I am stuck on the following problem that is as follows:
The integral equation $\quad \varphi(x)-\lambda \displaystyle\int_{-1}^{1}\cos[\pi(x-t)]\varphi(t) dt= g(x)$ has
1.a unique solution for $ \lambda \ne 1$ when $g(x)=x$
2.no solution for $\lambda \ne 1$ when $g(x)=1$
3.no solution for $\lambda =1$ when $g(x)=x$
4.infinite number of solutions for $\lambda =1$ when $\quad g(x)=1.$
What I did:
$$\begin{align}\quad g(x)&= \varphi(x)-\lambda \displaystyle\int_{-1}^{1}\cos[\pi(x-t)]\varphi(t)dt \\ \implies g(x) &=\varphi(x)-\lambda \displaystyle \int_{-1}^{1}\cos[\pi x-\pi t]\varphi(t)dt\\ &= \varphi(x) -\lambda \displaystyle\int_{-1}^{1} [\cos \pi x \cos \pi t +\sin \pi x \sin \pi t]\varphi(t)dt \\&=\varphi(x)- \lambda\cos \pi x\displaystyle\int_{-1}^{1} \cos \pi t \varphi(t)dt -\sin \pi x \displaystyle\int_{-1}^{1} \sin \pi t \varphi (t) dt \\&=....??\end{align}$$
Can someone explain with some details about how to tackle these types of problem.Thanks in advance for your time.
MY REATTEMPT TO THE PROBLEM:
( Using @Glen O 's suggestion)From the given problem we see,$$\begin{align}\quad \varphi(x)&=g(x)+\lambda \{A \cos (\pi x)+B \sin (\pi x)\} \end{align}$$,where $A=\displaystyle \int_{-1}^1 \cos(\pi t)\varphi(t)dt \longrightarrow \clubsuit$ and
$B=\displaystyle \int_{-1}^1 \sin(\pi t)\varphi(t)dt \longrightarrow \spadesuit$ .
After calculations,we get from $ \clubsuit \space \text{and}\space \spadesuit \space , A(1-\lambda)=\displaystyle \int_{-1}^1 \cos(\pi t)g(t)dt \space \text{and} \space B(1-\lambda)=\displaystyle \int_{-1}^1 \sin(\pi t)g(t)dt$ .
Now,(1) for $\lambda \ne 1,g(t)=t$,we get $A(1-\lambda)=0,B(1-\lambda)=\frac{2}{\pi}$ and so we can have a unique value of $A$ and $B$ and hence option $1$ is true.
(2)For $\lambda \ne 1,g(t)=1,$ we get $A(1-\lambda)=0,B(1-\lambda)=0$ and so we have $A=B=0$ and hence equation turns out to be $\varphi(x)=g(x)$ and hence no solution and so statement $2$ holds good.
(3)For $\lambda= 1,g(t) =t,$ we get $ 0 \times B=\frac{2}{\pi}$ which is impossible and hence statement $3$ does not hold.
(4)For $\lambda= 1,g(t) =t,$ we get $A(1-\lambda)=0,B(1-\lambda)=0$ and infinite number of values for $A,B$ is posssible. And hence statement $4$ holds true.
Am I missing something in my argument? Please feel free to comment.
Hint: As you correctly note, you may write the integral equation in the form
$$ \varphi(x) - \lambda \cos(\pi x)\int_{-1}^1\cos(\pi t)\varphi(t)dt-\lambda \sin(\pi x)\int_{-1}^1\sin(\pi t)\varphi(t)dt=g(x) $$ Notice that the integrals evaluate to simple numbers. So we have a solution of the form $$ \varphi(x) = g(x)+\lambda\big(A\cos(\pi x)+B\sin(\pi x)\big) $$ Now, $A=\int_{-1}^1 \cos(\pi t)\varphi(t)dt$. What do you get if you substitute our solution in for $\varphi(t)$, given $g(x)$? Similarly, what do you get for $B$?