1 dimensional heat conduction

dreamz · November 12, 2013, 11:37

We have one dimensional steady heat conduction with heat generation term. I use this to find temperature distribution within a rod. Now, i will require two boundary conditions in space co-ordinates. I specify, one end as insulated and other end with a specified heat flux. Physically this is not correct. However, mathematically the differential equation requires two boundary conditions. So, can we conclude that, we cannot specify two Neumann boundary conditions in this case?

Rami · November 13, 2013, 03:30

Hi dreamz,

Let us try to naively solve your energy eqaution
q,x = -(kT,x),x = S
assuming for simplicity k and S are constant.
integration:
q = -kT,x = Sx + A
-kT = 0.5*S*x^2 + Ax + B
To determine the integration constants A and B, use your BCs:
q(0) = 0 => A=0
q(L) = Q = SL (where Q is the prescribed value at x=L).

Now you have two problems:
(1) A contradiction: if both S and Q are given, you cannot satisfy the last equation.
(2) You cannot determine B.

Conclusion: the problem is ill-posed.
On the other hand, if your BCs were both Dirichlet or one Dirichlet and one Neumann, then the problem is well-posed, as you may see.

November 12, 2013, 11:37	1 dimensional heat conduction	#1
dreamz Member Ashutosh Join Date: Jul 2013 Posts: 98 Rep Power: 12	We have one dimensional steady heat conduction with heat generation term. I use this to find temperature distribution within a rod. Now, i will require two boundary conditions in space co-ordinates. I specify, one end as insulated and other end with a specified heat flux. Physically this is not correct. However, mathematically the differential equation requires two boundary conditions. So, can we conclude that, we cannot specify two Neumann boundary conditions in this case?

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November 13, 2013, 03:30		#2
Rami Senior Member Rami Ben-Zvi Join Date: Mar 2009 Posts: 155 Rep Power: 17	Hi dreamz, Let us try to naively solve your energy eqaution q,x = -(kT,x),x = S assuming for simplicity k and S are constant. integration: q = -kT,x = Sx + A -kT = 0.5Sx^2 + Ax + B To determine the integration constants A and B, use your BCs: q(0) = 0 => A=0 q(L) = Q = SL (where Q is the prescribed value at x=L). Now you have two problems: (1) A contradiction: if both S and Q are given, you cannot satisfy the last equation. (2) You cannot determine B. Conclusion: the problem is ill-posed. On the other hand, if your BCs were both Dirichlet or one Dirichlet and one Neumann, then the problem is well-posed, as you may see.