Can anyone help me solve a Non-Homogenous PDE for my assignment?

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In summary, the conversation discusses a problem with solving a differential equation, specifically the DE of T1' + cλnTn = (2q/a2J12(μm))*∫ rertJ02((μ1r)/a)dr. The conversation goes through various steps, using separation of variables and Bessel's equation, to arrive at a final solution of u(r,t) = J0((μ0r)/a)[C0cos((√c)(μ0/a)) + D0sin((√c)(μ0/a))] + (2 to infinity)∑J0((μnr)/a)[Cncos((√c
  • #1
Pauly Man
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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 separation 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:
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  • #2
Sit tight--I'm printing this out and taking it home.

Till tomorrow...
 
  • #3
I realized 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
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


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
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 let's 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
sweet jesus!
 
  • #8
I'll check it and get back to you tomorrow.
 
  • #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:
 
  • #10
Originally posted by Another God
sweet jesus!

LOL. Looks daunting I agree. :wink:
 
  • #11
Looks good to me.
 
  • #12
Thanx for all the help Tom.
 

Related to Can anyone help me solve a Non-Homogenous PDE for my assignment?

What is a non-homogenous PDE?

A non-homogenous PDE (partial differential equation) is a type of mathematical equation that involves multiple variables and their partial derivatives. It is called non-homogenous because it includes a term that is not equal to zero, unlike a homogenous PDE where all terms are equal to zero.

What is the difference between a non-homogenous PDE and a homogenous PDE?

The main difference between a non-homogenous PDE and a homogenous PDE is that a non-homogenous PDE contains a non-zero term, while a homogenous PDE has all terms equal to zero. This leads to different methods of solving and analyzing the equations.

What are some real-world applications of non-homogenous PDEs?

Non-homogenous PDEs have various applications in physics, engineering, and mathematical modeling. They are commonly used to describe phenomena such as heat transfer, fluid dynamics, and electromagnetic fields. They are also used in financial mathematics to model stock prices and options.

How are non-homogenous PDEs solved?

Solving a non-homogenous PDE involves finding a solution that satisfies the equation and any given boundary conditions. This can be done using various techniques such as separation of variables, Fourier transforms, and numerical methods. The choice of method depends on the specific equation and its complexity.

Why are non-homogenous PDEs important in science and engineering?

Non-homogenous PDEs are important in science and engineering because they provide a powerful tool for modeling and understanding complex systems. They allow scientists and engineers to describe and predict the behavior of various phenomena, leading to advancements in technology and scientific understanding.

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