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Given the problem:

$$y′ = (t − 1)\sin(y),\;\;\;y(1) = 1$$ find an approximation for $y(2)$.

Give an upper bound for the global error taking $n = 4$ (i.e., $h = \frac{1}{4}$)

The goal is to find an upper bound, not actually find an approximation for $y(2)$.

However I don't think it is possible to find an upper bound as we need to use the equation

$$|E_n| \leq \frac{T}{L}[(e^{(t_n - t_0)L} - 1)]$$

where $T$ is the max truncation error and $L$ is the Lipschitz constant.

But I am unable to get a Lipschitz constant from $y′ = (t − 1)\sin(y)$, I don't think it's possible without being given a bound on $t$.

$$|f(t, u) - f(t,v)| = |(t-1)\sin(u) - (t-1)\sin(v)|= |t-1||\sin(u) - \sin(v)|$$

$\sin\;$ has Lipschitz constant of $2$ so we can say

$$\leq 2|t-1||u - v|$$

And where can I go from here...?

Is the question just not well defined or am I missing something?

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Yes, you need to bound $t$. But this should not be a problem, since you are interested in $y(2)$. –  Siminore Nov 17 '12 at 17:55
    
what method are you using to find an approximate value for $y$ at $t = 2?$ –  abel Jan 22 at 15:42

1 Answer 1

Letting $f(t,y) = (t-1)\sin(y)$, you can use that an upper bound on $|\partial_y f(t,y)|$ will be a Lipschitz constant (this is an application of the mean value theorem). To see this, note, as the commenter mentioned, you only need to consider values of $1\leq t \leq 2$. It should be clear that a bound on this partial exists for any $t$ in this interval, and any value of $y$.

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