Solving sine-Gordon equation with boundary condition

Question

I need to numerically solve the sine-Gordon equation $$ \partial_{x,x} u(x,t) - \partial_{t,t} u(x,t) - \sin(u(x,t)) -\alpha \partial_t u(x,t) + \gamma =0 $$ for $x\in [0,15]$ and $t\in [0,2]$ with two boundary conditions $$\partial_x u(x,t)\vert_{x=0} = h, \hspace{0.5cm} \partial_x u(x,t)|_{x=15} = h + a_{ext} \sin(\omega_{ext} t). $$

My non-working Mathematica code is

al = 0.08; ga = 0.01; h = 5; aext = 3; omegaext = 1.4;
NDSolve[{D[u[x,t],x,x] - D[u[x,t],t,t] - Sin[u[x,t]] -al*D[u[x,t],t] + ga ==0, 
(D[u[x, t],x]/. {x -> 0}) == h, (D[u[x, t], x] /. {x -> 15}) == h + aext Sin[omegaext t] }, u, {x,0,15},{t,0,2}]

I get an error message like this

NDSolve::fembdnl: The dependent variable in (u^(1,0))[0,t]==5 in the boundary condition DirichletCondition[(u^(1,0))[0,t]==5,x==0.] needs to be linear.

Can anyone suggest me how to fix this?

Answer 1

There are several issues. First there need to be sufficient initial conditions for this telegraph type equation. Also you do not specify a Dirichlet type boundary condition. I have added arbitrary (wrong) information to get you started but you'd need to fix that.

Here is a tensor product version - which you should use if you are not familiar with the finite element method given below:

al = 0.08; ga = 0.01; h = 5; aext = 3; omegaext = 1.4;
sol1 = NDSolveValue[{D[u[x, t], x, x] - D[u[x, t], t, t] - 
     Sin[u[x, t]] - al*D[u[x, t], t] + ga == 0,(*(D[u[x,t],
   x]/. {x\[Rule]0})\[Equal]h,*)(D[u[x, t], x] /. {x -> 15}) == 
    h + aext Sin[omegaext t]
   (* fix these *)
   , u[0, t] == 0
   , u[x, 0] == 0
   , Derivative[0, 1][u][x, 0] == 0
   }, u, {x, 0, 15}, {t, 0, 2}]

FEM version:

al = 0.08; ga = 0.01; h = 5; aext = 3; omegaext = 1.4;
sol2 = NDSolveValue[{D[u[x, t], x, x] - D[u[x, t], t, t] - 
     Sin[u[x, t]] - al*D[u[x, t], t] + ga == 
    NeumannValue[h + aext Sin[omegaext t], x == 15]
   , u[x, 0] == 0
   , Derivative[0, 1][u][x, 0] == 0
   , DirichletCondition[u[x, t] == 0, x == 0]
   }, u, {x, 0, 15}, {t, 0, 2}]

For this one you probably want to think about the sign in the NeumannValue

Plot3D[{sol1[x, t], sol2[x, t]}, {x, 0, 15}, {t, 0, 2}, 
 PlotRange -> All]

Also note this coupled nonlinear Sine-Gordon example.