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Extension of Solution

解的延拓 | Extension of Solution

For Cauchy problem

\[ \begin{equation} \frac{dy}{dx} = f(x, y), \quad y(x_0) = y_0. \label{eq-cauchy} \end{equation} \]

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 \(\ref{eq-cauchy}\) can extend to its boundary.

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 \(\Gamma\) can extend to \(G\) \ \(G_1\).

We only consider positive extension, i.e. \(x\geq x_0\).

Consider closed region \(G_1\subset G\) that has point \((x_0,y_0)\) with the solution denoted by \(\varphi(x)\) 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, \(\exists \delta_0>0\)., 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 \(\delta_0'\)

For any closed region \(G_1\subset G\), Denote

\[ M=\max_{(x,y)\in G_1}|f(x_y)|+1 < \infty, \quad R_{\delta_0}(x',y')=\{(x,y): |x-x'|\leq \delta_0, |y-y'|\leq \delta_0\} \]

where \((x',y')\in G_1\).

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 \(\varphi_0(x)\) on interval \([x_0-\delta',x_0+\delta']\), where \(\delta'=\min{\delta_0, \delta_0/M}\). If there exists point on the curve \((x,\varphi_0(x)) \in G\)\\(G_1\), then we prove it.

If not, then the furthest point on the right \((x_0+\delta', \varphi_0(x_0+\delta'))\) must be in \(G_1\). So consider cauchy problem with initial condition \(y(x_0+\delta') = \varphi_0(x_0+\delta')\) in region \(R_{\delta_0} (x_0+\delta',\varphi_0(x_0+\delta'))\), according to Peano Theorem, we can get another solution \(\varphi_1(x)\) on interval \(x_0,x_0+2\delta'\).

Repeat the above procedure, we can say the interval can be extended to \([x_0, x_0+n\delta']\). 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 \(n\delta'>D_1\) and in the same time \(n\delta'<D\).

And we are done.