How we can show integral equation has a unique solution and how we can solve it?
We use the Banach fixedpoint theorem. First we recast the equation in a form that allows its application: \begin{align} x(t) = 0.1\int_{0}^{1} e^{ts}x(s) \, \mathrm{d}s + 1 =: Bx(t) + 1 =: Ax(t) \end{align} I'll assume $(C[0,1], \left\\cdot\right\_{\infty})$ is the Banach space of interest, but ostensibly you could use others ($L_1$ might be better). Now we attempt to prove $A$ is contractive on this space: \begin{align} \left\Ax Ay \right\_{\infty} &= 0.1\sup_{t \in [0,1]}\left \int_{0}^{1} e^{st}(x(s)  y(s)) \mathrm{d}s \right \\ &\le 0.1\sup_{t \in [0,1]}e^{t} \int_{0}^{1} \leftx(s)  y(s)\right \mathrm{d}s \\ &\le 0.1\left\xy\right\_{\infty} \end{align} So $A$ is contractive with $\left\B\right\_{\infty} = 0.1$ and hence by the Banach fixed point theorem this equation has a unique solution.
Using the sequence $x_{n+1}:= Ax_{n}$, we have the estimate \begin{align} \left\x_{n}  x\right\ \le \frac{0.1^{n}}{10.1}\left\x_{1}  x_{0}\right\_{\infty}. \end{align} Take $x_{0} = 0$, so that $x_{1} = 1$. Then \begin{align} x_{2}(t) = 0.1\int_{0}^{1}e^{st} \, \mathrm{d}s +1 = \frac{0.1}{t}(e^{t} 1) + 1 \end{align} This has the estimate \begin{align} \left\x_{2}  x\right\_{\infty} \le 0.011111 \end{align} I'll leave it to you to calculate $x_{3}$.
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Qaher
Updated on January 09, 2020Comments

Qaher over 2 years
My question is on the title, e.g. how we can show that this integral equation has a unique solution and how we can estimate what the approximate answer near $0.1$ can be?
$$ x(t) + 0.1 \int_0^1 e^{ts}x(s) ds = 1, \quad (0 \leqslant t \leqslant 1). $$
Thanks in advance.
What I've done so far:By using the Binomial of $e^{ts}=(e^{t})^s$, we have:
$$ (e^s)^t = (\sum_{n=1}^{\infty}(\frac{s^n}{n!}))^t \approx (1+s)^t $$ then, $$ (1+s)^t = \sum_{i=0}^t {t \choose i}s^i = 1 +ts+\frac{t(t1)}{2!}s^2+\cdots +s^t \tag{$*$} $$ by substituting $(*)$ in the integral we can estimate the solution, but i can not go any further.

Qaher over 7 years@RobinGoodfellow What I've done so far: By using the Binomial of $e^{ts}=(e^{t})^s$, we have: $$ (e^s)^s = (\sum_{n=1}^{\infty}(\frac{s^n}{n!}))^t \approx (1+s)^t $$ then, $$ (1+s)^t = \sum_{i=0}^t {t \choose i}s^i = 1 +ts+\frac{t(t1)}{2!}s^2+\cdots +s^t \tag{$*$} $$ by substituting $(*)$ in the integral we can estimate the solution, but i can not go any further.

Robin Goodfellow over 7 yearsWonderful. Please add that to the original post; you are supposed to demonstrate effort on your part to solve the problem before coming to us.


Qaher over 7 yearsWhat a beautiful application this method has! Thanks for the solution. The $e$ is dropped in the second inequality.

user14717 over 7 yearsThe $e$ is dropped (replaced by 1) because we wish to maximize $e^{t}$ over $t \in [0,1]$.

Merkh about 6 yearsWouldn't the maximum of $e^t$ over $t \in [0,1]$ be $e^{1} = e?$ Not that it changes the proof significantly, just the value of $\lambda \approx 0.27$ instead of $0.1.$