Nonhomogeneous Boundary Value Problem

In summary, the conversation discusses a nonhomogeneous boundary value problem of the heat equation in a disk with no angle dependence. The homogeneous case is solved using variables separation and Bessel functions, but the solution for the nonhomogeneous case is still unclear due to the constant term Q not being a function of space or time. The conversation also touches on the use of Green's function and the requirement for the source term and eigenfunction to be orthogonal for a nonhomogeneous equation to have a solution.
  • #1
Grogs
151
0
I've got a nonhomogeneous BVP I'm trying to solve. Both my book and my professor tend to focus on the really hard cases and completely skipp over the easier ones like this, so I'm not really sure how to solve it. It's the heat equation in a disk (polar coordinates) with no angle dependence.

[tex]\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + Q = \frac{\partial u}{\partial t}[/tex]

Q is a constant.

0 < r < 1
t > 0

Subject To:

BC1) u(0, t) bounded
BC2) u(1, t) = 0
IC) u(r, 0) = 0

I start with the homogeneous (Q=0) case and separate variables to get ODE's for T(t) and R(r):

[tex]rR'' + R' + \lambda^2 r R = 0 [/tex]

[tex]
T' + \lambda^2 T = 0
[/tex]

Solving and applying the BC's, I get

[tex]
R_{n}(r)=c_{2}J_{0}(j_{n0}r), n=1,2,3,...[/tex]

[tex]
T_{n}(t)=c_{4}e^{-j_{n0}^2 t}, n=1,2,3,...[/tex]

Where [itex]j_{n0}[/itex] represents the nth zero of the [itex]J_{0}[/itex] Bessel function. Putting them together,[tex]u_{hom}(r,t) = \sum \limits_{n=1} ^ {\infty} A_{n} e^{-j_{n0}^2 t} J_{0}(j_{n0}r) [/tex]

This is where I'm stuck at. In Previous problems, our inhomogeneous portion was a function of both space and time {Q(r,t)} and we would model both u(r,t) and Q(r,t) as infinite series of Bessel functions in r with time-dependent coefficients and substitute them back into the PDE, which would result in the sums and Bessel functions falling out, leaving an ODE we could solve to find the coefficients. That's not an option here since Q isn't a function of space OR time, so I really don't know where to go from here to get the partial portion of the solution. Thanks in advance for your help,

Grogs

EDIT: Corrected homogeneous solution
 
Last edited:
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  • #2
Of course it's a "function of space and time", it's just a constant function! Any function of x and t can be written, as you say, as
[tex]A_{0} + \sum \limits_{n=1} ^ {\infty} A_{n} e^{-j_{n0}^2 t} J_{0}(j_{n0}r) [/tex]
What must An, for n> 0, be for that to be a constant? What is A0?
 
  • #3
Halls,

The [itex]A_{n}[/itex]'s are zero in that case.

I should point that I goofed up when I wrote the homogeneous solution above. The n=0 case shouldn't be an eigenvalue (so no constants in the solution.) I must have been working on too many problems at once and gotten them mixed up. I edited my original post to reflect that correctly.

So where do I go from here? I've tried substituting the homogeneous solution back into the PDE, but nothing, including the sums, drops out so it's messier than the original PDE and doesn't seem to get me closer to a solution.

Thanks,

Grogs
 
  • #4
An= 0 for n>0 and obviously A0= C. Just include that when writing your equation as a sum of Bessel functions.
 
  • #5
Are you saying I should model [itex]u_{hom}[/itex] and [itex]u_{part}[/itex] as:

[tex]u_{hom} = \sum \limits_{n=1} ^ {\infty} A_{n} \exp(-j_{n0}^2 t) J_{0}(j_{n0}r)[/tex]

(There is no n=0 solution for the homogeneous case)

and

[tex]u_{part} = c + \sum \limits_{n=1} ^ {\infty} A_{n} \exp(-j_{n0}^2 t) J_{0}(j_{n0}r) = c[/tex]

(there's only a constant for the partial case)

resulting in

[tex]u(r,t) = c + \sum \limits_{n=1} ^ {\infty} A_{n} \exp(-j_{n0}^2 t) J_{0}(j_{n0}r) [/tex]

How do I solve for c though? If I substitute u(r,t) back into the PDE c gets differentiated out. If I substitute just the partial solution [itex]u_{part}=c{/itex] in, I get Q = 0, which obviously can't be true.
 
  • #6
I'm working on the exact same problem, any update?
 
  • #7
we can write in the solution of a nonhomogenuos equation as

u(x)=integral(g(x.x1)*f(x)dx...where f(x) is the sourse term...if sourse term is zero.ie homogenous equation,what will be the solution?zero?what it means?homogenous equation has no solution?

then in the series form of greens function if lambda is one of the eigenvalue of the differential operator,den greens function become infinte.then we require that for the nonhomogemous eqn to have solution sourse term and eigenfunction shuold be orhogonal...can anyone explain why it should be orthogonal?

thanku
 

Related to Nonhomogeneous Boundary Value Problem

1. What is a nonhomogeneous boundary value problem?

A nonhomogeneous boundary value problem is a mathematical problem that involves finding a solution to a differential equation subject to boundary conditions, where the boundary conditions are not equal to zero. In other words, the equation does not have a trivial solution and requires non-zero boundary values to be specified.

2. How is a nonhomogeneous boundary value problem solved?

The solution to a nonhomogeneous boundary value problem is typically found by using techniques such as separation of variables, variation of parameters, or the method of undetermined coefficients. These methods involve manipulating the given equation and applying the specified boundary conditions to find the solution.

3. What is the difference between a homogeneous and nonhomogeneous boundary value problem?

In a homogeneous boundary value problem, the boundary conditions are equal to zero, which means the equation has a trivial solution. In contrast, a nonhomogeneous boundary value problem has non-zero boundary conditions, and the equation does not have a trivial solution.

4. Are nonhomogeneous boundary value problems common in scientific research?

Yes, nonhomogeneous boundary value problems are common in scientific research, particularly in fields such as physics, engineering, and mathematics. These problems arise in many real-world applications, such as heat transfer, fluid mechanics, and quantum mechanics.

5. Can nonhomogeneous boundary value problems have multiple solutions?

Yes, it is possible for a nonhomogeneous boundary value problem to have multiple solutions. This can occur when the equation has multiple possible solutions that satisfy the given boundary conditions. In some cases, the solution may also depend on the choice of boundary conditions.

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