0range
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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
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Gregorius
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(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.
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0range
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(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.
Name:  CodeCogsEqn.gif
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and then

Name:  CodeCogsEqn (1).gif
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So then you're left with

Name:  CodeCogsEqn (2).gif
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Size:  804 Bytes



Would that be right? I think I've done something wrong
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Gregorius
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(Original post by 0range)
Name:  CodeCogsEqn.gif
Views: 88
Size:  1.6 KB

and then

Name:  CodeCogsEqn (1).gif
Views: 88
Size:  1.8 KB

So then you're left with

Name:  CodeCogsEqn (2).gif
Views: 84
Size:  804 Bytes



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

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

and write it as

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

Can you see where to go now?
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0range
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(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

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

and write it as

 \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

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

and integrate both sides so I'd have a

 \displaystyle \frac{du}{dx}

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

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

and integrate both sides so I'd have a

 \displaystyle \frac{du}{dx}

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

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

Now ponder what you get from

 \displaystyle \frac{d}{du} \left[ \left(\frac{du}{dx}\right)^2 \right]
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0range
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(Original post by Gregorius)
We've got to

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

Now ponder what you get from

 \displaystyle \frac{d}{du} \left[ \left(\frac{du}{dx}\right)^2 \right]
I really still can't see it 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??
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Gregorius
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(Original post by 0range)
I really still can't see it 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

 \displaystyle \frac{d}{du} \left[ \left(\frac{du}{dx}\right)^2 \right]= 2 \frac{du}{dx} \frac{d}{du}\left( \frac{du}{dx} \right)
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0range
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(Original post by Gregorius)
That expression evaluates to

 \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?
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(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

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

is the same as the equation

 \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  \frac{du}{dx}
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