1/100 Applications of Differential EquationsThe Heat Equationwith partial differential equationsDavid Richards and Adam Abrahamsen2/100 Introduction• The purpose of our presentation is to derive the heat equation sothat we are able to model the flow of heat through an object.• But before we begin our discussion of the mathematics of the heatequation, we must first determine what is meant by the term heat?A common example of the misunderstanding of the term heat isthe classic physics question of what contains more heat, a bathtubof warm water or a boiling cup of water?We all know the boiling cup of water has a higher temperaturebut contains less heat than the bathtub.3/100 • Therefore, in calculating problems concerned with heat we must dis-tinguish between the two types of measurements, the measurementof temperature and the measurement of the quantity of heatcontained in an object. In this discussion when we mention theterm heat we will be talking about the quantity of heat in anobject.4/100 Derivation of the heat equation• There are two methods used to solve for the rate of heat flow throughan object. The firs t method is derived from the properties of theobject. The second method is derived by measuring the rate of heatflow through the boundaries of the object.Method One• Experimental calculations show that the heat (Q) in ∆V at time (t)can be defined by,∆Q = cρu∆V. (2.1)• Where c is the specific heat, ρ is the density, u is the temperature,and ∆V is a small volume.5/100 • Consider this thin rod, made of a homogenous material and perfectlyinsulated along its length so that heat can only flow through its ends.Any p os ition along the rod is denoted as x, and the length of therod is denoted as L such that 0 ≤ x ≤ L.0Lx• Therefore the temperature u is the only condition that depends onposition(x) and time(t). Thus,∆Q = cρu(x, t)∆V (2.2)6/100 • Consider this thin rod that is perfectly insulated along its length sothat heat can only escape from its ends. Any position along the rodis denoted as x, and the length of the rod is denoted as L such that0 ≤ x ≤ L.0LUa b• Therefore the temperature u is the only condition that depends onposition(x) and time(t). Thus,∆Q = cρu(x, t)∆V (2.3)• Now consider a small section of the rod U defined as the intervalfrom x = a to x = b.7/100 • Consider this thin rod that is perfectly insulated along its length sothat heat can only escape from its ends. Any position along the rodis denoted as x, and the length of the rod is denoted as L such that0 ≤ x ≤ L.0LUa∆xb• Therefore the temperature u is the only condition that depends onposition(x) and time(t). Thus,∆Q = cρu(x, t)∆V (2.4)• Now consider a small section of the rod U defined as the intervalfrom x = a to x = b. The cross sectional area is defined as S, andthe width of this section is ∆x. This gives ∆V = S∆x.8/100 • We can now express the amount of heat in the cross-sectional areaas,∆Q = cρu(x, t)S∆x (2.5)• We find that the amount of heat in the section U is given by theintegral.Q(t) =Zbacρu(x, t)Sdx. (2.6)• Since the rod has unifor m thickness, S doesn’t change with res pectto time, and because we are dealing with homogenous materials cand ρ do not change with respect to time. Thus, by differentiatingwe take the partial of u to find the change in heat with res pect totime equaling,dQdt=Zbacρ∂u∂tdxS. (2.7)9/100 Method Two• Our second method of finding the change in heat Q with respect totime is also determined experimentally, with a rod similar to that inmethod one.0LUa∆xb10/100 Method Two• Our second method of finding the change in heat (Q) with respectto time is also determined experimentally, with a rod similar to thatin method one.• The rate of heat flow through U is inversely proportional to thewidth of U, and directly proportional to the cross-sectional area.0LUa∆xb11/100 Method Two• Our second method of finding the change in heat (Q) with respectto time is also determined experimentally, with a rod similar to thatin method one.• The rate of heat flow through U is inversely proportional to thewidth of U, and directly proportional to the cross-sectional area.• It is known that when two objects of different temperature are placedtogether (touching) heat will flow from the hotter object to thecooler one. If the temperature at b > a then heat will flow fromb → a.0LUa∆xb12/100 • Combining these properties we find,∆Q = −Cu(a + ∆x, t) − u(a, t)∆xS (2.8)• The proportionality constant C is known as the thermal conductivity.This varies depending on the type of material being evaluated.0LUa∆xb13/100 ∆Q = −Cu(a + ∆x, t) − u(a, t)∆xS (2.9)• To show this is true, we need to show the flow of heat through thesection U where the boundaries of the section are defined asa = x = 0;b = a + ∆x0LUa∆xb• If the temperature at u(a + ∆x, t) > u(a, t), then ∆Q would benegative which is logical because the temperature at a + ∆x isgreater than the temperature at a, so the heat would be flowingfrom a + ∆x to a, thus heat would be flowing out of the rod.14/100 • Letting ∆x → 0 in equation (??), the difference quotient ap-proaches∂u∂xthen the rate of heat flow through (U) at x = ais given by,⇒ −C∂u∂x(a, t)S (2.10)• Following the same argument we can show that the rate of heat flowthrough U at b is defined as,⇒ C∂u∂x(b, t)S (2.11)15/100 • Therefore, the amount of heat that U obtains at time t can be givenby,dQdt= C[rate in − rate out]S= C∂u∂x(b, t) −∂u∂x(a, t)S(2.12)• Using similar techniques as we used in method o ne we find thechange in heat with respect to time to be,dQdt= CZba∂2u∂x2dxS (2.13)16/100 Combining the two Methods• Now that we have two equations for the rate of heat flow into andout of the section U,Method 1dQdt=Zbacρ∂u∂tdxS (3.1)and,Method 2dQdt= CZba∂2u∂x2dxS. (3.2)17/100 We set these equations equal to each other.cρZba∂u∂tdxS = CZba∂2u∂x2dxS (3.3)• Doing some calculations we arrive at,∂u∂t− K∂2u∂x2= 0 (3.4)• More popularly written as,ut= Kuxx(3.5)• Where K is defined to be