Results 1 to 6 of 6

Math Help - Mixed Partial Derivatives

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    5

    Mixed Partial Derivatives

    Solve the equation u_xx - 3u_xt - 4u_tt = 0 subject to the initial conditions u(x,0) = x^2 and u_t(x,t) = e^x.

    Hint: consider a change of coordinates.

    Having spent a large amount of time getting wrong solutions to the above question, any help that can be offered would be very much appreciated!

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    show your working and we'll tell you what's wrong (although I don't think that'll be me ^^)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    5
    Ok no problem!

    I have seen something similar where you factor out the operator: (d_xx - 3d_xt - 4d_tt) = 0

    This gives us: (d_t + d_x)(4d_t + d_x)u = 0

    I then have attempted to define new coordinates. I was led to believe that I should try, for example: a = x +t and b = -4x + t (though this does not appear to work so perhaps this is where I have gone wrong?).

    We then want to find a v(a,b) such that it solves the equation.

    We then get;

    u_x = v_a - 4v_b
    u_t = v_a + v_b

    I then continue...but will avoid writing out the rest of it now in case my coordinates are wrong! If anybody can advise me if they are or are not correct then I will post the rest of my solution on that basis!

    Many thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Rocky View Post
    I have seen something similar where you factor out the operator: (d_xx - 3d_xt - 4d_tt) = 0

    This gives us: (d_t + d_x)(4d_t + d_x)u = 0 I think you mean (d_x + d_t)(d_x 4d_t)u = 0.

    I then have attempted to define new coordinates. I was led to believe that I should try, for example: a = x +t and b = -4x + t (though this does not appear to work so perhaps this is where I have gone wrong?).
    I get a = x t and b = 4x + t.

    We then want to find a v(a,b) such that it solves the equation.

    We then get;

    u_x = v_a - 4v_b
    u_t = v_a + v_b

    I then continue...
    Keep going, you're on the right lines. Now use the chain rule to find u_{xx},\; u_{xt} and u_{tt} in terms of v_{aa},\ v_{ab} and v_{bb}. For example, u_{xx} = \frac{\partial}{\partial a}(u_x)\frac{\partial a}{\partial x} + \frac{\partial}{\partial b}(u_x)\frac{\partial b}{\partial x}.

    You should then find that u_{xx}-3u_{xt}-4u_{tt} is some constant times v_{ab} (the terms in v_{aa} and v_{bb} all cancel out). So your equation is equivalent to v_{ab} = 0, whose general solution is v(a,b) = f(a)+g(b) for arbitrary (twice-differentiable) functions f, g. So the solution for u(x,t) is u(x,t) = f(x-t) + g(4x+t).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2009
    Posts
    5
    That is a brilliant help thanks very much! I have now got to this stage...but am now a little confused as to how I take into the account the initial conditions.

    I have attempted to somehow adapt d'Alambert's solution to the wave equation (or something along these lines) but the solution I attain does not seem to satisfy the original equation!

    In particular I obtained a solution along the lines of:

    (1/2) (x-t)^2 + (1/2) (4x+t)^2 + the integral of e^y dy (between the limits (x-t) and (4x+t))

    Again any help in finishing this question would be much appreciated! Thanks!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Rocky View Post
    Solve the equation u_xx - 3u_xt - 4u_tt = 0 subject to the initial conditions u(x,0) = x^2 and u_t(x,t) = e^x.
    Is that last equation correct? I would expect it to be u_t(x,0) = e^x. (if it is an initial condition, then it ought to tell you what happens at time 0.)
    In that case, if the general solution is u(x,t) = f(x-t) + g(4x+t) then the initial conditions are

    f(x) + g(4x) = x^2,\quad -f'(x) + g'(4x) = e^x.

    Integrate the second one to get -f(x) + \frac14g(4x) = e^x. (There should be a constant of integration somewhere there, but it will cancel out in the eventual solution, so let's ignore it.)

    Solve those two simultaneous equations for f(x) and g(4x) to get

    f(x) = \frac15(x^2-4e^x),\quad g(4x) = \frac45(x^2+e^x).

    Therefore

    f(x-t) = \frac15\bigl((x-t)^2-4e^{x-t}\bigr),\quad g(4x+t) = g\bigl(4(x+\frac t4)\bigr) = \frac45\bigl((x+\frac t4)^2+e^{x+(t/4)}).

    Add those together to get the solution for u(x,t).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial derivatives.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 30th 2011, 01:33 PM
  2. Partial Derivatives
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 16th 2010, 08:03 PM
  3. Higher Order Mixed Partial Derivatives
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 20th 2009, 05:58 PM
  4. Higher Order Mixed Partial Derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 18th 2009, 03:36 PM
  5. mixed partial derivatives
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 5th 2009, 07:14 AM

Search Tags


/mathhelpforum @mathhelpforum