1. ## Leibniz rule

Hi guys I am currently stuck at a question.

i) Show that if $\displaystyle x(t)$ satisfies the integral equation

$\displaystyle x(t)=a+bt+ \int_0^t (t-s)f(x(s)) ds$

then $\displaystyle x(t)$ is a solution to the initial value problem

$\displaystyle x''(t)=f(x(t))$ for $\displaystyle t>0$, with $\displaystyle x(0)=a , \ x'(0)=b.$

ii) Prove the converse of the result in part (i).

(Hint: You will need to do a change of order of integration in a double integral.)

Part (i) I differentiated $\displaystyle x(t)$ twice and got the result by applying Leibniz rule and the fundamental theorem of calculus.

For part (ii) I don't even know where to start. Where would I be using a double integral? I tried proving it just by integrating $\displaystyle x''(t)$ but I wasn't sure if i could do that.

Any help would be appreciated.

Thanks.

2. Originally Posted by sakodo
For part (ii) I don't even know where to start. Where would I be using a double integral? I tried proving it just by integrating $\displaystyle x''(t)$ but I wasn't sure if i could do that.

Necessarily

$\displaystyle x'(t)=\displaystyle\int_{0}^{t}f(x(s))\;ds +b$

Now, you'll have to integrate again.

3. Originally Posted by sakodo
Hi guys I am currently stuck at a question.

i) Show that if $\displaystyle x(t)$ satisfies the integral equation

$\displaystyle x(t)=a+bt+ \int_0^t (t-s)f(x(s)) ds$

then $\displaystyle x(t)$ is a solution to the initial value problem

$\displaystyle x''(t)=f(x(t))$ for $\displaystyle t>0$, with $\displaystyle x(0)=a , \ x'(0)=b.$

ii) Prove the converse of the result in part (i).

(Hint: You will need to do a change of order of integration in a double integral.)

Part (i) I differentiated $\displaystyle x(t)$ twice and got the result by applying Leibniz rule and the fundamental theorem of calculus.

For part (ii) I don't even know where to start. Where would I be using a double integral? I tried proving it just by integrating $\displaystyle x''(t)$ but I wasn't sure if i could do that.

Any help would be appreciated.

Thanks.
Written in term of Laplace-transform the DE is...

$\displaystyle \displaystyle s^{2}\ X(s) + s\ x(0) - x^{'} (0) = \mathcal{L} \{f[x(t)]\}$

... and from (1), performing the inverse Laplace trasform, You obtain...

$\displaystyle \displaystyle x(t)= (a+b\ t)\ \mathcal {U} (t) + \int_{0}^{t} (t-\tau) f[x(\tau)] \ d \tau$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$