# How we can show integral equation has a unique solution and how we can solve it?

1,902

We use the Banach fixed-point 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} \left|x(s) - y(s)\right| \mathrm{d}s \\ &\le 0.1\left\|x-y\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}}{1-0.1}\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, 2020

• 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?

$$x(t) + 0.1 \int_0^1 e^{ts}x(s) ds = 1, \quad (0 \leqslant t \leqslant 1).$$

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, $$(1+s)^t = \sum_{i=0}^t {t \choose i}s^i = 1 +ts+\frac{t(t-1)}{2!}s^2+\cdots +s^t \tag{*}$$ 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, $$(1+s)^t = \sum_{i=0}^t {t \choose i}s^i = 1 +ts+\frac{t(t-1)}{2!}s^2+\cdots +s^t \tag{*}$$ by substituting $(*)$ in the integral we can estimate the solution, but i can not go any further.
• Robin Goodfellow over 7 years
Wonderful. 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 years
What a beautiful application this method has! Thanks for the solution. The $e$ is dropped in the second inequality.
• user14717 over 7 years
The $e$ is dropped (replaced by 1) because we wish to maximize $e^{t}$ over $t \in [0,1]$.
• Wouldn'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 $0.1.$