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## Analysis of the eigenvalues and eigenfunctions for for all possible homogeneous boundary conditions

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

### Contents

The eigenvalues and eigenfunctions for over 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 .

### 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 is given and a plot of the first few normalized eigenfunctions.

#### 1.1 case 1: boundary conditions

 eigenvalues eigenfunctions None None None None

Normalized eigenfunctions: For

For

List of eigenvalues

List of numerical eigenvalues when

This is a plot showing how the eigenvalues change in value

This is a plot showing the corresponding normalized eigenfunctions for the first eigenvalues. We see that the number of zeros for is inside the interval . (not counting the end points). Hence which correspond to in this case, will have no zeros inside the interval. While which correspond to in this case, will have one zero and so on.

#### 1.2 case 2: boundary conditions

 eigenvalues eigenfunctions None None None None

Normalized eigenfunctions: For

For

List of eigenvalues

List of numerical eigenvalues when

This is a plot showing how the eigenvalues change in value

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

#### 1.3 case 3: boundary conditions

 eigenvalues eigenfunctions None None None None roots of

Normalized eigenfunctions: For

The normalization constant in this case depends on the eigenvalue.

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

This is a plot showing how the eigenvalues change in value

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

#### 1.4 case 4: boundary conditions

 eigenvalues eigenfunctions None None None None

Normalized eigenfunctions for

When

List of eigenvalues

List of numerical eigenvalues when

This is a plot showing how the eigenvalues change in value

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

#### 1.5 case 5: boundary conditions

 eigenvalues eigenfunctions None None Yes constant say

Normalized eigenfunction when

When

For , When

When

List of eigenvalues

List of numerical eigenvalues when

This is a plot showing how the eigenvalues change in value

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

#### 1.6 case 6: boundary conditions

 eigenvalues eigenfunctions None None None None Roots of

Normalized eigenfunctions for are

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

This is a plot showing how the eigenvalues change in value

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

#### 1.7 case 7: boundary conditions

 eigenvalues eigenfunctions Root of (one root) None None Roots of

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

This is a plot showing how the eigenvalues change in value

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

#### 1.8 case 8: boundary conditions

 eigenvalues eigenfunctions Root of (one root) None None Roots of

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

This is a plot showing how the eigenvalues change in value

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

#### 1.9 case 9: boundary conditions

 eigenvalues eigenfunctions None None

List of eigenvalues

List of numerical eigenvalues when

This is a plot showing how the eigenvalues change in value

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

### 2 Derivations

#### 2.1 case 1: boundary conditions

Let the solution be . This leads to the characteristic equation

Let

In this case is positive and hence is also positive. Let where . Hence the roots are . This gives the solution

First B.C. gives

The solution becomes

The second B.C. results in

But since , hence Leading to trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Applying the second B.C. gives

Therefore leading to trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

Non-trivial solution implies or for . Therefore

The corresponding eigenfunctions are

The normalized eigenfunctions are now found. In this problem the weight function is , therefore solving for from

Hence

For example, when the normalization constant becomes (since now )

For , the normalization constant becomes (since now )

The normalization value depends on the length. When

When

#### 2.2 case 2: boundary conditions

Let the solution be . This leads to the characteristic equation

Let

In this case is positive and hence is also positive. Let where . Hence the roots are . This gives the solution

First B.C. gives

Hence solution becomes

Second B.C. gives

But can not be zero, hence only other choice is , leading to trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

Hence solution becomes

Second B.C. gives

Leading to trivial solution. Therefore is not eigenvalue.

Let , the solution is

First B.C. gives

Hence solution becomes

Second B.C. gives

Non-trivial solution implies or for . Therefore

The eigenvalues are

The corresponding eigenfunctions are

The normalized eigenfunctions are now found. Since the weight function is , therefore solving for from

As was done earlier, the above results in

For the normalization constant becomes (since now )

For , the normalization constant becomes (since now )

Therefore, for

For

#### 2.3 case 3: boundary conditions

Let the solution be . This leads to the characteristic equation

Let

In this case is positive and hence is also positive. Let where . Hence the roots are . This gives the solution

First B.C. gives

Hence solution becomes

Second B.C. gives

But since and can not be zero, hence Leading to trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

Therefore leading to trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

For non-trivial solution, we want or Therefore the eigenvalues are given by the solution to

And the corresponding eigenfunction is

The normalized eigenfunctions are now found. Since the weight function is , therefore solving for from

As was done earlier, the above results in

Since there is no closed form solution to as it is a root of nonlinear equation , the normalized constant is found numerically. For , the first few roots are

In this case, the normalization constants depends on and are not the same as in earlier cases. The following small program was written to find the first normalization constants and to verify that each will make

The normalized constants are found to be (for )

The above implies that the first normalized eigenfunction is

And the second one is

And so on.

#### 2.4 case 4: boundary conditions

Let the solution be . This leads to the characteristic equation

Let

In this case is positive and hence is also positive. Let where . Hence the roots are . This gives the solution

First B.C. gives

Hence solution becomes

Second B.C. gives

But can not be zero, hence Leading to trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

Therefore leading to trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

For non-trivial solution, we want or for odd Therefore

The corresponding eigenfunctions are

The normalized eigenfunctions are now found. In this problem the weight function is , therefore solving for from

Hence

For example, when the normalization constant becomes (since now )

Which is the same when the eigenfunction was . For , the normalization constant becomes (since now )

The normalization value depends on the length. When

When

#### 2.5 case 5: boundary conditions

Let the solution be . This leads to the characteristic equation

Let

In this case is positive and hence is also positive. Let where . Hence the roots are . This gives the solution

First B.C. gives

Hence solution becomes

Second B.C. gives

But can not be zero since , hence Leading to trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

Therefore can be any value. Therefore is an eigenvalue and the corresponding eigenfunction is any constant, say .

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

For non-trivial solution, we want or for Therefore

And the corresponding eigenfunctions are

The normalized eigenfunctions are now found. In this problem the weight function is , therefore solving for from

As before, the above simplifies to

For example, when the normalization constant becomes (since now )

For , the normalization constant becomes (since now )

The normalization value depends on the length. When

When

For , corresponding to the eigenvalue, since the eigenfunction is taken as the constant , then

Therefore, When

When

#### 2.6 case 6: boundary conditions

Let the solution be . This leads to the characteristic equation

Let

In this case is positive and hence is also positive. Let where . Hence the roots are . This gives the solution

First B.C. gives

Hence solution becomes

Second B.C. gives

But can not be negative since its argument is positive here. And is always positive. In addition can not be zero since can not be zero as and is not zero. Therefore Leading to trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

This gives trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

For non-trivial solution, we want or Therefore the eigenvalues are the solution to

And the corresponding eigenfunctions are

Where are the roots of .

The normalized eigenfunctions are now found. Since the weight function is , therefore solving for from

As was done earlier, the above results in

Since there is no closed form solution to as it is a root of nonlinear equation , the normalized constant is found numerically. For , the first few roots are

In this case, the normalization constants depends on and are not the same as in earlier cases. The following small program was written to find the first normalization constants and to verify that each will make

The normalized constants are found to be (for )

The above implies that the first normalized eigenfunction is

And the second one is

And so on.

#### 2.7 case 7: boundary conditions

Let the solution be . This leads to the characteristic equation

Let

In this case is positive and is positive. Let where . Hence the roots are . This gives the solution

First B.C. gives

 (1)

Second B.C. gives

From (1) and the above now becomes

For non-trivial solution, we want . This means . Therefore is an eigenvalue and these are given by , where is the solution to

Or equivalently, the roots of

There is only one negative root when solving the above numerically, which is The corresponding eigenfunction is

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

This gives trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution now becomes

Second B.C. the above becomes

For non-trivial solution, we want or or

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

for each root .

#### 2.8 case 8: boundary conditions

Let the solution be . This leads to the characteristic equation

Let

In this case is positive and hence is also positive. Let where . Hence the roots are . This gives the solution

First B.C. gives

 (1)

Second B.C. gives

From (1) and the above becomes

For non-trivial solution, we want . This means . Or , therefore is eigenvalues and these are given by , where is the solution to

This has one root, found numerically which is . Hence . The corresponding eigenfunction is

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

This gives trivial solution. Therefore is not eigenvalue.

Let , The solution is

First B.C. gives

The solution becomes

Second B.C. gives

For non-trivial solution, we want or Therefore the eigenvalues are the solution to

And the corresponding eigenfunction is

#### 2.9 case 9: boundary conditions

Let the solution be . This leads to the characteristic equation

Let

In this case is positive and hence is also positive. Let where . Hence the roots are . This gives the solution

Hence

Left B.C. gives

 (1)

Right B.C. gives

Using (1) in the above, it simplifies to

But from (1) again, we see that and the above becomes

But since and so either or . results in trivial solution, therefore or but , hence is the eigenvalue. Corresponding eigenfunction is

Using (1) the above simplifies to

But , hence the eigenfunction is

Let Solution now is

Therefore

Left B.C.  gives

 (2)

Right B.C. gives

But from (2) and the above becomes

Which means and therefore the trivial solution. Therefore is not an eigenvalue.

Assuming Solution is

 (A)

Hence

Left B.C. gives

 (3)

Right B.C. gives

Using (3) in the above, it simplifies to

But from (3), we see that . Therefore the above becomes

Only choice for non trivial solution is either or . But implies but we said that . Hence other choice is

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

But and the above becomes