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 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 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!!!