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Change of variables in Heat Equation (and Fourier Series)

  1. Feb 5, 2017 #1
    Q: Suppose ##u(x,t)## satisfies the heat equation for ##0<x<a## with the usual initial condition ##u(x,0)=f(x)##, and the temperature given to be a non-zero constant C on the surfaces ##x=0## and ##x=a##.
    We have BCs ##u(0,t) = u(a,t) = C.## Our standard method for finding u doesn't work here, since ##e^{-k(\frac{n\pi}a)^2t}sin(\frac{n\pi}a)## does not satisfy these BCs.
    Make a change of variable from ##u## to ##v=u-C.## Show that ##v## satisfies the heat equation with BCs ##v=0## at ##x=0## and ##x=a.##
    Write down the solution for ##v(x,t).##Deduce an expression for ##u(x,t)## in terms of constants ##c_1,c_2,\ldots,## and write down a formula for ##c_n.##
    [Harder] Now suppose the BCs are ##u(0,t) = C##, ##u(a,t)=D## for constants ##C,D.## How could you solve the case?

    My question: These are extensions to homework which I'd like try to attempt, but I don't know where to start with the change of variable
     
  2. jcsd
  3. Feb 5, 2017 #2

    pasmith

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    Substitute [itex]u(x,t) = v(x,t) + C[/itex] into the PDE and boundary and initial conditions you are given for [itex]u[/itex] to obtain a PDE and boundary and initial conditions satisfied by [itex]v[/itex].
     
  4. Feb 5, 2017 #3
    From the given BCs for ##u##, am I right in saying that BCs for ##v## is ##v(0,t)=v(a,t)=u(0,t)+C=2C##? Also by substituting ##u(x,t)=v(x,t)+C## into the PDE do you mean partially differentiate it then substitute in like $$v=u+C$$ $$ \frac{\partial{v}}{\partial{t}}=\frac{\partial{u}}{\partial{t}}$$ $$\frac{\partial^2{v}}{\partial^2{x}}=\frac{\partial^2{u}}{\partial^2{x}}$$ so $$\frac{\partial{u}}{\partial{t}}=K\frac{\partial^2{u}}{\partial{x}^2}$$ becomes $$\frac{\partial{v}}{\partial{t}}=K\frac{\partial^2{v}}{\partial{x}^2}$$
    Did I get even the slightest of that right or am I going to a complete different direction??
     
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