Find the function \(f(x)\) that satisfies:

\(f(x) = 1 - \int_0^x [f(t)]^2 \ dt\),

for all \(x \geq 0.\)

Thanks!

jsaddern Dec 24, 2021

#2**+1 **

Answer #1 is not correct.

Best is to differentiate wrt x, turning the integral equation into a differential equation.

\(\displaystyle \frac{d}{dx}f(x)= -\{1.[f(x)]^{2} - 0.[f(0)]^{2} +\int^{x}_{0}\frac{\partial}{\partial x}[f(t)]^{2}dt , \)

\(\displaystyle \frac{d}{dx}f(x)= - [f(x)]^{2}.\)

Now integrate that.

There will be a constant of integration, so substitute back into the original equation to determine its value.

Tiggsy Dec 25, 2021

#4**+1 **

Hi Melody, the RHS comes from a standard result usually referred to as ' differentiation under the integral sign '.

For an indefinite integral the result takes the form

\(\displaystyle \frac{\partial}{\partial \alpha}\int f(\alpha,x)dx = \int\frac{\partial}{\partial \alpha}f(\alpha,x)dx .\)

It says that the order of the integral and the differentiation thingy are interchangeable, alpha is a parameter.

That's not always true, but it will be true for all of the functions you or I are ever going to meet.

As a simple example, since

\(\displaystyle \int e^{\alpha x} dx = \frac{1}{\alpha}e^{\alpha x}, \text{then}\\ \frac{\partial}{\partial \alpha} \int e^{\alpha x}dx=\int\frac{\partial}{\partial \alpha}e^{\alpha x} dx= \int xe^{\alpha x}dx =\frac{\partial}{\partial \alpha}(\frac{1}{\alpha}e^{\alpha x})= (\frac{x}{\alpha}-\frac{1}{\alpha^{2}})e^{\alpha x}.\)

(You can check the result by integrating by parts).

The result also works for a definite integral, providing that the limits of integration are constants.

If one or both of the limits of integration are functions of the parameter then the result takes the form

\(\displaystyle \frac{\partial}{\partial \alpha} \int^{b}_{a}f(\alpha,x)dx =\int^{b}_{a}\frac{\partial}{\partial \alpha}f(\alpha,x)dx+ \frac{db}{d \alpha}f(\alpha,b)-\frac{da}{d \alpha}f(\alpha,a).\)

For the example in the question, the parameter is x and the integration is wrt t, so we have (for the integral),

\(\displaystyle \frac{\partial}{\partial x}\int^{x}_{0}[f(t)]^{2}dt= \int^{x}_{0}\frac{\partial}{\partial x}[f(t)]^{2} dt +\frac{d}{d x}(x).[f(x)]^{2}-\frac{d}{dx}(0).[f(0)]^{2} \\ =[f(x)]^{2}.\)

Differentiate a function of t partially wrt x and you get zero, differentiate zero wrt x and you get zero so the first and third terms disappear.

***************************************************************************************************************************************

I hope that Christmas has gone well for you, and my very best wishes for the new year.

Tiggsy

Guest Dec 30, 2021

#6**0 **

I followed what you did so far Tiggsy BUT I still don't know how to finish it.

Maybe I have brain freeze.

\(f(x) = 1 - \int_0^x [f(t)]^2 \ dt\\ \)

\(\displaystyle \frac{d}{dx}f(x)= - [f(x)]^{2}\\ \int\displaystyle \frac{d}{dx}f(x)dx= -\int [f(x)]^{2}dx\\ f(x)= -\displaystyle \int [f(x)]^{2}dx\\ \)

???

Melody Jan 2, 2022

#7**0 **

Hi Melody

Second line down, let y = f(x) and you have the equation

\(\displaystyle \frac{dy}{dx}=-y^{2}, \text{ from which } \int \frac{dy}{y^{2}}=-\int dx. \)

Solve that for y ( = f(x) ) and substitute the result back into the original integral equation to determine the value of the constant of integration.

You should finish up with the result f(x) = 1/(x + 1).

Tiggsy

Guest Jan 2, 2022