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let $\lambda_1, \lambda_2$ be the charactieristic numbers and $f_1, f_2 $be the corresponding eigenfunctions for the corresponding eigenfunctions for the homogeneous integral equation $\varphi(x)-\lambda\int_0^1(2xt+4x^2)\varphi(t)dt=0 \; then \\1. \lambda_1\ne\lambda_2 \\2.\lambda_1=\lambda_2 \\3.\int_0^1f_1(x)f_2(x)dx=0 \\ 4.\int_0^1f_1(x)f_2(x)dx=1 $

My attempt: $\varphi(x)=\lambda\int_0^1(2xt+4x^2)\varphi(t)dt \\ consider \;\varphi(x)=2\lambda xc_1+4\lambda x^2c_2 \\ where\; c_1=\int_0^1t\varphi(t)dt \;and \;c_2=\int_0^1\varphi(t)dt\\ c_1=\int_0^1t(2\lambda tc_1 + 4\lambda t^2c_2)dt \\c_1=[2\lambda \frac{t^3}{3}c_1+4\lambda \frac{t^4}{4} c_2]_0^1\\c_1=2\lambda\frac{1}{3}c_1+4\lambda \frac{1}{4} c_2\\ c_1[1-\frac{2\lambda}{3}]=\lambda c_2.........(1)\\similarly\; c_2[1-\frac{4\lambda}{3}]=\lambda c_1........(2) \\ \begin{array}\| 1-\frac{2\lambda}{3} \;\;\; -\lambda \\-\lambda\;\;\;\;\;1-\frac{4\lambda}{3} \end{array}=0\\ \Rightarrow \lambda=-9\pm 3\sqrt10 \Rightarrow \lambda_1=-9+3\sqrt10 \;\;and\;\; \lambda_2=-9-3\sqrt10 \\ eigen functions are \\ f_1=2x[1-\frac{4\lambda}{3}]c_2+4\lambda x^2c_2 \\ f_1=2x[1-\frac{4(-9+3\sqrt10)}{3}]c_2+4(-9+3\sqrt10)x^2c_2 \\ f_1=(26-8\sqrt10)xc_2+(-36 +12\sqrt10)x^2c_2\\ f_2=2x[1-\frac{4(-9-3\sqrt10)}{3}]c_2+4(-9-3\sqrt10)x^2c_2\\ f_2=(26+8\sqrt10)xc_2'-(36+12\sqrt10)x^2c_2'\\ \int_0^1{f_1(x)f_2(x)}dx=c_2c_2'\int_0^1(-144x^4+48x^3+36x^2)dx \\ \int_0^1{f_1(x)f_2(x)}dx=-\frac{24}{5}c_2c_2'\\ \lambda_1\ne \lambda_2$
Option 1 is correct. I have a confusion that if I choose $c_2c_2'=0$ then I get $\int_0^1{f_1(x)f_2(x)}dx=0$ option 3 and if I choose $c_2c_2'=-\frac{5}{24}$ then i get $\int_0^1{f_1(x)f_2(x)}dx=1$ option 4 then which is the right answer among 3 and 4? please help me to understand this.

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1 Answer 1

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We have that $c_2c_2'=0 \iff c_2=0 $ or $c_2'=0 \iff f_1=0$ or $f_2=0$. But the zero function is not an eigenfunction.

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  • $\begingroup$ That means c2c2' can't be zero. But in the key answer it is given that the answer is 0 $\endgroup$
    – Priyanka
    Commented Sep 17, 2017 at 5:53
  • $\begingroup$ As per the solution i got the answers are option 1 and option 4. but in the key answer it is given that option 1 and option 3 is correct. how is this possible? please help me to understand the concept. Thanks in advance $\endgroup$
    – Priyanka
    Commented Sep 17, 2017 at 6:20
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    $\begingroup$ As I understand it, option $3$ is never correct. Option $4$ is correct for an appropriate choice of $c_2$ and $c_2'$, but not in general. $\endgroup$ Commented Sep 17, 2017 at 6:39

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