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Heat conduction in bars: varying the boundary condition. Harmonics problem

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data

    A problem with odd harmonics only. Show that the solution of the heat equation du/dt=c2*(d2u)/(dx2), subject to boundary conditions u(0,t)=0 and ux(L,t)=0, and the initial condition u(x,0)=f(x) , is

    u(x,t)= [tex]\sum[/tex] Bnsin[([tex]\pi[/tex]/2L)(2n+1)x]e-((c*[tex]\pi[/tex]/2L)*(2n+1))^2

    where n extends from 0 to [tex]\infty[/tex]

    Bn=2/L[tex]\int[/tex] f(x) sin[([tex]\pi[/tex]/2L)*(2n+1)x]dx

    where the limits extend from 0 to L

    2. Relevant equations

    Seperation of variables

    3. The attempt at a solution

    if u(0,t)=0 then u2(0,t)=0

    u2(0,t)=0= u2(L,t)

    not sure how to finish this derivation.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 13, 2008 #2


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    Gold Member

    This part is correct. What does that make the general solution for T(t) and X(t) and hence u(x,t)? Remember, you will need to examine 3 different cases: (1)k=0, (2)k<0 and (3)k>0.

    Huh? What is this supposed to mean?

    Where are you getting these boundary conditions? They are not equal to the conditions given in the question.
  4. Oct 13, 2008 #3
    3 cases : you mean when

    case 1: lambda =alpha^2 ==> X-alpha^2*X


    (r-alpha)(r+alpha)=0 ==> r= +-alpha

    X=c1e^alpha*x + c2e^alphae^-alpha
    X'(0)=0 ==> c1*alpha-c2*alpha
    X'(L)=alpha(e^alpha*L - e^(-alpha*L))c1=0 ==> c1=0

    case 2 lamda =0



    X'(0)=0 ==> c1=0

    X'(L) =0 ==>0=0


    case 3

    lambda = alpha^2 >0

    r^2+alpha^2 =0 ==> r=+- alpha*i

    X1=cos(alpha*x) X2=sin(alpha*x)

    X=X1+X2= cos(alpha*x)+sin(alpha*x)

    Ignore all the functions with u2. Those are all mistakes. ; I was trying to say that if u(0,t)=0 then u(0,t)=0==> ux(0,t)=0

    Therefore, ux(0,t)=0= ux(L,t)
    Last edited: Oct 13, 2008
  5. Oct 13, 2008 #4
    1. After SOV, you'll get 2 ODEs. One for spatial the other for time, then you select proper eigenvalue such that the PDE has nontrivial and physically possible solution considering the given BCs. In the case of
    [tex] \dfrac {T'}{T}= k \dfrac {X''}{X}=- \lambda [/tex] with Dirichlet boundary conditions, the possible eigenvalue is [tex]\lambda >0[/tex] (easy to find out by plugging the BCs in).

    2. Solve the spatial ODE, determine the coefficients and eigenvalues to satisfy the BCs.
    3. Solve the T, and plug the eigenvalue into the T.
    4. Write the u(x,t) in series form by applying the principle of superposition.
    3. Apply initial condition to u(x,t) by setting t=0 such that u(x,0)=f(x).
    4. determine the Fourier coeffs.
  6. Oct 15, 2008 #5
    Should I let X= c1 cos ([tex]\mu[/tex] x) + c2 sin ([tex]\mu[/tex]x)
    and k=-[tex]\mu[/tex]2 ; should I write T/T' in terms of c and k ?

    What is the spatial ODE?
    Not sure how to determined the eigenvalues
  7. Oct 15, 2008 #6
    Let's say you have a 1D heat diffusion problem
    \dfrac{\partial u}{\partial t} = k \dfrac{\partial ^2u}{\partial x^2}
    where 0<=x<=L, and with Dirichlet boundary conditions
    and initial temperature distribution

    To SOV set:
    We could set the separation constant [tex]-\lambda[/tex]
    you get
    \dfrac{\phi^''}{\phi}=\dfrac{1}{k} \dfrac {T^'}{t}=- \lambda
    you end up having 2 ODEs
    \phi^{''}+\lambda \phi = 0
    T^{'}+\lambda kt = 0

    [tex]\phi^{''}+\lambda \phi = 0[/tex] is the spatial ODE. it's the simplest form of Sturm-Liouville Eigenvalue problem.
    we can prove that the only way to get nontrivial solution is when [tex]\lambda>0[/tex], and the general solution to the spatial problem is
    \phi(x)=c1 sin(\sqrt{\lambda}x)+c2 cos(\sqrt{\lambda}x)
    To satisfy the Dirichlet BC, it requires c2=0 and
    \lambda=(\dfrac{n \pi}{L})^2
    now you have
    \phi(x)= cn \sin(\dfrac{n\pi x}{L})
    the solution to T is
    T(t)=Bne^{-k \lambda t}
    Plug lambda into T(t) and apply principle of superposition you have
    u(x,t)=\sum _{n=1} ^{\infty} A_{n}sin \dfrac{n \pi x}{L}e^{-k(n \pi/L)^2t}

    Now you need to make u(x,t) satisfy the initial condition and determine the A_{n}. Let us know if you have further problems.

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