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Math Help - \varphi'''-\lambda\varphi=0

  1. #1
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    \varphi'''-\lambda\varphi=0

    \varphi'''-\lambda\varphi=0

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

    m^3-\lambda=0\Rightarrow m^3=\lambda

    \displaystyle w_k=\lambda^{1/3}\left[\cos\left(\frac{2\pi k}{3}\right)+i\sin\left(\frac{2\pi k}{3}\right)\right]

    Let \lambda^{1/3}=s

    \displaystyle\varphi=Ae^{xs}+\exp\left(\frac{-xs}{2}\right)\left[B\cos\left(\frac{xs\sqrt{3}}{2}\right)+C\sin\left(  \frac{xs\sqrt{3}}{2}\right)\right]

    \displaystyle\varphi'=Ase^{xs}-\frac{\exp\left(\frac{-xs}{2}\right)}{2}\left[B\cos\left(\frac{xs\sqrt{3}}{2}\right)+C\sin\left(  \frac{xs\sqrt{3}}{2}\right)\right]+\exp\left(\frac{-xs}{2}\right)\left[\frac{Cs\sqrt{3}}{2}\cos\left(\frac{xs\sqrt{3}}{2}  \right)-\frac{Bs\sqrt{3}}{2}\sin\left(\frac{xs\sqrt{3}}{2}  \right)\right]

    \displaystyle\varphi''=As^2e^{xs}+\frac{\exp\left(  \frac{-xs}{2}\right)}{4}\left[B\cos\left(\frac{xs\sqrt{3}}{2}\right)+C\sin\left(  \frac{xs\sqrt{3}}{2}\right)\right]-s\exp\left(\frac{-xs}{2}\right)\left[\frac{Cs\sqrt{3}}{2}\cos\left(\frac{xs\sqrt{3}}{2}  \right)-\frac{Bs\sqrt{3}}{2}\sin\left(\frac{xs\sqrt{3}}{2}  \right)\right] \displaystyle +\exp\left(\frac{-xs}{2}\right)\left[\frac{-3Bs^2}{4}\cos\left(\frac{xs\sqrt{3}}{2}\right)-\frac{3Cs^2}{4}\sin\left(\frac{xs\sqrt{3}}{2}\righ  t)\right]

    \varphi_1(0)=A+B=1, \ \varphi_2(0)=A+B=0, \ \varphi_3(0)=A+B=0
    \displaystyle\varphi_1'(0)=sA-\frac{sB}{2}+\frac{Cs\sqrt{3}}{2}=0, \ \varphi_2'(0)=sA-\frac{sB}{2}+\frac{Cs\sqrt{3}}{2}=1, \ \varphi_3'(0)=sA-\frac{sB}{2}+\frac{Cs\sqrt{3}}{2}=0
    \displaystyle\varphi_1''(0)=s^2A+s^2B-\frac{s^2\sqrt{3}C}{2}=0, \ \varphi_2''(0)=s^2A+s^2B-\frac{s^2\sqrt{3}C}{2}=0, \ \varphi_3''(0)=s^2A+s^2B-\frac{s^2\sqrt{3}C}{2}=1

    \displaystyle\varphi_1\text{coefficient matrix}=\begin{bmatrix}1&1&0&1\\s&\frac{s}{2}&\fra  c{s\sqrt{3}}{2}&0\\s^2&s^2&-\frac{s^2\sqrt{3}}{2}&0\end{bmatrix}\Rightarrow\te  xt{rref}=\begin{bmatrix}1&0&0&-\frac{1}{3}\\0&1&0&\frac{4}{3}\\0&0&1&\frac{2\sqrt  {3}}{3}\end{bmatrix}

    \displaystyle\varphi_2\text{coefficient  matrix}=\begin{bmatrix}1&1&0&0\\s&\frac{s}{2}&\fra  c{s\sqrt{3}}{2}&1\\s^2&s^2&-\frac{s^2\sqrt{3}}{2}&0\end{bmatrix}\Rightarrow\te  xt{rref}=\begin{bmatrix}1&0&0&\frac{2}{3s}\\0&1&0&-\frac{2}{3s}\\0&0&1&0\end{bmatrix}

    \displaystyle\varphi_3\text{coefficient  matrix}=\begin{bmatrix}1&1&0&0\\s&\frac{s}{2}&\fra  c{s\sqrt{3}}{2}&0\\s^2&s^2&-\frac{s^2\sqrt{3}}{2}&1\end{bmatrix}\Rightarrow\te  xt{rref}=\begin{bmatrix}1&0&0&\frac{2}{3s^2}\\0&1&  0&-\frac{2}{3s^2}\\0&0&1&-\frac{2\sqrt{3}}{3s^2}\end{bmatrix}

    However, there is an error some where because the book provides the coefficients for \varphi_3 which are:

    \displaystyle A=\frac{1}{3s}, \ B=-\frac{1}{3s}, \ C=-\frac{\sqrt{3}}{3s}

    Therefore, I am not sure about my other coefficients for \varphi_1, \ \varphi_2
    Last edited by dwsmith; February 17th 2011 at 05:52 PM. Reason: Forgot the negative
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  2. #2
    MHF Contributor chisigma's Avatar
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    The DE is linear with constant coefficients, so that \varphi(*) is analythic and we can write...

    \displaystyle \varphi(t)= \sum_{n=0}^{\infty} \frac{\varphi^{(n)} (0)}{n!}\ t^{n} (1)

    But is...

    \varphi^{(n)} (0) = \lambda\ \varphi^{(n-3)} (0) (2)

    ... so that if we set...

    \varphi(0)=a_{0}

    \varphi^{'}(0)=a_{1}

    \varphi^{''}(0)=a_{2} (3)

    ... the solution is...

    \displaystyle \varphi(t)= a_{0}\ \sum_{k=0}^{\infty} \frac{\lambda^{k}}{(3k)!}\ t^{3k} + a_{1}\ \sum_{k=0}^{\infty} \frac{\lambda^{k}}{(3k+1)!}\ t^{3k+1} + a_{2}\ \sum_{k=0}^{\infty} \frac{\lambda^{k}}{(3k+2)!}\ t^{3k+2} (4)

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post
    The DE is linear with constant coefficients, so that \varphi(*) is analythic and we can write...

    \displaystyle \varphi(t)= \sum_{n=0}^{\infty} \frac{\varphi^{(n)} (0)}{n!}\ t^{n} (1)

    But is...

    \varphi^{(n)} (0) = \lambda\ \varphi^{(n-3)} (0) (2)

    ... so that if we set...

    \varphi(0)=a_{0}

    \varphi^{'}(0)=a_{1}

    \varphi^{''}(0)=a_{2} (3)

    ... the solution is...

    \displaystyle \varphi(t)= a_{0}\ \sum_{k=0}^{\infty} \frac{\lambda^{k}}{(3k)!}\ t^{3k} + a_{1}\ \sum_{k=0}^{\infty} \frac{\lambda^{k}}{(3k+1)!}\ t^{3k+1} + a_{2}\ \sum_{k=0}^{\infty} \frac{\lambda^{k}}{(3k+2)!}\ t^{3k+2} (4)

    Kind regards

    \chi \sigma
    This is probably a great response, but, unfortunately, I don't know how to apply this to the DE.
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  4. #4
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    Problem found. \displaystyle\frac{1}{4}+\frac{-3}{4}=\frac{-1}{2}\neq 1
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  5. #5
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    Follow up:

    \displaystyle\varphi_1\text{coefficient matrix}=\begin{bmatrix}1&1&0&1\\s&-\frac{s}{2}&\frac{s\sqrt{3}}{2}&0\\s^2&-\frac{s^2}{2}&-\frac{s^2\sqrt{3}}{2}&0\end{bmatrix}\Rightarrow\te  xt{rref}=\begin{bmatrix}1&0&0&\frac{1}{3}\\0&1&0&\  frac{2}{3}\\0&0&1&0\end{bmatrix}

    \displaystyle\varphi_1=\frac{e^{xs}}{3}+\frac{2}{3  }\exp\left(\frac{-xs}{2}\right)\cos\left(\frac{xs\sqrt{3}}{2}\right)

    \displaystyle\varphi_2\text{coefficient matrix}=\begin{bmatrix}1&1&0&0\\s&-\frac{s}{2}&\frac{s\sqrt{3}}{2}&1\\s^2&-\frac{s^2}{2}&-\frac{s^2\sqrt{3}}{2}&0\end{bmatrix}\Rightarrow\te  xt{rref}=\begin{bmatrix}1&0&0&\frac{1}{3s}\\0&1&0&-\frac{1}{3s}\\0&0&1&\frac{1}{s\sqrt{3}}\end{bmatri  x}

    \displaystyle\varphi_2=\frac{e^{xs}}{3s}+\frac{\ex  p\left(\frac{-xs}{2}\right)}{3s}\left[-\cos\left(\frac{xs\sqrt{3}}{2}\right)+\sqrt{3}\sin  \left(\frac{xs\sqrt{3}}{2}\right)\right]

    \displaystyle\varphi_3\text{coefficient matrix}=\begin{bmatrix}1&1&0&0\\s&-\frac{s}{2}&\frac{s\sqrt{3}}{2}&0\\s^2&-\frac{s^2}{2}&-\frac{s^2\sqrt{3}}{2}&1\end{bmatrix}\Rightarrow\te  xt{rref}=\begin{bmatrix}1&0&0&\frac{1}{3s^2}\\0&1&  0&-\frac{1}{3s^2}\\0&0&1&-\frac{\sqrt{3}}{3s^2}\end{bmatrix}

    \displaystyle\varphi_3=\frac{e^{xs}}{3s^2}-\frac{\exp\left(\frac{-xs}{2}\right)}{3s^2}\left[\cos\left(\frac{xs\sqrt{3}}{2}\right)+\sqrt{3}\sin  \left(\frac{xs\sqrt{3}}{2}\right)\right]

    Find \varphi(0)=5, \ \varphi'(0)=-3, \ \varphi''(0)=9

    \displaystyle \varphi(x)=A\left[\frac{e^{xs}}{3}+\frac{2}{3}\exp\left(\frac{-xs}{2}\right)\cos\left(\frac{xs\sqrt{3}}{2}\right)  \right] \displaystyle +B\left[\frac{e^{xs}}{3s}+\frac{\exp\left(\frac{-xs}{2}\right)}{3s}\left[-\cos\left(\frac{xs\sqrt{3}}{2}\right)+\sqrt{3}\sin  \left(\frac{xs\sqrt{3}}{2}\right)\right]\right] \displaystyle +C\left[\varphi_3=\frac{e^{xs}}{3s^2}-\frac{\exp\left(\frac{-xs}{2}\right)}{3s^2}\left[\cos\left(\frac{xs\sqrt{3}}{2}\right)+\sqrt{3}\sin  \left(\frac{xs\sqrt{3}}{2}\right)\right]\right]

    \varphi(0):A=5

    \displaystyle\varphi'(0):B=\frac{-54(s-2)}{17s}

    \displaystyle\varphi''(0):C=\frac{-9(s-36)}{34}

    Are my coefficients correct?
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