|Aug22-11, 07:20 AM||#1|
First order linear PDE, need help understanding solution/method
1. The problem statement, all variables and given/known data
Solve the initial boundary value problem:
u_t + cu_x = -ku
u is a function of x,t
u(x,0) = 0, x > 0
u(0,t) = g(t), t > 0
treat the domains x > ct and x < ct differently in this problem. the boundary condition affects the solution in the region x < ct, while the IC affects it in the region x > ct.
3. The attempt at a solution
The question previous question walks you through transforming the coordinates to get the general solution:
A = x - ct B = t
So by the chain rule:
u_x = (dA/dx)u_A + (dB/dx)u_B
u_t = (dA/dt)u_A + (dB/dt)u_B
dA/dx = 1 dB/dt = 0
dA/dt = 1 dB/dt = 1
u_x = u_A
u_t = -cu_A + u_B
the new equation is now:
-cu_A + u_B + cu_A + ku
= u_B + ku
u_B = -ku
--> u = exp(-kB).f(A)
--> u(x,t) = exp(-kt).f(x-ct)
the justification given here is: "since the intergration constant is only constant in B, it may depend on A". I don't understand this step, what does that justification mean? I see that all they did was swap from B and A back to x-ct and t though I dont quite understand whats happening.
Anyway, now we have the general solution, so we can work on the main part of the question:
u(x,t) = exp(-kt).f(x-ct)
Since u(x,0) = 0 then f(x) = 0 for x > 0
Since u(0,t) = g(t) then exp(-kt).f(-ct) = g(t) for t > 0
Here is where I get confused...
=> f(z) = g(-z/c) exp(-(kz/c)) for z < 0
then the next step says
=> u(x,t) = 0 for x > ct
and u(x,t) = exp(-kt).exp(-(k/c)*(x-ct)).g(-(x-ct)/c) for x < ct
=> u(x,t) = exp(-kx/c).g(t-(x/c)) for x < ct
No idea what happens once they substitute z, and why they are doing the things they've done. Any help is appreciated. Sorry for the difficult notation and thank you in advance!!!
|Aug22-11, 11:39 AM||#2|
With four first equation, it's just s trick. The change of variables is perfectly valid, because if you calculate the Jacobian you obtain a non-zero result, that means the co-ordinate system (A,B) is perfectly allowable. Or that I think you mean that if f(x,y)=A(x) + B(y) say, then [itex]\partial f/\partial x=A'(x)[/itex], remember these are partial differentials that you're dealing with, not full ones.
For the second part of your question, you need to understand that f is a function of a single variable only, for the second part that are introducing a new variable z=-ct, as [itex]t\geqslant 0[/itex], we can say that [itex]z\leqslant 0[/itex], so we have found f(B) in two regions, when [itex]B\leqslant 0[/itex] and when [itex]B\geqslant 0[/itex]. The variable in the equation is x-ct, so for x-ct>0, the solution is zero, i.e. for x>ct and for x-ct<0, the solution is as you mentioned it, .e. for x<ct.
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