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Solitons - Static Kinks

Hi guys, I'm working through a sheet to help me learn about Solitons. I'm on section 3 now, which is about static kinks, and I can't really seem to start this exercise. Can someone give me a hint, show me what I'd have to do generally?





http://imgur.com/ywh8nQq

Thanks guys
Original post by 0range
Hi guys, I'm working through a sheet to help me learn about Solitons. I'm on section 3 now, which is about static kinks, and I can't really seem to start this exercise. Can someone give me a hint, show me what I'd have to do generally?

http://imgur.com/ywh8nQq

Thanks guys


Isn't this just an application of the chain rule? Differentiating equation (11) w.r.t. x is the same as differentiating w.r.t. u and multiplying by du/dx. The square roots cancel leaving you with the V'(u) term.
Reply 2
Original post by Gregorius
Isn't this just an application of the chain rule? Differentiating equation (11) w.r.t. x is the same as differentiating w.r.t. u and multiplying by du/dx. The square roots cancel leaving you with the V'(u) term.


CodeCogsEqn.gif

and then

CodeCogsEqn (1).gif

So then you're left with

CodeCogsEqn (2).gif



Would that be right? I think I've done something wrong
(edited 8 years ago)
Original post by 0range
CodeCogsEqn.gif

and then

CodeCogsEqn (1).gif

So then you're left with

CodeCogsEqn (2).gif



Would that be right? I think I've done something wrong

That's right, you've got the "if" part. Now see if you can do the "only if" part. Take that equation

d2udx2=V(u) \displaystyle \frac{d^2u}{dx^2} = V'(u)

and write it as

dudxddu(dudx)=V(u) \displaystyle \frac{du}{dx} \frac{d}{du}\left( \frac{du}{dx} \right) = V'(u)

Can you see where to go now?
Reply 4
Original post by Gregorius
That's right, you've got the "if" part. Now see if you can do the "only if" part. Take that equation

d2udx2=V(u) \displaystyle \frac{d^2u}{dx^2} = V'(u)

and write it as

dudxddu(dudx)=V(u) \displaystyle \frac{du}{dx} \frac{d}{du}\left( \frac{du}{dx} \right) = V'(u)

Can you see where to go now?


Hmm I may need a tiny bit more help. What I was thinking was that I'd need to integrate from here somehow. Like move the

Unparseable latex formula:

\displaystyle \frac{du}{dx} \frac{du}{dx} \right) = V'(u) {du}\left



and integrate both sides so I'd have a

dudx \displaystyle \frac{du}{dx}

on the left after integration, but I'm not sure what the other side would be..
Original post by 0range
Hmm I may need a tiny bit more help. What I was thinking was that I'd need to integrate from here somehow. Like move the

Unparseable latex formula:

\displaystyle \frac{du}{dx} \frac{du}{dx} \right) = V'(u) {du}\left



and integrate both sides so I'd have a

dudx \displaystyle \frac{du}{dx}

on the left after integration, but I'm not sure what the other side would be..


We've got to

dudxddu(dudx)=V(u) \displaystyle \frac{du}{dx} \frac{d}{du}\left( \frac{du}{dx} \right) = V'(u)

Now ponder what you get from

ddu[(dudx)2] \displaystyle \frac{d}{du} \left[ \left(\frac{du}{dx}\right)^2 \right]
Reply 6
Original post by Gregorius
We've got to

dudxddu(dudx)=V(u) \displaystyle \frac{du}{dx} \frac{d}{du}\left( \frac{du}{dx} \right) = V'(u)

Now ponder what you get from

ddu[(dudx)2] \displaystyle \frac{d}{du} \left[ \left(\frac{du}{dx}\right)^2 \right]


I really still can't see it :frown: I was thinking of multiplying both sides with d/du and then integrating both sides ?

Would the bit to ponder not just turn into 0??
Original post by 0range
I really still can't see it :frown: I was thinking of multiplying both sides with d/du and then integrating both sides ?

Would the bit to ponder not just turn into 0??


That expression evaluates to

ddu[(dudx)2]=2dudxddu(dudx) \displaystyle \frac{d}{du} \left[ \left(\frac{du}{dx}\right)^2 \right]= 2 \frac{du}{dx} \frac{d}{du}\left( \frac{du}{dx} \right)
Reply 8
Original post by Gregorius
That expression evaluates to

ddu[(dudx)2]=2dudxddu(dudx) \displaystyle \frac{d}{du} \left[ \left(\frac{du}{dx}\right)^2 \right]= 2 \frac{du}{dx} \frac{d}{du}\left( \frac{du}{dx} \right)


Right I can see how you got that (I'm confusing myself a lot here)
To be able to do that, would you not have to differentiate the rhs as well? (so the v'(u))
and would that not be going in the wrong direction?
Original post by 0range
Right I can see how you got that (I'm confusing myself a lot here)
To be able to do that, would you not have to differentiate the rhs as well? (so the v'(u))
and would that not be going in the wrong direction?


We've simply re-written the left hand side of one of our equations...So now we can say

dudxddu(dudx)=V(u) \displaystyle \frac{du}{dx} \frac{d}{du}\left( \frac{du}{dx} \right) = V'(u)

is the same as the equation

(12)ddu[(dudx)2]=V(u) \displaystyle \left( \frac{1}{2} \right) \frac{d}{du} \left[ \left(\frac{du}{dx}\right)^2 \right] = V'(u)

which you can integrate straight away. The end result will be the required equation for dudx \frac{du}{dx}

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