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Comparison Theorem

Based on the theorem on Extension of solution, we can only deduce that solution of an ODE can extend to its boundary Once the right function \(f(x,y)\) is continous on an open set \(G\).

But it does not tell us the existing interval of independent variable \(x\). In particular, for two dimension problem, \(y'=f(x,y)\in C(\mathbb{R}^2)\), it could either be

\[ x\rightarrow \beta <\infty, y\rightarrow \infty \]

or

\[ x\rightarrow \infty, y\rightarrow \beta <\infty \]

So to better know the existing interval of \(x\), we have to give a more detailed tool for analysis.

Note: The theorem presented below are only appliable to two dimension problem of ODE, cause in high dimension problem, we cannot compare the magnitude of \(\pmb{y}\) but under some metric.

To unify the form, in this part we talk about ODE

\[ \begin{equation} \frac{dy}{dx}=f(x,y),\quad y(x_0)=y_0 \label{eq1} \end{equation} \]

First Comparison Theorem

First Comparison Theorem

Assume \(f(x,y), F(x,y)\in C(G)\), where \(G\) is an open set, and

\[ f(x,y) <F(x,y),\quad \forall(x,y)\in G \]

And we assume \(y=\phi(x)\) and \(y=\varPhi(x)\) on \((a,b)\) are solution to ODE

\[ \begin{cases} y'=f(x,y)\\ y(x_0)=y_0 \end{cases}, \quad \begin{cases} y'=F(x,y)\\ y(x_0)=y_0 \end{cases} \]

respectively, where \((x_0,y_0)\in G\). Then

\[ \phi(x)<\varPhi(x), \quad x_0<x<b \]

Formulate a function

\[ \psi(x)=\varPhi(x)-\phi(x) \]

then prove by contradiction.

Let

\[ \psi(x)=\varPhi(x)-\phi(x) \]

then by initial condition

\[ \psi(x_0)=0,\quad \psi'(x_0)=F(x_0,y_0)-f(x_0,y_0)>0 \]

So we can see that there exists \(\delta>0\), such that

\[ \psi(x)=\varPhi(x)-\phi(x)>0,\quad x_0<x\leq x_0+\delta \]

If proposition

\[ \varPhi(x)>\phi(x),\quad \forall x\in[x_0,b) \]

does not hold, then \(\exists x_0+\delta<x_1<b\) such that

\[ \varPhi(x_1)<\phi(x_1) \]

Let

\[ \alpha=\min\{x\in (x_0+\delta,b): \psi(x)=0\} \]

so we can deduce that

\[ \psi(\alpha)=0,\quad \psi(x)>0,\quad \forall x\in (x_0,\alpha) \]

easy to see that \(\psi(\alpha)\leq 0\). But

\[ \psi'(\alpha)= F(\alpha,\varPhi(\alpha))-f(\alpha,\phi(\alpha))>0 \]

which contradicts!

Upper and Lower Solution

To simplify the notation, we have to introduce upper and lower solution on the right side.

Definition of upper and lower solution

(i) If function \(v(x)\) on \([x_0,b)\) satisfies

\[ \frac{d v(x)}{dx}<f(x,v(x)), \quad v(x_0)\leq y_0 \]

then it is called the lower solution of ODE \(\ref{eq1}\) on the right side.

(ii) If function \(w(x)\) on \([x_0,b)\) satisfies

\[ \frac{d w(x)}{dx}>f(x,v(x)), \quad w(x_0)\leq y_0 \]

then it is called the upper solution of ODE \(\ref{eq1}\) on the right side.

From image point of view, we can say that the tangent of points on the upper solution on the right side must be bigger than the vector field on the same points. They are shown in the following graph.

Theorem for upper and lower solution

Assume \(v(x)\) and \(w(x)\) are defined in the above definition, then

\[ v(x)<\phi(x)<w(x),\quad \forall x\in [x_0,b) \]
  • Define an assistant function
\[ g(x)=\frac{dw(x)}{dx}-f(x,w(x)),\quad F(x,y)=f(x,y)+g(x) \]

which is larger than zero on the line \(y=w(x)\), then can deduce the result just by First Comparison Theorem

\(w(x)\) satisfies ODE

\[ \begin{align*} \frac{dy}{dx}&=F(x,y)\\ &=f(x,y)+g(x)\\ &=f(x,y)+\frac{dw(x)}{dx}-f(x,w(x))\\ &>f(x,y) \end{align*} \]

So by First Comparison Theorem we have \(w(x)>\phi(x)\) on interval \([x_0,b)\).

It is similar to prove \(v(x)<\phi(x)\).

Maximum and Minimum solution

Maximum and Minimum solution

If there are two solution \(\varPsi(x)\), \(\varPhi(x)\) to the ODE \(\ref{eq1}\), such that for all solution \(y(x)\) to the same ODE,

\[ \varPsi(x)\leq y(x)\leq \varPhi(x),\quad \forall x\in[x_0,b) \]

then we call \(\varPsi(x)\), \(\varPhi(x)\) the minimum and maximum solution of ODE on interval \([x_0,b)\) respectively.

Apparently, the minimum and maximum solution are unique. Here we prove their existence.

Theorem of existence for minimum and maximum solution

Assume \(f(x,y)\in C(R)\) where

\[ R=\{(x,y):x_0\leq x\leq x_0+a, |y-y_0|\leq b\} \]

Then initial problem \(\ref{eq1}\) has maximum and minimum solution on \([x_0,x_0+h)\), where

\[ h<\alpha=\min\{a,b/M\}, \quad M=\max_{(x,y)\in R}{|f(x,y)|} \]

Formulate assistant initial problems

\[ \frac{dy}{dx}=f(x,y)+\varepsilon_n, \quad y(x_0)=y_0 \]

where \(\varepsilon_n>0\), \((n=1,2,\cdots)\) which is monotonically decreasing to \(0\), e.g. \(\varepsilon_n=1/n\). Then show its limit of solution sequence is the maxmimum solution.

It is similar to use \(-\varepsilon_n\) to prove the existence of minimum solution.

Formulate assistant initial problems

\[ \begin{align} \frac{dy}{dx}=f(x,y)+\varepsilon_n, \quad y(x_0)=y_0 \label{eq2} \end{align} \]

where \(\varepsilon_n>0\), \((n=1,2,\cdots)\) which is monotonically decreasing to \(0\), e.g. \(\varepsilon_n=1/n\).

By Peano Theorem, ODE \(\ref{eq2}\) has solution on interval \(|x-x_0|\leq \alpha_n\), where

\[ \alpha_n=\min\{a,b/M_n\},\quad M_n=\max_{(x,y)\in D}\{|f(x,y)+\varepsilon_n|\} \]

Notice that

\[ \lim_{n\rightarrow \infty}\alpha_n = \alpha = \min\{a,b/M\} \]

so we choose \(h<\alpha\) such that all ODE \(\ref{eq2}\) can have a solution \(y=\phi_n(x)\) on interval \(|x-x_0|\leq h\).

  • Show that \(\{\phi_n(x)\}\) has a subsequence that converges to the solution of the original ODE \(\ref{eq1}\).

This is similar as we prove in Peano Theorem.

  • Show that the limit function \(y=\varPhi(x)\) is the maximum solution of ODE \(\ref{eq1}\).

By First Comparison Theorem we have \(\phi_n(x)>y(x)\), for all solution \(y(x)\) of the original ODE \(\ref{eq1}\). Then replace \(n\) with \(n_j\) and let \(n\rightarrow \infty\) we have

\[ \varPhi(x)\geq y(x),\quad \forall x\in [x_0,h] \]

which shows that \(\varPhi(x)\) is the maximum solution of ODE \(\ref{eq1}\).

Corollary

Initial problem \(\ref{eq1}\) has unique solution iff its maximum solution equals to minimum solution for all independent variable \(x\) on a given interval.

Second Comparison Theorem

In First Comparison Theorem, we need to find \(F(x,y)\) which is strictly larger than \(f(x,y)\). What if we happen to find \(F(x,y)\) also equals \(f(x,y)\)?

Apparently, we need to strengthen our condition to get a corresponding answer. This gives us the Second Comparison Theorem as below.

Second Comparison Theorem

Assume \(f(x,y), F(x,y)\in C(G)\), where \(G\) is an open set, and

\[ f(x,y) \leq F(x,y),\quad \forall(x,y)\in G \]

And we assume \(y=\phi(x)\) and \(y=\varPhi(x)\) on \((a,b)\) are solution to ODE

\[ \begin{cases} y'=f(x,y)\\ y(x_0)=y_0 \end{cases}, \quad \begin{cases} y'=F(x,y)\\ y(x_0)=y_0 \end{cases} \]

respectively, where \((x_0,y_0)\in G\). And \(y=\phi(x)\) is the minimum solution on the right side. Then

\[ \phi(x)\leq \varPhi(x), \quad x_0<x<b \]

Also by formulating assistant initial problems.

We still consider assistant problems

\[ \frac{dy}{dx}=f(x,y)-\varepsilon_n,\quad y(x_0)=y_0 \]

So by the proof of Theorem for Maximum and Minimum Solution, we can have function subsequence

\[ \phi_{n_j}\rightrightarrows \phi \]

which satisfies

\[ \phi(x)\leq \varPhi(x) \]

Result for estimation

Application for Comparison Theorem

Assume \(f(x,y)\in C(G)\) where

\[ G=\{(x,y): x_0<x<b,-\infty<y<+\infty\} \]

We have \((x_0,y_0)\in G\), denote \([x_0,\alpha)\) is the maximum existence interval on the right side for solution to ODE \(\ref{eq1}\).

(i) If ODE \(\ref{eq1}\) has upper and lower solution \(w(x)\), \(v(x)\) with their public existing interval \([x_0,\beta)\), then \(\beta\leq \alpha\).

(ii) If ODE \(\ref{eq1}\) has upper solution \(w(x)\) (or lower solution \(v(x)\)) with its maximum existing interval \([x_0,\beta)\), and satisfies

\[ \lim_{x\rightarrow \beta^-}w(x)=-\infty (\text{or } \lim_{x\rightarrow \beta^-}v(x)=+\infty ) \]

then \(\alpha\leq \beta\).