# A Fredholm integral equation

• Apr 18th 2011, 05:13 AM
mathlovely
A Fredholm integral equation
solve fredholm integral equation
integral from 0 to 1 of exp(x+t)u(t)dt =x
(Thinking)
• Apr 18th 2011, 05:47 AM
Ackbeet
I don't know whether your x is a function of t which you're trying to find, or whether x is a variable.

Case 1: x is a variable independent of t. Then exp(x) comes out of the integral. If you let C = int_0^1 u(t) dt, then you must solve

C exp(x) = x, or

C = x exp(-x).

Use Lambert's W function to solve.

Case 2: x is a function of t. Then the LHS is a constant (because you've integrated out the only independent variable in the LHS), which implies that the RHS is a constant, and hence x is a constant. Again, pull exp(x) out of the integral, and solve as before.
• Apr 18th 2011, 11:30 PM
mathlovely
thank u toooooooo mutch BUT
(Crying)(Crying)
Quote:

Originally Posted by mathlovely
solve fredholm integral equation
integral from 0 to 1 of exp(x+t)u(t)dt =x
(Thinking)

THIS FIE with first kind and homogenous equation
Is there any solution for it
• Apr 19th 2011, 03:25 AM
Ackbeet
I have an error in my definition for $C.$ It should be

http://quicklatex.com/cache3/ql_2312...9c87e83_l3.png

Was there a problem with my solution? Your second post is enigmatic; it also is bordering on bumping, which is forbidden on MHF.
• Apr 19th 2011, 05:28 AM
mathlovely
your C is ok but ????
I solve it like this
C=integral from 0 to 1 of exp (t)*( C *exp(t)) after I subtitute u(t) by C*exp(t)
but I DON'T know if
there is any solution
• Apr 19th 2011, 05:59 AM
Ackbeet
Your posts are still unintelligible to me. Let's back up a bit. For what are you trying to solve? Is the following your integral equation?

http://quicklatex.com/cache3/ql_fd70...a200dc3_l3.png

If so, then we note that u = u(t), and the independent variable t is integrated out on the LHS. If x is dependent on t, then the LHS is just a constant, right? You'd have the following:

http://quicklatex.com/cache3/ql_096e...4bdc9b9_l3.png

The LHS is a constant because the only independent variable in sight has been integrated out. Therefore, x(t) is constant, and you can pull out the exponential the same way as in the second case, which follows.

Otherwise, if x is independent of t, you can factor out as follows:

http://quicklatex.com/cache3/ql_7138...6c313da_l3.png

I've been using QuickLaTeX to do the LaTeX rendering here, which you can embed as images. It's not bad, actually.

Does this make any sense?
• Apr 19th 2011, 10:18 AM
ivalmian
I think he dislikes the fact it's a transcendental equation... which is something you can't really deal with, unless LHC in Ackbeet's solution is a particularly nice constant (eg 0). In that case x is either 0 or some some sort of complex infinity. If LHC is non 0 but small and you want a small x solution you can approximate the RHS as

http://quicklatex.com/cache3/ql_76c5...8cfb1dc_l3.png

But generally there are no closed form exact solutions (ie something that looks like x = blah).
• Apr 19th 2011, 10:49 AM
Ackbeet
Quote:

Originally Posted by ivalmian
I think he dislikes the fact it's a transcendental equation... which is something you can't really deal with, unless LHC in Ackbeet's solution is a particularly nice constant (eg 0). In that case x is either 0 or some some sort of complex infinity. If LHC is non 0 but small and you want a small x solution you can approximate the RHS as

http://quicklatex.com/cache3/ql_76c5...8cfb1dc_l3.png

But generally there are no closed form exact solutions (ie something that looks like x = blah).

Other than the Lambert W function. I suppose it's debatable whether that's really a closed-form solution, though.