h(x, t) of paint at the wall at position x and time t
is described by the the quasi-linear PDE
d h(x, t) + h^2(x, t)d h(x,t) = s(x, t)
We have added a source term s(x, t) which
describes the rate with which we add paint to the wall. In order to add paint where at the moment is little paint, we choose
s(x, t) = 1
The Cauchy data are given in implicit form:
t0(s) = 0; x0(s) = s; h0(s) = 1 - tanh(s)
Use the method of characteristics to find the solution in the form
x = h^2t - t^2 + arctan(1-sqrt(h^2-2t))
i have managed to get the h^2t from the required answer but as far as the arctan bit goes i dont have a clue.
Any help please!!!!