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And so I want to solve the following equation, subject to these initial conditions:

$\ u_{tt} - u_{xx} = 6u^5+(8+4a)u^3-(2+4a)u$

$\ u(0,x)=\tanh(x), u_t(0,x)=0$

When I use NDSolve to solve within the intervals $\ [0,10] \times [-5,5]$, I tried this as a code:

NDSolve[{
   D[u[t, x], t, t] - D[u[t, x], x, x] ==
     6 u[t, x]^5 + (8+4a) u[t, x]^3 - (2+4a) u[t, x],
   u[0, x] == Tanh[x], D[u[0, x], t] == 0},
 u[t, x], {t, 0, 10}, {x, -5, 5}]

But Mathematica then returns the following line:

NDSolve::deqn: Equation or list of equations expected instead of True in the first argument

where the True affirmation seemingly refers to the second initial condition. What is it that I did wrong? Is there anything wrong with the code? Is there anything wrong with the problem?

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  • $\begingroup$ D[u[0, x], t] == 0 is trivially true because u[0,x] does not depend on t. To write the equation you mention, write it as Derivative[1, 0][u][0, x]. As a beginner as yet unfamiliar with Derivative, you could have constructed this expression as D[u[t, x], t] /. t -> 0, i.e. take the derivative first, and substitute the value t=0 only afterwards. $\endgroup$ – Szabolcs Feb 4 '14 at 16:26
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Since u[x,0] does not depend on t differentiating leads to 0==0. It should be Derivative[1, 0][u][0, x] == 0 Also for your problem one needs some boundary conditions.

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  • $\begingroup$ Now I'd love to be able to compute a numerical integral of the solution, or a function thereof... $\endgroup$ – NSERC Protester Feb 6 '14 at 15:41

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