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Math Help - \varphi^{(4)}-\lambda\varphi=0

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

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

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

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

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

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

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

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

    \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

    \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

    \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

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

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

    \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\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?
    Last edited by dwsmith; March 7th 2011 at 10:29 AM. Reason: Differentiation versus exponentiation notation.
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  2. #2
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    I wonder if you're switching problems mid-way here? I get the following:

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

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

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

    \varphi'(L)=0\quad\Rightarrow\quad sC_{1}\,e^{sL}-sC_{2}\,e^{-sL}-sC_{3}\sin(sL)+sC_{4}\cos(sL)=0.
    Last edited by Ackbeet; March 8th 2011 at 11:13 AM. Reason: Wrong subscript in third equation, third term.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    I wonder if you're switching problems mid-way here? I get the following:

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

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

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

    \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 \varphi(x)=A\varphi_1+B\varphi_2+C\varphi_3+D\varp  hi_4\text{?}

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

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

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

    \varphi_1'''(0)=0, \ \varphi_2'''(0)=0, \ \varphi_3'''(0)=0, \ \varphi_4'''(0)=1
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  4. #4
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    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?
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    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 \varphi_t, t=1,2,3,4 in post 1.

    For

    \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]

    \varphi(0): \ A=0

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

    \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\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\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 s_n represents the eigenvalues.

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

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

    y(0) = y_0, y'(0) = y_1, y''(0) = y_2, y'''(0) = y_3

    but you're not. You're given

    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 \lambda or s_n.
    Last edited by Jester; March 8th 2011 at 04:39 AM. Reason: forgot a D.
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  7. #7
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    Quote Originally Posted by Danny View Post
    Just a comment @dw - your method would work fine if you were given IC's

    y(0) = y_0, y'(0) = y_1, y''(0) = y_2, y'''(0) = y_3

    but you're not. You're given

    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 \lambda or s_n.
    I am not sure how to eliminate C and D. Do you have any suggestions or ideas?
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  8. #8
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    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.
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  9. #9
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    Quote Originally Posted by Danny View Post
    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 \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)
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  10. #10
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    \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\Rightarrow \frac{C}{D}=-s\frac{\sinh(Ls)-\sin(Ls)}{\cosh(Ls)-\cos(Ls)}

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

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

    \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 \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.
    Last edited by dwsmith; March 10th 2011 at 12:19 PM. Reason: Added book's solution.
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  11. #11
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    \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)

    \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

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

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

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

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

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