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# Class Note for MATH 456 at UMass(3)

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Date Created: 02/06/15
Math 4567Spring 2006 Dr Alexandros Sopasakis page 30 37 Heat Equation We will now look at a model for describing the distribution of temperature in a solid material as a function of time and space It is probably easy to see the need for such a description and applications of it to our everyday life are abound As usual there are several simpli cations and assumptions behind such a model Among them is the assumption that the body7 which we wish to measure the temperature of7 is homogeneous ie it is composed of the exact same material and no foreign bodies are in it There do exist more complicated models however which can describe the temperature distribution in non homogeneous bodies Similarly we will assume that our object is perfectly insulated from surrounding sources or sinks of heat and in fact that it can not generate heat of each own To simplify the mathematical modeling aspect further we will rst attempt to solve this problem in just one spatial dimension So we consider 0 a homogeneous material rod of length L o a constant cross section S 0 further assume that the rod is completely insulated everywhere around The well known mathematical model which we will derive below that makes use of these assumptions and describes the temperature distribution in three dimensions is the famous heat or di usion equation7 ut V um uyy um 318 where V is known thermal dz usz m tg and is related to7 as well as calculated from7 properties of the material The heat equation is in fact another realization of the conservation of energy which we saw a version of in the Euler equations We present below the derivation of this model 371 Derivation of the heat equation There are several approaches in order to obtain a mathematical model for temperature through a body Mainly these approaches are characterized into two main categories 0 Relativistic o Newtonian Although much more interesting mathematically the relativistic approach7 which produces the exact same equation 3187 is too complicated to present here in detail This method relies on atomistic properties and generates a system of equations which produce 318 We will instead present the classical Newton approach In this case we represent our variables of interest through classical conservation laws which we know that any model relating them ought to adhere to Thus in a similar fashion as we did for the derivation of our linear advection equation we now consider an in nitesimal element in our rod and write the conservation of energy law for it We rst provide some necessary background from physics It is well known from the Fourier heat conduction law that heat Q is transported in direction opposite to the temperature gradient of u and is proportional to it7 319 mm w an an an E aiy Math 4567Spring 2006 Dr Alexandros Sopasakis page 31 where n is the proportionality constant also known as thermal conductivity It is also known that heat Q is related to mass m and temperature u via the following formula Qmt Amumt 320 where A is the known speci c heat for our material Let us complete this derivation for now in just one dimension m Note however that this result can be very easily generalized to all three dimensions Consider an in nitesimal piece from our rod with length mm Am Then if the rod has cross section S this piece has volume SAm Further assuming that the density of the material for the body under consideration is c then the in nitesimal mass for our in nitesimal volume element is simply given by Am cSAm Therefore based on 320 the equivalent heat for our volume element is described by Q Amu AcSAmu 321 Let us now take into account physical properties In other words that the energy or heat in any piece of the insulated rod ought to be conserved Considering again our in nitesimal volume piece with length m m Am we can state the following rate of change of energy heat rate heat owing in rate heat owing out or mathematically written as 887 QmmtS 7 qutm AmtS 322 Note however that the rate of change of energy heat can be found by di erentiating our equation in 321 8Q Bu 7 A SA 7 323 at C 5 8t We substitute this for the left hand side of 322 and obtain Bu AcSAmE SQmt 7 Qm Amt Let us rewrite the above by dividing with Am and S 8 QltA7U 7Q7t V07 7 at Am Note however that the right hand side of the above is simply the derivative of Q with respect to m aQam if we let Am 7 00 Thus we obtain A Bu 8Q c7 77 at am Finally using the Fourier law of heat conduction 319 in one dimension Q 7ndudm we obtain our heat equation Bu 8211 7 V7 at am2 where V nAc groups all parameters together This derivation can be easily generalized to higher dimensions Math 456Spring 2006 Dr Alexandros Sopasakis page 32 372 Solutions of the Heat equation There are several different methods which allow knowledge of exact or numerical solutions for the heat equation Before however we even start discussing the possible solutions of this equation we should get acquainted with some amazing properties that these solutions tend to have To shed light into this subject we use the well known maximum principle which in fact can be applied to more general problems with similar success We give below a version of the maximum principle as applied to the heat equation Theorem 15 Suppose the heat equation is de ned on the following time space domain 0 lt t lt T and a S a S h Then the temperature u attains both its maximum and minimum values on the boundary of this domain Further that temperature is unique Theoretical solutions separation of variables method Let us examine a concrete example of a rod length L with temperature ut We keep the ends of the rod a 0 and at a L at a constant temperature u 0 while we will set the initial temperature in the rod to be increasing with distance via u0 a In other words we propose the following heat equation model for this example7 37 3275 for 0ltltL7 tgt0 u0t 0 for t gt 0 uL7t L for t gt 0 u0 a 0 lt a lt L We will apply a very general procedure called separation of variables in order to obtain the solution to this model The idea behind this procedure is general enough that can be successfully applied to several other problems with similar results The separation of variables method assumes that the solution um t which we are looking for can be separated into a product of solutions each of which depends only on one variable In other words we assume ut XTt We then obtain the time and 2nd space derivative of the above7 ut XT t and um XTt Substituting into our heat equation above we obtain the following7 XvTt uX cTt Let us now separate variables by moving all terms with a t to the left and all the terms with an a to the right7 Tt XHW VTt X It is important to note here that if we pick a speci c time t t1 then our equation above becomes Ttl 7 COHS an 7 XMltgt VTt1 t t Xa Math 456Spring 2006 Dr Alexandros Sopasakis page 33 for all m Similarly if we pick a speci c space value 1 then it becomes7 Tit Mm 7 t t VTlttgt Xml cons an for all All in all the following information has been revealed7 T t X constant and constant VTt X Note that we have successfully reduced the original PDE problem into two very simple ODE problems We will solve both of the above ODEs and produce the values of Tt and Once this is done then we can reassemble the solution of our original PDE from the fact that ut XTt It is customary to denote the general constant above by 7A and we therefore have the following two ODEs to solve7 which we can rewrite as7 TKU tAVT 0 X c AXv 0 It is easy to obtain the solutions for each of those equations In general any linear second order homogeneous equation such as XH IXm bX 0 is rst transformed to a quadratic formula 102 aw b 0 We solve it using the quadratic formula Depending on the two solutions of this quadratic 101 and LU2 we have the following solutions for the ODE7 General X 01 expw1 02 expw2 if 101 a LU2 and both are real 2nd order X 01 expw1 02 expw2 if 101 LU2 and both are real ODE Solutions X expom01 cos 02 sin if 10 is complex 10 or 15 324 Similarly we obtain the solutions for our rst order ODE Tt AVTOS 0 with one of the methods which we have already learned previously If for instance we use the separation of variables method for this ODE we obtain the solution to be Tt Tt0e t t0gt As a result the total solution to our heat equation model is just the product of both the rst order ODE and the second order ODE together um 75 XTt

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