The solution of the symmetric integral equation below: $$g(s) = f(s) + \lambda \int_{-1}^{1} (st +s^2t^2)g(t)dt \tag{$*$}$$ with separable kernels method is $$g(s) = f(s) + \lambda \int_{-1}^{1} (\frac{st}{(1-\frac{2}{3}\lambda)}+\frac{s^2t^2}{(1-\frac{2}{5}\lambda)} )f(t) dt$$ let $f(s)=(s+1)^2$ and $\lambda = 1$ in $(*)$, i.e let: $$g(s) = (s+1)^2 + \int_{-1}^{1} (st +s^2t^2)g(t)dt \tag{$**$}$$ The question is, How I can (maybe using of the Hilbert-Schmidt theorem to) find the eigenfunctions corresponding to eigenvalues $\lambda_1 = \frac{3}{2}, \lambda_2 = \frac{5}{2}$ for the symmetric kernel in $(**)$?

The issue in the question has also briefly been surveyed in the first example appeared on the 153th page of the book titled "Linear Integral Equations" by Ram P. Kanwal.

The picture blow is an extraction from the second edition of aforementioned reference. enter image description here


Late answer, still, for the sake of those who might land here looking for detailed explanation here:

You can solve for resolvent kernel as in Example 4 in Section 2.2 of Ram P Kanwal, to get $$\Gamma(s,t;\lambda)=\frac{st}{1-\frac23\lambda}+\frac{s^2t^2}{1-\frac25\lambda}$$

Putting $\lambda=1$ as in question, we get $$\Gamma(s,t;\lambda)=\frac{st}{1-\frac23}+\frac{s^2t^2}{1-\frac25}$$

Taking $\dfrac32$ common in the first term, and $\dfrac25$ common in second term, we get $$\Gamma(s,t;\lambda)=\frac{st}{\frac23(\frac32-\lambda)}+\frac{s^2t^2}{\frac25(\frac52-\lambda)}=\frac{\frac32st}{\frac32-\lambda}+\frac{\frac52s^2t^2}{\frac52-\lambda}\tag{1}$$

By preceding theory, we have $$\Gamma(s,t;\lambda)=\sum_{k=1}^\infty{\phi_k(s)\phi^*_k(t)\over\lambda_k-\lambda}$$ where

  1. $\lambda=1$
  2. $\lambda_1=\dfrac32$
  3. $\lambda_2=\dfrac52$ and
  4. $\phi^*_k=\phi_k$ since integral is over real numbers.

Hence $$\Gamma(s,t;\lambda)=\frac{\phi_1(s)\phi_1(t)}{\frac32-1}+\frac{\phi_2(s)\phi_2(t)}{\frac52-1}\tag{2}$$

Comparing equations (1) and (2), $$\phi_1(s)=\sqrt\frac32s,\ \phi_2(s)=\sqrt\frac52s^2$$

Where multiplying with $\dfrac{\sqrt2}{\sqrt2}$ in denominator and numerator gives the desired result.

(Note: there is a mistake in print; $\phi_2(s)=\sqrt{\dfrac52}s^2$, and not $\phi_2(s)=\sqrt{\dfrac{10}2}s^2$ - corrected in the very next step in textbook.)


If you solve the eigenvalue problem $g(s) = \lambda \int_{-1}^{1} (st +s^2t^2)g(t)dt $, you will find the two eigenvalues $\lambda_1=3/2$ and $\lambda_2=5/2$, and their corresponding eigenfunctions. The coefficients of the eigenfunctions are determined by the normalization condition $\int_{-1}^1 \phi(x)^2 dx=1$.

  • $\begingroup$ A little late question, can you please elaborate? $\endgroup$ – Jesse P Francis Oct 13 '18 at 4:16

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