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Modified Heat Equation

  1. Aug 8, 2007 #1
    Hi folks,

    Given the following heat equation

    [tex]u_t = u_{xx} + t - x^2,[/tex]

    I'd like to find all solutions [tex]u(x,t)\in C^2(\mathbb{R}^2)[/tex] such that the quotient

    [tex]|u(x,t)| / (|x|^5 + |t|^5)[/tex]

    goes to zero as the sum [tex]|x| + |t|[/tex] goes to infinity.

    I know how to do the same problem with the usual heat equation [tex]u_t = u_{xx}[/tex], but I'm not entirely sure how to deal with this extra [tex]t - x^2[/tex] term. I suspect I can still start by taking the Fourier transform (in x) of each side to get something like:

    [tex]\partial_t \hat{u}(\xi,t) = \widehat{\partial_t u}(\xi, t) = -|\xi|^2 \hat{u}(\xi,t) + \widehat{t-x^2}(\xi) = -|\xi|^2\hat{u}(\xi, t) + t\delta(\xi) - \delta^{\prime\prime}(\xi),[/tex]

    which gives me an ODE in t that is easy enough to solve. The issue is that I think this method only gives me smooth (by which I mean infinitely differentiable) solutions. Are there other [tex]C^2[/tex] solutions that I am missing with this approach?

    Thanks in advance for any help.
  2. jcsd
  3. Aug 8, 2007 #2
    This solves "nicely" by a modified techinique using Seperation of Variables. If the boundary value problem is on a string of finite length I can post all the steps where are required to solve this analytically.
  4. Aug 9, 2007 #3
    Your equation is linear, so, as long as you have nice boundary conditions, the solution is unique. Check your notes.
  5. Aug 9, 2007 #4
    Why is it linear? Perhaps you mean to say it is quasi-linear. It cannot be linear for if u_1 and u_2 are solutions does not mean that u_1 + u_2 are solutions.

    @bob321. I will post complete steps, but I am unable to since you do not provide a boundary and initial value problems.
  6. Aug 9, 2007 #5
    The problem is over the entire real line so there are no boundary conditions, and the initial condition [tex]u(x,0)[\tex] can be an arbitrary [tex]C^2[\tex] function. I've actually since worked out the general solution using the Fourier transform, as I started to do in my original post.

  7. Aug 10, 2007 #6
    It is linear. If [itex]L=u_t+u_{xx},[/itex] then [itex]L[u_1+u_2]=L[u_1]+L[u_2][/itex]. What is not is homogeneous. A PDE is said to be quasilinear when is linear in the higher derivative term, but not necesarily in the terms of lower order.

    That is a boundary condition. You want your solutions to converge at [itex]\pm \infty[/itex], so [itex]\lim_{x\rightarrow \pm \infty}u(x,t)=0[/itex].
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