The One-Dimensional Heat Equation

1 The heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsThe One-Dimensional heat EquationR. C. DailedaTrinity UniversityPartial Differential EquationsFebruary 25, 2014 Daileda1-D heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsIntroductionGoal:Model heat flow in a One-Dimensional object (thin rod).Set up:Place rod of lengthLalongx-axis, one end at origin:xL0heated rodLetu(x,t) = temperature in rod at positionx, timet.(Ideal) Assumptions:Rod is given some initial temperature distributionf(x) alongits is perfectly insulated, heat only moves internal heat sources or heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsThe heat EquationOne can show thatusatisfies theone-dimensional heat equationut= :This can be derived via conservation of energy and Fourier slaw of heat conduction (see textbook pp.)

2 143-144).The constantc2is thethermal diffusivity:K0= thermal conductivity,c2=K0s ,s= specific heat , = heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsInitial and Boundary ConditionsTo completely determineuwe must also specify:Initial conditions:The initial temperature profileu(x,0) =f(x) for 0<x< conditions:Specific behavior atx0= 0, temperature:u(x0,t) =Tfort> end:ux(x0,t) = 0 fort> end:ux(x0,t) =Au(x0,t) fort> heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsSolving the heat EquationCase 1: homogeneous Dirichlet boundary conditionsWe now apply separation of variables to the heat problemut=c2uxx(0<x<L,t>0),u(0,t) =u(L,t) = 0(t>0),u(x,0) =f(x)(0<x<L).

3 We seek separated solutions of the formu(x,t) =X(x)T(t). Inthis caseut=XT uxx=X T XT =c2X T X X=T c2T= with the boundary conditions we obtain the systemX kX= 0,X(0) =X(L) = 0,T c2kT= heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsAlready know: up to constant multiples, the only solutions totheBVP inXarek= 2n= n L 2,X=Xn= sin ( nx) = sin n xL ,n satisfyT c2kT=T + cn L |{z} n2T= 0T = 2nT T=Tn=bne thus have thenormal modesof the heat Equation :un(x,t) =Xn(x)Tn(t) =bne 2ntsin( nx),n heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsSuperposition and initial conditionApplying the principle of superposition gives the general solutionu(x,t) = Xn=1un(x,t) = Xn=1bne 2ntsin( nx).

4 If we now impose our initial condition we find thatf(x) =u(x,0) = Xn=1bnsin n xL ,which is the sine series expansion off(x). Hencebn=2 LZL0f(x) sin n xL heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsRemarksAs before, if the sine series off(x) is already known, solutioncan be built by simply including exponential can show that this is theonlysolution to the heatequation with the given initial of the decaying exponential factors: The normal modes tend to zero (exponentially) ast . Overall,u(x,t) 0 (exponentially)uniformly in xast . Ascincreases,u(x,t) 0 more agrees with heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsExampleSolve the heat problemut= 3uxx(0<x<2,t>0),u(0,t) =u(2,t) = 0(t>0),u(x,0) = 50(0<x<2).

5 We havec= 3,L= 2 and, by exercise (withp=L= 2)f(x) = 50 =200 Xk=012k+ 1sin (2k+ 1) x2 .Since 2k+1=c(2k+ 1) L= 3(2k+ 1) 2, we obtainu(x,t) =200 Xk=012k+ 1e 3(2k+1)2 2t/4sin (2k+ 1) x2 .Daileda1-D heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsSolving the heat EquationCase 2a: steady state solutionsDefinition:We say thatu(x,t) is asteady state solutionifut 0( time-independent).Ifu(x,t) is a steady state solution to the heat Equation thenut 0 c2uxx=ut= 0 uxx= 0 u=Ax+ state solutions can help us deal with inhomogeneousDirichlet boundary conditions. Note thatu(0,t) =T1u(L,t) =T2 B=T1AL+B=T2 u= T2 T1L x+ heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsSolving the heat EquationCase 2b: inhomogeneous Dirichlet boundary conditionsNow consider the heat problemut=c2uxx(0<x<L,t>0),u(0,t) =T1,u(L,t) =T2(t>0),u(x,0) =f(x)(0<x<L).

6 Step 1:Letu1denote the steady state solution from above:u1= T2 T1L x+ 2:Letu2=u :By superposition,u2still solves the heat heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsThe boundary and initial conditions satisfied byu2areu2(0,t) =u(0,t) u1(0) =T1 T1= 0,u2(L,t) =u(L,t) u1(L) =T2 T2= 0,u2(x,0) =f(x) u1(x).Step 3:Solve the heat Equation with homogeneous Dirichletboundary conditions and initial conditions above. This 4:Assembleu(x,t) =u1(x) +u2(x,t).Remark:According to our earlier work, limt u2(x,t) = callu2(x,t) thetransientportion of the haveu(x,t) u1(x) ast , the solution tends tothe steady heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsExampleSolve the heat 3uxx(0<x<2,t>0),u(0,t) = 100,u(2,t) = 0(t>0),u(x,0) = 50(0<x<2).

7 We havec= 3,L= 2,T1= 100,T2= 0 andf(x) = steady state solution isu1= 0 1002 x+ 100 = 100 corresponding homogeneous problem foru2is thusut= 3uxx(0<x<2,t>0),u(0,t) =u(2,t) = 0(t>0),u(x,0) = 50 (100 50x) = 50(x 1)(0<x<2).Daileda1-D heat EquationThe heat equationHomogeneous Dirichlet conditionsInhomogeneous Dirichlet conditionsAccording to exercise (withp=L= 2), the sine series for50(x 1) is 100 Xk=11ksin 2k x2 , onlyevenmodes occur. Since 2k=c2k L= 3k ,u2(x,t) = 100 Xk=11ke 3k2 2tsin (k x).Henceu(x,t) =u1(x)+u2(x,t) = 100 50x 100 Xk=11ke 3k2 2tsin (k x).Daileda1-D heat Equatio