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August 31st, 2017, 04:34 PM   #1
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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?
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August 31st, 2017, 05:18 PM   #2
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Are a and b non-zero constants that have the same sign?
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September 12th, 2017, 03:21 PM   #3
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Quote:
Originally Posted by nbehlman View Post
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?
Personally, I have never liked the "Laplace transform"- and it isn't necessary here. The differential equation aT''- bt= 0 has "characteristic equation" as^2- b= 0 or s^2= b/a. Now, here is where skipjack's question is important.

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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