Finding Constants to Solution Given 2nd Order, Nonlinear DE

  • #1
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Homework Statement



Okay, here's the deal:


I have been given a second order nonlinear differential equation, and I have also been given the general solution with constants A and B. I am supposed to find the constants A and B. The solution represents a fermion at rest, since the solution does not vary with time.

Homework Equations



The differential equation is as follows:


(d^2phi/dt^2)-(d^2phi/dx^2)-phi+phi^3=0

The soliton solution is:


Phi(x)=A(tanh(Bx))

The Attempt at a Solution


I have five pages of attempts to solve this. One professor told me it was impossible, but the professor that I am doing research with told me that it is possible. First, I found the first and second derivatives of phi, then, I input them into the differential equation. The first term goes out because the particle is at rest. Then, I tried different methods.
I tried using hyperbolic trig identities.
I tried writing the equation in terms of sinh and cosh and then tried to eliminate some things. He told me that I need to end up with a polynomial that has A and B and no x, and that I must have hyperbolic tangent on both sides of the equation to find the constants.
I also tried getting a polynomial with tanh(Bx) representing a variable p and then I got factors A, p, and the polynomial. If the equation equals zero, then at least one of those factors must be zero, so I set them equal to zero, but I still have x in there. I tried setting tanh (Bx) (aka p) equal to zero and setting the polynomial to zero and then I solved the polynomial for p and made the p equation equal to zero, however, it looked very complicated and the professor said it was a simple calculation, and even then, that would mean that phi is zero, and that is not the case because I am supposed to use phi to get energy density.

Please help.
 

Answers and Replies

  • #2
phyzguy
Science Advisor
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I think you're making this harder than it is. The solution is just a 2-3 lines. Can your write down d^2phi/dx^2? Do you remember that 1-tanh[x]^2 = sech[x]^2. Also remember that you are trying to find an A and B that solve the equation, so you are free to choose A and B to make the problem simplify.
 
  • #3
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d^2phi/dx^2= -2AB^2tanh(Bx), correct?

Now I have gotten the equation

tanh(Bx)[2AB^2-A+A^3tanh^2(Bx)]=0

I set 2AB^2-A+A^3tanh^2(Bx)=0

tanh(Bx)= sqrt[2B-1]/A

But now what? Was I supposed to change the tanh^2(Bx) into 1-sech^2(Bx)?
 
  • #4
phyzguy
Science Advisor
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d^2phi/dx^2= -2AB^2tanh(Bx), correct?
No. You're not differentiating this correctly. The first derivative of tanh(x) is sech^2(x). The second derivative is more complicated than just tanh[x]. Try again.
 
  • #5
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The second derivative is

phi''=-2AB^2tanh(Bx)sech^2(Bx)= -2AB^2tanh(Bx)+2AB^2tanh^3(Bx)

and so

tanh(Bx)[2AB^2-A]+tanh^3(Bx)[-2AB^2+A^3]=0
 
  • #6
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I keep getting that phi is zero.
 
  • #7
phyzguy
Science Advisor
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OK, better. Of course phi = 0 is a solution, but there is another solution. Suppose those two coefficients (2AB^2 -A), and (-2AB^2+A^3) are both zero. Is there a set of A and B that can make that happen?
 
  • #8
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Ahhh yes. I understand!
Thank you very much :)
 
  • #9
75
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Alright, I ended up with A=+/- 1 and B=1/sqrt(2).
 
  • #10
phyzguy
Science Advisor
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Sounds good to me. See, it wasn't that hard after all!
 

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