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Lecture05

# Lecture05 - Note to myself redraw 1D wave equation...

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Note to myself: redraw 1D wave equation characteristics with correct slopes. 12.5 Heat Equation: Solution by Fourier Series Mathematical model Consider a body of homogeneous material and de- note its temperature by u ( x , t ) [K], x Ω R 3 , t 0. Let V Ω denote a bounded control volume with surface S := ∂V . The amount of heat trans- ferred per unit time from the control volume to the neighboring material is given by the integral of the heat flux q [Wm 2 ] over the boundary, ∂Q ∂t = integraldisplay S q · n ds = integraldisplay V div q d x , (144) if no work is done and if there are no heat sources or sinks inside of V . In ( 144 ), n denotes the outward unit normal vector field on S . We used the divergence theorem in the second equation of ( 144 ). Quantities on both sides of ( 144 ) have the physical unit of power [W]. Fourier’s law states that the heat flux q is proportional to the negative temperature gradient, q = k u, (145) with the thermal conductivity k [Wm 1 K 1 ], i. e. heat flows from warmer to colder regions. Assuming a constant k , the right-hand side of ( 144 ) becomes integraldisplay V div q d x = integraldisplay V k Δ u d x (146) For the left-hand side of ( 144 ) we write ∂Q ∂t = ∂Q ∂u ∂u ∂t = integraldisplay V c p ρ ∂u ∂t d x , (147) with the specific heat capacity c p [Jkg 1 K 1 ] and with the mass density ρ [kgm 3 ]. Therefore we obtain integraldisplay V ( c p ρu t k Δ u ) d x = 0 , (148) 29

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and because the control volume V Ω was chosen arbitrarily, we obtain the PDE u t c 2 Δ u = 0 , in Ω × (0 , ) , c 2 := k c p ρ [m 2 s 1 ] . (149) Equation ( 149 ) is called the heat equation, with the thermal diffusivity c 2 . It is a second-order, linear, parabolic PDE. The modeling approach used here is very common. You might have seen it already in a lecture on mathematical modeling, such as MATH 564. We use separation of variables to obtain solutions in the one-dimensional case. For that purpose, we consider the initial-boundary value problem u t = c 2 u xx , in (0 , L ) × (0 , ) , (150) u (0 , t ) = u ( L, t ) = 0 , t 0 , (151) u ( x, 0) = f ( x ) , x [0 , L ] . (152) A physical interpretation of the solution is the temperature distribution in a thin bar of length L > 0, which is insulated in the lateral direction, kept at a constant temperature at the endpoints, and has a prescribed initial temperature distribution.
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