Solution on nonlinear pde with intial condtion

Question

It produces an error. please suggest an alternative solution to the above-coupled pde problem.

 ClearAll[U, V]
System = {D[U[x, t], t] == - U[x, t] D[U[x, t], x]-9.8 D[V[x, t], x], 
   D[V[x,t],t] == -V[x,t] D[U[x,t],x]-U[x,t] D[V[x,t],x]-(0.2x-20) D[U[x,t],x]-0.2 U[x, t],
    V[x, 0] == 2 (Sech[0.0136931 x])^2, 
   U[x, 0] == 1.4 (Sech[0.0136931 x])^2 };
{U, V} = NDSolveValue[System, {U, V}, {x, 0, 1}, {t, 0, 1}]

Answer 1

It produces an error.

You need boundary conditions. You only have initial conditions. For numerical solver, both are needed.

I put some below, but you need to put the correct ones as you know the problem better.

ClearAll[U, V, x, t]
g = 98/10;
a = 986691840920751/72057594037927936;
pde1 = D[U[x, t], t] == -U[x, t] *D[U[x, t], x] - g *D[V[x, t], x]
pde2 = D[V[x, t], t] == -V[x, t]* D[U[x, t], x] - 
    U[x, t]* D[V[x, t], x] - (2/10*x - 20) D[U[x, t], x] - 2/10 *U[x, t];
ic = {V[x, 0] == 2 *Sech[a*x]^2, U[x, 0] == 14/10 (Sech[a* x])^2}
bc = {V[0, t] == 0, U[0, t] == 0}
System = {pde1, pde2, bc, ic};
NDSolveValue[System, {U, V}, {x, 0, 1}, {t, 0, 1}]

Answer 2

There are initial values missing.

For an example I assume the following initial values:

U[0, t] == 1.4, V[0, t] == 2

Then there should be a comma instead of a semicolon.

Then you must decrease the initial step size not to run into trouble.

With all these changes:

ClearAll[U, V]
System = {D[U[x, t], t] == -U[x, t] D[U[x, t], x] - 9.8 D[V[x, t], x],
    D[V[x, t], t] == -V[x, t] D[U[x, t], x] - 
     U[x, t] D[V[x, t], x] - (0.2 x - 20) D[U[x, t], x] - 0.2 U[x, t],
    V[x, 0] == 2 (Sech[0.0136931 x])^2, 
   U[x, 0] == 1.4 (Sech[0.0136931 x])^2, U[0, t] == 1.4, V[0, t] == 2};
{U, V} = NDSolveValue[System, {U, V}, {x, 0, 1}, {t, 0, 1}, 
  StartingStepSize -> 0.01]