home
PDF (letter size
PDF (legal size

Analysis of the eigenvalues and eigenfunctions for $$y''(x)+\lambda y(x)=0$$ for all possible homogeneous boundary conditions

May 23, 2020   Compiled on May 23, 2020 at 5:10am

Contents

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 $$y(0)=0,y(L)=0$$

 eigenvalues eigenfunctions $$\lambda <0$$ None None $$\lambda =0$$ None None $$\lambda >0$$ $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots$$ $$\Phi _{n}(x)=c_{n}\sin \left ( \sqrt{\lambda _{n}}\,x\right )$$

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 $$y(0)=0,y^{\prime }(L)=0$$

 eigenvalues eigenfunctions $$\lambda <0$$ None None $$\lambda =0$$ None None $$\lambda >0$$ $$\lambda _{n}=\left ( \frac{n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots$$ $$\Phi _{n}(x)=c_{n}\sin \left ( \sqrt{\lambda _{n}}\,x\right )$$

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 $$y(0)=0,y(L)+y^{\prime }(L)=0$$

 eigenvalues eigenfunctions $$\lambda <0$$ None None $$\lambda =0$$ None None $$\lambda >0$$ roots of $$\tan \left ( \sqrt{\lambda }L\right ) +\sqrt{\lambda }=0$$ $$\Phi _{n}(x)=c_{n}\sin \left ( \sqrt{\lambda _{n}}\,x\right )$$

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 $$y^{\prime }(0)=0,y(L)=0$$

 eigenvalues eigenfunctions $$\lambda <0$$ None None $$\lambda =0$$ None None $$\lambda >0$$ $$\lambda _{n}=\left ( \frac{n\pi }{2L}\right ) ^{2}\qquad n=1,3,5,\cdots$$ $$\Phi _{n}(x)=c_{n}\cos \left ( \sqrt{\lambda _{n}}\,x\right )$$

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 $$y^{\prime }(0)=0,y^{\prime }(L)=0$$

 eigenvalues eigenfunctions $$\lambda <0$$ None None $$\lambda =0$$ Yes constant say $$1$$ $$\lambda >0$$ $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots$$ $$\Phi _{n}(x)=c_{n}\cos \left ( \sqrt{\lambda _{n}}\,x\right )$$

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 $$y^{\prime }(0)=0,y(L)+y^{\prime }(L)=0$$

 eigenvalues eigenfunctions $$\lambda <0$$ None None $$\lambda =0$$ None None $$\lambda >0$$ Roots of $$\sqrt{\lambda }\tan \left ( \sqrt{\lambda }L\right ) =1$$ $$\Phi _{n}(x)=c_{n}\cos \left ( \sqrt{\lambda _{n}}\,x\right )$$

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 $$y(0)+$$$$y^{\prime }(0)=0,y(L)=0$$

 eigenvalues eigenfunctions $$\lambda <0$$ Root of $$\tanh \left ( \sqrt{-\lambda }L\right ) =\sqrt{-\lambda }$$(one root) $$\Phi (x)=\sinh \left ({\sqrt{-\lambda }x}\right ) -\sqrt{-\lambda }\cosh \left ({\sqrt{-\lambda }x}\right )$$ $$\lambda =0$$ None None $$\lambda >0$$ Roots of $$\tan \left ( \sqrt{\lambda }L\right ) =\sqrt{\lambda }$$ $$\Phi _{n}(x)=\sin \left ({\sqrt{\lambda }x}\right ) -\sqrt{\lambda }\cos \left ({\sqrt{\lambda }x}\right )$$

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 $$y(0)+$$$$y^{\prime }(0)=0,y^{\prime }(L)=0$$

 eigenvalues eigenfunctions $$\lambda <0$$ Root of $$\tanh \left ( \sqrt{-\lambda }L\right ) =\frac{1}{\sqrt{-\lambda }}$$(one root) $$\Phi _{-1}(x)=\sinh \left ( \sqrt{-\lambda }x\right ) -\sqrt{-\lambda }\cosh \left ( \sqrt{-\lambda }x\right )$$ $$\lambda =0$$ None None $$\lambda >0$$ Roots of $$\tan \left ( \sqrt{\lambda }L\right ) =\frac{-1}{\sqrt{\lambda }}$$ $$\Phi _{n}(x)=\sin \left ({\sqrt{\lambda }x}\right ) -\sqrt{\lambda }\cos \left ({\sqrt{\lambda }x}\right )$$

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 $$y(0)+$$$$y^{\prime }(0)=0,y(L)+y^{\prime }(L)=0$$

 eigenvalues eigenfunctions $$\lambda <0$$ $$-1$$ $$\Phi _{-1}(x)=\sinh \left ({x}\right ) -\cosh \left ({x}\right )$$ $$\lambda =0$$ None None $$\lambda >0$$ $$\lambda _{n}=\left ( \frac{n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots$$ $$\Phi _{n}(x)=\sin \left ({\sqrt{\lambda _{n}}x}\right ) -\sqrt{\lambda _{n}}\cos \left ({\sqrt{\lambda _{n}}x}\right )$$

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$$0=c_{1}+c_{2}\mu \tag{1}$$ 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$$0=c_{1}+c_{2}\mu \tag{1}$$ 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$$0=c_{1}+\mu c_{2} \tag{1}$$ 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$$0=c_{2}+c_{1} \tag{2}$$ 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 $$y=c_{1}\cos \left ( \sqrt{\lambda }x\right ) +c_{2}\sin \left ( \sqrt{\lambda }x\right ) \tag{A}$$ 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$$0=c_{1}+\sqrt{\lambda }c_{2} \tag{3}$$ 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*}