# Fredholm Integral of Second Kind, Eigenvalues

1. Jan 5, 2006

### beautiful1

I need help with an integral eigenvalue equation...I am lost on how to handle this:

$$\int_{-\infty}^{\infty} dy K(x,y) \psi_n(y) = \lambda_n \psi_n(x)$$

The kernel, $$K(x,y)$$ is a 2D, correlated Gaussian. I have read that for this case an analytic solution exist for the eigenvalues, $$\lambda_n$$, and the eigenfunctions, $$\psi_n(x)$$, are given in terms of the Hermite functions (polynomials?).

Any suggestions on starting this solution would be appreciated.

p.s. dear moderator, perhaps you know if this should be posted in the differential equations subforum. I wasn't sure.

Last edited: Jan 6, 2006
2. Jan 6, 2006

### HallsofIvy

Staff Emeritus
Hey, I'm a moderator and I'm not sure either! It might get more responses in differential equation than calculus so I will move it there.

"2d Gaussian"? That's $Const e^{-x^2-y^2}= Const e^{-x^2}e^{-y^2}$ isn't it? If so then do this: write the equation as
$$Const e^{-x^2}\int_{-\infty}^{\infty}e^{-y^2}\psi_n(y)dy= \lambda_n\psi_n(x)[/itex]. Notice that the integral on the left is a definite integral: it is a constant: Let $X_n= \int_{-\infty}{\infty}e^{-y^2}\psi_n(y)dy$. Then the equation says $Const X_ne^{-x^2}= \lamda_n\psi_n(x)$. Multiply both sides by $e^{-x^2}$ to get $Const X_n e^{-2x^2}= \lambda_n \psi_n(x)e^{-x^2}$. Now integrate both sides, with respect to x, from -infinity to infinity: $Const X_n \int_{-\infty}^{/infty}e^{-2x^2}dx= \lamba_n X_n$. What value of $\lamba_n$ makes that true for any Xn? Last edited: Jan 6, 2006 3. Jan 6, 2006 ### saltydog Hall you mind if I summarize your work here: $$\mathcal{F}\left(\Psi_n\right)=\lambda_n\Psi_n$$ where: $$\mathcal{F}\left\{f\right\}=\int_{-\infty}^{\infty}K(x,y)f(y)dy$$ with: $$K(x,y)=Ce^{-(x^2+y^2)}$$ so that we have: $$\lambda_n\Psi_n(x)=Ce^{-x^2}\int_{-\infty}^{\infty}e^{-y^2}\Psi_n(y)dy$$ Representing the definite integral as the constant $X_n$ as Hall indicated above: $$X_n=\int_{-\infty}^{\infty}e^{-y^2}\Psi_n(y)dy$$ we obtain: $$\lambda_n\Psi_n(x)=CX_ne^{-x^2}$$ Multiplying both sides by $e^{-x^2}$$ and integrating: $$\int_{-\infty}^{\infty}\lambda_n\Psi_n(x)e^{-x^2}dx=\int_{-\infty}^{\infty}CX_ne^{-2x^2}dx$$ but from above: $$\int_{-\infty}^{\infty}\Psi_n(v)e^{-v^2}dv=X_n$$ so that we're left with: $$\lambda_n X_n=CX_n\int_{-\infty}^{\infty}e^{-2x^2}dx$$ If I've incorrectly interpreted Hall's analysis above, I'm sure he'll . . . indicate such. Last edited: Jan 6, 2006 4. Jan 6, 2006 ### beautiful1 Thank you both for your response. I will think about this approach as it looks helpful. But there is a slight complication, which is that the Guassian is correlated, i.e. not separable in x and y. The example I have in mind is $$K(x,y) = C \exp \{-\sigma_1^2 (c x + sy)^2-\sigma_2^2 (cy - sx)^2 \}$$ where $$C$$, $$\sigma_1$$, and $$\sigma_2$$ are real constants and $$c = \cos \theta$$ and $$s = \sin\theta$$ for some angle $$\theta$$. Plotted, such a function would be a 2D, squeezed Gaussian rotated w.r.t to the x-y axes. But your solution may also work in this case and I am pursuing that. Sorry for not being more explicit earlier. Last edited: Jan 6, 2006 5. Jan 7, 2006 ### saltydog How about expanding the kernel in a power series and then solving (approximating) it as per above? For example: $$e^{-((x+y)^2-(x-y)^2)}= 1-4xy+8x^2y^2-\frac{32x^3y^3}{3}+\frac{32x^4y^4}{3}-...$$ 6. Jan 7, 2006 ### beautiful1 Thanks for the second reply saltydog; your suggestions mirrors another approach I found in "Methods of Theoretical Physics" by Morse and Feschbach. There, the suggestions is to assume the kernel is an expansion of the form $$K(x,y)= \sum_{n=0}^{\infty} h_{n}(x) g_{n}(y)$$ where [itex] h_n$ is a complete set of functions and $g_n(y)$ is the corresponding coefficient. Substituting this forumal in yields

$$\psi_n(x) = \lambda^{-1} \sum_{n} A_n h_n(x)$$

with

$$A_n = \int dy g_n(y) \psi_n(y)$$

Inserting the series expansion for $\psi_n(y)$ yields

$$A_n = \sum_p \alpha_{np} A_p$$

which is a set of simultaneous equations for the $A$'s with coefficients
$$\alpha_{np} = \int dy g_n(y) h_p(y)$$

This can then converted into the usual eigenvalue problem.
And for a judicious selection of the original functions, the problem can be made relatively easy. For example, in the case of the Gaussian, a diagonal basis using the Hermite functions (Hermite polynomials times a Gaussian) is a good choice, since

$$\alpha_{np} = a_{nn} \delta_{pn}$$

and

$$\psi_n(x) = h_n(x)$$

I think your suggestions of a power series would be somewhat similar.

Thanks to everyones help.

BTW, there may be some mistakes in the above.

Last edited: Jan 7, 2006
7. Jan 7, 2006

### saltydog

I used the wrong equation last night. Consider a kernel of the form:

$$K(x,y)=e^{-((x+y)^2+(y-x)^2)}$$

and expand it out to 6 terms in a Taylor series:

\begin{align*} e^{-((x+y)^2+(y-x)^2)}&=1-2x^2+2x^4-\frac{4x^6}{3} \\ &+\left(-2+4x^2-4x^4+\frac{8x^6}{3}\right)y^2 \\ &+\left(2-4x^2+4x^4-\frac{8x^6}{3}\right)y^4 \end{align}

I've attached plots of the kernel and its 6-term Taylor equivalent. As you can see they are similar in a region about the origin; the kernel rapidly decays beyond this. Can one then solve the related homogeneous Fredholm equation:

$$u(x)=\int_{-a}^{a}T(x,y)u(y)dy$$

where T(x,y) is the Taylor expansion of K(x,y)

and obtain an approximate solution to the unbounded case?

Would the accuracy improve as more terms are added and the limits of integration are expanded. No?

I suspect all of this can be analyzed from the perspective of integral operators in Hilbert Space and it's completeness thereof.

Anyway Beau (how about I just call you that?), I've read your post above and will look into more. Thanks.

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8. Jan 7, 2006

### beautiful1

Thanks saltydog, I really like your approach of expanding the kernel. I would think that it is a very general and useful method for use with many other kernels too.

I'm not sure how to solve the remaining integrals, perhaps by using a Taylor series expansion of u(x)? If this was truncated at the same order as $K$ then maybe coefficients of like powers could be equated. Maybe. I'll check my new favorite book by Morse and Feschbach (which incidently is selling for almost \$1000 on Amazon!)

Thanks for your help. I'll let you know how things turn out.
beau

9. Jan 8, 2006

### krab

Just define new variables:
$$u=cx+sy,\ v=sx-cy$$
It's separable in those variables. This just represents a rotation of the axes by an angle of theta.

10. Jan 8, 2006

### saltydog

Hello Krab.

I don't understand how to make that substitution for the rest of the equation: the u(y)dy part of the integral in particular. Also, how would the left-hand side change as well? Might you explain a little further please?