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NDSolve::ivone: Boundary values may only be specified for one independent variable. Initial values may only be specified at one value of the other independent variable. >>

I keep getting this, although I wish to use NDSolveValue to solve the following differential equation: $\ u_{xx}-\frac{u_t}{\sqrt{t^2+x^2}}=-6u^5+(8+4e)u^3-(2+4e)u $ where e is a parameter, with the initial conditions being $\ u(0.001,x)=tanh(x), u_t(0.001,x)=0$ and the boundary conditions being $\ u(t,0)=0,u(t,6)=1 $. I have loads of operations I wish to perform on the solution, but every single operation depends on getting the solution in the first place. Here's the code I intend to run in order to perform everything I want on it:

eps4 = 0;
phi6m4 = NDSolveValue[{
   D[u[t, x], x, x] - 
   D[u[t, x], t]*1/Sqrt[t^2 + x^2] == -6 u[t, x]^5 +
         (8 + 4 eps4) u[t, x]^3 - (2 + 4 eps4)u[t, x],
   u[0.001, x] == Tanh[x],
   Derivative[1, 0][u][0.001, x] == 0,
   u[t, 0] == 0, u[t, 7] == 1}
     , u, {t, 0.001, 6}, {x, 0, 7}];

NDSolveValue::ivone: Boundary values may only be specified for one independent variable. Initial values may only be specified at one value of the other independent variable. >>

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    $\begingroup$ the error is in NDSolveValue, why post all that other stuff? $\endgroup$ – george2079 Mar 5 '14 at 20:19
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    $\begingroup$ The error arises due to specifying initial condition on both u and D[u,t] which is not permissable since the system is first order in t. $\endgroup$ – george2079 Mar 5 '14 at 20:29
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As it is already mentioned in the comments, the initial condition Derivative[1, 0][u][0.001, x] == 0, should not appear there. After it removal the equation is solved as expected:

    eps4 = 0;
phi6m4V = 
 NDSolveValue[{D[u[t, x], x, x] - 
     D[u[t, x], t]/
      Sqrt[t^2 + x^2] == -6 u[t, x]^5 + (8 + 4 eps4) u[t, x]^3 - (2 + 
        4 eps4) u[t, x], u[0.001, x] == Tanh[x], u[t, 0] == 0, 
   u[t, 7] == 1}, u, {t, 0.001, 6}, {x, 0, 7}]

This

 Plot3D[phi6m4V[t, x], {t, 0.001, 6}, {x, 0, 7}]

Shows the solution: enter image description here

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