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I am stuck with the following problem:

For a continuous function $f (t ), 0 ≤ t ≤1,$ the integral equation $y(t)=f(t)+3 \displaystyle \int_{0}^{1}tsy(s)ds \,$ has

(a) a unique solution if $\displaystyle \int_{0}^{1}sf(s)ds \ne 0$

(b) no solution if $\displaystyle \int_{0}^{1}sf(s)ds = 0$

(c) infinitely many solutions if $\displaystyle \int_{0}^{1}sf(s)ds = 0$

(d) infinitely many solutions if $\displaystyle \int_{0}^{1}sf(s)ds \ne 0 $

My Attempt:

$y(t)=f(t)+3 \displaystyle \int_{0}^{1}tsy(s)ds \, \implies y(t)=f(t)+3tC$ where $\,C=\displaystyle \int_{0}^{1}s y(s)ds=\displaystyle \int_{0}^{1}s(f(s)+3sC)ds \implies....C=\displaystyle \int_{0}^{1}sf(s)ds+C$ [I have skipped some steps].

Now,looking at the options ,I think option (b) is correct as the given condition in (b) implies $C=C$.

Am I right? Please help.

learner
  • 6,726

1 Answers1

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Note that you can write the equation as $Ly = f$, where the linear operator $L$ is defined by $(Ly)(t) = 3t \int_0^1 s y(s) ds - y(t)$.

Since $L$ is linear, checking the kernel is a generally good idea.

Then $y \in \ker L$ iff $Ly = 0$ iff $y(t) = 3t \int_0^1 s y(s) ds$ for $t \in [0,1]$. It follows that $\ker L = \{ t \mapsto k t \}_{k \in \mathbb{R}}$ (ie, all functions of the form $y(t) = k t$). In particular, the kernel is non-trivial.

Now consider the equation $L y = f$. It is straightforward to verify that if $\int_0^1 s f(s) ds = 0$, then $Lf = f$, that is, $y=f$ solves the integral equation.

Hence if $\int_0^1 s f(s) ds = 0$, then all solutions to the equation $Ly = f$ are given by $f+y$, with $y \in \ker L$, or more explicitly, $y$ solves the equation iff $y(t) = f(t) + kt$ for some constant $k$.

Hence (c) is the correct choice.

copper.hat
  • 172,524
  • $\ker L \supseteq { t \mapsto k t }_{k \in \mathbb{R}}$ is clear and sufficient for your argument, although out of curiosity, how did you deduce equality? – davin Jun 04 '13 at 18:09
  • If $Ly = 0$, then let $k=3\int_0^1 s y(s) ds$. Then $y(t) = kt$ for all $t$. – copper.hat Jun 04 '13 at 18:17
  • It's still strange logic, since $k$ is a function of $y$, and you're then defining $y$ in terms of $k$. I don't see how that exhausts all possible functions. – davin Jun 04 '13 at 18:24
  • There is no strange logic. I just need to show that if $y \in \ker L$, then $y$ has the form $y(t) = k t$ for some $k$. It doesn't matter how I find the $k$ as long as $y$ has that form. – copper.hat Jun 04 '13 at 18:25
  • Indeed. Cheers. – davin Jun 04 '13 at 18:29
  • Why the downvote? I mean why after almost 4 years? – copper.hat Mar 20 '17 at 05:06