home

PDF (letter size

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

Normalized eigenfunctions: For $L=1,$$${\mathrm{\Phi }}_n(x)=\sqrt{2}\sin \left(\sqrt{{\lambda }_n}x\right)$$ For $L=\pi ,$$${\mathrm{\Phi }}_n(x)=\sqrt{\frac{2}{\pi }}\sin \left(\sqrt{{\lambda }_n}x\right)$$

List of eigenvalues $$\{\frac{{\pi }^2}{L^2},\frac{4{\pi }^2}{L^2},\frac{9{\pi }^2}{L^2},\frac{16{\pi }^2}{L^2},\cdots \}$$ 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 ${\mathrm{\Phi }}_n(x)$ is $n-1$ inside the interval $0<x<\pi$. (not counting the end points). Hence ${\mathrm{\Phi }}_1(x)$ which correspond to ${\lambda }_1=1$ in this case, will have no zeros inside the interval. While ${\mathrm{\Phi }}_2(x)$ 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,$$${\mathrm{\Phi }}_n(x)=\sqrt{2}\sin \left(\sqrt{{\lambda }_n}x\right)$$ For $L=\pi ,$$${\mathrm{\Phi }}_n(x)=\sqrt{\frac{2}{\pi }}\sin \left(\sqrt{{\lambda }_n}x\right)$$

List of eigenvalues $$\{\frac{{\pi }^2}{4L^2},\frac{9{\pi }^2}{4L^2},\frac{25{\pi }^2}{4L^2},\frac{49{\pi }^2}{4L^2},\cdots \}$$ 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 $L=\pi ,$

$$\begin{array}{rl}{\mathrm{\Phi }}_1& =\left(0.729448\right)\sin \left(\sqrt{0.620}x\right)\\ {\mathrm{\Phi }}_2& =\left(0.766385\right)\sin \left(\sqrt{2.794}x\right)\\ & ⋮\end{array}$$

The normalization constant in this case depends on the eigenvalue.

List of numerical eigenvalues when $L=\pi$ (since there is no analytical solution) $$\{0.620,2.794,6.845,12.865,20.879,\cdots \}$$ 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.

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

Normalized eigenfunctions for $L=1$

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{2}\cos \left(\sqrt{{\lambda }_n}x\right)n=1,3,5,\cdots$$

When $L=\pi$

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{\frac{2}{\pi }}\cos \left(\sqrt{{\lambda }_n}x\right)n=1,3,5,\cdots$$

List of eigenvalues$$\{\frac{{\pi }^2}{4L^2},\frac{9{\pi }^2}{4L^2},\frac{25{\pi }^2}{4L^2},\frac{49{\pi }^2}{4L^2},\cdots \}$$ 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.

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

Normalized eigenfunction when $L=1$

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{2}\cos \left(\sqrt{{\lambda }_n}x\right)n=1,2,3,\cdots$$

When $L=\pi$

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{\frac{2}{\pi }}\cos \left(\sqrt{{\lambda }_n}x\right)n=1,2,3,\cdots$$

For ${\tilde{\mathrm{\Phi }}}_0$, When $L=1$

$${\tilde{\mathrm{\Phi }}}_0=1$$

When $L=\pi$

$${\tilde{\mathrm{\Phi }}}_0=\sqrt{\frac{1}{\pi }}$$

List of eigenvalues$$\{0,\frac{{\pi }^2}{L^2},\frac{4{\pi }^2}{L^2},\frac{9{\pi }^2}{L^2},\frac{16{\pi }^2}{L^2},\cdots \}$$ List of numerical eigenvalues when $L=\pi$$$\{0,1,4,9,16,\cdots \}$$ 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.

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

Normalized eigenfunctions for $L=\pi$ are

$$\begin{array}{rl}{\mathrm{\Phi }}_1& =\left(0.705925\right)\cos \left(\sqrt{0.147033}x\right)\\ {\mathrm{\Phi }}_2& =\left(0.751226\right)\cos \left(\sqrt{1.48528}x\right)\\ & ⋮\end{array}$$

List of numerical eigenvalues when $L=\pi$ (There is no analytical solution for the roots)$$\{0.147033,1.48528,4.576,9.606,16.622,\cdots \}$$ 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.

1.7 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)$$\{-0.992,1.664,5.631,11.623,\cdots \}$$ 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)$$\{-1.007,0.480,3.392,8.376,24,368,\cdots \}$$ 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$$\{-1,\frac{{\pi }^2}{L^2},\frac{4{\pi }^2}{L^2},\frac{9{\pi }^2}{L^2},\frac{16{\pi }^2}{L^2},\cdots \}$$ List of numerical eigenvalues when $L=\pi$$$\{-1,1,4,9,16,\cdots \}$$ 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 $y=A{e}^{rx}$. This leads to the characteristic equation

$$\begin{array}{rl}{r}^2+\lambda & =0\\ r& =\pm \sqrt{-\lambda }\end{array}$$

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 $\sinh \left(\mu L\right)\ne 0$ since $\mu L\ne 0$, 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$$\begin{array}{rl}\sqrt{{\lambda }_n}& =\frac{n\pi }{L}n=1,2,3,\cdots \\ {\lambda }_n& ={\left(\frac{n\pi }{L}\right)}^{2}n=1,2,3,\cdots \end{array}$$

The corresponding eigenfunctions are$${\mathrm{\Phi }}_n=c_n\sin \left(\sqrt{{\lambda }_n}x\right)n=1,2,3,\cdots$$

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

$$\begin{array}{rl}{\int }_0^{L}r\left(x\right){\mathrm{\Phi }}_n^{2}dx& =1\\ {\int }_0^L{c}_n^2{\sin }^2\left(\sqrt{{\lambda }_n}x\right)dx& =1\\ c_n^2{\int }_0^L(\frac{1}{2}-\frac{1}{2}\cos \left(2\sqrt{{\lambda }_n}x\right))dx& =1\\ {\int }_0^L\frac{1}{2}dx-{\int }_0^L\frac{1}{2}\cos \left(2\sqrt{{\lambda }_n}x\right)dx& =\frac{1}{c_n^2}\\ \frac{1}{2}L-\frac{1}{2}{\left(\frac{\sin \left(2\sqrt{{\lambda }_n}x\right)}{2\sqrt{{\lambda }_n}}\right)}_0^L& =\frac{1}{c_n^2}\\ \frac{1}{2}L-\frac{1}{4\sqrt{{\lambda }_n}}\sin \left(2\sqrt{{\lambda }_n}L\right)& =\frac{1}{c_n^2}\\ 2\sqrt{{\lambda }_n}L-\sin \left(2\sqrt{{\lambda }_n}L\right)& =\frac{4\sqrt{{\lambda }_n}}{c_n^2}\end{array}$$

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 $\sqrt{{\lambda }_n}=\frac{n\pi }{L}=n\pi$)

$$\begin{array}{rl}{c}_n& =\sqrt{\frac{4n\pi }{2n\pi -\sin \left(2n\pi \right)}}\\ & =\sqrt{\frac{4n\pi }{2n\pi }}\\ c_n& =\sqrt{2}\end{array}$$

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

$$\begin{array}{rl}{c}_n& =\sqrt{\frac{4n}{2n\pi -\sin \left(2n\pi \right)}}\\ & =\sqrt{\frac{4n}{2n\pi }}\\ c_n& =\sqrt{\frac{2}{\pi }}\end{array}$$

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

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{2}\sin \left(\sqrt{{\lambda }_n}x\right)n=1,2,3,\cdots$$

When $L=\pi$

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{\frac{2}{\pi }}\sin \left(\sqrt{{\lambda }_n}x\right)n=1,2,3,\cdots$$

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

Let the solution be $y=A{e}^{rx}$. This leads to the characteristic equation$$\begin{array}{rl}{r}^2+\lambda & =0\\ r& =\pm \sqrt{-\lambda }\end{array}$$

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$$\begin{array}{rl}{y}^{\prime }\left(x\right)& =\mu c_2\cosh \left(\mu x\right)\\ 0& =\mu c_2\cosh \left(\mu L\right)\end{array}$$

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$$\begin{array}{rl}{y}^{\prime }\left(x\right)& =c_2\\ 0& =c_2\end{array}$$

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$$\begin{array}{rl}{y}^{\prime }\left(x\right)& =\sqrt{\lambda }{c}_2\cos \left(\sqrt{\lambda }x\right)\\ 0& =\sqrt{\lambda }{c}_2\cos \left(\sqrt{\lambda }L\right)\end{array}$$

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$$\begin{array}{rl}\sqrt{{\lambda }_n}L& =\frac{n\pi }{2}\\ \sqrt{{\lambda }_n}& =\frac{n\pi }{2L}n=1,3,5,\cdots \end{array}$$

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

$${\mathrm{\Phi }}_n=c_n\sin \left(\sqrt{{\lambda }_n}x\right)n=1,3,5,\cdots$$

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

$$\begin{array}{rl}{\int }_0^{L}r\left(x\right){\mathrm{\Phi }}_n^{2}dx& =1\\ {\int }_0^L{c}_n^2{\sin }^2\left(\sqrt{{\lambda }_n}x\right)dx& =1\end{array}$$

As was done earlier, the above results in

$$c_n=\sqrt{\frac{4\sqrt{{\lambda }_n}}{2\sqrt{{\lambda }_n}L-\sin \left(2\sqrt{{\lambda }_n}L\right)}}n=1,3,5,\cdots$$

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

$$\begin{array}{rl}{c}_n& =\sqrt{\frac{4\frac{n\pi }{2}}{2\frac{n\pi }{2}-\sin \left(2\frac{n\pi }{2}\right)}}\\ & =\sqrt{\frac{2n\pi }{n\pi }}\\ c_n& =\sqrt{2}\end{array}$$

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

$$\begin{array}{rl}{c}_n& =\sqrt{\frac{4\frac{n}{2}}{2\frac{n}{2}\pi -\sin \left(2\frac{n}{2}\pi \right)}}\\ & =\sqrt{\frac{2n}{n\pi }}\\ c_n& =\sqrt{\frac{2}{\pi }}\end{array}$$

Therefore, for $L=1$

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{2}\sin \left(\sqrt{{\lambda }_n}x\right)n=1,3,5,\cdots$$

For $L=\pi$

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{\frac{2}{\pi }}\sin \left(\sqrt{{\lambda }_n}x\right)n=1,3,5,\cdots$$

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

Let the solution be $y=A{e}^{rx}$. This leads to the characteristic equation

$$\begin{array}{rl}{r}^2+\lambda & =0\\ r& =\pm \sqrt{-\lambda }\end{array}$$

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$$0=c_2(\sinh \left(\mu L\right)+\mu \cosh \left(\mu x\right))$$ But $\sinh \left(\mu L\right)\ne 0$ since $\mu L\ne 0$ 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$$\begin{array}{rl}0& =c_{2}L+c_2\\ & =c_2(1+L)\end{array}$$

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$$0=c_2(\sin \left(\sqrt{\lambda }L\right)+\sqrt{\lambda }\cos \left(\sqrt{\lambda }L\right))$$ 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 $${\mathrm{\Phi }}_n=c_n\sin \left(\sqrt{{\lambda }_n}x\right)n=1,2,3,\cdots$$

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

$$\begin{array}{rl}{\int }_0^{L}r\left(x\right){\mathrm{\Phi }}_n^{2}dx& =1\\ {\int }_0^L{c}_n^2{\sin }^2\left(\sqrt{{\lambda }_n}x\right)dx& =1\end{array}$$

As was done earlier, the above results in

$$c_n=\sqrt{\frac{4\sqrt{{\lambda }_n}}{2\sqrt{{\lambda }_n}L-\sin \left(2\sqrt{{\lambda }_n}L\right)}}n=1,2,3,\cdots$$

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 $${\lambda }_n=\{0.620,2.794,6.845,12.865,20.879,\cdots \}$$

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

$${\mathrm{\Phi }}_1=\left(0.729448\right)\sin \left(\sqrt{0.620}x\right)$$

And the second one is

$${\mathrm{\Phi }}_2=\left(0.766385\right)\sin \left(\sqrt{2.794}x\right)$$

And so on.

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

Let the solution be $y=A{e}^{rx}$. This leads to the characteristic equation$$\begin{array}{rl}{r}^2+\lambda & =0\\ r& =\pm \sqrt{-\lambda }\end{array}$$

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$$\begin{array}{rl}y& =c_1\cosh \left(\mu x\right)+c_2\sinh \left(\mu x\right)\\ y^{\prime }& =c_1\mu \sinh \left(\mu x\right)+c_2\mu \cosh \left(\mu x\right)\end{array}$$

First B.C. $y^{\prime }\left(0\right)=0$ gives$$\begin{array}{rl}0& =c_2\mu \\ c_2& =0\end{array}$$

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$$\begin{array}{rl}y\left(x\right)& =c_1\cos \left(\sqrt{\lambda }x\right)+c_2\sin \left(\sqrt{\lambda }x\right)\\ y^{\prime }\left(x\right)& =-c_1\sqrt{\lambda }\sin \left(\sqrt{\lambda }x\right)+c_2\sqrt{\lambda }\cos \left(\sqrt{\lambda }x\right)\end{array}$$

First B.C. $y^{\prime }\left(0\right)=0$ gives$$\begin{array}{rl}0& =c_2\sqrt{\lambda }\\ c_2& =0\end{array}$$

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 $${\lambda }_n={\left(\frac{n\pi }{2L}\right)}^{2}n=1,3,5,\cdots$$

The corresponding eigenfunctions are

$${\mathrm{\Phi }}_n=c_n\cos \left(\sqrt{{\lambda }_n}x\right)n=1,3,5,\cdots$$

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

$$\begin{array}{rl}{\int }_0^{L}r\left(x\right){\mathrm{\Phi }}_n^{2}dx& =1\\ {\int }_0^L{c}_n^2{\cos }^2\left(\sqrt{{\lambda }_n}x\right)dx& =1\\ c_n^2{\int }_0^L(\frac{1}{2}+\frac{1}{2}\cos \left(2\sqrt{{\lambda }_n}x\right))dx& =1\\ {\int }_0^L\frac{1}{2}dx+{\int }_0^L\frac{1}{2}\cos \left(2\sqrt{{\lambda }_n}x\right)dx& =\frac{1}{c_n^2}\\ \frac{1}{2}L+\frac{1}{2}{\left(\frac{\sin \left(2\sqrt{{\lambda }_n}x\right)}{2\sqrt{{\lambda }_n}}\right)}_0^L& =\frac{1}{c_n^2}\\ \frac{1}{2}L+\frac{1}{4\sqrt{{\lambda }_n}}\sin \left(2\sqrt{{\lambda }_n}L\right)& =\frac{1}{c_n^2}\\ 2\sqrt{{\lambda }_n}L+\sin \left(2\sqrt{{\lambda }_n}L\right)& =\frac{4\sqrt{{\lambda }_n}}{c_n^2}\end{array}$$

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 $\sqrt{{\lambda }_n}=\frac{n\pi }{2L}=\frac{n\pi }{2}$)

$$\begin{array}{rl}{c}_n& =\sqrt{\frac{4\frac{n\pi }{2}}{2\frac{n\pi }{2}+\sin \left(2\frac{n\pi }{2}\right)}}\\ & =\sqrt{\frac{2n\pi }{n\pi }}\\ c_n& =\sqrt{2}\end{array}$$

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 $\sqrt{{\lambda }_n}=\frac{n\pi }{2L}=\frac{n}{2}$)

$$\begin{array}{rl}{c}_n& =\sqrt{\frac{4\frac{n}{2}}{2\frac{n}{2}\pi +\sin \left(2\frac{n}{2}\pi \right)}}\\ & =\sqrt{\frac{2n}{2n\pi }}\\ c_n& =\sqrt{\frac{2}{\pi }}\end{array}$$

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

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{2}\cos \left(\sqrt{{\lambda }_n}x\right)n=1,3,5,\cdots$$

When $L=\pi$

$${\tilde{\mathrm{\Phi }}}_n=\sqrt{\frac{2}{\pi }}\cos \left(\sqrt{{\lambda }_n}x\right)n=1,3,5,\cdots$$

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

Let the solution be $y=A{e}^{rx}$. This leads to the characteristic equation$$\begin{array}{rl}{r}^2+\lambda & =0\\ r& =\pm \sqrt{-\lambda }\end{array}$$

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$$\begin{array}{rl}y& =c_1\cosh \left(\mu x\right)+c_2\sinh \left(\mu x\right)\\ y^{\prime }& =c_1\mu \sinh \left(\mu x\right)+c_2\mu \cosh \left(\mu x\right)\end{array}$$

First B.C. $y^{\prime }\left(0\right)=0$ gives$$\begin{array}{rl}0& =c_2\mu \\ c_2& =0\end{array}$$

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 $\mu L\ne 0$, 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$$\begin{array}{rl}y\left(x\right)& =c_1\cos \left(\sqrt{\lambda }x\right)+c_2\sin \left(\sqrt{\lambda }x\right)\\ y^{\prime }\left(x\right)& =-c_1\sqrt{\lambda }\sin \left(\sqrt{\lambda }x\right)+c_2\sqrt{\lambda }\cos \left(\sqrt{\lambda }x\right)\end{array}$$