Suppose that $f$ satisfies the Lipschitz condition.

For given initial values $y_0, z_0\in \mathbb{R}$, we consider the IVPs: $$y'=f(t,y), \ y(a)=y_0 \\ z'=f(t,z), \ z(a)=z_0$$ with $a\leq t \leq b$.


1) Show that the problem has unique solution, i.e. if $y_0=z_0$ then $y(t)=z(t)$ for all $t$. $$\max_{1\leq t\leq b}|y(t)-z(t)|\leq c|y_0-z_0|$$

2) Let $\{y^n\}$ and $\{z^n\}$ be aproximation of the problem that we get by the mean value method, let $$\epsilon^n:=y^n-z^n \\ \epsilon^{n+1/2}:=y^{n+1/2}-z^{n+1/2} \\ y^{n+1/2}=\frac{1}{2}(y^{n+1}+y^n)$$ show that $$(\epsilon^{n+1}-\epsilon^n)\epsilon^{n+1/2}=h[f(t^{n+1/2}, \frac{1}{2}(y^{n+1}+y^n))-f(t^{n+1/2}, \frac{1}{2}(z^{n+1}+z^n))]\epsilon^{n+1/2}$$


I have done the following:

1) I have shown that $e^{-2Lt}(y(t)-z(t))^2$ is descreasing for $t\in [a,b]$.

Then we have the following: $$0\leq e^{-2Lt}(y(t)-z(t))^2\leq e^{-2Lt_0}(y(t_0)-z(t_0))^2=e^{-2Lt_0}(y_0-z_0)^2$$ If $y_0=z_0$ we get $e^{-2Lt}(y(t)-z(t))^2=0 \Rightarrow y(t)-z(t)=0\Rightarrow y(t)=z(t)$$ Therefore the problem has a unique solution.

For the inequality we can show that $y(t)-z(t)$ is decreasing, but then it follows that $|y(t)-z(t)|\leq |y_0-z_0|$, but how do we get that constant $c$ at the right side?

2) For this one I don't really have an idea. Could yyou give me a hint?

  • 1
    $\begingroup$ In 2), what is the meaning of "approximation", is it a sequence of functions or the result of some numerical method? And what is the "mean value method"? Is it the implicit midpoint method? $\endgroup$ Aug 29, 2019 at 12:54
  • $\begingroup$ It is meant the result of a numerical method.. @LutzL $\endgroup$
    – Mary Star
    Aug 29, 2019 at 13:24

1 Answer 1


By the definition of the implicit midpoint method, you directly have \begin{align} y^{n+1}-y^n&=hf(t^{n+1/2},y^{n+1/2})\\ z^{n+1}-z^n&=hf(t^{n+1/2},z^{n+1/2})\\ \end{align} so that in the difference of both equations you get without any further transformation $$ ϵ^{n+1}-ϵ^n=h[f(t^{n+1/2},y^{n+1/2})-f(t^{n+1/2},z^{n+1/2})]. $$ Now simply multiply with $ϵ^{n+1/2}$ to get the claimed equation.


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