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Math Help - Eigenfunction Expansion

  1. #1
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    Eigenfunction Expansion

    Find the basic function at x_0=0 for

    \varphi''+4\varphi'+\lambda\varphi=0

    Also, find the solution which satisfies \varphi(0)=3, \ \varphi'(0)=-1

    m^2+4m+\lambda=0

    \displaystyle m=\frac{-4\pm\sqrt{16-4\lambda}}{2}=-2\pm\sqrt{4-\lambda}

    \varphi=\exp\left(z(-2+\sqrt{4-\lambda})\right)

    What I don't understand is why the book multiplied through by x instead of of z = x + yi, because looking at the first solution, the imaginary constant is being used in relations to sqrt{4-\lambda}.

    How was \varphi_2 acquired and is the coefficient 2 associated with sine in the first solutions due to solving for the conditions?

    The solution to the problem is:

    \displaystyle\varphi_1=e^{-2x}\left[\cos(x\sqrt{4-\lambda})+\frac{2\sin(x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}\right]

    \displaystyle\varphi_2=e^{-2x}\frac{\sin (x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}
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  2. #2
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    \varphi=\exp\left(x(-2\pm\sqrt{4-\lambda})\right)\Rightarrow \varphi=C_1\exp(-2x+x\sqrt{4-\lambda})+C_2\exp(-2x-x\sqrt{4-\lambda})

    Then solve for the initial conditions?

    What about taking into account if \lambda>4 \ \ \ \text{or} \ \ \ \lambda\in\left[ [4,0)\cup (0,-\infty)\right]\text{?}

    When \lambda=0, the DE is reduced to \varphi''+4\varphi'=0, and then we have

    t^2+4t=0\Rightarrow t=0, -4

    \varphi=C_1+C_2e^{-4x}

    \varphi_1(0)=C_1+C_2=3

    \displaystyle\varphi_1'(0)=C_2=\frac{1}{4}

    \displaystyle\varphi_1=\frac{1}{4}\left(11+e^{-4x}\right)

    \varphi_2(0)=C_1+C_2=-1

    \displaystyle\varphi_2'(0)=C_2=-\frac{3}{4}

    \displaystyle\varphi_2=-\frac{1}{4}\left(1+3e^{-4x}\right)

    \varphi(x)=A\left[\frac{1}{4}\left(11+e^{-4x}\right)\right]+B\left[-\frac{1}{4}\left(1+3e^{-4x}\right)\right]
    Last edited by dwsmith; February 14th 2011 at 03:37 PM. Reason: Added when lambda = 0 and fixed some errors
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  3. #3
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    Now, if \lambda\in\left[ [4,0)\cup (0,-\infty)], then we obtain:

    \displaystyle \sqrt{4-\lambda}\in\mathbb{R}, \ \ \ \sqrt{4-\lambda}=\omega

    \displaystyle \varphi=C_1\exp(-2x+x\omega)+C_2\exp(-2x-x\omega)

    \varphi(0)=C_1+C_2=3

    \varphi'(0)=C_1(\omega-2)+C_2(-2-\omega)=-1

    \displaystyle C_1=\frac{5+3\omega}{2\omega}, \ \ C_2=\frac{-5+3\omega}{2\omega}

    \displaystyle\varphi_1(x)=\frac{5+3\omega}{2\omega  }\exp(-2x+x\omega)+\frac{-5+3\omega}{2\omega}\exp(-2x-x\omega)

    So omega can't be 0; therefore, lambda can't equal 4.

    \lambda\in\left[ [4,0)\cup (0,-\infty)]\Rightarrow \lambda\in\left[ (4,0)\cup (0,-\infty)]

    \varphi_2(0)=C_1+C_2=-1

    \varphi_2'(0)=C_1(\omega-2)+C_2(-2-\omega)=3

    \displaystyle C_1=\frac{1-\omega}{2\omega}, \ \ C_2=\frac{-1-\omega}{2\omega}

    \displaystyle\varphi_2(x)=\frac{1-\omega}{2\omega}\exp(-2x+x\omega)+\frac{-1-\omega}{2\omega}\exp(-2x-x\omega)

    \displaystyle\varphi(x)=A\left[\frac{5+3\omega}{2\omega}\exp(-2x+x\omega)+\frac{-5+3\omega}{2\omega}\exp(-2x-x\omega)\right]+B\left[\frac{1-\omega}{2\omega}\exp(-2x+x\omega)+\frac{-1-\omega}{2\omega}\exp(-2x-x\omega)\right]

    Correct?
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  4. #4
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    I don't understand how the book obtained its solution which would be for \lambda>4

    \displaystyle\varphi_1=e^{-2x}\left[\cos(x\sqrt{4-\lambda})+\frac{2\sin(x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}\right]

    \displaystyle\varphi_2=e^{-2x}\frac{\sin (x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}

    When I do the problem, I obtain:

    \varphi=e^{-2x}\left(C_1\cos(x\sqrt{4-\lambda})+C_2i\sin(x\sqrt{4-\lambda})\right)

    \varphi(0)=C_1=3

    \varphi=e^{-2x}\left(3\cos(x\sqrt{4-\lambda})+C_2i\sin(x\sqrt{4-\lambda})\right)

    \varphi'=-2e^{-2x}\left(3\cos(x\sqrt{4-\lambda})+C_2i\sin(x\sqrt{4-\lambda})\right)+e^{-2x}\left(C_2i\sqrt{4-\lambda}\cos(x\sqrt{4-\lambda})-3\sqrt{4-\lambda}\sin(x\sqrt{4-\lambda})\right)

    \displaystyle\varphi'(0)=-2\cdot 3+C_2i\sqrt{4-\lambda}=-1\Rightarrow C_2=\frac{-5i}{\sqrt{4-\lambda}}

    Before I attempt to solve \varphi_2, why am I so off with \varphi_1\text{?}

    \displaystyle\varphi_1(x)=e^{-2x}\left(3\cos(x\sqrt{4-\lambda})-i\frac{5\sin(x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}\right)\neq e^{-2x}\left[\cos(x\sqrt{4-\lambda})+\frac{2\sin(x\sqrt{4-\lambda})}{\sqrt{4-\lambda}}\right]
    Last edited by dwsmith; February 14th 2011 at 05:05 PM. Reason: forgot the i
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