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Non-Homogenous PDE

  1. Apr 27, 2003 #1
    I have an assignment due at the end of the week, and I was wondering if someone could check my working for me, as I am prone to making errors. Also, in Step Five I am unsure how to solve for T(t), can anyone point me in the right direction?

    ∂u/∂t = (c/r)*(r(∂u/∂r)) + Q(r,t)
    0 < r < a ; c > 0

    u(0,t) is finite
    u(a,t) = 0
    u(r,0) = 0

    Q(r,t) = qJ0((&mu;1r)/a)

    Where &mu;1 is the first crossing of J0(x), which is a Bessel Function of the first kind of order zero, q is a constant.

    Solve for u(r,t).

    Step One-

    Use seperation of variables on the homogenous equation, (ie. Set Q(r,t) = 0):

    u(r,t) = R(r)T(t)

    &rArr; &part;u/&part;t = RT'
    &rArr; &part;u/&part;r = R'T
    &rArr; &part;2u/&part;r2 = R"T

    &rArr; (1/c)*(T'/T) = (R"/R) + (1/r)*(R'/R) = -&lambda;

    Step Two-

    Solve for R(r) first:

    r2R" + rR' + &lambda;r2R = 0
    let &lambda; = k2
    and &rho; = kr

    &rArr; dR/r = (dR/d&rho;)*(d&rho;/dr) = k(dR/d&rho;)
    &rArr; d2R/dr2 = k2(d2R/d&rho;2)
    &rArr; &rho;2R"(&rho;) + &rho;R'(&rho;) + &rho;2R(&rho;) = 0

    which is Bessels equation, with &nu; = 0.

    &rArr; R(&rho;) = AJ0(&rho;) + BY0(&rho;)
    &rArr; R(r) = AJ0(kr) + BY0(kr)
    where Y(x) is a Bessel Function of the second kind.

    Now Y(x) approaches infinity as x approaches zero, therefore if R(r) is to satisfy the first boundaryb condition;

    u(0,t) finite;

    then B must equal zero.

    &rArr; R(r) = AJ0(kr)

    The second boundary condition is;

    u(a,t) = R(a)T(t) = 0;

    &rArr; R(a) = 0 and/or T(t) = 0
    Now T(t) = 0 is a trivial solution, as this implies u(r,t) = 0. So we look at R(a) = 0;

    &rArr; R(a) = AJ0(ka) = 0
    A = 0 is another trivial solution, so we look at;

    J0(ka) = 0

    We define &mu;n to be the nth zero of J0, then

    kn = &mu;n/a

    Setting A = 1 we get;
    &rArr; R(r) = J0((&mu;nr)/a)

    Step Three-

    Expand Q(r,t) as a Fourier series of Rn(r):

    Let Q(r,t) = &sum; bn(t)Rn(r)

    Use the orthogonality condition:

    &int; rJ0(knr)J0(kmr)dr = 0 ; n &ne; m

    &int; rJ0(knr)J0(kmr)dr = (a2/2)*J12(kma) ; n = m

    &rArr; bm(t) = 0 ; m &ne; 1

    &rArr; bm(t) = (2q/a2J12(&mu;m))*&int; rertJ02((&mu;1r)/a)dr ; m = 1

    (So far so good, I think. I have a solution for R(r) with no unknowns, and have expanded Q(r,t) out in terms of the eigenfunctions R(r), and have "solved" for the coefficient bm).

    Step Four-

    Substitute the solution;

    u(r,t) = &sum;Rn(r)Tn(t);

    into the original PDE;

    &sum;RnT'n = c&sum;(R"n + (1/r)R'n)Tn + &sum;bnRn

    Now R' + (1/r)R' = -&lambda;R

    &rArr; &sum;(T'n + c&lambdanTn)Rn = &sum;bnRn

    Use the orthogonality condition for Rn to get;

    &sum;T'n + c&lambdanTn = &sum;bn

    &rArr; &sum;T'n + c&lambdanTn = 0 ; n &ne; 1

    &rArr; &sum;T'n + c&lambdanTn = (2q/a2J12(&mu;m))*&int; rertJ02((&mu;1r)/a)dr ; n = 1

    Step Five-

    Now I have to solve for T(t) using the equations above. For n > 1 the solution is easy to find;

    T'n + c&lambda;nTn = 0

    &rArr; Tn(t) = Cncos((&radic;c)(&mu;n/a)) + Dnsin((&radic;c)(&mu;n/a))

    For n = 1 the I am unsure how to find the solution. The DE to solve is given below, and I've never had to solve a DE like it before:

    T1' + c&lambdanTn = (2q/a2J12(&mu;m))*&int; rertJ02((&mu;1r)/a)dr

    I can find the homogenous solution, however I can't find the particular solution. HELP!

    The final solution-

    The final solution is:

    u(r,t) = J0((&mu;0r)/a)[C0cos((&radic;c)(&mu;0/a)) + D0sin((&radic;c)(&mu;0/a))] + (2 to infinity)&sum;J0((&mu;nr)/a)[Cncos((&radic;c)(&mu;n/a)) + Dnsin((&radic;c)(&mu;n/a))] + J0((&mu;1r)/a)[(Particular Solution of T1(t))
    Last edited: Apr 27, 2003
  2. jcsd
  3. Apr 27, 2003 #2

    Tom Mattson

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    Sit tight--I'm printing this out and taking it home.

    Till tomorrow...
  4. Apr 28, 2003 #3
    I realised last nite that I for some reason solved the equation for T(t) as a second order DE, when it is clearly a first order DE. So ignore the final solution and part of step five.

    For n &ge; 1-

    T'n + c&lambda;nTn = 0

    T(t) = e-(c&lambda;nt)
  5. Apr 28, 2003 #4

    Tom Mattson

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    I had some trouble getting through your initial post because of what I am convinced are typos.

    For instance, think that this:

    Should be this:


    At least, that would be consistent with what you write later.

    Also this:

    Should presumably be this:

    &rArr; dR/dr = (dR/d&rho;)*(d&rho;/dr) = k(dR/d&rho;)

    Also, I don't know what


    is supposed to be. It shows up as a rectangle.

    If I'm right about those typos, then you are OK through Steps 1 and 2. I am still checking Step 3.
  6. Apr 29, 2003 #5
    Re: Re: Non-Homogenous PDE

    Yep, there was a typo there. Although it should still be c/r, and not c:


    Also this:

    Yep, that is correct.

    That is supposed to be a "implies" symbol, it doesn't seem to work on all broswers and OS's. Sorry.

    Cool. :smile:
  7. Apr 30, 2003 #6
    I think I've solved it!

    I found out today that the reason I was having so much difficulty with the problem was that the assignment had a "typo". Q(r,t) is not given by:

    Q(r,t) = qJ0((&mu;1r)/a)ert

    But by:

    Q(r,t) = qJ0((&mu;1r)/a)e&nu;t

    This makes the problem a heck of a lot easier to solve.

    Step one and step two remain unchanged, step three changes only with the final answer:

    bm(t) = 0 ; m &ne; 1
    bm(t) = (qe&nu;tJ02(&mu;1))/(J02(&mu;m)) ; m =1

    Step four reamins unchanged until the end result:

    where T'n + ckn2Tn = bn(t) , where bn is given above.

    Step five becomes:

    Step Five-

    For n &ge; 1-

    Tn(t) = Dne-ckn2t

    For n = 1-

    Let &alpha; = ckn2
    Let &beta; = q(J02(&mu;1)/J02(&mu;n))


    T'1 + &alpha;T1 = &beta;e&nu;t

    The integrating factor is:

    I.F- e&alpha;t


    e&alpha;tT1(t) - T1(0) = (&beta;/(&alpha; + &nu;))(e(&alpha; + &nu;)t - 1)

    Now, Tn(0) = 0 from the initial condition. SO the above simplifies. Now we add the T(t)'s together:

    Tn(t) = (&beta;/(&alpha; + &nu;))(e&nu;t - e-&alpha;t) + (2 to infinity)&sum;Dne-ckn2t

    Now lets look at the initial condition to determine Dn-

    Tn(0) = (&beta;/(&alpha; + &nu;))(1 - 1) + (2 to infinity)&sum;Dn = 0

    Tn(0) = (2 to infinity)&sum;Dn = 0

    So Dn = 0

    Therefore Tn(t) is given by:

    Tn(t) = T1(t) = (q(J02(&mu;1)/J02(&mu;1))
    /(ckn2 + &nu;))(e&nu;t - e-ckn2t)

    Which upon simplification becomes:

    Tn(t) = T1(t) = (q/(ckn2 + &nu;))(e&nu;t - e-ckn2t)

    Therefore the final solution is given by:

    u(r,t) = &sum;Rn(r)Tn(t) = R1(r)T1(t)

    u(r,t) = J0((&mu;1r)/a)(q/(c(&mu;1/a)2 + &nu;))(e&nu;t - e-c(&mu;1/a)2t)

    This solution satifsies the boundary and initial conditions, and so I hopw that it is right.
  8. Apr 30, 2003 #7

    Another God

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    sweet jesus!
  9. Apr 30, 2003 #8

    Tom Mattson

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    I'll check it and get back to you tomorrow.
  10. Apr 30, 2003 #9
    Thanx for all the help Tom, I really appreciate it.

    Most of the class has the same result, so I'm crossing fingers. :smile:
  11. Apr 30, 2003 #10
    LOL. Looks daunting I agree. :wink:
  12. May 1, 2003 #11

    Tom Mattson

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    Looks good to me.
  13. May 4, 2003 #12
    Thanx for all the help Tom.
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