0

Suppose I have

v[x_] = Square[1.453 Sech[x + 1]] + I  Sech[x + 1] Tanh[x + 1]

And I have to solve the equation:

mu1 u1[x] - u1''[x] - v[x] u1[x] == 0

for u1[x]. The conditions that are given are:

u1[-2] == 1, u1'[-2] == 0 .

I have tried DSolve but it shows errors:

Solve::inex: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help.

How can I Solve this equation in Mathematical symbolically?

sarnold
  • 102,305
  • 22
  • 181
  • 238
Mashriq Ahmed
  • 7
  • 3
  • after giving the input : sol = DSolve[{mu1 u1[x] - u1''[x] - v[x] u1[x] == 0, u1[-2] == 1, u1'[-2] == 0}, u1, x] i got : " Solve::inex: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >>" and the output shows exactly the input – Mashriq Ahmed Jun 14 '12 at 04:06
  • @MashriqAhmed In the definition of `v` what do you mean by `Square` ? Is it the second power (`^2`) or the square root (`Sqrt`) ? – b.gatessucks Jun 14 '12 at 09:45
  • Square is not a function. Use ^2. – Chris Degnen Jun 14 '12 at 09:59
  • i used ^2. But its shows the same error: "Solve::inex: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >>" i am so stuck with this problem. – Mashriq Ahmed Jun 14 '12 at 17:30
  • @ b.gatessucks. its not Square, its (^2) – Mashriq Ahmed Jun 14 '12 at 17:31

1 Answers1

1

This simpler version solves:

sol = DSolve[{mu1*u1[x] - u1''[x] - Cos[x]* u1[x] == 0,
    u1[-2] == 1,
    u1'[-2] == 0},
   u1, x];

GraphicsRow[Table[Plot[Evaluate[u1[x] /. sol], {x, 0, 20},
   PlotRange -> All], {mu1, 1, 3}]]

enter image description here

But your equation with v included has problems, presumably mathematical:

DSolve[{mu1*u1[x] - u1''[x] -
    ((1.453*Sech[x + 1])^2 + I*Sech[x + 1]*Tanh[x + 1])* u1[x] == 0,
  u1[-2] == 1,
  u1'[-2] == 0},
 u1, x]
Chris Degnen
  • 8,443
  • 2
  • 23
  • 40
  • The equation of v[x] is V[x]=(A sech(x ±D/2 ))^2+i B sech(x ±D/2 )tanh(x ±D/2), where A=sqrt(2+(B^2) /9) and D is the separation between the two potentials. This is the equation of a Potential. Can this differential equation be solved for u1[x]??? – Mashriq Ahmed Jun 14 '12 at 17:42