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Integral equations.

Alan

Member
Jul 21, 2012
58
Hi, I have the next question from Pipkin's textbook on Integral Equations:

question 9, page 10:
Find the values of [tex]k[/tex] for which the following equation has solutions that aren't identically zero. If [tex]k\neq 0[/tex], find representative solutions:


[tex]\int_{-\pi}^{\pi} \sin(x+y)u(y)dy = ku(x)[/tex]




What I have done so far is the following:


denote by [tex]c_1=\int \cos(y)u(y)dy,\ c_2=\int \sin(y)u(y)dy[/tex]
so we have:[tex]c_1\sin(x)+c_2 \cos(x)=ku(x)[/tex], so I found represntaive solutions, but how do I find the values of k?


Thanks in advance.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I think something is missing maybe the value of u(x) at some points ?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hi, I have the next question from Pipkin's textbook on Integral Equations:

question 9, page 10:
Find the values of [tex]k[/tex] for which the following equation has solutions that aren't identically zero. If [tex]k\neq 0[/tex], find representative solutions:


[tex]\int_{-\pi}^{\pi} \sin(x+y)u(y)dy = ku(x)[/tex]




What I have done so far is the following:


denote by [tex]c_1=\int \cos(y)u(y)dy,\ c_2=\int \sin(y)u(y)dy[/tex]
so we have:[tex]c_1\sin(x)+c_2 \cos(x)=ku(x)[/tex], so I found represntaive solutions, but how do I find the values of k?


Thanks in advance.
Hi Alan! :)

You have [tex]c_1\sin(x)+c_2 \cos(x)=ku(x)[/tex]
So [tex]u(x) = \frac 1 k (c_1\sin(x)+c_2 \cos(x))[/tex]
Have you tried to substitute that into your original equation?
 

Alan

Member
Jul 21, 2012
58
Hi Alan! :)

You have [tex]c_1\sin(x)+c_2 \cos(x)=ku(x)[/tex]
So [tex]u(x) = \frac 1 k (c_1\sin(x)+c_2 \cos(x))[/tex]
Have you tried to substitute that into your original equation?
And then what do I get? I get the following:

$$\int_{-\pi}^{\pi}\sin(x+y)(c_1\sin y+c_2\cos y)/k dy = c_1\sin x + c_2 \cos x = ((c_1\cos x )\pi+(c_2\sin x)\pi)/k$$

How to procceed from there?

Edit:
We should have: $c_1 \pi/k = c_2 , c_2\pi /k= c_1$, which means that: $(\pi/k)^2 =1$, i.e $k=\pm \pi$.

Thanks I solved it now... :-D