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Pde with boundary conditions

  1. Nov 1, 2009 #1
    [tex]u_t=u_{xx}+2u_x[/tex]
    0<=x<=L, t>=0, u(x,0)=f(x), u_x(0,t)=u_x(L,t)=0

    How to do this?
     
  2. jcsd
  3. Nov 1, 2009 #2

    HallsofIvy

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    Just a standard "separation of variables" will work.

    Assume u is of the form u(x,t)= A(x)B(t). Then ux= A'B, ut= AB', and uxx= A"B. Putting those into the equation, AB'= A"B+ 2A'B.

    Divide both sides by AB:
    [tex]\frac{B'}{B}= \frac{A"+2A'}{A}[/tex]

    Since the left side depends upon t only while the right side depends upon x only, in oder to be equal they must both be equal to the same constant. Call that constant "[itex]\lambda[/itex]".

    Then we have [itex]B'/B= \lambda[/itex] or [itex]B'= \lambda B[/itex] and [itex](A"+ 2A')/A= \lambda[/itex] or [itex]A"+ 2A'= \lambda A[/itex].

    Now look at the boundary conditions: [itex]u_x(0,t)= A(0)B'(t)= 0[/itex] for all t so we must have A(0)= 0. Also, [itex]u_x(L,t)= A(0)B'(t)= 0[/itex] for all t so A(L)= 0.

    Solve the boundary problem [itex]A"+ 2A'- \lamba A= 0[/itex], A(0)= 0, A'(0)= 0 to determine what [itex]\lambda[/itex] can be. You should see that this is an eigenvalue problem. [itex]A"+ 2A'= \lambda A[/itex] is trivially satisfied for A(x) identically equal to 0 but then we can't satisfy "u(x,0)= f(x)". The values of [itex]\lambda[/itex] for which there are not trivial solutions are the "eigenvalues" for this problem and the non-trivial solutions are the "eigenvectors" or, in this case, "eigenfunctions". Once you have determined what the eigenvalues are, you can put them into B"= \lamba B to solve that.

    As you may have learned in Linear Algebra (which should be a prerequisite for both differential equations and partial differential equations) the set of all vectors satisfying [itex]Lv= \lambda v[/itex] for a subspace so there will be an infinite number of eigenfunctions. Generally you will need to write the entire solution as a sum of eigenfunctions.

    Once you have all of that theory, you can start talking about Fourier series. Solutions to linear differential equations with constant coefficients can be written as Fourier series
    [tex]\sum P_n(t)sin(\frac{n\pi}{L}x)+ Q_n(t)cos(\frac{n\pi}{L}x)[/tex]

    And since the derivative of sin is never 0 at multiples of [itex]\pi[/itex], you could see immediately that the coefficients of [itex]sin((n\pi/L)x) must be 0. The simplest way to do this problem, jumping over a lot of preliminary theory, is to start by assuming a function of the form
    [tex]u(x,t)= \sum_{n=0}^\infty A_n(t)sin(\frac{n\pi}{L}x)[/tex]
     
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