Extension of Solution

解的延拓 | Extension of Solution

For Cauchy problem

$$\begin{array}{}\text{(1)}& \frac{dy}{dx}=f(x,y),y(x_0)=y_0.\end{array}$$

it is clear that we are satisfied with the result of intervals in the previous chapter, especially when we have to shrink the interval of solution using Contraction Mapping Method.

So a naive idea is, can we use the Peano/Picard theorem repeatedly to extend the interval of parameter $t$? If so, in what cases can we extend it and to where can we extend?

Here comes the following theorem.

解的延拓定理 | Theorem of extension of solution

Assume $f(x,y)\in C(G)$, where $G$ is an open set(region). Then the solution of Cauchy problem $\text{1}$ can extend to its boundary.

HintsProof

The proof is equivalent to prove that, for each closed set $G_1\subset G$ and $(x_0,y_0)\in G_1$, the solution $\mathrm{\Gamma }$ can extend to $G$ \ $G_1$.

We only consider positive extension, i.e. $x\ge x_0$.

Consider closed region $G_1\subset G$ that has point $(x_0,y_0)$ with the solution denoted by $\phi (x)$ that satisfies initial condition $y(x_0)=y_0$.

• Use the property of Open Set.

Because $G$ is an open set, the distance between $G$ and $G_1$ can not be zero, that is, $\mathrm{\exists }{\delta }_0>0$., s.t.

$$\{(x,y):|x-a|<{\delta }_0,|y-b|<{\delta }_0,(a,b)\in \partial G_1\}\subset G$$

Here ${\delta }_0$ is a constaint condition that guarantees the extended interval of solution will not exceed the boundary of $G_1$.

• Using ${\delta }_0$ to generate extended intervals with length ${\delta }^{\prime }$

For any closed region $G_1\subset G$, Denote

$$M=\underset{(x,y)\in G_1}{max}|f(x_y)|+1<\mathrm{\infty },R_{{\delta }_0}(x^{\prime },y^{\prime })=\{(x,y):|x-x^{\prime }|\le {\delta }_0,|y-y^{\prime }|\le {\delta }_0\}$$

where $(x^{\prime },y^{\prime })\in G_1$.

For Cauchy problem with initial condition $y(x_0)=y_0$ in region $R_{{\delta }_0}(x_0,y_0)$, according to Peano Theorem, we can have a solution ${\phi }_0(x)$ on interval $[x_0-{\delta }^{\prime },x_0+{\delta }^{\prime }]$, where ${\delta }^{\prime }=min\{{\delta }_0,{\delta }_0/M\}$. If there exists point on the curve $(x,{\phi }_0(x))\in G$\$G_1$, then we prove it.

If not, then the furthest point on the right $(x_0+{\delta }^{\prime },{\phi }_0(x_0+{\delta }^{\prime }))$ must be in $G_1$. So consider cauchy problem with initial condition $y(x_0+{\delta }^{\prime })={\phi }_0(x_0+{\delta }^{\prime })$ in region $R_{{\delta }_0}(x_0+{\delta }^{\prime },{\phi }_0(x_0+{\delta }^{\prime }))$, according to Peano Theorem, we can get another solution ${\phi }_1(x)$ on interval $x_0,x_0+2{\delta }^{\prime }$.

Repeat the above procedure, we can say the interval can be extended to $[x_0,x_0+n{\delta }^{\prime }]$. If we denote the distance between $(x_0,y_0)$ and $\partial G_1$ as $D_1$, distance between $(x_0,y_0)$ and $\partial G$ as $D$, then choose $n$ such that $n{\delta }^{\prime }>D_1$ and in the same time $n{\delta }^{\prime }<D$.

And we are done.