The document discusses using the Laplace transform to solve the heat equation for transient conduction. It presents a problem of calculating the temperature over time within a semi-infinite slab where the initial temperature is suddenly changed at the boundary. By taking the Laplace transform, the PDE is converted to an ODE, making the solution easier to obtain. The solution is found in the transform domain and then inverted back to the time domain to give the temperature as a function of position and time within the slab. Transforming the time variable rather than the space variable is preferred because it avoids issues with evaluating boundary terms involving derivatives.

1. Solving the Heat Equation for Transient Conduction using Laplace Transform March 20 2010 Desmond Adair School of Engineering University of Tasmania School of Engineering University of Tasmania

3. To show how it is implemented in MathematicaSchool of Engineering University of Tasmania

4. Problem formulation Consider a semi-infinite slab where the distance variable, y goes from 0 to ∞. The temperature is initially uniform within the slab and can be considered to be 0. At t = 0, the temperature at y = 0 is suddenly increased to 1. It is required to calculate the temperature as a function of time, t, within the slab. This problem is commonly solved by a similarity variabletechnique that arises because of the absence of a physical length scale. Here the Laplace transform is used, which is an integral transform that effectively converts (linear) PDEs (if we wish to use it on say time) to a set of ODEs in frequency space. This is a set of ODEs because the transform creates an explicit parameter, s, which is effectively a frequency and the equations need to be solved for all positive s. School of Engineering University of Tasmania

5. Problem formulation equation boundary conditions the boundary conditions are School of Engineering University of Tasmania

6. Problem analysis If a transform technique is used the intention is to simplify the problem by transforming the pde to an ode (or an algebraic equation from an ode). Once the transform is done, the need is to evaluate the integral that arises at the boundaries. So the boundary conditions and the domain of the problem must be in a form conducive to this. The Laplace transform is defined from 0 to ∞. In this problem both of the domains are from 0 to ∞, however the first try is to do the transform in time. In Mathematica this command is LaplaceTransform[heateq, t, s] and the new parameter is s. School of Engineering University of Tasmania

7. Problem analysis The first two terms make an ode in the transformed θ(t, y). Let’s call this The last term, which arose from integration by parts of the term must be evaluated from the boundary conditions. The temperature for all y at t = 0 is zero, thus this term is 0. Note that the t -> ∞ boundary term is usually zero ( this is not in this calculation) because it is multiplied by Exp[-s t] (and thus Mathematicaautomatically makes it 0). This results an ode in (y) and have generated an explicit parameter, s. The LaplaceTransform [t,s] does not affect any y derivatives. We can write School of Engineering University of Tasmania

8. Problem analysis This is easily by doing Now get the solution out of the {{}}’s We see that we cannot stand an exponentially increasing part so that c2 = 0. Now we need to evaluate (y, s) at y= 0 or some place that we know it. Well, we know θ [t, y] at y = 0 (= 1). School of Engineering University of Tasmania

9. Problem analysis Thus we can transform this to get a value for θ [y = 0, s]. Now find the value C[1] after setting C[2] = 0 and evaluating the expression at y = 0. Thus our solution in transformed space is School of Engineering University of Tasmania

10. Plot solution in frequency space to see what s does The top plot is for s = 10. It is a difficult to tell for sure, but the different values s represent different modes. The term s is roughly a frequency so that higher s’s decay faster!! School of Engineering University of Tasmania

11. Examine the issue of time to “frequency” We can more of what s means by transforming tn- - > s So, we see the inverse relation of t and s – which that is a little justification for calling s the “frequency”. School of Engineering University of Tasmania

12. Put solution in frequency space to see what α does Now if we make plots for different values of α we get: The top curve is for α =10. it is seen that the value of α gives the rate of “diffusion” which we also expect once the solution is transformed back. School of Engineering University of Tasmania

13. Transform back to time Now we must transform back to get the solution in physical space. There are three ways. One is to get a table of transforms and inverse transforms. A good table is in Spiegel’s mathematics handbook (M.R. Spiegel, Mathematical Handbook, Schaum’s Outline Series, McGraw-Hill, 1968). The second way is to use Mathematica and the problem would be, do you believe the answer. The third way is to do the complex integration yourself, the problem here is, would you believe the answer? However, Mathematicaagrees with the tables. School of Engineering University of Tasmania

14. Plot the solution in physical space Note that the bottom curve is for t = .02

15. Plot the solution in physical space Again the bottom curve is for t = 0.02. By comparing the two plots for different α, we can see that the value of the diffusivity controls how fast heat transfer occurs.

16. Why transform time variable instead of space variable? Now suppose that initially the y variable had been transformed Now we need to evaluate the boundary terms. The first one sθ[t,0] = s for all t of interest. However, the second one presents a bit of a problem. We don’t really know the heat flux at the boundary so we don’t know the derivative. Consequently, transforming y to s does not help solve the problem! School of Engineering University of Tasmania