (μL ))
      1

But sinh (μL )  can not be negative since its argument is positive here. And coshμL  is always positive. In addition cosh (μL ) + μ sinh (μL )  can not be zero since sinh (μL )  can not be zero as μL ⁄=  0  and cosh(μL )  is not zero. Therefore c1 = 0,  Leading to trivial solution. Therefore λ <  0  is not eigenvalue.

Let λ = 0  , The solution is

y (x) = c + c x
         1   2

First B.C. y′(0) = 0  gives

0 = c
     2

The solution becomes

y (x) = c1

Second B.C.          ′
y(L ) + y (L ) = 0  gives

0 = c1

This gives trivial solution. Therefore λ =  0  is not eigenvalue.

Let λ > 0  , The solution is

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First B.C. y′(0) = 0  gives

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The solution becomes

             (√ -- )
y(x) = c1cos    λx

Second B.C.          ′
y(L ) + y (L ) = 0  gives

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For non-trivial solution, we want    (√ -- )   √ --   (√ -- )
cos   λL   −   λ sin    λL   = 0  or √ --    (√ -- )
  λ tan    λL   = 1  Therefore the eigenvalues are the solution to

        (     )
√ --     √ --
  λ tan    λL   = 1

And the corresponding eigenfunctions are

         (∘  ---)
Φn =  cos    λnx      n =  1,2,3,⋅⋅⋅

Where λn  are the roots of        (     )
√--     √ --
 λ tan    λL   = 1  .

The normalized ˜Φn  eigenfunctions are now found. Since the weight function is r (x) = 1  , therefore solving for c
 n  from

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As was done earlier, the above results in

     ∘ ----------√------------
        --------4--λn---------
cn =    2√ λ-L + sin(2 √ λ-L)     n =  1,2,3,⋅⋅⋅
            n             n

Since there is no closed form solution to λn  as it is a root of nonlinear equation   --   (   -- )
√ λ tan  √ λL   = 1  , the normalized constant is found numerically. For L = π  , the first few roots are

λn = {0.147033, 1.48528,4.57614, 9.60594, 25.6247,36.6282, 64.6318, 81.6328,100.634, 121.634,⋅⋅⋅}

In this case, the normalization constants depends on n  and are not the same as in earlier cases. The following small program was written to find the first 10  normalization constants and to verify that each will make ∫L        (√ ---)
 0 c2ncos2    λnx  dx = 1

The normalized constants are found to be (for L = π  )

cn = {0.705925, 0.751226, 0.776042, 0.786174, 0.790773, 0.793157, 0.794531, ⋅⋅⋅}


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The above implies that the first normalized eigenfunction is

                   (            )
                     √ ---------
Φ1 = (0.705925 )cos    0.147033x

And the second one is

                    (√ -------- )
Φ2  = (0.751226) cos   1.48528x

And so on.

2.7 case 7: boundary conditions        ′
y(0) + y (0) = 0,y(L ) = 0

Let the solution be        rx
y = Ae  . This leads to the characteristic equation

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Let λ < 0

In this case − λ  is positive and √----
 − λ  is positive. Let √ ----
  − λ = μ  where μ > 0  . Hence the roots are ± μ  . This gives the solution

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First B.C. y (0) + y ′(0 ) = 0  gives

0 = c1 + c2μ
(1)

Second B.C. y (L ) = 0  gives

0 = c1cosh (μL ) + c2 sinh (μL )

From (1) c1 = − c2μ  and the above now becomes

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For non-trivial solution, we want sinh (μL ) − μ cosh(μL ) = 0  . This means tanh(μL ) = μ  . Therefore λ <  0  is an eigenvalue and these are given by         2
λn = − μn  , where μn  is the solution to

tanh (μL ) = μ

Or equivalently, the roots of

     (√ ---- )   √ ----
tanh    − λL   =   − λ

There is only one negative root when solving the above numerically, which is λ −1 = 0.992.  The corresponding eigenfunction is

          (     (∘  ------)   ∘  ------    (∘  ------))
Φ −1 = c−1  sinh     − λ −1x −    − λ −1cosh    − λ −1x

Let λ = 0  , The solution is

y (x) = c1 + c2x

First B.C.         ′
y(0) + y (0) = 0  gives

0 = c + c
     1   2

The solution becomes

y (x) = c (1 − x)
         1

Second B.C. y(L )  gives

0 = c1(1 − L )

This gives trivial solution. Therefore λ =  0  is not eigenvalue.

Let λ > 0  , The solution is

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First B.C. y (0) + y ′(0 ) = 0  gives

           √--
0 = c +  c  λ
     1    2

The solution now becomes

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Second B.C. y (L) = 0  the above becomes

       (   ( √ --)    √--   ( √ -- ))
0 =  c2  sin    λL  −   λ cos    λL

For non-trivial solution, we want    (√ -- )   √ --   ( √ --)
sin    λL   −   λ cos    λL  =  0  or     (√ -- )   √ --
tan    λL   −   λ = 0  or

√ --      (√ -- )
  λ = tan    λL

Therefore the eigenvalues are the solution to the above (must be done numerically)  And the corresponding eigenfunctions are

           (    (∘  ---)   ∘  ---   (∘ --- ))
Φn (x) = cn  sin     λnx  −    λncos    λnx

for each root λn  .

2.8 case 8: boundary conditions        ′         ′
y(0) + y (0) = 0,y (L) = 0

Let the solution be y = Aerx  . This leads to the characteristic equation

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Let λ < 0

In this case − λ  is positive and hence √ ----
  − λ  is also positive. Let √ ----
  − λ = μ  where μ >  0  . Hence the roots are ± μ  . This gives the solution

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First B.C.          ′
y (0) + y (0 ) = 0  gives

0 = c1 + c2μ
(1)

Second B.C. y′(L ) = 0  gives

0 = c1μsinh (μL ) + c2μ cosh(μL )

From (1) c1 = − c2μ  and the above becomes

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For non-trivial solution, we want −  μsinh (μL ) + cosh (μL ) = 0  . This means − μ tanh (μL ) + 1 = 0  . Or             -1
tanh (μL) = μ  , therefore λ < 0  is eigenvalues and these are given by λn =  − μ2n  , where μn  is the solution to

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This has one root, found numerically which is λ −1 = − 1  . Hence √ ----
  − λ = 1  . The corresponding eigenfunction is

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Let λ = 0  , The solution is

y (x) = c1 + c2x

First B.C. y(0) + y′(0) = 0  gives

0 = c1 + c2

The solution becomes

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Second B.C.   ′
y  (L )  gives

0 = − c1

This gives trivial solution. Therefore λ =  0  is not eigenvalue.

Let λ > 0  , The solution is

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First B.C. y (0) + y ′(0 ) = 0  gives

           √--
0 = c1 + c2 λ

The solution becomes

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Second B.C. y ′(L ) = 0  gives

      (√ --   ( √ --)        ( √ --) )
0 = c2   λ cos    λL  +  λsin    λL

For non-trivial solution, we want     (     )           (     )
      √ --     √--     √ --
λsin    λL  +   λ cos    λL   = 0  or      (     )
       √ --       √--
λ tan    λL  =  −  λ  Therefore the eigenvalues are the solution to

   ( √ --)    − √ λ-   − 1
tan    λL  =  ----- =  √---
                λ       λ

And the corresponding eigenfunction is

            (   (√ -- )   √ --   (√ -- ))
Φn (x) = cn  sin    λx   −   λ cos   λx

2.9 case 9: boundary conditions y(0) + y′(0) = 0,y(L ) + y′(L) = 0

Let the solution be y = Aerx  . This leads to the characteristic equation

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Let λ < 0

In this case − λ  is positive and hence √ ----
  − λ  is also positive. Let √ ----
  − λ = μ  where μ >  0  . Hence the roots are ± μ  . This gives the solution

y = c1 cosh(μx ) + c2 sinh (μx )

Hence

 ′
y = μc1 sinh (μx ) + μc2 cosh (μx)

Left B.C. gives

0 = c1 + μc2
(1)

Right B.C. gives

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Using (1) in the above, it simplifies to

0 = sinh (μL )(c2 + μc1 )

But from (1) again, we see that c1 = − μc2   and the above becomes

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But sinh (μ2L ) ⁄= 0  since μ2L  ⁄= 0  and so either c =  0
 2  or (1 − μ2) = 0  . c =  0
 2  results in trivial solution, therefore       2
(1 − μ ) = 0  or   2
μ  = 1  but  2
μ  = − λ  , hence λ =  − 1  is the eigenvalue. Corresponding eigenfunction is

y = c1cosh (x) + c2 sinh (x)

Using (1) the above simplifies to

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But     √ ----
μ =   − λ = 1  , hence the eigenfunction is

|---------------------------------|
|y (x) = c2(− cosh (x ) + sinh (x))|
----------------------------------|

Let λ = 0  Solution now is

y = c1x + c2

Therefore

y′ = c1

Left B.C. 0 = y (0 ) + y′(0)   gives

0 = c2 + c1
(2)

Right B.C. 0 = y(L ) + y ′(L )  gives

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But from (2) c1 = − c2   and the above becomes

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Which means c2 = 0  and therefore the trivial solution. Therefore λ =  0  is not an eigenvalue.

Assuming λ > 0  Solution is

         ( √ --)         (√ -- )
y = c1cos    λx  + c2 sin    λx
(A)

Hence

       √ --    ( √ --)   √ --     ( √-- )
y′ = −   λc1sin    λx  +   λc2 cos   λx

Left B.C. gives

         √ --
0 = c1 +   λc2
(3)

Right B.C. gives

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Using (3) in the above, it simplifies to

       ( √ --) (     √ -- )
0 = sin    λL    c2 −  λc1

But from (3), we see that c =  − √ λc
 1         2   . Therefore the above becomes

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Only choice for non trivial solution is either (1 + λ) = 0  or     (  -- )
sin  √ λL   = 0  . But (1 + λ) = 0  implies λ = − 1  but we said that λ > 0  . Hence other choice is

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The above are the eigenvalues. The corresponding eigenfunction is from (A)

               (   --- )         (   --- )
Φ  (x) = c  cos  ∘ λ  x  + c  sin  ∘ λ x
 n        1n        n       2n        n

But         √ ---
c1n = −   λnc2n   and the above becomes

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