1. ## \varphi^{(4)}-\lambda\varphi=0

$\displaystyle \varphi^{(4)}-\lambda\varphi=0, \ \ 0<x<L, \ \ \varphi(0)=\varphi'(0)=\varphi(L)=\varphi'(L)=0$

$\displaystyle m^4=\lambda, \ \ r=\lambda, \ \ \theta=0$

$\displaystyle \displaystyle m=r^{1/n}e^{\frac{ik\theta}{n}}\Rightarrow m=\pm s, \ \pm si, \ \ \text{where} \ s=\lambda^{1/4}$

$\displaystyle \varphi=C_1e^{xs}+C_2e^{-xs}+C_3\cos(xs)+C_4\sin(xs)$

$\displaystyle \varphi'=sC_1e^{xs}-sC_2e^{-xs}-sC_3\sin(xs)+sC_4\cos(xs)$

$\displaystyle \varphi''=s^2C_1e^{xs}+s^2C_2e^{-xs}-s^2C_3\cos(xs)-s^2C_4\sin(xs)$

$\displaystyle \varphi'''=s^3C_1e^{xs}-s^3C_2e^{-xs}+s^3C_3\sin(xs)-s^3C_4\cos(xs)$

$\displaystyle \varphi_1(0): \ C_1+C_2+C_3=1, \ \varphi_1'(0): \ sC_1-sC_2+sC_4=0, \ \varphi_1''(0): \ s^2C_1+s^2C_2-s^2C_3=0, \ \varphi_1'''(0): \ s^3C_1-s^3C_2-s^3C_4=0$

$\displaystyle \varphi_2(0): \ C_1+C_2+C_3=0, \ \varphi_2'(0): \ sC_1-sC_2+sC_4=1, \ \varphi_2''(0): \ s^2C_1+s^2C_2-s^2C_3=0, \ \varphi_2'''(0): \ s^3C_1-s^3C_2-s^3C_4=0$

$\displaystyle \varphi_3(0): \ C_1+C_2+C_3=0, \ \varphi_3'(0): \ sC_1-sC_2+sC_4=0, \ \varphi_3''(0): \ s^2C_1+s^2C_2-s^2C_3=1, \ \varphi_3'''(0): \ s^3C_1-s^3C_2-s^3C_4=0$

$\displaystyle \varphi_4(0): \ C_1+C_2+C_3=0, \ \varphi_4'(0): \ sC_1-sC_2+sC_4=0, \ \varphi_4''(0): \ s^2C_1+s^2C_2-s^2C_3=0, \ \varphi_4'''(0): \ s^3C_1-s^3C_2-s^3C_4=1$

$\displaystyle \displaystyle\varphi_1=\frac{1}{4}\left(e^{xs}+e^{-xs}\right)+\frac{\cos(xs)}{2}=\frac{\cosh(xs)+\cos (xs)}{2}$

$\displaystyle \displaystyle\varphi_2=\frac{1}{4s}\left(e^{xs}-e^{-xs}\right)+\frac{\sin(xs)}{2s}=\frac{\sinh(xs)+\si n(xs)}{2s}$

$\displaystyle \displaystyle\varphi_3=\frac{1}{4s^2}\left(e^{xs}+ e^{-xs}\right)-\frac{\cos(xs)}{2s^2}=\frac{\cosh(xs)-\cos(xs)}{2s^s}$

$\displaystyle \displaystyle\varphi_4=\frac{1}{4s^3}\left(e^{xs}-e^{-xs}\right)-\frac{\sin(xs)}{2s^3}=\frac{\sinh(xs)-\sin(xs)}{2s^3}$

Is this correct so far?

2. I wonder if you're switching problems mid-way here? I get the following:

$\displaystyle \varphi(0)=0\quad\Rightarrow\quad C_{1}+C_{2}+C_{3}=0,$

$\displaystyle \varphi'(0)=0\quad\Rightarrow\quad sC_{1}-sC_{2}+sC_{4}=0,$

$\displaystyle \varphi(L)=0\quad\Rightarrow\quad C_{1}\,e^{sL}+C_{2}\,e^{-sL}+C_{3}\cos(sL)+C_{4}\sin(sL)=0,$ and

$\displaystyle \varphi'(L)=0\quad\Rightarrow\quad sC_{1}\,e^{sL}-sC_{2}\,e^{-sL}-sC_{3}\sin(sL)+sC_{4}\cos(sL)=0.$

3. Originally Posted by Ackbeet
I wonder if you're switching problems mid-way here? I get the following:

$\displaystyle \varphi(0)=0\quad\Rightarrow\quad C_{1}+C_{2}+C_{3}=0,$

$\displaystyle \varphi'(0)=0\quad\Rightarrow\quad sC_{1}-sC_{2}+sC_{4}=0,$

$\displaystyle \varphi(L)=0\quad\Rightarrow\quad C_{1}\,e^{sL}+C_{2}\,e^{-sL}+C_{2}\cos(sL)+C_{4}\sin(sL)=0,$ and

$\displaystyle \varphi'(L)=0\quad\Rightarrow\quad sC_{1}\,e^{sL}-sC_{2}\,e^{-sL}-sC_{3}\sin(sL)+sC_{4}\cos(sL)=0.$
First, don't I have to obtain the general $\displaystyle \varphi(x)=A\varphi_1+B\varphi_2+C\varphi_3+D\varp hi_4\text{?}$

$\displaystyle \varphi_1(0)=1, \ \varphi_2(0)=0, \ \varphi_3(0)=0, \ \varphi_4(0)=0$

$\displaystyle \varphi_1'(0)=0, \ \varphi_2'(0)=1, \ \varphi_3'(0)=0, \ \varphi_4'(0)=0$

$\displaystyle \varphi_1''(0)=0, \ \varphi_2''(0)=0, \ \varphi_3''(0)=1, \ \varphi_4''(0)=0$

$\displaystyle \varphi_1'''(0)=0, \ \varphi_2'''(0)=0, \ \varphi_3'''(0)=0, \ \varphi_4'''(0)=1$

4. Hmm. You're trying to find a fundamental matrix? If so, I'm afraid this problem is out of my league. Maybe Danny could take a look?

5. Originally Posted by Ackbeet
Hmm. You're trying to find a fundamental matrix? If so, I'm afraid this problem is out of my league. Maybe Danny could take a look?
I have already solved for the coefficients of $\displaystyle \varphi_t, t=1,2,3,4$ in post 1.

For

$\displaystyle \displaystyle\varphi(x)=\frac{1}{2}\left[A(\cosh(xs)+\cos(xs))+\frac{B}{s}\left(\sinh(xs)+\ sin(xs)\right)+\frac{C}{s^2}\left(\cosh(xs)-\cos(xs)\right)+\frac{1}{s^3}\left(\sinh(xs)-\sin(xs)\right)\right]$

$\displaystyle \varphi(0): \ A=0$

$\displaystyle \varphi'(0): \ sB=0\Rightarrow B=0$

$\displaystyle \displaystyle\varphi(x)=\frac{1}{2}\left[\frac{C}{s^2}\left(\cosh(xs)-\cos(xs)\right)+\frac{D}{s^3}\left(\sinh(xs)-\sin(xs)\right)\right]$

$\displaystyle \displaystyle\varphi(L): \ \frac{C}{2{s^2}_n}\left(\cosh(Ls_n)-\cos(Ls_n)\right)+\frac{D}{2{s^3}_n}\left(\sinh(Ls _n)-\sin(Ls_n)\right)=0$

$\displaystyle \displaystyle\varphi'(L): \ \frac{C}{2{s_n}}\left(\sinh(Ls_n)+\sin(Ls_n)\right )+\frac{D}{2{s^2}_n}\left(\cosh(Ls_n)-\cos(Ls_n)\right)=0$

Where $\displaystyle s_n$ represents the eigenvalues.

Now, if everything is correct, I am struggling with solving for the eigenvalues.

$\displaystyle \displaystyle\frac{C}{2{s^2}_n}\left(\cosh(Ls_n)-\cos(Ls_n)\right)+\frac{D}{2{s^3}_n}\left(\sinh(Ls _n)-\sin(Ls_n)\right)-\frac{C}{2{s_n}}\left(\sinh(Ls_n)+\sin(Ls_n)\right )-\frac{D}{2{s^2}_n}\left(\cosh(Ls_n)-\cos(Ls_n)\right)=0$

6. Just a comment @dw - your method would work fine if you were given IC's

$\displaystyle y(0) = y_0, y'(0) = y_1, y''(0) = y_2, y'''(0) = y_3$

but you're not. You're given

$\displaystyle y(0) = y_0, y'(0) = y_1, y(L) = y_2, y'(L) = y_3.$

Some of it works but you're still left with two equations to solve. Adrian's method is much more natural and direct.

Next, dw, with your two equations for C and D, eliminate C and D and get a single equation for $\displaystyle \lambda$ or $\displaystyle s_n$.

7. Originally Posted by Danny
Just a comment @dw - your method would work fine if you were given IC's

$\displaystyle y(0) = y_0, y'(0) = y_1, y''(0) = y_2, y'''(0) = y_3$

but you're not. You're given

$\displaystyle y(0) = y_0, y'(0) = y_1, y(L) = y_2, y'(L) = y_3.$

Some of it works but you're still left with two equations to solve. Adrian's method is much more natural and direct.

Next, dw, with your two equations for C and D, eliminate C and D and get a single equation for $\displaystyle \lambda$ or $\displaystyle s_n$.
I am not sure how to eliminate C and D. Do you have any suggestions or ideas?

8. Yes, in post 5, you have two equations for C and D. Move the D terms to the right side of both equations and then divide the left side and right sides. The C and D terms will cancel. Then simplify.

9. Originally Posted by Danny
Yes, in post 5, you have two equations for C and D. Move the D terms to the right side of both equations and then divide the left side and right sides. The C and D terms will cancel. Then simplify.
I don't see how division is going to cancel the C and D terms. Here is what I have after simplifying:

$\displaystyle \displaystyle \frac{C}{s_n}\left(\cosh(Ls_n)-\cos(Ls_n)\right)-C\left(\sinh(Ls_n)+\sin(Ls_n)\right)=\frac{D}{s_n} \left(\cosh(Ls_n)-\cos(Ls_n)\right)-\frac{D}{{s_n}^2}\left(\sinh(Ls_n)-\sin(Ls_n)\right)$

10. $\displaystyle \displaystyle \varphi(L)=\frac{C}{2s^2}\left[\cosh(Ls)-\cos(Ls)\right]+\frac{D}{2s^3}\left[\sinh(Ls)-\sin(Ls)\right]=0$

$\displaystyle \displaystyle\Rightarrow \frac{C}{D}=-s\frac{\sinh(Ls)-\sin(Ls)}{\cosh(Ls)-\cos(Ls)}$

$\displaystyle \displaystyle\varphi'(L)=\frac{C}{2s}\left[\sinh(Ls)+\sin(Ls)\right]+\frac{D}{2s^2}\left[\cosh(Ls)-\cos(Ls)\right]=0$

$\displaystyle \displaystyle\Rightarrow \frac{C}{D}=-s\frac{\cosh(Ls)-\cos(Ls)}{\sinh(Ls)+\sin(Ls)}$

$\displaystyle \displaystyle \frac{\sinh(Ls)-\sin(Ls)}{\cosh(Ls)-\cos(Ls)}=\frac{\cosh(Ls)-\cos(Ls)}{\sinh(Ls)+\sin(Ls)}\Rightarrow \frac{\left(\cosh(Ls)-\cos(Ls)\right)^2}{\sinh^2(Ls)-\sin^2(Ls)}=1$

Is this much correct?

The solution is:

$\displaystyle \displaystyle \frac{1-\cosh(Ls)\cos(Ls)}{2s^4}=0$

I have been unable to manipulate my solution in order to obtain the books. I am guessing I am wrong, or I am unable to make a slick substitution with an identity.

11. $\displaystyle \displaystyle\frac{(\cosh(Ls)-\cos(Ls))^2}{\sinh^2(Ls)-\sin^2(Ls)}=1\Rightarrow\cosh^2(Ls)-2\cosh(Ls)\cos(Ls)+\cos^2(Ls)=\sinh^2(Ls)-\sin^2(Ls)$

$\displaystyle \displaysytle\Rightarrow\underbrace{\cosh^2(Ls)-\sinh^2(Ls)}_{1}-2\cosh(Ls)\cos(Ls)+\underbrace{\cos^2(Ls)+\sin^2(L s)}_{1}=0$

$\displaystyle \Rightarrow 2(1-\cosh(Ls)\cos(Ls))=0$

Now, let $\displaystyle Ls=x$ and looking at the Taylor expansion of cosh and cos because 0 isn't eigenvalue.

$\displaystyle \displaystyle \left(1-\sum_{x=0}^{\infty}\frac{x^{2n}}{(2n)!}\sum_{x=0}^ {\infty}\frac{(-1)^{n}}{(2n)!}x^{2n}\right)$

$\displaystyle \displaystyle\sum_{x=0}^{\infty}\frac{x^{2n}}{(2n) !}\sum_{x=0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac {x^4}{4!}-\frac{x^4}{2!2!}+\cdots$

$\displaystyle \displaystyle 1-\left(1-\frac{x^2}{2!}+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac {x^4}{4!}-\frac{x^4}{2!2!}+\cdots\right)=\frac{x^4}{6}-\frac{x^8}{2\cdot 6!}+\cdots\pm\cdots=0$

$\displaystyle \displaystyle\frac{1-\cosh(Ls_n)\cos(Ls_n)}{{s_n}^4}=0$

Why the book multiplied through by $\displaystyle \frac{1}{2}$ beats me.