I have some problems here. The first involves what I believe is called Burger's Equation, but in any case, I have a partial differential equation to solve, and all I would like to find out is the step by step process of going about it, so I can work it out myself. We have u times δu/δx + δu/δt = 2. The initial conditions are given as u(x, 0 ) = -x. We are first asked to find the Characteristic Curves. What I did was use the Chain Rule, and went about as follows :
To begin with, let us look at the Chain Rule to see if we can make the equation above fit the pattern it gives. For this we have :
du/dt=δu/δx times dx/dt + δu/δt times dt/dt, and so, to give it a solution more conducive to this, we make sure that the u in the problem is analogous to the dx/dt in this variation of Burger's Equation, both being the coefficients of δu/δx , and the 2 from the Right Hand Side of the problem is like the du/dt in the Chain Rule expression, and it is obvious that dt/dt=1.
So dx/dt times δu/δx + δu/δt = du/dt compared to u times δu/δx + δu/δt = 2 makes u=dx/dt, and du/dt=2. For a start, this means that along the characteristic lines, u changes for each second with a slope of two, where if we integrate du/dt=2, and get ∫(du/dt) dt=∫2dt, we end up with u=2t+A , with A some constant, or function not dependent upon t, and at t=0, because the initial condition states that u(x,0)= -x, then A= -x, while for u=dx/dt, this implies that x=ut+B, with B a constant of integration or function not depending on t, since dx/dt= d/dt(ut)=u.
I also went on to add : From the information above, since u=2t+A,∀t∈R∶t ≥0, and it was found that A= -x, then u=2t-x. But since it is also known that x=ut+B, and if we rearrange u=2t-x, we can also shew that x=2t-u, then ut+B= 2t-u.
Now out of this, which are the characteristic curves ? Are they the lines x=ut+B, and u=2t-x, or have I totally mucked this up ? At what stage also should I have put in the change of coordinate variables with t=tau and x=xi ?
We are required next to find the solution u in terms of the original variables x and t, but I am fuzzy on how to relate this to the chain rule equation. We are also to go on to sketch this, and I assume these are lines in the plane.
We are also asked to consider this same initial value problem but with a new initial condition that u(x, 0) = f(x), where f'(x) is bounded, then to find those characteristic curves and from the Jacobian, determine the earliest possible time at which the solution will break down.
We are also informed that it is not possible to find an explicit form of u as a function of x and t but it is possible to find an implicit solution F(x, t, u) = 0, and we are asked to find it, and I have been looking at YouTube videos all weekend to work out the procedures for solving PDE'S like these, but I am still not sure how to proceed.
Finally, we are asked to consider the initial value problem δu/δt - u squared times δu/δx = 0, with initial conditions in a piecewise, Heaviside type set up, u(x, 0) = g(x) = { negative half, if x less than or equal to zero ; 1 if x between zero and 1 exclusive ; and one half when x is greater than or equal to one }
From this we are asked to sketch or plot those base characteristics, then find the solution, as well as determine in which region it is single-valued. Next we sketch or plot the solution for t = 0, 2, 4, and finally find the shock solution for this system.
I have some ideas, but I just need some hints to put it all together, in the way of some step by step procedures. Thank You.
In your integration of u you are confosing the initial value (at t=0) of x and its vakue at a later time. From $u=dx/dt$ and $u=2t-x_o$ one gets $x(t) = t^2 -x_0t + x_0$, i.e.: $u(t^2 -x_0t + x_0, t) = 2t-x_0$. Now, given $x,t$, you can solve for $x_0$ and write the solution $u$ in terms of $x,t$ to get $u(x,t) = 2t - (x-t)/(1-t)$ for small $t$.