First order linear PDE, understanding solution/method

[complex000]

Homework Statement

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.

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

Therefore

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

therefore

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 don't quite understand what's 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!

 

[hunt_mat]

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 $\partial f/\partial x=A'(x)$, 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 $t\geqslant 0$, we can say that $z\leqslant 0$, so we have found f(B) in two regions, when $B\leqslant 0$ and when $B\geqslant 0$. 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.