Finding an upper bound for the local error with the Euler method

Question

This is not about a particular problem but more a question about how to best approach this kind of problem. I'll give three examples and my approach to them.

Problem 1

$$ y' = 2y - 5 \sin(t)$$ $$ y(0) = 1 $$

For this I understand that I need to find an upper bound for $|y'(t)|$ but since the straight approach doesn't get me anywhere I tried to find an upper bound for $ y(t) $ so using the mean value theorem I get:

$$ | y(t) - y(0) | = | y(t) - 1 | = |t - 0| |y'(\alpha)| = |t| |2y(\alpha) - 5 \sin(\alpha)| \leq 2|t||y(\alpha)| + 5|t|$$

Where $\alpha \in [0, t]$. Again I find myself need an upper bound for $|y(t)|$ that I don't know how to find.

Problem 2

$$y' = -2ty$$ $$y(0) = 1$$ $$ t \geq 0$$

Trying to find an upper bound for $|y(t)|$ I get:

$$ |y(t) - y(0)|=|y(t)-1|=|t| |y'(\alpha)|= 2 \alpha t |y(\alpha)|$$

Again not particularly useful. I can find the Lipschitz "constant" for $F(t,y) = -2ty$ which is $K = 2t$ but it doesn't really help m in any way I can see.

Problem 3

$$y'(t) = \frac{e^{-y}}{t}$$ $$y(1)=0$$ $$ t \geq 1$$ This one starts asking me to prove that $0 \leq y(t) \leq t$ so with the same approach as before:

$$|y(t) - y(1)| = |y(t)| = |t-1|\frac{e^{-y(\alpha)}}{\alpha}$$

Now I looked at the euler method approximation which is $y(t_{i+1}) = y(t_{i}) + h \frac{e^{-y(t_{i})}}{t_{i}}$. Looking at a few iterations we can see that the method is decreasing so:

$$|y(t)| = |t-1|\frac{e^{-y(\alpha)}}{\alpha} \leq |t-1|\frac{e^{-y(1)}}{\alpha}$$

Since $\alpha \geq 1$ and $t > 1$ we have:

$$ |y(t)| \leq |t-1|\frac{e^{-y(1)}}{\alpha} \leq |t|\frac{e^{0}}{1} = |t|$$

Curiously the second item on this problem asks me to prove that the method is a decreasing one, so I imagine I'm not supposed to be using that in my answer to the first one.

Now if I want to find an upper bound for $|y'(t)|$:

$$ |y'(t)| = |\frac{e^{-y(t)}}{t}| \leq |\frac{e^{-y(1)}}{t}| = \frac{1}{t} $$

Which I guess given the $t$ would work.

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What is the correct way to approach this kind of problem then? I've looked at my class notes, but all examples are more along the lines of $y' = 3\sin(y(t)) + 4t$ which are almost trivial to find bounds for. My guess is that since I'm quite a bit rusty (haven't done any math in a few years) I'm missing something pretty basic that I can't identify.

Any help would be greatly appreciated.

Answer 1

Consider continuous IVP is $$\left\{\begin{array}{ll} y' = f(t,y), & t\in[0,T],\\ y(0) = y_0. \end{array}\right. $$ I guess you mean local truncation error of Forward Euler Method. $$ y_{n+1}=y_{n}+h f\left(t_{n}, y_{n}\right) $$ This is a simple case where the method is based on a first order Taylor expansion. Assuming $y_{n}=y\left(t_{n}\right)$ we have $$ \begin{aligned} y_{n+1} &=y\left(t_{n}\right)+h f\left(t_{n}, y\left(t_{n}\right)\right) \\ &=y\left(t_{n}\right)+h y^{\prime}\left(t_{n}\right) \end{aligned} $$ since we are solving the ODE $y^{\prime}(t)=f(t, y(t)) .$ Taylor's expansion of the exact value is $$ y\left(t_{n+1}\right)=y\left(t_{n}\right)+h y^{\prime}\left(t_{n}\right)+O\left(h^{2}\right) $$ So the local truncation error is $$ y\left(t_{n+1}\right)-y_{n+1}=O\left(h^{2}\right). $$