Existence of a unique solution.

Question

Consider the integral equation for some given function $y(t)\in C[0,1]$ and a given constant $\lambda$ with $|\lambda|<1$, $$x(t)-\lambda\int_0^1 e^{t-s}x(s)ds=y(t).$$

Show that there exists a unique solution $x(t)\in C[0,1]$.

Answer 1

First, set $\displaystyle x_0(t) = y(t) + \frac{λ}{1 - λ} \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s$. Because$$\begin{align*} y(t) + λ \int_0^1 \mathrm{e}^{t - s} x_0(s) \,\mathrm{d}s &= y(t) + λ \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s + λ \int_0^1 \mathrm{e}^{t - s} \left(\frac{λ}{1 - λ}\int_0^1 \mathrm{e}^{s- u} y(u) \,\mathrm{d}u\right) \,\mathrm{d}s\\ &= y(t) + λ \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s + \frac{λ^2}{1 - λ} \int_0^1 \int_0^1 \mathrm{e}^{t - u} y(u) \,\mathrm{d}u\mathrm{d}s\\ &= y(t) + λ \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s + \frac{λ^2}{1 - λ} \int_0^1 \mathrm{e}^{t - u} y(u) \,\mathrm{d}u \int_0^1 \mathrm{d}s\\ &= y(t) + λ \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s + \frac{λ^2}{1 - λ} \int_0^1 \mathrm{e}^{t - u} y(u) \,\mathrm{d}u\\ &= y(t) + \frac{λ}{1 - λ} \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s = x_0(t), \end{align*}$$ then $x_0(t)$ is a solution to the given equation.

Now suppose $x(t)$ is a solution to the given equation, then$$ x(t) = y(t) + λ \int_0^1 \mathrm{e}^{t - s} x(s) \,\mathrm{d}s = y(t) + λ \mathrm{e}^t \int_0^1 \mathrm{e}^{-s} x(s) \,\mathrm{d}s. $$ Denote $\displaystyle c = λ \int_0^1 \mathrm{e}^{-s} x(s) \,\mathrm{d}s$, then $x(t) = y(t) + c\mathrm{e}^t$. Therefore,$$\begin{align*} y(t) + c\mathrm{e}^t &= x(t) = y(t) + λ \mathrm{e}^t \int_0^1 \mathrm{e}^{-s} x(s) \,\mathrm{d}s = y(t) + λ \mathrm{e}^t \int_0^1 \mathrm{e}^{-s} (y(s) + c\mathrm{e}^s) \,\mathrm{d}s\\ &= y(t) + λ \mathrm{e}^t \left(\int_0^1 \mathrm{e}^{-s} y(s) \,\mathrm{d}s + c\right), \end{align*}$$ which implies$$ c = λ \left(\int_0^1 \mathrm{e}^{-s} y(s) \,\mathrm{d}s + c\right), $$ i.e.$$ c = \frac{λ}{1 - λ} \int_0^1 \mathrm{e}^{-s} y(s) \,\mathrm{d}s. $$ Hence $x(t) = y(t) + c\mathrm{e}^t \equiv x_0(t)$, so the solution to the given equation is unique.