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The eigenvalues and eigenfunctions for \(y^{\prime \prime }+\lambda y=0\) 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 \(L=\pi \).

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 \(L=\pi \) is given and a plot of the first few normalized eigenfunctions.

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

Normalized eigenfunctions: For \(L=1,\)\[ \Phi _{n}(x)=\sqrt {2}\sin \left ( \sqrt {\lambda _{n}}\,x\right ) \] For \(L=\pi ,\)\[ \Phi _{n}(x)=\sqrt {\frac {2}{\pi }}\sin \left ( \sqrt {\lambda _{n}}\,x\right ) \]

List of eigenvalues \[ \left \{ \frac {\pi ^{2}}{L^{2}},\frac {4\pi ^{2}}{L^{2}},\frac {9\pi ^{2}}{L^{2}},\frac {16\pi ^{2}}{L^{2}},\cdots \right \} \] List of numerical eigenvalues when \(L=\pi \) \[ \{1,4,8,16,25,\cdots \} \] This is a plot showing how the eigenvalues change in value

This is a plot showing the corresponding normalized eigenfunctions for the first \(4\) eigenvalues. We see that the number of zeros for \(\Phi _{n}(x)\) is \(n-1\) inside the interval \(0<x<\pi \). (not counting the end points). Hence \(\Phi _{1}(x)\) which correspond to \(\lambda _{1}=1\) in this case, will have no zeros inside the interval. While \(\Phi _{2}(x)\) which correspond to \(\lambda _{2}=4\) in this case, will have one zero and so on.

1.2 case 2: boundary conditions \(y(0)=0,y^{\prime }(L)=0\)

Normalized eigenfunctions: For \(L=1,\)\[ \Phi _{n}(x)=\sqrt {2}\sin \left ( \sqrt {\lambda _{n}}\,x\right ) \] For \(L=\pi ,\)\[ \Phi _{n}(x)=\sqrt {\frac {2}{\pi }}\sin \left ( \sqrt {\lambda _{n}}\,x\right ) \]

List of eigenvalues \[ \left \{ \frac {\pi ^{2}}{4L^{2}},\frac {9\pi ^{2}}{4L^{2}},\frac {25\pi ^{2}}{4L^{2}},\frac {49\pi ^{2}}{4L^{2}},\cdots \right \} \] List of numerical eigenvalues when \(L=\pi \) \[ \{0.25,2.25,6.25,12.25,20.25,\cdots \} \] This is a plot showing how the eigenvalues change in value

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

1.3 case 3: boundary conditions \(y(0)=0,y(L)+y^{\prime }(L)=0\)

Normalized eigenfunctions: For \(L=\pi ,\)

\begin {align*} \Phi _{1} & =\left ( 0.729448\right ) \sin \left ( \sqrt {0.620}x\right ) \\ \Phi _{2} & =\left ( 0.766385\right ) \sin \left ( \sqrt {2.794}x\right ) \\ & \vdots \end {align*}

The normalization constant in this case depends on the eigenvalue.

List of numerical eigenvalues when \(L=\pi \) (since there is no analytical solution) \[ \left \{ 0.620,2.794,6.845,12.865,20.879,\cdots \right \} \] This is a plot showing how the eigenvalues change in value

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

1.4 case 4: boundary conditions \(y^{\prime }(0)=0,y(L)=0\)

Normalized eigenfunctions for \(L=1\)

\[ \tilde {\Phi }_{n}=\sqrt {2}\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \]

When \(L=\pi \)

\[ \tilde {\Phi }_{n}=\sqrt {\frac {2}{\pi }}\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \]

List of eigenvalues\[ \left \{ \frac {\pi ^{2}}{4L^{2}},\frac {9\pi ^{2}}{4L^{2}},\frac {25\pi ^{2}}{4L^{2}},\frac {49\pi ^{2}}{4L^{2}},\cdots \right \} \] List of numerical eigenvalues when \(L=\pi \)\[ \left \{ 0.25,2.25,6.25,12.25,20.25,\cdots \right \} \] This is a plot showing how the eigenvalues change in value

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

1.5 case 5: boundary conditions \(y^{\prime }(0)=0,y^{\prime }(L)=0\)

Normalized eigenfunction when \(L=1\)

\[ \tilde {\Phi }_{n}=\sqrt {2}\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \]

When \(L=\pi \)

\[ \tilde {\Phi }_{n}=\sqrt {\frac {2}{\pi }}\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \]

For \(\tilde {\Phi }_{0}\), When \(L=1\)

\[ \tilde {\Phi }_{0}=1 \]

When \(L=\pi \)

\[ \tilde {\Phi }_{0}=\sqrt {\frac {1}{\pi }}\]

List of eigenvalues\[ \left \{ 0,\frac {\pi ^{2}}{L^{2}},\frac {4\pi ^{2}}{L^{2}},\frac {9\pi ^{2}}{L^{2}},\frac {16\pi ^{2}}{L^{2}},\cdots \right \} \] List of numerical eigenvalues when \(L=\pi \)\[ \left \{ 0,1,4,9,16,\cdots \right \} \] This is a plot showing how the eigenvalues change in value

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

1.6 case 6: boundary conditions \(y^{\prime }(0)=0,y(L)+y^{\prime }(L)=0\)

Normalized eigenfunctions for \(L=\pi \) are

\begin {align*} \Phi _{1} & =\left ( 0.705925\right ) \cos \left ( \sqrt {0.147033}x\right ) \\ \Phi _{2} & =\left ( 0.751226\right ) \cos \left ( \sqrt {1.48528}x\right ) \\ & \vdots \end {align*}

List of numerical eigenvalues when \(L=\pi \) (There is no analytical solution for the roots)\[ \left \{ 0.147033,1.48528,4.576,9.606,16.622,\cdots \right \} \] This is a plot showing how the eigenvalues change in value

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

1.7 case 7: boundary conditions \(y(0)+\)\(y^{\prime }(0)=0,y(L)=0\)

List of numerical eigenvalues when \(L=\pi \) (There is no analytical solution for the roots)\[ \left \{ -0.992,1.664,5.631,11.623,\cdots \right \} \] This is a plot showing how the eigenvalues change in value

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

1.8 case 8: boundary conditions \(y(0)+\)\(y^{\prime }(0)=0,y^{\prime }(L)=0\)

List of numerical eigenvalues when \(L=\pi \) (There is no analytical solution for the roots)\[ \left \{ -1.007,0.480,3.392,8.376,24,368,\cdots \right \} \] This is a plot showing how the eigenvalues change in value

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

1.9 case 9: boundary conditions \(y(0)+\)\(y^{\prime }(0)=0,y(L)+y^{\prime }(L)=0\)

List of eigenvalues\[ \left \{ -1,\frac {\pi ^{2}}{L^{2}},\frac {4\pi ^{2}}{L^{2}},\frac {9\pi ^{2}}{L^{2}},\frac {16\pi ^{2}}{L^{2}},\cdots \right \} \] List of numerical eigenvalues when \(L=\pi \)\[ \left \{ -1,1,4,9,16,\cdots \right \} \] This is a plot showing how the eigenvalues change in value

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

2 Derivations

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

Let the solution be \(y=Ae^{rx}\). This leads to the characteristic equation

\begin {align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt {-\lambda } \end {align*}

Let \(\lambda <0\)

In this case \(-\lambda \) is positive and hence \(\sqrt {-\lambda }\) is also positive. Let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). Hence the roots are \(\pm \mu \). This gives the solution\[ y=c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \] First B.C. \(y\left ( 0\right ) =0\) gives\[ 0=c_{1}\] The solution becomes\[ y\left ( x\right ) =c_{2}\sinh \left ( \mu x\right ) \] The second B.C. \(y\left ( L\right ) =0\) results in\[ 0=c_{2}\sinh \left ( \mu L\right ) \] But \(\sinh \left ( \mu L\right ) \neq 0\) since \(\mu L\neq 0\), hence \(c_{2}=0,\) Leading to trivial solution. Therefore \(\lambda <0\) is not eigenvalue.

Let \(\lambda =0\), The solution is\[ y\left ( x\right ) =c_{1}+c_{2}x \] First B.C. \(y\left ( 0\right ) =0\) gives\[ 0=c_{1}\] The solution becomes \[ y\left ( x\right ) =c_{2}x \] Applying the second B.C. \(y\left ( L\right ) =0\) gives\[ 0=c_{2}L \] Therefore \(c_{2}=0,\) leading to trivial solution. Therefore \(\lambda =0\) is not eigenvalue.

Let \(\lambda >0\), The solution is\[ y\left ( x\right ) =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \] First B.C. \(y\left ( 0\right ) =0\) gives\[ 0=c_{1}\] The solution becomes \[ y\left ( x\right ) =c_{2}\sin \left ( \sqrt {\lambda }x\right ) \] Second B.C. \(y\left ( L\right ) =0\) gives\[ 0=c_{2}\sin \left ( \sqrt {\lambda }L\right ) \] Non-trivial solution implies \(\sin \left ( \sqrt {\lambda }L\right ) =0\) or \(\sqrt {\lambda }L=n\pi \) for \(n=1,2,3,\cdots \). Therefore\begin {align*} \sqrt {\lambda _{n}} & =\frac {n\pi }{L}\qquad n=1,2,3,\cdots \\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}

The corresponding eigenfunctions are\[ \Phi _{n}=c_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \]

The normalized \(\tilde {\Phi }_{n}\) eigenfunctions are now found. In this problem the weight function is \(r\left ( x\right ) =1\), therefore solving for \(c_{n}\) from

\begin {align*} \int _{0}^{L}r\left ( x\right ) \Phi _{n}^{2}dx & =1\\ \int _{0}^{L}c_{n}^{2}\sin ^{2}\left ( \sqrt {\lambda _{n}}x\right ) dx & =1\\ c_{n}^{2}\int _{0}^{L}\left ( \frac {1}{2}-\frac {1}{2}\cos \left ( 2\sqrt {\lambda _{n}}x\right ) \right ) dx & =1\\ \int _{0}^{L}\frac {1}{2}dx-\int _{0}^{L}\frac {1}{2}\cos \left ( 2\sqrt {\lambda _{n}}x\right ) dx & =\frac {1}{c_{n}^{2}}\\ \frac {1}{2}L-\frac {1}{2}\left ( \frac {\sin \left ( 2\sqrt {\lambda _{n}}x\right ) }{2\sqrt {\lambda _{n}}}\right ) _{0}^{L} & =\frac {1}{c_{n}^{2}}\\ \frac {1}{2}L-\frac {1}{4\sqrt {\lambda _{n}}}\sin \left ( 2\sqrt {\lambda _{n}}L\right ) & =\frac {1}{c_{n}^{2}}\\ 2\sqrt {\lambda _{n}}L-\sin \left ( 2\sqrt {\lambda _{n}}L\right ) & =\frac {4\sqrt {\lambda _{n}}}{c_{n}^{2}} \end {align*}

Hence

\[ c_{n}=\sqrt {\frac {4\sqrt {\lambda _{n}}}{2\sqrt {\lambda _{n}}L-\sin \left ( 2\sqrt {\lambda _{n}}L\right ) }}\]

For example, when \(L=1\) the normalization constant becomes (since now \(\sqrt {\lambda _{n}}=\frac {n\pi }{L}=n\pi \))

\begin {align*} c_{n} & =\sqrt {\frac {4n\pi }{2n\pi -\sin \left ( 2n\pi \right ) }}\\ & =\sqrt {\frac {4n\pi }{2n\pi }}\\ c_{n} & =\sqrt {2} \end {align*}

For \(L=\pi \), the normalization constant becomes (since now \(\sqrt {\lambda _{n}}=\frac {n\pi }{\pi }=n\))

\begin {align*} c_{n} & =\sqrt {\frac {4n}{2n\pi -\sin \left ( 2n\pi \right ) }}\\ & =\sqrt {\frac {4n}{2n\pi }}\\ c_{n} & =\sqrt {\frac {2}{\pi }} \end {align*}

The normalization \(c_{n}\) value depends on the length. When \(L=1\)

\[ \tilde {\Phi }_{n}=\sqrt {2}\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \]

When \(L=\pi \)

\[ \tilde {\Phi }_{n}=\sqrt {\frac {2}{\pi }}\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \]

2.2 case 2: boundary conditions \(y(0)=0,y^{\prime }(L)=0\)

Let the solution be \(y=Ae^{rx}\). This leads to the characteristic equation\begin {align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt {-\lambda } \end {align*}

Let \(\lambda <0\)

In this case \(-\lambda \) is positive and hence \(\sqrt {-\lambda }\) is also positive. Let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). Hence the roots are \(\pm \mu \). This gives the solution\[ y=c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \] First B.C. gives\[ 0=c_{1}\] Hence solution becomes \[ y\left ( x\right ) =c_{2}\sinh \left ( \mu x\right ) \] Second B.C. gives\begin {align*} y^{\prime }\left ( x\right ) & =\mu c_{2}\cosh \left ( \mu x\right ) \\ 0 & =\mu c_{2}\cosh \left ( \mu L\right ) \end {align*}

But \(\cosh \left ( \mu L\right ) \) can not be zero, hence only other choice is \(c_{2}=0\), leading to trivial solution. Therefore \(\lambda <0\) is not eigenvalue.

Let \(\lambda =0\), The solution is\[ y\left ( x\right ) =c_{1}+c_{2}x \] First B.C. gives\[ 0=c_{1}\] Hence solution becomes \[ y\left ( x\right ) =c_{2}x \] Second B.C. gives\begin {align*} y^{\prime }\left ( x\right ) & =c_{2}\\ 0 & =c_{2} \end {align*}

Leading to trivial solution. Therefore \(\lambda =0\) is not eigenvalue.

Let \(\lambda >0\), the solution is\[ y\left ( x\right ) =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \] First B.C. gives\[ 0=c_{1}\] Hence solution becomes \[ y\left ( x\right ) =c_{2}\sin \left ( \sqrt {\lambda }x\right ) \] Second B.C. gives\begin {align*} y^{\prime }\left ( x\right ) & =\sqrt {\lambda }c_{2}\cos \left ( \sqrt {\lambda }x\right ) \\ 0 & =\sqrt {\lambda }c_{2}\cos \left ( \sqrt {\lambda }L\right ) \end {align*}

Non-trivial solution implies \(\cos \left ( \sqrt {\lambda }L\right ) =0\) or \(\sqrt {\lambda }L=\frac {n\pi }{2}\) for \(n=1,3,5,\cdots \). Therefore\begin {align*} \sqrt {\lambda _{n}}L & =\frac {n\pi }{2}\\ \sqrt {\lambda _{n}} & =\frac {n\pi }{2L}\qquad n=1,3,5,\cdots \end {align*}

The eigenvalues are\[ \lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots \] The corresponding eigenfunctions are

\[ \Phi _{n}=c_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \]

The normalized \(\tilde {\Phi }_{n}\) eigenfunctions are now found. Since the weight function is \(r\left ( x\right ) =1\), therefore solving for \(c_{n}\) from

\begin {align*} \int _{0}^{L}r\left ( x\right ) \Phi _{n}^{2}dx & =1\\ \int _{0}^{L}c_{n}^{2}\sin ^{2}\left ( \sqrt {\lambda _{n}}x\right ) dx & =1 \end {align*}

As was done earlier, the above results in

\[ c_{n}=\sqrt {\frac {4\sqrt {\lambda _{n}}}{2\sqrt {\lambda _{n}}L-\sin \left ( 2\sqrt {\lambda _{n}}L\right ) }}\qquad n=1,3,5,\cdots \]

For \(L=1\) the normalization constant becomes (since now \(\sqrt {\lambda _{n}}=\frac {n\pi }{2L}=\frac {n\pi }{2}\))

\begin {align*} c_{n} & =\sqrt {\frac {4\frac {n\pi }{2}}{2\frac {n\pi }{2}-\sin \left ( 2\frac {n\pi }{2}\right ) }}\\ & =\sqrt {\frac {2n\pi }{n\pi }}\\ c_{n} & =\sqrt {2} \end {align*}

For \(L=\pi \), the normalization constant becomes (since now \(\sqrt {\lambda _{n}}=\frac {n\pi }{2\pi }=\frac {n}{2}\))

\begin {align*} c_{n} & =\sqrt {\frac {4\frac {n}{2}}{2\frac {n}{2}\pi -\sin \left ( 2\frac {n}{2}\pi \right ) }}\\ & =\sqrt {\frac {2n}{n\pi }}\\ c_{n} & =\sqrt {\frac {2}{\pi }} \end {align*}

Therefore, for \(L=1\)

\[ \tilde {\Phi }_{n}=\sqrt {2}\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \]

For \(L=\pi \)

\[ \tilde {\Phi }_{n}=\sqrt {\frac {2}{\pi }}\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \]

2.3 case 3: boundary conditions \(y(0)=0,y(L)+y^{\prime }(L)=0\)

Let the solution be \(y=Ae^{rx}\). This leads to the characteristic equation

\begin {align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt {-\lambda } \end {align*}

Let \(\lambda <0\)

In this case \(-\lambda \) is positive and hence \(\sqrt {-\lambda }\) is also positive. Let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). Hence the roots are \(\pm \mu \). This gives the solution\[ y=c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \] First B.C. \(y\left ( 0\right ) =0\) gives\[ 0=c_{1}\] Hence solution becomes \[ y\left ( x\right ) =c_{2}\sinh \left ( \mu x\right ) \] Second B.C. \(y\left ( L\right ) +y^{\prime }\left ( L\right ) =0\) gives\[ 0=c_{2}\left ( \sinh \left ( \mu L\right ) +\mu \cosh \left ( \mu x\right ) \right ) \] But \(\sinh \left ( \mu L\right ) \neq 0\) since \(\mu L\neq 0\) and \(\cosh \left ( \mu x\right ) \) can not be zero, hence \(c_{2}=0,\) Leading to trivial solution. Therefore \(\lambda <0\) is not eigenvalue.

Let \(\lambda =0\), The solution is\[ y\left ( x\right ) =c_{1}+c_{2}x \] First B.C. \(y\left ( 0\right ) =0\) gives\[ 0=c_{1}\] The solution becomes \[ y\left ( x\right ) =c_{2}x \] Second B.C. \(y\left ( L\right ) +y^{\prime }\left ( L\right ) =0\) gives\begin {align*} 0 & =c_{2}L+c_{2}\\ & =c_{2}\left ( 1+L\right ) \end {align*}

Therefore \(c_{2}=0,\) leading to trivial solution. Therefore \(\lambda =0\) is not eigenvalue.

Let \(\lambda >0\), The solution is

\[ y\left ( x\right ) =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \] First B.C. \(y\left ( 0\right ) =0\) gives

\[ 0=c_{1}\] The solution becomes \[ y\left ( x\right ) =c_{2}\sin \left ( \sqrt {\lambda }x\right ) \] Second B.C. \(y\left ( L\right ) +y^{\prime }\left ( L\right ) =0\) gives\[ 0=c_{2}\left ( \sin \left ( \sqrt {\lambda }L\right ) +\sqrt {\lambda }\cos \left ( \sqrt {\lambda }L\right ) \right ) \] For non-trivial solution, we want \(\sin \left ( \sqrt {\lambda }L\right ) +\sqrt {\lambda }\cos \left ( \sqrt {\lambda }L\right ) =0\) or \(\tan \left ( \sqrt {\lambda }L\right ) +\sqrt {\lambda }=0\) Therefore the eigenvalues are given by the solution to \[ \tan \left ( \sqrt {\lambda }L\right ) +\sqrt {\lambda }=0 \]

And the corresponding eigenfunction is \[ \Phi _{n}=c_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \]

The normalized \(\tilde {\Phi }_{n}\) eigenfunctions are now found. Since the weight function is \(r\left ( x\right ) =1\), therefore solving for \(c_{n}\) from

\begin {align*} \int _{0}^{L}r\left ( x\right ) \Phi _{n}^{2}dx & =1\\ \int _{0}^{L}c_{n}^{2}\sin ^{2}\left ( \sqrt {\lambda _{n}}x\right ) dx & =1 \end {align*}

As was done earlier, the above results in

\[ c_{n}=\sqrt {\frac {4\sqrt {\lambda _{n}}}{2\sqrt {\lambda _{n}}L-\sin \left ( 2\sqrt {\lambda _{n}}L\right ) }}\qquad n=1,2,3,\cdots \]

Since there is no closed form solution to \(\lambda _{n}\) as it is a root of nonlinear equation \(\tan \left ( \sqrt {\lambda }L\right ) +\sqrt {\lambda }=0\), the normalized constant is found numerically. For \(L=\pi \), the first few roots are \[ \lambda _{n}=\left \{ 0.620,2.794,6.845,12.865,20.879,\cdots \right \} \]

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 \(\int _{0}^{L}c_{n}^{2}\sin ^{2}\left ( \sqrt {\lambda _{n}}x\right ) dx=1\)

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

\[ c_{n}=\{0.729448,0.766385,0.782173,0.788879,0.792141,0.79393,0.795006,0.7957,0.796171,0.796506\} \]

The above implies that the first normalized eigenfunction is

\[ \Phi _{1}=\left ( 0.729448\right ) \sin \left ( \sqrt {0.620}x\right ) \]

And the second one is

\[ \Phi _{2}=\left ( 0.766385\right ) \sin \left ( \sqrt {2.794}x\right ) \]

And so on.

2.4 case 4: boundary conditions \(y^{\prime }(0)=0,y(L)=0\)

Let the solution be \(y=Ae^{rx}\). This leads to the characteristic equation\begin {align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt {-\lambda } \end {align*}

Let \(\lambda <0\)

In this case \(-\lambda \) is positive and hence \(\sqrt {-\lambda }\) is also positive. Let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). Hence the roots are \(\pm \mu \). This gives the solution\begin {align*} y & =c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \\ y^{\prime } & =c_{1}\mu \sinh \left ( \mu x\right ) +c_{2}\mu \cosh \left ( \mu x\right ) \end {align*}

First B.C. \(y^{\prime }\left ( 0\right ) =0\) gives\begin {align*} 0 & =c_{2}\mu \\ c_{2} & =0 \end {align*}

Hence solution becomes \[ y\left ( x\right ) =c_{1}\cosh \left ( \mu x\right ) \] Second B.C. \(y\left ( L\right ) =0\) gives\[ 0=c_{1}\cosh \left ( \mu L\right ) \] But \(\cosh \left ( \mu L\right ) \) can not be zero, hence \(c_{1}=0,\) Leading to trivial solution. Therefore \(\lambda <0\) is not eigenvalue.

Let \(\lambda =0\), The solution is\[ y\left ( x\right ) =c_{1}+c_{2}x \] First B.C. \(y^{\prime }\left ( 0\right ) =0\) gives\[ 0=c_{2}\] The solution becomes \[ y\left ( x\right ) =c_{1}\] Second B.C. \(y\left ( L\right ) =0\) gives\[ 0=c_{1}\] Therefore \(c_{1}=0,\) leading to trivial solution. Therefore \(\lambda =0\) is not eigenvalue.

Let \(\lambda >0\), The solution is\begin {align*} y\left ( x\right ) & =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \\ y^{\prime }\left ( x\right ) & =-c_{1}\sqrt {\lambda }\sin \left ( \sqrt {\lambda }x\right ) +c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) \end {align*}

First B.C. \(y^{\prime }\left ( 0\right ) =0\) gives\begin {align*} 0 & =c_{2}\sqrt {\lambda }\\ c_{2} & =0 \end {align*}

The solution becomes \[ y\left ( x\right ) =c_{1}\cos \left ( \sqrt {\lambda }x\right ) \] Second B.C. \(y\left ( L\right ) =0\) gives\[ 0=c_{1}\cos \left ( \sqrt {\lambda }L\right ) \] For non-trivial solution, we want \(\cos \left ( \sqrt {\lambda }L\right ) =0\) or \(\sqrt {\lambda }L=\frac {n\pi }{2}\) for odd \(n=1,3,5,\cdots \) Therefore \[ \lambda _{n}=\left ( \frac {n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots \]

The corresponding eigenfunctions are

\[ \Phi _{n}=c_{n}\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \]

The normalized \(\tilde {\Phi }_{n}\) eigenfunctions are now found. In this problem the weight function is \(r\left ( x\right ) =1\), therefore solving for \(c_{n}\) from

\begin {align*} \int _{0}^{L}r\left ( x\right ) \Phi _{n}^{2}dx & =1\\ \int _{0}^{L}c_{n}^{2}\cos ^{2}\left ( \sqrt {\lambda _{n}}x\right ) dx & =1\\ c_{n}^{2}\int _{0}^{L}\left ( \frac {1}{2}+\frac {1}{2}\cos \left ( 2\sqrt {\lambda _{n}}x\right ) \right ) dx & =1\\ \int _{0}^{L}\frac {1}{2}dx+\int _{0}^{L}\frac {1}{2}\cos \left ( 2\sqrt {\lambda _{n}}x\right ) dx & =\frac {1}{c_{n}^{2}}\\ \frac {1}{2}L+\frac {1}{2}\left ( \frac {\sin \left ( 2\sqrt {\lambda _{n}}x\right ) }{2\sqrt {\lambda _{n}}}\right ) _{0}^{L} & =\frac {1}{c_{n}^{2}}\\ \frac {1}{2}L+\frac {1}{4\sqrt {\lambda _{n}}}\sin \left ( 2\sqrt {\lambda _{n}}L\right ) & =\frac {1}{c_{n}^{2}}\\ 2\sqrt {\lambda _{n}}L+\sin \left ( 2\sqrt {\lambda _{n}}L\right ) & =\frac {4\sqrt {\lambda _{n}}}{c_{n}^{2}} \end {align*}

Hence

\[ c_{n}=\sqrt {\frac {4\sqrt {\lambda _{n}}}{2\sqrt {\lambda _{n}}L+\sin \left ( 2\sqrt {\lambda _{n}}L\right ) }}\]

For example, when \(L=1\) the normalization constant becomes (since now \(\sqrt {\lambda _{n}}=\frac {n\pi }{2L}=\frac {n\pi }{2}\))

\begin {align*} c_{n} & =\sqrt {\frac {4\frac {n\pi }{2}}{2\frac {n\pi }{2}+\sin \left ( 2\frac {n\pi }{2}\right ) }}\\ & =\sqrt {\frac {2n\pi }{n\pi }}\\ c_{n} & =\sqrt {2} \end {align*}

Which is the same when the eigenfunction was \(\sin \left ( \frac {n\pi }{2L}x\right ) \). For \(L=\pi \), the normalization constant becomes (since now \(\sqrt {\lambda _{n}}=\frac {n\pi }{2L}=\frac {n}{2}\))

\begin {align*} c_{n} & =\sqrt {\frac {4\frac {n}{2}}{2\frac {n}{2}\pi +\sin \left ( 2\frac {n}{2}\pi \right ) }}\\ & =\sqrt {\frac {2n}{2n\pi }}\\ c_{n} & =\sqrt {\frac {2}{\pi }} \end {align*}

The normalization \(c_{n}\) value depends on the length. When \(L=1\)

\[ \tilde {\Phi }_{n}=\sqrt {2}\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \]

When \(L=\pi \)

\[ \tilde {\Phi }_{n}=\sqrt {\frac {2}{\pi }}\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,3,5,\cdots \]

2.5 case 5: boundary conditions \(y^{\prime }(0)=0,y^{\prime }(L)=0\)

Let the solution be \(y=Ae^{rx}\). This leads to the characteristic equation\begin {align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt {-\lambda } \end {align*}

Let \(\lambda <0\)

In this case \(-\lambda \) is positive and hence \(\sqrt {-\lambda }\) is also positive. Let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). Hence the roots are \(\pm \mu \). This gives the solution\begin {align*} y & =c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \\ y^{\prime } & =c_{1}\mu \sinh \left ( \mu x\right ) +c_{2}\mu \cosh \left ( \mu x\right ) \end {align*}

First B.C. \(y^{\prime }\left ( 0\right ) =0\) gives\begin {align*} 0 & =c_{2}\mu \\ c_{2} & =0 \end {align*}

Hence solution becomes \[ y\left ( x\right ) =c_{1}\cosh \left ( \mu x\right ) \] Second B.C. \(y^{\prime }\left ( L\right ) =0\) gives\[ 0=c_{1}\mu \sinh \left ( \mu L\right ) \] But \(\sinh \left ( \mu L\right ) \) can not be zero since \(\mu L\neq 0\), hence \(c_{1}=0,\) Leading to trivial solution. Therefore \(\lambda <0\) is not eigenvalue.

Let \(\lambda =0\), The solution is

\[ y\left ( x\right ) =c_{1}+c_{2}x \] First B.C. \(y^{\prime }\left ( 0\right ) =0\) gives\[ 0=c_{2}\] The solution becomes \[ y\left ( x\right ) =c_{1}\] Second B.C. \(y^{\prime }\left ( L\right ) =0\) gives\[ 0=0 \] Therefore \(c_{1}\) can be any value. Therefore \(\lambda =0\) is an eigenvalue and the corresponding eigenfunction is any constant, say \(1\).

Let \(\lambda >0\), The solution is\begin {align*} y\left ( x\right ) & =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \\ y^{\prime }\left ( x\right ) & =-c_{1}\sqrt {\lambda }\sin \left ( \sqrt {\lambda }x\right ) +c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) \end {align*}

First B.C. \(y^{\prime }\left ( 0\right ) =0\) gives\begin {align*} 0 & =c_{2}\sqrt {\lambda }\\ c_{2} & =0 \end {align*}

The solution becomes \[ y\left ( x\right ) =c_{1}\cos \left ( \sqrt {\lambda }x\right ) \] Second B.C. \(y^{\prime }\left ( L\right ) =0\) gives\[ 0=-c_{1}\sqrt {\lambda }\sin \left ( \sqrt {\lambda }L\right ) \] For non-trivial solution, we want \(\sin \left ( \sqrt {\lambda }L\right ) =0\) or \(\sqrt {\lambda }L=n\pi \) for \(n=1,2,3,\cdots \) Therefore \[ \lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \] And the corresponding eigenfunctions are \[ \Phi _{n}\left ( x\right ) =c_{n}\cos \left ( \sqrt {\lambda }x\right ) \qquad n=1,2,3,\cdots \]

The normalized \(\tilde {\Phi }_{n}\) eigenfunctions are now found. In this problem the weight function is \(r\left ( x\right ) =1\), therefore solving for \(c_{n}\) from

\begin {align*} \int _{0}^{L}r\left ( x\right ) \Phi _{n}^{2}dx & =1\\ \int _{0}^{L}c_{n}^{2}\cos ^{2}\left ( \sqrt {\lambda _{n}}x\right ) dx & =1 \end {align*}

As before, the above simplifies to

\[ c_{n}=\sqrt {\frac {4\sqrt {\lambda _{n}}}{2\sqrt {\lambda _{n}}L+\sin \left ( 2\sqrt {\lambda _{n}}L\right ) }}\]

For example, when \(L=1\) the normalization constant becomes (since now \(\sqrt {\lambda _{n}}=\frac {n\pi }{L}=n\pi \))

\begin {align*} c_{n} & =\sqrt {\frac {4n\pi }{2n\pi +\sin \left ( 2n\pi \right ) }}\\ c_{n} & =\sqrt {2} \end {align*}

For \(L=\pi \), the normalization constant becomes (since now \(\sqrt {\lambda _{n}}=\frac {n\pi }{L}=n\))

\begin {align*} c_{n} & =\sqrt {\frac {4n}{2n\pi +\sin \left ( 2n\pi \right ) }}\\ c_{n} & =\sqrt {\frac {2}{\pi }} \end {align*}

The normalization \(c_{n}\) value depends on the length. When \(L=1\)

\[ \tilde {\Phi }_{n}=\sqrt {2}\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \]

When \(L=\pi \)

\[ \tilde {\Phi }_{n}=\sqrt {\frac {2}{\pi }}\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \]

For \(n=0\), corresponding to the \(\lambda _{0}\) eigenvalue, since the eigenfunction is taken as the constant \(1\), then

\begin {align*} \int _{0}^{L}c_{0}^{2}dx & =1\\ c_{0} & =\sqrt {\frac {1}{L}} \end {align*}

Therefore, When \(L=1\)

\[ \tilde {\Phi }_{0}=1 \]

When \(L=\pi \)

\[ \tilde {\Phi }_{0}=\sqrt {\frac {1}{\pi }}\]

2.6 case 6: boundary conditions \(y^{\prime }(0)=0,y(L)+y^{\prime }(L)=0\)

Let the solution be \(y=Ae^{rx}\). This leads to the characteristic equation\begin {align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt {-\lambda } \end {align*}

Let \(\lambda <0\)

In this case \(-\lambda \) is positive and hence \(\sqrt {-\lambda }\) is also positive. Let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). Hence the roots are \(\pm \mu \). This gives the solution\begin {align*} y & =c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \\ y^{\prime } & =c_{1}\mu \sinh \left ( \mu x\right ) +c_{2}\mu \cosh \left ( \mu x\right ) \end {align*}

First B.C. \(y^{\prime }\left ( 0\right ) =0\) gives\begin {align*} 0 & =c_{2}\mu \\ c_{2} & =0 \end {align*}

Hence solution becomes \[ y\left ( x\right ) =c_{1}\cosh \left ( \mu x\right ) \] Second B.C. \(y\left ( L\right ) +y^{\prime }\left ( L\right ) =0\) gives\[ 0=c_{1}\left ( \cosh \left ( \mu L\right ) +\mu \sinh \left ( \mu L\right ) \right ) \] But \(\sinh \left ( \mu L\right ) \) can not be negative since its argument is positive here. And \(\cosh \mu L\) is always positive. In addition \(\cosh \left ( \mu L\right ) +\mu \sinh \left ( \mu L\right ) \) can not be zero since \(\sinh \left ( \mu L\right ) \) can not be zero as \(\mu L\neq 0\) and \(\cosh \left ( \mu L\right ) \) is not zero. Therefore \(c_{1}=0,\) Leading to trivial solution. Therefore \(\lambda <0\) is not eigenvalue.

Let \(\lambda =0\), The solution is\[ y\left ( x\right ) =c_{1}+c_{2}x \] First B.C. \(y^{\prime }\left ( 0\right ) =0\) gives\[ 0=c_{2}\] The solution becomes \[ y\left ( x\right ) =c_{1}\] Second B.C. \(y\left ( L\right ) +y^{\prime }\left ( L\right ) =0\) gives\[ 0=c_{1}\] This gives trivial solution. Therefore \(\lambda =0\) is not eigenvalue.

Let \(\lambda >0\), The solution is\begin {align*} y\left ( x\right ) & =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \\ y^{\prime }\left ( x\right ) & =-c_{1}\sqrt {\lambda }\sin \left ( \sqrt {\lambda }x\right ) +c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) \end {align*}

First B.C. \(y^{\prime }\left ( 0\right ) =0\) gives\begin {align*} 0 & =c_{2}\sqrt {\lambda }\\ c_{2} & =0 \end {align*}

The solution becomes \[ y\left ( x\right ) =c_{1}\cos \left ( \sqrt {\lambda }x\right ) \] Second B.C. \(y\left ( L\right ) +y^{\prime }\left ( L\right ) =0\) gives\begin {align*} 0 & =c_{1}\cos \left ( \sqrt {\lambda }L\right ) -c_{1}\sqrt {\lambda }\sin \left ( \sqrt {\lambda }L\right ) \\ & =c_{1}\left ( \cos \left ( \sqrt {\lambda }L\right ) -\sqrt {\lambda }\sin \left ( \sqrt {\lambda }L\right ) \right ) \end {align*}

For non-trivial solution, we want \(\cos \left ( \sqrt {\lambda }L\right ) -\sqrt {\lambda }\sin \left ( \sqrt {\lambda }L\right ) =0\) or \(\sqrt {\lambda }\tan \left ( \sqrt {\lambda }L\right ) =1\) Therefore the eigenvalues are the solution to \[ \sqrt {\lambda }\tan \left ( \sqrt {\lambda }L\right ) =1 \] And the corresponding eigenfunctions are \[ \Phi _{n}=\cos \left ( \sqrt {\lambda _{n}}x\right ) \qquad n=1,2,3,\cdots \] Where \(\lambda _{n}\) are the roots of \(\sqrt {\lambda }\tan \left ( \sqrt {\lambda }L\right ) =1\).

The normalized \(\tilde {\Phi }_{n}\) eigenfunctions are now found. Since the weight function is \(r\left ( x\right ) =1\), therefore solving for \(c_{n}\) from

\begin {align*} \int _{0}^{L}r\left ( x\right ) \Phi _{n}^{2}dx & =1\\ \int _{0}^{L}c_{n}^{2}\cos ^{2}\left ( \sqrt {\lambda _{n}}x\right ) dx & =1 \end {align*}

As was done earlier, the above results in

\[ c_{n}=\sqrt {\frac {4\sqrt {\lambda _{n}}}{2\sqrt {\lambda _{n}}L+\sin \left ( 2\sqrt {\lambda _{n}}L\right ) }}\qquad n=1,2,3,\cdots \]

Since there is no closed form solution to \(\lambda _{n}\) as it is a root of nonlinear equation \(\sqrt {\lambda }\tan \left ( \sqrt {\lambda }L\right ) =1\), the normalized constant is found numerically. For \(L=\pi \), the first few roots are

\[ \lambda _{n}=\left \{ 0.147033,1.48528,4.57614,9.60594,25.6247,36.6282,64.6318,81.6328,100.634,121.634,\cdots \right \} \]

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 \(\int _{0}^{L}c_{n}^{2}\cos ^{2}\left ( \sqrt {\lambda _{n}}x\right ) dx=1\)

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

\[ c_{n}=\{0.705925,0.751226,0.776042,0.786174,0.790773,0.793157,0.794531,\cdots \} \]

The above implies that the first normalized eigenfunction is

\[ \Phi _{1}=\left ( 0.705925\right ) \cos \left ( \sqrt {0.147033}x\right ) \]

And the second one is

\[ \Phi _{2}=\left ( 0.751226\right ) \cos \left ( \sqrt {1.48528}x\right ) \]

And so on.

2.7 case 7: boundary conditions \(y(0)+y^{\prime }(0)=0,y(L)=0\)

Let the solution be \(y=Ae^{rx}\). This leads to the characteristic equation\begin {align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt {-\lambda } \end {align*}

Let \(\lambda <0\)

In this case \(-\lambda \) is positive and \(\sqrt {-\lambda }\) is positive. Let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). Hence the roots are \(\pm \mu \). This gives the solution\begin {align*} y & =c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \\ y^{\prime } & =c_{1}\mu \sinh \left ( \mu x\right ) +c_{2}\mu \cosh \left ( \mu x\right ) \end {align*}

First B.C. \(y\left ( 0\right ) +y^{\prime }\left ( 0\right ) =0\) gives\begin {equation} 0=c_{1}+c_{2}\mu \tag {1} \end {equation} Second B.C. \(y\left ( L\right ) =0\) gives\[ 0=c_{1}\cosh \left ( \mu L\right ) +c_{2}\sinh \left ( \mu L\right ) \] From (1) \(c_{1}=-c_{2}\mu \) and the above now becomes\begin {align*} 0 & =-c_{2}\mu \cosh \left ( \mu L\right ) +c_{2}\sinh \left ( \mu L\right ) \\ & =c_{2}\left ( \sinh \left ( \mu L\right ) -\mu \cosh \left ( \mu L\right ) \right ) \end {align*}

For non-trivial solution, we want \(\sinh \left ( \mu L\right ) -\mu \cosh \left ( \mu L\right ) =0\). This means \(\tanh \left ( \mu L\right ) =\mu \). Therefore \(\lambda <0\) is an eigenvalue and these are given by \(\lambda _{n}=-\mu _{n}^{2}\), where \(\mu _{n}\) is the solution to \[ \tanh \left ( \mu L\right ) =\mu \] Or equivalently, the roots of\[ \tanh \left ( \sqrt {-\lambda }L\right ) =\sqrt {-\lambda }\]

There is only one negative root when solving the above numerically, which is \(\lambda _{-1}=0.992.\)The corresponding eigenfunction is\[ \Phi _{-1}=c_{-1}\left ( \sinh \left ( \sqrt {-\lambda _{-1}}x\right ) -\sqrt {-\lambda _{-1}}\cosh \left ( \sqrt {-\lambda _{-1}}x\right ) \right ) \] Let \(\lambda =0\), The solution is\[ y\left ( x\right ) =c_{1}+c_{2}x \] First B.C. \(y\left ( 0\right ) +y^{\prime }\left ( 0\right ) =0\) gives\[ 0=c_{1}+c_{2}\] The solution becomes \[ y\left ( x\right ) =c_{1}\left ( 1-x\right ) \] Second B.C. \(y\left ( L\right ) \) gives\[ 0=c_{1}\left ( 1-L\right ) \] This gives trivial solution. Therefore \(\lambda =0\) is not eigenvalue.

Let \(\lambda >0\), The solution is\begin {align*} y\left ( x\right ) & =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \\ y^{\prime }\left ( x\right ) & =-c_{1}\sqrt {\lambda }\sin \left ( \sqrt {\lambda }x\right ) +c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) \end {align*}

First B.C. \(y\left ( 0\right ) +y^{\prime }\left ( 0\right ) =0\) gives\[ 0=c_{1}+c_{2}\sqrt {\lambda }\] The solution now becomes \begin {align*} y\left ( x\right ) & =-c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \\ & =c_{2}\left ( \sin \left ( \sqrt {\lambda }x\right ) -\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) \right ) \end {align*}

Second B.C. \(y\left ( L\right ) =0\) the above becomes \[ 0=c_{2}\left ( \sin \left ( \sqrt {\lambda }L\right ) -\sqrt {\lambda }\cos \left ( \sqrt {\lambda }L\right ) \right ) \] For non-trivial solution, we want \(\sin \left ( \sqrt {\lambda }L\right ) -\sqrt {\lambda }\cos \left ( \sqrt {\lambda }L\right ) =0\) or \(\tan \left ( \sqrt {\lambda }L\right ) -\sqrt {\lambda }=0\) or \[ \sqrt {\lambda }=\tan \left ( \sqrt {\lambda }L\right ) \] Therefore the eigenvalues are the solution to the above (must be done numerically)  And the corresponding eigenfunctions are \[ \Phi _{n}\left ( x\right ) =c_{n}\left ( \sin \left ( \sqrt {\lambda _{n}}x\right ) -\sqrt {\lambda _{n}}\cos \left ( \sqrt {\lambda _{n}}x\right ) \right ) \] for each root \(\lambda _{n}\).

2.8 case 8: boundary conditions \(y(0)+y^{\prime }(0)=0,y^{\prime }(L)=0\)

Let the solution be \(y=Ae^{rx}\). This leads to the characteristic equation\begin {align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt {-\lambda } \end {align*}

Let \(\lambda <0\)

In this case \(-\lambda \) is positive and hence \(\sqrt {-\lambda }\) is also positive. Let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). Hence the roots are \(\pm \mu \). This gives the solution\begin {align*} y & =c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \\ y^{\prime } & =c_{1}\mu \sinh \left ( \mu x\right ) +c_{2}\mu \cosh \left ( \mu x\right ) \end {align*}

First B.C. \(y\left ( 0\right ) +y^{\prime }\left ( 0\right ) =0\) gives\begin {equation} 0=c_{1}+c_{2}\mu \tag {1} \end {equation} Second B.C. \(y^{\prime }\left ( L\right ) =0\) gives\[ 0=c_{1}\mu \sinh \left ( \mu L\right ) +c_{2}\mu \cosh \left ( \mu L\right ) \] From (1) \(c_{1}=-c_{2}\mu \) and the above becomes\begin {align*} 0 & =-c_{2}\mu ^{2}\sinh \left ( \mu L\right ) +c_{2}\mu \cosh \left ( \mu L\right ) \\ & =c_{2}\mu \left ( -\mu \sinh \left ( \mu L\right ) +\cosh \left ( \mu L\right ) \right ) \end {align*}

For non-trivial solution, we want \(-\mu \sinh \left ( \mu L\right ) +\cosh \left ( \mu L\right ) =0\). This means \(-\mu \tanh \left ( \mu L\right ) +1=0\). Or \(\tanh \left ( \mu L\right ) =\frac {1}{\mu }\), therefore \(\lambda <0\) is eigenvalues and these are given by \(\lambda _{n}=-\mu _{n}^{2}\), where \(\mu _{n}\) is the solution to \begin {align*} \tanh \left ( \mu L\right ) & =\frac {1}{\mu }\\ \tanh \left ( \sqrt {-\lambda }L\right ) & =\frac {1}{\sqrt {-\lambda }} \end {align*}

This has one root, found numerically which is \(\lambda _{-1}=-1\). Hence \(\sqrt {-\lambda }=1\). The corresponding eigenfunction is

\begin {align*} \Phi _{-1}\left ( x\right ) & =c_{-1}\left ( -\mu \cosh \left ( \mu x\right ) +\sinh \left ( \mu x\right ) \right ) \\ & =c_{-1}\left ( -\cosh \left ( x\right ) +\sinh \left ( x\right ) \right ) \end {align*}

Let \(\lambda =0\), The solution is\[ y\left ( x\right ) =c_{1}+c_{2}x \] First B.C. \(y\left ( 0\right ) +y^{\prime }\left ( 0\right ) =0\) gives\[ 0=c_{1}+c_{2}\] The solution becomes \begin {align*} y\left ( x\right ) & =c_{1}\left ( 1-x\right ) \\ y^{\prime } & =-c_{1} \end {align*}

Second B.C. \(y^{\prime }\left ( L\right ) \) gives\[ 0=-c_{1}\] This gives trivial solution. Therefore \(\lambda =0\) is not eigenvalue.

Let \(\lambda >0\), The solution is\begin {align*} y\left ( x\right ) & =c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \\ y^{\prime }\left ( x\right ) & =-c_{1}\sqrt {\lambda }\sin \left ( \sqrt {\lambda }x\right ) +c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) \end {align*}

First B.C. \(y\left ( 0\right ) +y^{\prime }\left ( 0\right ) =0\) gives\[ 0=c_{1}+c_{2}\sqrt {\lambda }\] The solution becomes \begin {align*} y\left ( x\right ) & =-c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \\ & =c_{2}\left ( \sin \left ( \sqrt {\lambda }x\right ) -\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) \right ) \end {align*}

Second B.C. \(y^{\prime }\left ( L\right ) =0\) gives\[ 0=c_{2}\left ( \sqrt {\lambda }\cos \left ( \sqrt {\lambda }L\right ) +\lambda \sin \left ( \sqrt {\lambda }L\right ) \right ) \] For non-trivial solution, we want \(\lambda \sin \left ( \sqrt {\lambda }L\right ) +\sqrt {\lambda }\cos \left ( \sqrt {\lambda }L\right ) =0\) or \(\lambda \tan \left ( \sqrt {\lambda }L\right ) =-\sqrt {\lambda }\) Therefore the eigenvalues are the solution to \[ \tan \left ( \sqrt {\lambda }L\right ) =\frac {-\sqrt {\lambda }}{\lambda }=\frac {-1}{\sqrt {\lambda }}\] And the corresponding eigenfunction is \[ \Phi _{n}\left ( x\right ) =c_{n}\left ( \sin \left ( \sqrt {\lambda }x\right ) -\sqrt {\lambda }\cos \left ( \sqrt {\lambda }x\right ) \right ) \]

2.9 case 9: boundary conditions \(y(0)+y^{\prime }(0)=0,y(L)+y^{\prime }(L)=0\)

Let the solution be \(y=Ae^{rx}\). This leads to the characteristic equation\begin {align*} r^{2}+\lambda & =0\\ r & =\pm \sqrt {-\lambda } \end {align*}

Let \(\lambda <0\)

In this case \(-\lambda \) is positive and hence \(\sqrt {-\lambda }\) is also positive. Let \(\sqrt {-\lambda }=\mu \) where \(\mu >0\). Hence the roots are \(\pm \mu \). This gives the solution\[ y=c_{1}\cosh \left ( \mu x\right ) +c_{2}\sinh \left ( \mu x\right ) \] Hence\[ y^{\prime }=\mu c_{1}\sinh \left ( \mu x\right ) +\mu c_{2}\cosh \left ( \mu x\right ) \] Left B.C. gives\begin {equation} 0=c_{1}+\mu c_{2} \tag {1} \end {equation} Right B.C. gives\begin {align*} 0 & =c_{1}\cosh \left ( \mu L\right ) +c_{2}\sinh \left ( \mu L\right ) +\mu c_{1}\sinh \left ( \mu L\right ) +\mu c_{2}\cosh \left ( \mu L\right ) \\ & =\cosh \left ( \mu L\right ) \left ( c_{1}+\mu c_{2}\right ) +\sinh \left ( \mu L\right ) \left ( c_{2}+\mu c_{1}\right ) \end {align*}

Using (1) in the above, it simplifies to\[ 0=\sinh \left ( \mu L\right ) \left ( c_{2}+\mu c_{1}\right ) \] But from (1) again, we see that \(c_{1}=-\mu c_{2}\) and the above becomes\begin {align*} 0 & =\sinh \left ( \mu L\right ) \left ( c_{2}-\mu \left ( \mu c_{2}\right ) \right ) \\ & =\sinh \left ( \mu L\right ) \left ( c_{2}-\mu ^{2}c_{2}\right ) \\ & =c_{2}\sinh \left ( \mu L\right ) \left ( 1-\mu ^{2}\right ) \end {align*}

But \(\sinh \left ( \mu ^{2}L\right ) \neq 0\) since \(\mu ^{2}L\neq 0\) and so either \(c_{2}=0\) or \(\left ( 1-\mu ^{2}\right ) =0\). \(c_{2}=0\) results in trivial solution, therefore \(\left ( 1-\mu ^{2}\right ) =0\) or \(\mu ^{2}=1\) but \(\mu ^{2}=-\lambda \), hence \(\lambda =-1\) is the eigenvalue. Corresponding eigenfunction is\[ y=c_{1}\cosh \left ( x\right ) +c_{2}\sinh \left ( x\right ) \] Using (1) the above simplifies to\begin {align*} y & =-\mu c_{2}\cosh \left ( x\right ) +c_{2}\sinh \left ( x\right ) \\ & =c_{2}\left ( -\mu \cosh \left ( x\right ) +\sinh \left ( x\right ) \right ) \end {align*}

But \(\mu =\sqrt {-\lambda }=1\), hence the eigenfunction is\[ \fbox {$y\left ( x\right ) =c_2\left ( -\cosh \left ( x\right ) +\sinh \left ( x\right ) \right ) $}\] Let \(\lambda =0\) Solution now is\[ y=c_{1}x+c_{2}\] Therefore\[ y^{\prime }=c_{1}\] Left B.C. \(0=y\left ( 0\right ) +y^{\prime }\left ( 0\right ) \) gives\begin {equation} 0=c_{2}+c_{1} \tag {2} \end {equation} Right B.C. \(0=y\left ( L\right ) +y^{\prime }\left ( L\right ) \) gives\begin {align*} 0 & =\left ( c_{1}L+c_{2}\right ) +c_{1}\\ 0 & =c_{1}\left ( 1+L\right ) +c_{2} \end {align*}

But from (2) \(c_{1}=-c_{2}\) and the above becomes\begin {align*} 0 & =-c_{2}\left ( 1+L\right ) +c_{2}\\ 0 & =-c_{2}L \end {align*}

Which means \(c_{2}=0\) and therefore the trivial solution. Therefore \(\lambda =0\) is not an eigenvalue.

Assuming \(\lambda >0\) Solution is \begin {equation} y=c_{1}\cos \left ( \sqrt {\lambda }x\right ) +c_{2}\sin \left ( \sqrt {\lambda }x\right ) \tag {A} \end {equation} Hence\[ y^{\prime }=-\sqrt {\lambda }c_{1}\sin \left ( \sqrt {\lambda }x\right ) +\sqrt {\lambda }c_{2}\cos \left ( \sqrt {\lambda }x\right ) \] Left B.C. gives\begin {equation} 0=c_{1}+\sqrt {\lambda }c_{2} \tag {3} \end {equation} Right B.C. gives\begin {align*} 0 & =c_{1}\cos \left ( \sqrt {\lambda }L\right ) +c_{2}\sin \left ( \sqrt {\lambda }L\right ) -\sqrt {\lambda }c_{1}\sin \left ( \sqrt {\lambda }L\right ) +\sqrt {\lambda }c_{2}\cos \left ( \sqrt {\lambda }L\right ) \\ & =\cos \left ( \sqrt {\lambda }L\right ) \left ( c_{1}+\sqrt {\lambda }c_{2}\right ) +\sin \left ( \sqrt {\lambda }L\right ) \left ( c_{2}-\sqrt {\lambda }c_{1}\right ) \end {align*}

Using (3) in the above, it simplifies to\[ 0=\sin \left ( \sqrt {\lambda }L\right ) \left ( c_{2}-\sqrt {\lambda }c_{1}\right ) \] But from (3), we see that \(c_{1}=-\sqrt {\lambda }c_{2}\). Therefore the above becomes\begin {align*} 0 & =\sin \left ( \sqrt {\lambda }L\right ) \left ( c_{2}-\sqrt {\lambda }\left ( -\sqrt {\lambda }c_{2}\right ) \right ) \\ & =\sin \left ( \sqrt {\lambda }L\right ) \left ( c_{2}+\lambda c_{2}\right ) \\ & =c_{2}\sin \left ( \sqrt {\lambda }L\right ) \left ( 1+\lambda \right ) \end {align*}

Only choice for non trivial solution is either \(\left ( 1+\lambda \right ) =0\) or \(\sin \left ( \sqrt {\lambda }L\right ) =0\). But \(\left ( 1+\lambda \right ) =0\) implies \(\lambda =-1\) but we said that \(\lambda >0\). Hence other choice is \begin {align*} \sin \left ( \sqrt {\lambda }L\right ) & =0\\ \sqrt {\lambda }L & =n\pi \qquad n=1,2,3,\cdots \\ \lambda _{n} & =\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \end {align*}

The above are the eigenvalues. The corresponding eigenfunction is from (A)\[ \Phi _{n}\left ( x\right ) =c_{1_{n}}\cos \left ( \sqrt {\lambda _{n}}x\right ) +c_{2_{n}}\sin \left ( \sqrt {\lambda _{n}}x\right ) \] But \(c_{1_{n}}=-\sqrt {\lambda _{n}}c_{2_{n}}\) and the above becomes\begin {align*} \Phi _{n}\left ( x\right ) & =-\sqrt {\lambda _{n}}c_{2_{n}}\cos \left ( \sqrt {\lambda _{n}}x\right ) +c_{2}\sin \left ( \sqrt {\lambda _{n}}x\right ) \\ & =C_{n}\left ( -\sqrt {\lambda _{n}}\cos \left ( \sqrt {\lambda _{n}}x\right ) +\sin \left ( \sqrt {\lambda _{n}}x\right ) \right ) \end {align*}