# IVP using Leibniz formula

• May 10th 2011, 03:22 PM
morganfor
IVP using Leibniz formula
How does one formulate this integral equation as an initial value problem?

u(t)=1 + integral from 0 to t (s ln (s/t) u(s) ds)

we're supposed to use Leibniz formula but I'm not sure how to apply it. Thanks!
• May 10th 2011, 07:23 PM
Ackbeet

$u(t)=1+\int_{0}^{t}s\ln\left(\frac{s}{t}\right)u(s )\,ds,$

and you need to convert to an IVP using the Leibniz formula, which is, in this case,

$\frac{d}{dt}\int_{a(t)}^{b(t)}f(s,t)\,ds=\frac{db( t)}{dt}\,f(b(t),t)-\frac{da(t)}{dt}\,f(a(t),t)+\int_{a(t)}^{b(t)} \frac{\partial }{\partial t}\,f(s,t)\,ds.$

So, why not just differentiate the original integral equation with respect to t, and use the Leibniz formula on the integral. What do you get?
• May 10th 2011, 11:10 PM
Drexel28
Quote:

Originally Posted by Ackbeet

$u(t)=1+\int_{0}^{t}s\ln\left(\frac{s}{t}\right)u(s )\,ds,$

and you need to convert to an IVP using the Leibniz formula, which is, in this case,

$\frac{d}{dt}\int_{a(t)}^{b(t)}f(s,t)\,ds=\frac{db( t)}{dt}\,f(b(t),t)-\frac{da(t)}{dt}\,f(a(t),t)+\int_{a(t)}^{b(t)} \frac{\partial }{\partial t}\,f(s,t)\,ds.$

So, why not just differentiate the original integral equation with respect to t, and use the Leibniz formula on the integral. What do you get?

Why not just note that $\displaystyle \int _0^t s\log\left(\frac{s}{t}\right)u(s)\text{ }ds=\int_0^t s\log(s)u(s)\text{ }ds-\log(t)\int_0^t su(s)\text{ }ds$ from which all you need is the product rule and the FTC?
• May 11th 2011, 02:41 AM
Ackbeet
Quote:

Originally Posted by Drexel28
Why not just note that $\displaystyle \int _0^t s\log\left(\frac{s}{t}\right)u(s)\text{ }ds=\int_0^t s\log(s)u(s)\text{ }ds-\log(t)\int_0^t su(s)\text{ }ds$ from which all you need is the product rule and the FTC?

True enough, but if the OP'er must use the Leibniz formula because of class restraints, then we're back to square one.