1

I am just working through some exam practice problems, and I am a bit stuck with this one:

Consider the IVP:

$$ \frac{dy}{dt} = f(t,y), \space y(0)=y_{0} $$

(a) Expand solution $y = Y(t)$ of the IVP in a two term ts with r about $t= t_{n+1}$, to show that:

$$Y_{n} = Y_{n+1}-hf(t_{n+1},Y_{n+1})+\frac{1}{2}Y''(\psi_{n})h^2$$

where $Y_n = Y(t_n)$ and $\psi_{n}$ lies between $t_n$ and $t_{n+1}$

(b) minus from back-Euler method, show the global error $e_{n}$ in the back-Euler method:

$$e_{n+1} = \frac{e_{n}}{1-hJ}-\frac{Y''(\psi_{n})h^2}{2(1-hJ)}$$

where $J=f_{y}(t_{n+1},\eta_{n+1})$ and $\eta_{n+1}$ lies between $t_n$ and $t_{n+1}$

For part (a):

I have from my textbook that the ts expansion about $t_n$ is:

$$Y_{n+1} = Y_{n}+hf(t_{n},Y_{n})+\frac{1}{2}Y''(\psi_{n})h^2 $$

Hence, if I expand around $t_{n+1}$, then I have:

$$Y_{n+2} = Y_{n+1}+hf(t_{n+1},Y_{n+1})+\frac{1}{2}Y''(\psi_{n+1})h^2$$ I need to get rid of the $Y_{n+2}$ and bring in $Y_{n}$, so I substitute for $Y_{n+1}$ ,rearrange to isolate $Y_{n}$, and then substitute again for $Y_{n+2}$ this gives me:

$$Y_{n} = Y_{n+1}+hf(t_{n+1},Y_{n+1})+\frac{1}{2}Y''(\psi_{n+1})h^2-hf(t_{n},Y_{n})-\frac{1}{2}Y''(\psi_{n})h^2-hf(t_{n+1},Y_{n+1})-\frac{1}{2}Y''(\psi_{n+1})h^2 $$

I'm not really sure what I am doing here, because it seems like all I am doing is rearranging the taylor expansion about $t_n$, the sign is clearly wrong, and I am explicitly asked to use the expansion about $t_{n+1}$...Any help would be greatly appreciated here.

Attempting part(b):

I rearrange the solution from part(a) to isolate $Y_{n+1}$ and minus the back-E method:

$$Y_{n+1} = Y_{n}+hf(t_{n+1},Y_{n+1})$$

This gives me:

$$e_{n+1} = Y_{n+1}-y_{n+1} = Y_{n}-y_{n}+h[f(t_{n+1},Y_{n+1})-f(t_{n+1},y_{n+1})]+\frac{1}{2}Y''(\psi_{n})h^2 = (1-hJ(\psi_{n}))(Y_n-y_n)+\frac{1}{2}Y''(\psi_{n})h^2 = (1-hJ(\psi_{n}))e_{n}+\frac{1}{2}Y''(\psi_{n})h^2$$ By the intermediate value theorem.

Then I am stuck again. I am sure there is some trick to manipulate the last expression, but I have failed at multiple attempts.

Any help would be greatly appreciated!

JJJ
  • 874

1 Answers1

1

Recall the Taylor expansion about a point $a$ is given by

$$f(x) = f(a) + f^{\prime}(a) (x-a) + f^{\prime\prime}(\xi)(x-a)^{2}/2$$

for $\xi$ between $x$ and $a$.

For $Y_{n} = Y(t_{n})$ and $t_{n+k} = t_{n} +kh$, expand

$$Y_{n} = Y_{n+1} + f(t_{n+1}, Y_{n+1})(t_{n}-t_{n+1}) + Y^{\prime\prime}(\psi_{n})(t_{n}-t_{n+1})/2 = Y_{n+1}-hf(t_{n+1},Y_{n+1})+\frac{1}{2}Y''(\psi_{n})h^2$$

as desired.

For part b), we have, as you wrote,

$$e_{n+1}= e_{n}+h[f(t_{n+1},Y_{n+1})-f(t_{n+1},y_{n+1})]+\frac{1}{2}Y''(\psi_{n})h^2.$$

By the Mean Value Theorem, there exists a $\eta_{n+1}$ so that

$$f(t_{n+1},Y_{n+1})-f(t_{n+1},y_{n+1}) = (Y_{n+1}-y_{n+1})f_{y}(t_{n+1},\eta_{n+1}) = Je_{n+1}.$$

So

$$e_{n+1}= e_{n}+hJe_{n+1}+\frac{1}{2}Y''(\psi_{n})h^2$$

or

$$e_{n+1}(1-hJ)= e_{n}+\frac{1}{2}Y''(\psi_{n})h^2$$

and the result follows.

JessicaK
  • 7,655