Solving Integral Equation: sin(x)+∫_0^π sin(x-t)y(t)dt

[sara_87]

Homework Statement

solve the following integral equation

y(x)=sin(x)+\int_0^\pi sin(x-t)y(t)dt

Homework Equations

The Attempt at a Solution

If the limits of integration were from 0 to x then i could solve this using Laplace transfroms because the definition of the convolution is:
integral g(x-t)y(t) where the limits are from 0 to x.
But here, we have the limits from 0 to pi.
Does anyone have any ideas.
Thank you

 

[Dick]

How about if you differentiate with respect to x twice. What happens if you add that to the original equation?

 

[sara_87]

after I differentiate once, I get:

y'(x)=cos(x)(1+\int^\pi_0 (cos(t)-sin(t))y(t)dt)

differenitiating agin gives

y''(x)=-sin(x)(1+\int^\pi_0(cos(t)-sin(t))y(t)dt)

adding to the original gives

y''+y=\int^\pi_0 (sin(x)-cos(x))sin(t)y(t)dt

is that right?
What shall i do now?

 

[hisham.i]

y(x)=sin(x)+\int_0^\pi sin(x-t)y(t)dt
let's try to solve this integral by integration by parts twice:
take,
\int_0^pi \sin(x-t)y(t)dt
let u=y(t) then du=y(t)dt
dv=sin(x-t)dt then v=cos(x-t)
apply u*v -\int vdu
we will obtain two factors A+\int cos(x-t)y(t)dt
another time integrate the second integral by parts we will obtain the same integral as above we will get 2*the first integral and solve for it!

 

[hisham.i]

the first integral i wrote above i mean pi and not from 0--->p, writing mistake!

 

[sara_87]

I think you made a mistake,

if u=y(t) then u'=y'(t) because we are integrating with respect to t.
so in the intgrand we will have y'(t) and if we integrate by parts again, we will have y''(t) in the integrand.

 

[hisham.i]

yes you'r right i have made a mistake.
take u=y(t) du=y'(t)dt
dv=sin(x-t)dt v=-cos(x-t)
then after calculating it, calculate the second integral by the same method but take
u=cos(x-t) du=sin(x-t)dt
dv=y'(t)dt v=y(t)

 

[sara_87]

I don't think this is the right approach. after the integration by parts, we get that the integral =0 leaving us with
y(x) = sin(x)

is this right?

 

[Dick]

after I differentiate once, I get:

y'(x)=cos(x)(1+\int^\pi_0 (cos(t)-sin(t))y(t)dt)

differenitiating agin gives

y''(x)=-sin(x)(1+\int^\pi_0(cos(t)-sin(t))y(t)dt)

adding to the original gives

y''+y=\int^\pi_0 (sin(x)-cos(x))sin(t)y(t)dt

is that right?
What shall i do now?

I get that y''(x)=(-sin(x))+integral(-sin(x-t)*y(t)). So y(x)+y''(x)=0. I don't see how you are factoring a cos(x) out of that first derivative.

 

[vela]

after I differentiate once, I get:

y'(x)=cos(x)(1+\int^\pi_0 (cos(t)-sin(t))y(t)dt)

How did you pull the cos(x) out of the integral? You just want to differentiate sin(x-t) wrt x and leave it all inside the integral, i.e.

\int sin(x-t)y(t)dt \Rightarrow \int cos(x-t)y(t)dt

 

[sara_87]

thank you both for pointing out my mistake.
after differentiating twice and adding y and y'' I get
y''(x)+y(x)=0
so the solution is
y(x)=Acos(x)+Bsin(x)

to find A and B I need initial conditions.
How can I get the initial conditions?

 

[Dick]

thank you both for pointing out my mistake.
after differentiating twice and adding y and y'' I get
y''(x)+y(x)=0
so the solution is
y(x)=Acos(x)+Bsin(x)

to find A and B I need initial conditions.
How can I get the initial conditions?

Plug your y(x) into the integral equation. Now you can explicitly do the integral. If you move all of the terms to one side you will get something like c*sin(x)+d*cos(x)=0 (where c and d are linear functions of A and B). Now what?

 

[sara_87]

thanks,
i did this just now and got that:

c=A-1-pi*B/2
d=B+pi*A/2

A=4/(4+pi^2)
and
B=-2Pi/(4+pi^2)

is this right?

 

[Dick]

Right. You set c=d=0. Those look like the same values I got for A and B.

 

[sara_87]

wohoo, thanks.
Im doing another question and it is the same as this one but instead of sin(x-t)y(t) in the integrand, i have (x-t)y(t) in the integrand.
Do I also differentiate twice?

 

[Dick]

wohoo, thanks.
Im doing another question and it is the same as this one but instead of sin(x-t)y(t) in the integrand, i have (x-t)y(t) in the integrand.
Do I also differentiate twice?

It's certainly worth a try. What do you think will happen?

 

[hisham.i]

Why integration by parts is wrong in such a case?

 

[sara_87]

we have

y(x)=1+\int_0^1 (x-t)y(t)dt

y'(x) = \int_0^1 y(t)dt

y''=0

hmmm...do i add them?

 

[Dick]

we have

y(x)=1+\int_0^1 (x-t)y(t)dt

y'(x) = \int_0^1 y(t)dt

y''=0

hmmm...do i add them?

Well, you don't need to. y''=0 is already an ODE you can easily solve.

 

[Dick]

Why integration by parts is wrong in such a case?

For one thing, you did it wrong. If u=y then du=y'(x)dx. Now there's a y'(x) in your integral. The next time you integrate by parts you get y''(x). You aren't going to get back where you started. You are just making things more complicated.

 

[sara_87]

when I solve y''=0 and then substitute it into the original equation I get:

x(A*1/2+B)+(A*1/3+B*3/2-1)=0

how will this help to find A and B?

 

[Dick]

when I solve y''=0 and then substitute it into the original equation I get:

x(A*1/2+B)+(A*1/3+B*3/2-1)=0

how will this help to find A and B?

The same as the other one. If c*x+d=0 for all x, then c=0 and d=0.

 

[sara_87]

wohoo, thank you very much.
I get A=-12/5 and B=6/5

just one final question,
If now instead of (x-t)y(t) in the integrand, I have ln(x/t)y(t), do i still use differentiation?

I would then get:

y'(x)=\frac{1}{x}\int_0^1 y(t)dt
y''(x)=\frac{-1}{x^2}\int_0^1 y(t)dt

 

[Dick]

Step back and think about it. The whole game here is to eliminate the integral so you just have an ODE in y(x) left. Can you think of a way to use those two equations to eliminate the integral? I'll bet you can.

 

[sara_87]

ya, I can

I get:
y'' + 1/x y'=0
solving gives:
y(x) = Aln(x)+B
is this right?
then i do the same as before to find A and B, right?

 

[Dick]

ya, I can

I get:
y'' + 1/x y'=0
solving gives:
y(x) = Aln(x)+B
is this right?
then i do the same as before to find A and B, right?

I think you've got it.

 

[sara_87]

woohooo
Thanks ;)

 

[andylu224]

when I solve y''=0 and then substitute it into the original equation I get:

x(A*1/2+B)+(A*1/3+B*3/2-1)=0

how will this help to find A and B?

I'm just wondering how did get these combinations for the constants after you've plugged your solution into the integration equation.

 

[Dick]

I'm just wondering how did get these combinations for the constants after you've plugged your solution into the integration equation.

You calculate the integral.

 

[andylu224]

i know you calculate the integral to obtain the constants, but I am just at a loss in doing what you both did.

the solution to the ODE would be y = Ax + B. when inserted into the original equation,

Ax + B = 1 + (int 0->1) (x-t)y(t)dt

(A - (int 0->1)y(t)dt)x + (B - 1 + (int 0->1) ty(t)dt) = 0

I'm just wondering how you evaluated those integral terms into the constant terms.
I had a similar query about the question done before as well.

 

[Dick]

It's not Ax+B, it's f(x)=Aln(x)+B. You want to integrate ln(x/t)*f(t)=ln(x/t)*(Aln(t)+B)dt from t from 0 to 1. That breaks up into integrating ln(t)^2 and ln(t). Do them by integrating by parts. If you are having trouble with the individual integrations, it might be best to post a separate thread with your specific integration questions. This doesn't have much to do with integral equations.

 

[╔(σ_σ)╝]

Have you tried the laplace transform ... it applies very nicely here .:)

This equation is the Volterra integral equations .

 

[sara_87]

How does the Laplace trandform apply here?

Which integral equaiton is Volterra?
The ones are posted are Fredholm.