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Thread: Eigenfunction Expansion

  1. #1
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    Eigenfunction Expansion

    Find the basic function at $\displaystyle x_0=0$ for

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

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

    $\displaystyle m^2+4m+\lambda=0$

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

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

    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 $\displaystyle \varphi_2$ 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:

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

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

    Then solve for the initial conditions?

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

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

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

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

    $\displaystyle \varphi_1(0)=C_1+C_2=3$

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

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

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

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

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

    $\displaystyle \varphi(x)=A\left[\frac{1}{4}\left(11+e^{-4x}\right)\right]+B\left[-\frac{1}{4}\left(1+3e^{-4x}\right)\right]$
    Last edited by dwsmith; Feb 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 $\displaystyle \lambda\in\left[ [4,0)\cup (0,-\infty)]$, then we obtain:

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

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

    $\displaystyle \varphi(0)=C_1+C_2=3$

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

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

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

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

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

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

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

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

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

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

    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 $\displaystyle \lambda>4$

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

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

    When I do the problem, I obtain:

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

    $\displaystyle \varphi(0)=C_1=3$

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

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

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

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

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