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I'm trying to show that the eigenvalues of the following integral equation \begin{align*} \lambda \phi(t) = \int_{-T/2}^{T/2} dx \phi(x)e^{-\Gamma|t-x|} \end{align*}

are given by \begin{align*} \Gamma \lambda_k = \frac{2}{1+u_k^2} \end{align*}

where $u_k$ are the solutions to the transcendental equation \begin{align*} \tan(\Gamma T u_k) = \frac{2u_k}{u_k^2-1}. \end{align*}

My approach was to separate this into two integrals: \begin{align*} \int_{-T/2}^{T/2} dx \phi(x)e^{|t-x|} = e^{-\Gamma t}\int_{-T/2}^t dx \phi(x)e^{\Gamma x} + e^{\Gamma t} \int_t^{T/2} dx \phi(x) e^{-\Gamma x}. \end{align*}

Then I differentiated the eigenvalue equation twice with this modification to find that \begin{align*} \ddot{\phi} = \frac{\Gamma^2-2\Gamma}{\lambda}\phi \end{align*} indicating that $\phi(t)$ is a sum of exponentials $Ae^{\kappa t} + Be^{-\kappa t}$ where $\kappa^2$ is the coefficient in the previous equation. Can anyone confirm this is the correct approach? I don't think there's any way to factor the original kernel and invert the result. And I'm having trouble determining the initial conditions of the equation to set $A$ and $B$. Assuming I can find these my instinct would be to plug $\phi$ back into the original equation and explicitly integrate and solve for $\lambda$.

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I figured it out and thought I'd share the solution in case anyone else is plagued with this question. First of all there's a typo: it should read $\ddot{phi} = -\frac{2\Gamma - \lambda\Gamma^2}{\lambda}$. To solve this you plug this back in to the original equation and evaluate when $t = 0$. Then you do the same for the equation $\lambda \dot{\phi}$ at $t = 0$. After simplifying you get the two separate solutions $\tan(\kappa T/2)=\frac{\Gamma}{\kappa}$ and $\tan(\kappa T/2) = -\frac{\kappa}{\Gamma}$. By messing around and using the double angle formula for $\tan(\theta)$ it's doable. –  fatbox Nov 30 '11 at 9:28
    
In my previous comment I meant to say: $\ddot{\phi} = -\frac{2\Gamma - \lambda \Gamma^2}{\lambda}$ and I'm now defining $\kappa \equiv \frac{2\Gamma - \lambda \Gamma^2}{\lambda}$ and assuming an exponential solution: $\phi(t) = Ae^{i\kappa t} + Be^{-i\kappa t}$. When you plug in and check for $\phi(0)$ and $\dot{\phi}(0)$ you find the two equations I mentioned previously. You will find that one equation corresponds to $A = B$ and the other to $A = -B$. These are the symmetric and antisymmetric wavefunctions, which are analogous to the bound states of the finite square well potential. –  fatbox Nov 30 '11 at 9:33
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