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August 31st, 2017, 04:34 PM  #1 
Newbie Joined: Aug 2017 From: USA Posts: 1 Thanks: 0  Analytical Solution
I am trying to derive an analytical solution for a heat transfer problem. I have a conduction term and a heat sink term that is proportional to temperature. The conduction term alone would leave me with the Laplace equation, T''=0, When I add the heat sink term and the constants, I believe the equation I am trying to solve is aT''bT=0. I am doing this in 1D only. I know the boundary conditions for T and T' at x=0 and x=1. How do I solve this problem? Do I use a Laplace transform?

August 31st, 2017, 05:18 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1385 
Are a and b nonzero constants that have the same sign?

September 12th, 2017, 03:21 PM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,649 Thanks: 681  Quote:
If a= 0 s^2= b/a is impossible. The equation in that case is bT= 0 so that T= 0. If a and b are the same sign then b/a is positive so $\displaystyle s= \pm\sqrt{b/a}$ is real and the general solution is $\displaystyle Ae^{\sqrt{b/a}x}+ Be^{\sqrt{b/a}x}$. When x= 0, that is $\displaystyle A+ B$ and when x= 1, that is $\displaystyle Ae^{b/a}+ Be^{b/a}$ set those equal to the given values and you have two linear equations to solve for A and B. If a and b are of opposite sign then b/a is negative so $\displaystyle s= \pm\sqrt{b/a}$ is complex. If we write the those complex numbers as $\displaystyle \alpha\pm \beta$ then the general solution to the differential equation is $\displaystyle e^{\alpha x}(A \cos(\beta x)+ B\sin(\beta x)$. When x= 0, that is $\displaystyle A \cos(\beta)$ and when x= 1 $\displaystyle e^{\alpha}(A \cos(\beta)+ B \sin(\beta)$. Again, set those equal to the given values and solve for A and B. Last edited by skipjack; September 12th, 2017 at 03:29 PM.  

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