Heat Equation with insulated endpoints.

Click For Summary
SUMMARY

The discussion centers on solving the heat equation for a bar with insulated endpoints, described by the partial differential equation (PDE) u_{t} = k u_{xx} - h (u-T), where h > 0. Participants explore the transformation v = e^{ht}(u-T) to convert this equation into a more standard form, u_{t} = k u_{xx}. The challenge lies in correctly applying the transformation to derive the necessary relationships between the variables. The conversation highlights the importance of understanding the underlying principles of heat propagation and the specific forms of the equations involved.

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Familiarity with heat equation variants
  • Knowledge of boundary value problems (BVPs)
  • Experience with mathematical transformations and derivatives
NEXT STEPS
  • Study the derivation of solutions for the standard heat equation u_{t} = k u_{xx}
  • Learn about boundary conditions and their impact on heat equations
  • Explore the method of separation of variables for solving PDEs
  • Investigate the implications of insulated boundaries in thermal analysis
USEFUL FOR

Mathematicians, physics students, and engineers involved in thermal analysis or studying heat transfer phenomena will benefit from this discussion.

Kizaru
Messages
44
Reaction score
0

Homework Statement


Assume that a bar is insulated at the endpoints. If it loses heat through its lateral surface at a rate per unit length proportional to the difference u(x,t) - T, where T is the temperature of the medium surrounding the bar, the equation of heat propagation is now

[tex]u_{t} = k u_{xx} - h (u-T)[/tex]

where h > 0

Homework Equations


Use the function

[tex]v = e^{ht}(u-t)[/tex]

to reduce this BVP to one already solved.


The Attempt at a Solution



"To one already solved" is referring to heat equation variants in which the PDE is of form

[tex]u_{t} = k u_{xx}[/tex]

I can solve it from that form, I just need to convert into something of that form.

Some things I noticed, partial derivative of v with respect to t, and equated to the second partial derivative with respect to x yields u_t = u_xx - h(u-t)
This is off by the constant k which is in front of the u_xx in the original PDE. Not sure what I'm missing from here.
 
Physics news on Phys.org
Hi Kizaru! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)
Kizaru said:
Some things I noticed, partial derivative of v with respect to t, and equated to the second partial derivative with respect to x yields u_t = u_xx - h(u-t)

Yes, you have the correct basic technique, I can't see quite how you haven't got there. :confused:

If v = eht(u - t),

then vt = … and kvxx = … ? :smile:
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K