Integral equation with a derivative of the function inside the integral

[damoj]

f(x) = 2\int_{0}^{t} sin(8u)f'(t-u) du + 8sin(8t) , t\geq 0
is this problem solvable? I've never seen an integral equation like this with an f'(t-u)

i tried to solve it us the convolution theorem and laplace transforms but ended up with

s^{2} F(s) + 64F(s)- 16(F(s) - f(0)) =64
and i haven't been given f(0)

 

[obafgkmrns]

This is just a hunch, but if you assume f(t) = A sin(at + theta), you will wind up with an equation in terms of sines and cosines. Smart money says that "a" would be 8. Then it's a matter of solving algebraic equations to get the phase and amplitude. Another hunch: It may be easier to convert everything to exponentials and work it from there.

 

[vela]

f(x) = 2\int_{0}^{t} sin(8u)f'(t-u) du + 8sin(8t) , t\geq 0
is this problem solvable? I've never seen an integral equation like this with an f'(t-u)

I think you want f(t) on the LHS, right?

i tried to solve it us the convolution theorem and laplace transforms but ended up with

s^{2} F(s) + 64F(s)- 16(F(s) - f(0)) =64
and i haven'[/color]t been given f(0)

What do you get if you set t=0 in the original equation?

I didn't bother trying to reproduce what you did, but your approach sounds fine. If you can't figure it out, show your work and we'll be able to provide more guidance.

 

[Ray Vickson]

f(x) = 2\int_{0}^{t} sin(8u)f'(t-u) du + 8sin(8t) , t\geq 0
is this problem solvable? I've never seen an integral equation like this with an f'(t-u)

i tried to solve it us the convolution theorem and laplace transforms but ended up with

s^{2} F(s) + 64F(s)- 16(F(s) - f(0)) =64
and i haven't been given f(0)

You could always set f(0) = c (an unspecified constant) and get a solution in terms of t and c. However, if looks like you CAN get f(0) from the original integral equation, at least if f' does not have a singularity at zero.

RGV

 

[damoj]

thanks for the replies

yeah it was meant to be f(t) on the LHS

Im going to have another look at it and get back to you guys.

 

[damoj]

so i think i solved it, thanks guys.
heres the solution in case anyone has a similar one in the future.

f(t)=2∫t0sin(8u)f′(t−u)du+8sin(8t),t≥0

set t=0
then f(0)= 0

the convolution gives us

(sF(s) - f(0)) \cdot \frac{8}{s^{2}+64}

and the rest is just algebra
f(t) = 64te^{8t}