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Partial Differential Equations

  1. Oct 27, 2011 #1
    Can anyone help with these problems? I have no idea where to start. What is the general approach?

    Determine the solution of ∂ρ/∂t = (sin x)ρ which satisfies ρ(x,0) = cos x.
    Determine the solution of ∂ρ/∂t = ρ which satisfies ρ(x,t) = 1 + sin x along x =-2t.

    Relevant equations: ∂ρ/∂t + ∂/∂x(q(ρ)) = 0 or ∂ρ/∂t + ∂/∂x(ρu(ρ)) = 0 and q = ρu.
     
    Last edited: Oct 27, 2011
  2. jcsd
  3. Oct 28, 2011 #2

    Fredrik

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    I suck at differential equations, but I think it's pretty clear that the approach should be to solve these equations as if they were ordinary differential equations (since they contain no derivatives with respect to x). The "constants" that appear in the solutions can of course depend on x, so instead of writing e.g. A, you write A(x), where A is a function. Then you use the "which satisfies..." statements to find A.
     
  4. Oct 28, 2011 #3
    So how would I solve them? Which method should I use?
     
  5. Oct 28, 2011 #4

    Fredrik

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    You should look at them and immediately see the solution. If you can solve the ordinary differential equation y'=y, you can solve these.

    Edit: Uh, wait. What equations are you trying to solve? You said that you were looking for the solutions of ∂ρ/∂t = ρ and a similar equation. But did you perhaps mean something entirely different? I don't understand what the "relevant equations" have to do with anything, or what q and u are. If the "relevant" equations are the ones you're trying to solve, and the first things you said meant something entirely different than what I thought, then I don't know the answer.

    Edit 2: If I was right the first time about what equations you want to solve, then you can do this: Let x be arbitrary. Define a function f by f(t)=ρ(x,t). Plug this into the equation you want to solve. Solve it for f. Since x was arbitrary, the "constants" in the solution may have different values for different values of x. Figure out the rest.
     
    Last edited: Oct 28, 2011
  6. Oct 29, 2011 #5
    Thanks.
     
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