Diff. eqn + erf (error function)

[veronik]

[SOLVED] Diff. eqn + erf (error function)

I’m stacked with this problem for many days, someone can help me pleeeeease:

(a) $$f \left( x \right) =\int _{-\infty }^{{x}^{2}/2}\!{e^{x-1/2\,{t}^{2<br /> }}}{dt}$$

I foud the solution: $$f \left( x \right) =1/2\,{e^{x}}\sqrt {2\pi } \left( 1+{\it <br /> erf} \left( 1/4\,{x}^{2}\sqrt {2} \right) \right)$$

(b) Find the solution of the dfferential equatio:
$${\frac {d^{2}}{d{x}^{2}}}y \left( x \right) =f \left( x \right)$$ with y(0)=0 and dy(0)/dx = 0

In the form : $$y \left( x \right) =\int _{0}^{x}\! \left( x-t \right) f \left( t<br /> \right) {dt}$$

Veronica

 

[tiny-tim]

Welcome to PF!

(b) Find the solution of the dfferential equatio:
$${\frac {d^{2}}{d{x}^{2}}}y \left( x \right) =f \left( x \right)$$ with y(0)=0 and dy(0)/dx = 0

In the form : $$y \left( x \right) =\int _{0}^{x}\! \left( x-t \right) f \left( t<br /> \right) {dt}$$

Veronica

Hi veronica! Welcome to PF!

(… the f in (a) isn't supposed to be the same as the f in (b), is it? …)

I suppose you know what $$\frac{d}{dx}\int _{0}^{x}\! \left( x-t \right) f \left( t<br /> \right) {dt}$$ is?

Are you sure they don't mean something like $$\int _{0}^{x}\! \left( x-t \right)^2 f \left( t<br /> \right) {dt}$$ ?

 

[veronik]

Hi veronica! Welcome to PF!

(… the f in (a) isn't supposed to be the same as the f in (b), is it? …)

I suppose you know what $$\frac{d}{dx}\int _{0}^{x}\! \left( x-t \right) f \left( t<br /> \right) {dt}$$ is?

Are you sure they don't mean something like $$\int _{0}^{x}\! \left( x-t \right)^2 f \left( t<br /> \right) {dt}$$ ?

Thanks for your hospitality and your help tiny tim ,
1-the f in (a) is the same as in (b)...
2-I didnt calculate it..
3-it's x-t

its' (x-t).

I'm waiting for your help

 

[tiny-tim]

3-it's x-t

No, it's not.

Try a simpler one: what is $$\frac{d}{dx}\int _{0}^{x}\! f\left( t<br /> \right) {dt}$$ ?

Hint: put f(t) = g´(t).

 

[D H]

For (a), use the definition of the error function.

For (b), the solution is given. I'm a bit confused. What are you supposed to do?

EDIT:
You are given f(x). The information $d^2y/dx^2 = f(x)$ and $y(0) = y'(0) = 0$ are superfluous. Use the integral form.

 

[veronik]

I suppse ... the question is to demonstrate that the solution is in that form ( $$\int _{0}^{x}\! \left( x-t \right)^2 f \left( t<br /> \right) {dt}$$ )



tiny tin: the x-t was for you question about if it was in the form (x-t)f(t) or (x-t)^2f(t)

did u found the solution?

I'm getting tired :shy:... I'm from europe, it's already 2am here!

 

[D H]

So, not the same f(x), as I thought originally. Use the Leibniz integral rule.

 

[veronik]

I think that it's the same f (it was stated like that in the question!)

 

[D H]

The given equations are valid for any reasonable f(x). You need to show this is the case.

 

[tiny-tim]

did u found the solution?

I'm getting tired :shy:... I'm from europe, it's already 2am here!

Hi Veronica!

I'm from London - we're one hour behind you! :zzz:

Look, with f(t) = g´(t), $$\frac{d}{dx}\int _{0}^{x}\! f\left( t<br /> \right) {dt}$$
$$=\,\frac{d}{dx}\int _{0}^{x}\! g'\left( t<br /> \right) {dt}$$
$$=\,\frac{d}{dx}\left[\,g\left(t)\,\right] _{0}^{x}$$
$$=\,\frac{d}{dx}(g(x)\,-\,g(0))$$
= g´(x) = f(x).

Now you try - what is $$\frac{d}{dx}\int _{0}^{x}\! x f\left( t<br /> \right) {dt}$$ ?

And then what is $${\frac {d^{2}}{d{x}^{2}}}\int _{0}^{x}\! \left( x-t \right) f \left( t \right) {dt}$$ ?

 

[coomast]

I’m stacked with this problem for many days, someone can help me pleeeeease:

(a) $$f \left( x \right) =\int _{-\infty }^{{x}^{2}/2}\!{e^{x-1/2\,{t}^{2<br /> }}}{dt}$$

I foud the solution: $$f \left( x \right) =1/2\,{e^{x}}\sqrt {2\pi } \left( 1+{\it <br /> erf} \left( 1/4\,{x}^{2}\sqrt {2} \right) \right)$$

(b) Find the solution of the dfferential equatio:
$${\frac {d^{2}}{d{x}^{2}}}y \left( x \right) =f \left( x \right)$$ with y(0)=0 and dy(0)/dx = 0

In the form : $$y \left( x \right) =\int _{0}^{x}\! \left( x-t \right) f \left( t<br /> \right) {dt}$$

Veronica

The first part is indeed correct.

Hi Veronica!

I'm from London - we're one hour behind you! :zzz:

Look, with f(t) = g´(t), $$\frac{d}{dx}\int _{0}^{x}\! f\left( t<br /> \right) {dt}$$
$$=\,\frac{d}{dx}\int _{0}^{x}\! g'\left( t<br /> \right) {dt}$$
$$=\,\frac{d}{dx}\left[\,g\left(t)\,\right] _{0}^{x}$$
$$=\,\frac{d}{dx}(g(x)\,-\,g(0))$$
= g´(x) = f(x).

Now you try - what is $$\frac{d}{dx}\int _{0}^{x}\! x f\left( t<br /> \right) {dt}$$ ?

And then what is $${\frac {d^{2}}{d{x}^{2}}}\int _{0}^{x}\! \left( x-t \right) f \left( t \right) {dt}$$ ?

This is the way to show that the formula given is indeed the solution to the differential equation. However it does not give the derivation of the formula which (I assume) is to be found. The equation can be easily integrated to become:

$$y=\int \int f(x) dxdx+Ax+B$$

Naming now f(x)=g'(x) and g(x)=h'(x) and applying the boundary conditions you arrive at:

$$y(x)=h(x)-h(0)-xg(0)$$

This can be rewritten as:

[Begin Edit]
The solution has been removed.
I apologize if this caused any inconvenience, but I was under the assumption that it was not done to post the entire solution in the homework section, but is allowed for the other sections. Sorry about this. What if I was asked in a private message to give the solution?
[End Edit]

Which is the solution you are looking for. So, the boundary conditions are indeed necessary to arrive at the result.

 

[D H]

Please do not give out complete solutions. In fact, please edit the solution out of that last post.