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#### Thorra

##### Member
I'll look over your other posts tomorrow but I'd like to post the 3rd problem for now in case you got the time to answer somewhere between that time.

So the third and final (for now, anyway) problem.

Problem III

Approximate the differential equation $$\frac{d^3 u}{dx^3}=g(x)$$ on a model* that has no more than 5 points, and with a constant step $h$ with at least the third local approximation order ($\mathcal O(h^3)$) in this differential equation solution set of functions.

* When I say "model", I mean like a pattern, that is ready to repeat itself throughout the grid, but just taking the one basic vital thing from it that repeats throughout. I think.

And I can only hope to god I translated that well enough, especially the underlined part.

My Take?

Honestly, I don't really have a take on this problem. It's just more cosmos for me, one where I don't understand what's really asked of me and why is it only "up to" 5 points. Well mostly the underlined part is the confusing part (which is the original point I guess). Welp, anyway, off I go to sleep.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I'll look over your other posts tomorrow but I'd like to post the 3rd problem for now in case you got the time to answer somewhere between that time.

So the third and final (for now, anyway) problem.

Problem III

Approximate the differential equation $$\frac{d^3 u}{dx^3}=g(x)$$ on a model* that has no more than 5 points, and with a constant step $h$ with at least the third local approximation order ($\mathcal O(h^3)$) in this differential equation solution set of functions.

* When I say "model", I mean like a pattern, that is ready to repeat itself throughout the grid, but just taking the one basic vital thing from it that repeats throughout. I think.

And I can only hope to god I translated that well enough, especially the underlined part.

My Take?

Honestly, I don't really have a take on this problem. It's just more cosmos for me, one where I don't understand what's really asked of me and why is it only "up to" 5 points. Well mostly the underlined part is the confusing part (which is the original point I guess). Welp, anyway, off I go to sleep.

Let's pick a specific example.
Suppose you have:
$$u'''(x) = \sin(x)$$
Do you know how to solve it?
That is, can you tell what the set of solutions is for this differential equation?

Here's my interpretation of the problem.

You are given, say, g(x), g(x+h), g(x+2h), g(x+3h), and g(x+4h).
Furthermore, you have that u'''(x + ih) = g(x + ih), where i=0, ..., 4.
And you have to find u(x), u(x+h), u(x+2h), u(x+3h), and u(x+4h) with order $\mathcal O(h^3)$ such that they fit in the differential equation.
When doing so, I think you're supposed to treat u(x), u'(x), and u''(x) as constants, since a third order differential equation has multiple solutions that are fully determined by picking these 3 as given constants.

To be honest, I do not quite know yet how to solve the 3rd order differential equation without getting stuck in a morass of equations.
Say, u'(x)=g(x).
Can you get started with that?

Btw, I have changed the title of your thread to be a bit more descriptive than just "Problem III".

#### Thorra

##### Member
Let's pick a specific example.
Suppose you have:
$$u'''(x) = \sin(x)$$
Do you know how to solve it?
That is, can you tell what the set of solutions is for this differential equation?
Aw man I don't think I can actually. We're only just having the differential equation course now and AFAIK we're only looking at the first order derivative equations, maaybe 2nd order in some cases.

Here's my interpretation of the problem.

You are given, say, g(x), g(x+h), g(x+2h), g(x+3h), and g(x+4h).
Furthermore, you have that u'''(x + ih) = g(x + ih), where i=0, ..., 4.
And you have to find u(x), u(x+h), u(x+2h), u(x+3h), and u(x+4h) with order $\mathcal O(h^3)$ such that they fit in the differential equation.
When doing so, I think you're supposed to treat u(x), u'(x), and u''(x) as constants, since a third order differential equation has multiple solutions that are fully determined by picking these 3 as given constants.
so... I advance the g(x+ih) functions in the Taylor series again? And that's v'(x+ih), v''(x+ih) and eventually v'''(x+ih)? Or what?

To be honest, I do not quite know yet how to solve the 3rd order differential equation without getting stuck in a morass of equations.
Say, u'(x)=g(x).
Can you get started with that?
I dunno... maybe this problem is too difficult for me to understand in such a short time?

Btw, I have changed the title of your thread to be a bit more descriptive than just "Problem III".
Hehe, thanks. That was a testament to how sleepy I was I guess.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Aw man I don't think I can actually. We're only just having the differential equation course now and AFAIK we're only looking at the first order derivative equations, maaybe 2nd order in some cases.
In this case the process is fairly straight forward.
It's a matter of repeatedly taking the integral from both sides.
\begin{array}{}
u'''(x)&=&\sin(x) \\
\int u'''(x)dx&=&\int \sin(x)dx \\
u''(x)&=& -\cos(x) + C_1 \\
\int u''(x)dx&=& \int (-\cos(x) + C_1)dx \\
u'(x) &=& -\sin(x) + C_1 x + C_2 \\
\int u'(x) &=& \int (-\sin(x) + C_1 x + C_2)dx \\
u(x) &=& \cos(x) + \frac 1 2 C_1 x^2 + C_2 x + C_3
\end{array}
There you go! The solution!!

As a result, we find that:
\begin{array}{}
u(0) &=& \cos(0) + C_3 \\
u'(0) &=& -\sin(0) + C_2 \\
u''(0) &=& -\cos(0) + C_1
\end{array}

Hey!
These values are isolating the integration constants! so... I advance the g(x+ih) functions in the Taylor series again? And that's v'(x+ih), v''(x+ih) and eventually v'''(x+ih)? Or what?
Sounds good...

I dunno... maybe this problem is too difficult for me to understand in such a short time?
We won't know until you try at least something. #### Thorra

##### Member
In this case the process is fairly straight forward.
It's a matter of repeatedly taking the integral from both sides.
\begin{array}{}
u'''(x)&=&\sin(x) \\
\int u'''(x)dx&=&\int \sin(x)dx \\
u''(x)&=& -\cos(x) + C_1 \\
\int u''(x)dx&=& \int (-\cos(x) + C_1)dx \\
u'(x) &=& -\sin(x) + C_1 x + C_2 \\
\int u'(x) &=& \int (-\sin(x) + C_1 x + C_2)dx \\
u(x) &=& \cos(x) + \frac 1 2 C_1 x^2 + C_2 x + C_3
\end{array}
There you go! The solution!!
Oh well that don't seem that tough. Though... what is the function (set) we must derive in my problem? Looks to my like a non defined universal writing of any function.

As a result, we find that:
\begin{array}{}
u(0) &=& \cos(0) + C_3 \\
u'(0) &=& -\sin(0) + C_2 \\
u''(0) &=& -\cos(0) + C_1
\end{array}
Hey!
These values are isolating the integration constants! Haha, nice. Though,, shouldn't $u(0)=C_3 + 1$ and $u^{''}(0)=C_1-1$ ? Or you just merge them into one big constant again?

Sounds good...

We won't know until you try at least something. Well I just don't know where to begin. Why $g(x)$ and $u(x)$ are two different given functions and stuff.

Well here goes nothing. I just used some substituton, really.

\begin{array}{}
g(x)&=&\frac{du}{dx} \\
g(x+h)&=&\frac{du}{dx}+hg'(x)+\frac{h^2}{2}g''(x)+\mathcal O(h^3) \\
g(x+2h)&=&\frac{du}{dx}+2hg'(x)+\frac{4h^2}{2}g''(x)+\mathcal O(h^3) \\
g(x+3h)&=&\frac{du}{dx}+3hg'(x)+\frac{9h^2}{2}g''(x)+\mathcal O(h^3) \\
g(x+4h)&=&\frac{du}{dx}+4hg'(x)+\frac{16h^2}{2}g''(x)+\mathcal O(h^3) \\
\end{array}
and then just told out the u'(x) for all of them (except the first... that one's not really approximated to the 3rd degree..)

\begin{array}{}
\frac{du}{dx}&=&g(x+h)-hg'(x)-\frac{h^2}{2}g''(x)+\mathcal O(h^3) \\
\frac{du}{dx}&=&g(x+2h)-2hg'(x)-\frac{4h^2}{2}g''(x)+\mathcal O(h^3) \\
\frac{du}{dx}&=&g(x+3h)-3hg'(x)-\frac{8h^2}{2}g''(x)+\mathcal O(h^3) \\
\frac{du}{dx}&=&g(x+4h)-4hg'(x)-\frac{16h^2}{2}g''(x)+\mathcal O(h^3) \\
\end{array}

Now if I am even on the right road here, I can only imagine an extreemly long writing for a $\frac{d^3 u}{dx^3}$ solution with 3rd level approximation, as then I'd go up to like $g^{(4)}(x)+\mathcal O(h^5)$ and then make another Linear Algebraic Equation System where that and two previous derivatives and maybe some more would have to equal 0 (like in Task II) (so that the $\frac{d^3u}{dx^3}h^2$ can survive to the 3rd approximation when we'd tell out $\frac{d^3u}{dx^3}=(...) +\frac{\mathcal O(h^5)}{h^2}$). And that's just nutty.

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Oh well that don't seem that tough. Though... what is the function (set) we must derive in my problem? Looks to my like a non defined universal writing of any function.
Formally. the solution set is:
$$\{x \mapsto \cos(x)+C_1x^2+C_2x+C_3 \quad |\quad C_1, C_2, C_3 \in \mathbb R \}$$
This is hardly "any" function - its a cosine summed with an arbitrary parabola.

Haha, nice. Though,, shouldn't $u(0)=C_3 + 1$ and $u^{''}(0)=C_1-1$ ? Or you just merge them into one big constant again?
Yes. You can do both.
I left them like that because $\cos(0)$ actually represents $g^{(-3)}(0)$, where I mean the 3rd anti-derivative of g.

Well I just don't know where to begin. Why $g(x)$ and $u(x)$ are two different given functions and stuff.

Well here goes nothing. I just used some substituton, really.

\begin{array}{}
g(x)&=&\frac{du}{dx} \\
g(x+h)&=&\frac{du}{dx}+hg'(x)+\frac{h^2}{2}g''(x)+\mathcal O(h^3) \\
g(x+2h)&=&\frac{du}{dx}+2hg'(x)+\frac{4h^2}{2}g''(x)+\mathcal O(h^3) \\
g(x+3h)&=&\frac{du}{dx}+3hg'(x)+\frac{9h^2}{2}g''(x)+\mathcal O(h^3) \\
g(x+4h)&=&\frac{du}{dx}+4hg'(x)+\frac{16h^2}{2}g''(x)+\mathcal O(h^3) \\
\end{array}
and then just told out the u'(x) for all of them (except the first... that one's not really approximated to the 3rd degree..)

\begin{array}{}
\frac{du}{dx}&=&g(x+h)-hg'(x)-\frac{h^2}{2}g''(x)+\mathcal O(h^3) \\
\frac{du}{dx}&=&g(x+2h)-2hg'(x)-\frac{4h^2}{2}g''(x)+\mathcal O(h^3) \\
\frac{du}{dx}&=&g(x+3h)-3hg'(x)-\frac{8h^2}{2}g''(x)+\mathcal O(h^3) \\
\frac{du}{dx}&=&g(x+4h)-4hg'(x)-\frac{16h^2}{2}g''(x)+\mathcal O(h^3) \\
\end{array}

Now if I am even on the right road here, I can only imagine an extreemly long writing for a $\frac{d^3 u}{dx^3}$ solution with 3rd level approximation, as then I'd go up to like $g^{(4)}(x)+\mathcal O(h^5)$ and then make another Linear Algebraic Equation System where that and two previous derivatives and maybe some more would have to equal 0 (like in Task II) (so that the $\frac{d^3u}{dx^3}h^2$ can survive to the 3rd approximation when we'd tell out $\frac{d^3u}{dx^3}=(...) +\frac{\mathcal O(h^5)}{h^2}$). And that's just nutty.
Since you need to find:
$$u(x), u(x+h), u(x+2h), u(x+3h), u(x+4h)$$
given
$$g(x+ih)$$
$$u(x), u'(x), ..., u^{(j)}(x)$$

I would try to expand:
$$u(x+h), u((x+h)-h), u(x+2h), u((x+2h)-h), ...$$

But hey! I haven't figured it out yet. #### Thorra

##### Member
Hello, I'm back! I'll warm up with trying to understand this problem and then I'll post a big one. The final problem. I know you've probably forgotten most of the thoughts on sovling a problem like this by now. I know I have. But if you find it easy to remember this, then I'll appreciate a reply!

Here's my interpretation of the problem.

You are given, say, g(x), g(x+h), g(x+2h), g(x+3h), and g(x+4h).
Furthermore, you have that u'''(x + ih) = g(x + ih), where i=0, ..., 4.
And you have to find u(x), u(x+h), u(x+2h), u(x+3h), and u(x+4h) with order $\mathcal O(h^3)$ such that they fit in the differential equation.
When doing so, I think you're supposed to treat u(x), u'(x), and u''(x) as constants, since a third order differential equation has multiple solutions that are fully determined by picking these 3 as given constants.
Constants like $u''(0)=C_1$, $u'(0)=C_2$, $u(0)=C_3$? (I figured this from your next post when you tripple integrate the sin(x) step by step. Although cos(0) isn't 0 so nevermind that... Still not sure why x=0.

Since you need to find:
$$u(x), u(x+h), u(x+2h), u(x+3h), u(x+4h)$$
given
$$g(x+ih)$$
$$u(x), u'(x), ..., u^{(j)}(x)$$
j=3, right?

I would try to expand:
$$u(x+h), u((x+h)-h), u(x+2h), u((x+2h)-h), ...$$

But hey! I haven't figured it out yet. Hey... I don't quite get what you're doing there. Are you going one step forth, then one step back from the previous step? Doesn't really look that way and that would be pointless anyway, so I don't get it. Plus, u((x+h)-h) just looks like u(x) to me. And u(x+h) = u((x+2h)-h). I guess you mean the resulting function of u(x+h) then taken to a function of u(x-h)... hmm. But then why write u(x+2h) immidiately rather then u((x+h)+h) or something. Sorry, I don't see a larger connection, I know it was a big mistake to take this subject.

To be clear: do I have to write out the five solutions of u=(x+ih) in terms of h and g(x)? Like the righthandside of the equation only containing h and g(x)? That's one thing I didn't understand from the very beginning cause I comprehend this not.

So I have u'''(x+ih)=g(x+ih) so, u(x+ih) is the 3rd anti-derivative (or integral) of g(x-ih)? $u'''(x+ih)=g^{(-3)}(x+ih)$.
So this whole task is about integrating a function through approximation and numerical methods hell, rather than the precise mathematical integration hell? This task sucks...

Ok... Expanding
\begin{array}{}
u(x+h)=v(x)+hv'(x)+\frac{1}{2}h^2v''(x)+\mathcal O(h^3) \\
u((x+h)-h)=v(x)+hv'(x)+\frac{1}{2}h^2v''(x)+v(x)-v'(x)h+\frac{1}{2}v''(x)h^2+\mathcal O(h^3) \\
u(x+2h)=v(x)+2v'(x)h+\frac{4}{2}h^2v''(x)+\mathcal O(h^3) \\
u((x+2h)-h)=v(x)+2hv'(x)+\frac{4}{2}h^2v''(x)+v(x)-v'(x)h+\frac{1}{2}v''(x)h^2+\mathcal O(h^3) \\
u(x+3h)=v(x)+3v'(x)h+\frac{9}{2}h^2v''(x)+\mathcal O(h^3) \\
\end{array}
? These the 5 points? What next?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello, I'm back! I'll warm up with trying to understand this problem and then I'll post a big one. The final problem. I know you've probably forgotten most of the thoughts on sovling a problem like this by now. I know I have. But if you find it easy to remember this, then I'll appreciate a reply!
Welcome back! Constants like $u''(0)=C_1$, $u'(0)=C_2$, $u(0)=C_3$? (I figured this from your next post when you tripple integrate the sin(x) step by step. Although cos(0) isn't 0 so nevermind that... Still not sure why x=0.

j=3, right?
Yes and yes.

Hey... I don't quite get what you're doing there. Are you going one step forth, then one step back from the previous step? Doesn't really look that way and that would be pointless anyway, so I don't get it. Plus, u((x+h)-h) just looks like u(x) to me. And u(x+h) = u((x+2h)-h). I guess you mean the resulting function of u(x+h) then taken to a function of u(x-h)... hmm. But then why write u(x+2h) immidiately rather then u((x+h)+h) or something. Sorry, I don't see a larger connection, I know it was a big mistake to take this subject.
Yes, $u((x+h)-h) = u(x)$.
However, we can expand $u((x+h)-h)$ like:
$$u(x) = u((x+h)-h) = u(x+h) - h u'(x+h) +...$$

To be clear: do I have to write out the five solutions of u=(x+ih) in terms of h and g(x)? Like the righthandside of the equation only containing h and g(x)? That's one thing I didn't understand from the very beginning cause I comprehend this not.

So I have u'''(x+ih)=g(x+ih) so, u(x+ih) is the 3rd anti-derivative (or integral) of g(x-ih)? $u'''(x+ih)=g^{(-3)}(x+ih)$.
So this whole task is about integrating a function through approximation and numerical methods hell, rather than the precise mathematical integration hell? This task sucks...

Ok... Expanding
$$u(x+h)=v(x)+hv'(x)+\frac{1}{2}h^2v''(x)+\mathcal O(h^3)$$
Mixing up $u$ and $v$ again? $$u((x+h)-h)=v(x)+hv'(x)+\frac{1}{2}h^2v''(x)+v(x)-v'(x)h+\frac{1}{2}v''(x)h^2+\mathcal O(h^3)$$
This expansion should be:
$$u(x) = u((x+h)-h) = u(x+h) - h u'(x+h) +...$$
The term $u(x+h)$ is one of the unknowns that we want to find.
Let's call it $u_1 = u(x+h)$ for short.

\begin{array}{}
u(x+2h)=v(x)+2v'(x)h+\frac{4}{2}h^2v''(x)+\mathcal O(h^3) \\
u((x+2h)-h)=v(x)+2hv'(x)+\frac{4}{2}h^2v''(x)+v(x)-v'(x)h+\frac{1}{2}v''(x)h^2+\mathcal O(h^3) \\
u(x+3h)=v(x)+3v'(x)h+\frac{9}{2}h^2v''(x)+\mathcal O(h^3) \\
\end{array}
? These the 5 points? What next?
You would actually get more equations if you write out all combinations.
The target would be to find $u_i = u(x+ih)$ in terms of $C_1, C_2, C_3$ and $g_i = g(x+ih) = u'''(x+ih)$.