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Non-Homogenous PDE

  • Thread starter Pauly Man
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  • #1
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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:

Answers and Replies

  • #2
Tom Mattson
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Sit tight--I'm printing this out and taking it home.

Till tomorrow...
 
  • #3
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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)
 
  • #4
Tom Mattson
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OK.

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

For instance, think that this:

Originally posted by Pauly Man
&part;u/&part;t = (c/r)*(r(&part;u/&part;r)) + Q(r,t)
Should be this:

&part;u/&part;t=(c&part;/&part;r)*(r(&part;u/&part;r))+Q(r,t)

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

Also this:

&rArr; dR/r = (dR/d&rho;)*(d&rho;/dr) = k(dR/d&rho;)
Should presumably be this:

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

Also, I don't know what

&rArr;

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.
 
  • #5
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Originally posted by Tom

&part;u/&part;t=(c&part;/&part;r)*(r(&part;u/&part;r))+Q(r,t)

At least, that would be consistent with what you write later.
Yep, there was a typo there. Although it should still be c/r, and not c:

&part;u/&part;t=((c/r)&part;/&part;r)*(r(&part;u/&part;r))+Q(r,t)

Also this:

Originally posted by Tom

Should presumably be this:

&rArr; dR/dr = (dR/d&rho;)*(d&rho;/dr) = k(dR/d&rho;)
Yep, that is correct.

Originally posted by Tom

Also, I don't know what

&rArr;

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

Originally posted by Tom

If I'm right about those typos, then you are OK through Steps 1 and 2. I am still checking Step 3.
Cool. :smile:
 
  • #6
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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))

then;

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

The integrating factor is:

I.F- e&alpha;t

So;

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.
 
  • #7
Another God
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sweet jesus!
 
  • #8
Tom Mattson
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I'll check it and get back to you tomorrow.
 
  • #9
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Thanx for all the help Tom, I really appreciate it.

Most of the class has the same result, so I'm crossing fingers. :smile:
 
  • #10
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Originally posted by Another God
sweet jesus!
LOL. Looks daunting I agree. :wink:
 
  • #11
Tom Mattson
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Looks good to me.
 
  • #12
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Thanx for all the help Tom.
 

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