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

Question

Consider the integral equation: \begin{align*} x(t) - \lambda\displaystyle\int_0^1 e^{t-s}x(s)ds = \Gamma(t), \end{align*} where $\Gamma(t) \in C([0,1])$ is a given function, and $\lambda$ is a real number satisfying $|\lambda|<1$. Show that there exists a unique solution $x(t) \in C([0,1])$ to the integral equation.

My idea is that consider $(C([0,1]),\rho)$, where $\rho(x,y) = \max\limits_{a\leq t\leq b}|x(t)-y(t)|$, and for each $x \in C([0,1])$, define $g(x)(t) = \Gamma(t)+\lambda\displaystyle\int_0^1 e^{t-s}x(s)ds$. I am able to prove that $g$ is self-mapping. Then I need to prove that $g$ is contraction mapping. For $x,y \in C([0,1])$,

$|g(x)-g(y)|(t) = \left|\lambda\displaystyle\int_0^1 e^{t-s}(x(s)-y(s))ds\right|\leq |\lambda|\displaystyle\int_0^1 e^{t-s}ds.\rho(x,y)< (-e^{t-1}+e^t)\rho(x,y).$

However, when I take the maximum, I noticed that $\max\limits_{0\leq t \leq 1} (-e^{t-1}+e^t) = e-1>1$, so that I cannot prove that this is contraction mapping. Can somebody please help me to fix this?

Answer 1

Let $y(t)=e^{-t}x(t).$ The equation takes the form $$y(t)-\lambda \int\limits_0^1y(s)\,ds =e^{-t}\Gamma(t)\quad (*)$$ Now we can apply the contraction mapping theorem for $|\lambda|<1$ to the operator $$(Ty)(t)=e^{-t}\Gamma(t)+\lambda \int\limits_0^1y(s)\,ds$$ where $y\in C[0,1].$

Remark The equation $(*)$ has a very special form $$y(t)=\lambda\int\limits_0^1y(s)\,ds +f(t)\quad (**)$$ Hence the solution is equal $y(t)=c+f(t)$ for a constant $c.$ Substituting this in $(**)$ gives $$c+f(t)=\lambda c +\int\limits_0^1f(t)\,dt+f(t)$$ Therefore $$c={1\over 1-\lambda}\int\limits_0^1f(t)\,dt,\quad \lambda\neq 1$$ For $\lambda=1$ there is no solution if $\int\limits_0^1f(t)\,dt\neq 0.$ If $\int\limits_0^1f(t)\,dt= 0,$ the solutions are equal $c+f(t)$ for any constant $c.$

Answer 2

The equation indeed has a unique solution for any $\lambda \neq 1$ and Contraction Mapping Theorem is not the right approach.

Pulling out $e^{t}$ from the integral we see that the equation gives $x(t)=\Gamma (t)+ce^{t}$ for some constant $c$. To evaluate $c$ multiply this equation by $e^{-t}$ and integrate. You will get $c=\frac 1 {1-\lambda} \int_0^{1}e^{-s}\Gamma (s) ds$. Thus, this is the only possible solution. Now go back to the original equation and check that this is indeed a solution.