Existence and Uniqueness Theorem
Sufficient condition for existence of solution #
Consider a DE in two var: $$ \frac{d y}{d x}=f(x,y) $$
- Let f
- be continuous on D
- satisfies Lipschitz condition (wrt y) in D
- Let $(x_{0},y_{0})$ be an interior point in D and $a,b$ such that rectangle $|x-x_{0}|\le a,|y-y_{0}|\le b$ lies in D
- Let $M=\max(f(x,y))$ and $h=\min\left(a, \frac{b}{M}\right)$
If these are met, there exists a unique solution $\phi$ of the IVP on $|x-x_{0}|\le h$
uniqueness #
We assume there are $\phi(x),\psi(x)$ satisfying the IVP, and prove that they must be equal
we will bisect the interval and first prove for the right half $x_{0}\le x\le x_{0}+h$
$$ \begin{align*} |\phi(x)-\psi(x)|&\le \left|\int_{x_{0}}^{x}f[t,\phi(t)]-f[t,\psi(t)]dt\right| \\ &\le \int_{x_{0}}^{x}|f[t,\phi(t)]-f[t,\psi(t)]|dt\\ \\\ \text{because it follows Lipschitz condition}\\ f[t,\phi(t)]-f[t,\psi(t)]&\le K |\phi(t)-\psi(t)|\\ \\ |\phi(x)-\psi(x)|&\le K\int_{x_{0}}^{x}|\phi(t)-\psi(t)|dt\\ \\ \text{Let }U(x)&= \int_{x_{0}}^{x}|\phi(t)-\psi(t)|dt\\ \text{So, }U’(x)&\le kU(x)\\ U’(x)-kU(x)&\le 0\\ (U’(x)-kU(x))e^{-K(x-x_{0})}&\le 0\\ \frac{d}{dx}(U(x)e^{-K(x-x_{0})})\le 0\\ \text{By integrating, we get}\\ U(x)e^{-K(x-x_{0})}-U(x_{0})e^{-K(x_{0}-x_{0})}&\le 0\\ U(x_{0})=0, \text{so }U(x)e^{-K(x-x_{0})}&\le 0\\ U(x)&\le 0\\ \\ \text{But we know }U(x)&\ge 0\\ \therefore U(x)=0\\ \end{align*} $$
Since integrand is continuous, the only possible way for that to happen is $\phi(x)=\psi(x)$ for $x\ge x_{0}$
Similarly, we can prove for $x_{0}-h\le x\le x_{0}$
Picards iteration convergence #
Iteration on g, let us name that $T_{g}(x)=y_{0}+\int_{x_{0}}^{x}f[t,g(t)]dt$
Bounding iterations #
In the interval $|x-x_{0}|\le h$, if $|g(x)-y_{0}|\le b$, then $|T_{g}(x)-y_{0}|\le b$
$h\le \frac{b}{M}$ will be used here
$$ \begin{align*} |T_{g}(x)-y_{0}|&\le |\int_{x_{0}}^{x}f[t,g(t)]dt|\\ &\le \int_{x_{0}}^{x}|f[t,g(t)]|dt\\ &\le M\int_{x_{0}}^{x}dt\\ &\le M|x-x_{0}|\\ &\le Mh\\ &\le b \end{align*} $$
Convergence to solution #
$\phi_{n}$ converges to solution $\phi(x)$
$$ \begin{align*} M_{0}&= \max_{|t-x_{0}|\le h}|f(t,y_{0})|\\ \phi_{n}(x)&= y_{0}+\int_{x_{0}}^{x}f[t,\phi_{n-1}(t)]dt\\ \\ \text{The following estimate holds:}\\ |\phi_{n+1}(x)-\phi_{n}(x)|&\le M_{0}K^{n} \frac{|x-x_{0}|^{n+1}}{(n+1)!}\\ \\ |\phi(x)-\phi_{n}(x)|&\le M_{0}K^{n} \frac{|x-x_{0}|^{n+1}}{(n+1)!}e^{K|x-x_{0}|} \end{align*} $$
We can prove the first estimate using induction.