Definitions, Theorems, Examples, Topics for Asmar 2nd edition Chapter 3: 3.5 and 3.6 only 3.5 The One-Dimensional Heat Equation ==== The BVP for J.B. Fourier's insulated bar problem is u_t = c^2 u_xx u(0,t)=0, u(L,t)=0, u(x,0)=f(x), 0 < x < L, t > 0. SEPARATION OF VARIABLES Seek product solutions u(x,t)=X(x)T(t). The plan is to obtain the solution of Fourier's problem by superposition of these product solutions, hence The solution u(x,t) = sum of product solutions times constants THEOREM. A product solution u=XT satisfies for some constant k: X'' - kX = 0, X(0)=X(L)=0, X(x) nonzero, T' - (k c^2)T = 0, T(t) nonzero. THEOREM. The X-BVP has a solution X(x) nonzero if and only if -k=(n Pi/L)^2, n=1,2,3,..., in which case X(x)=c1 sin(n Pi x/L) for some nonzero constant c1. The corresponding factor T(t) in this case is given by T(t) = c2 exp(-(n Pi c/L)^2 t) for nonzero c2. THEOREM. A product solution u=XT must be a constant times one of the functions sin(n Pi x/L) exp(-(n Pi c/L)^2 t) THEOREM. The form of a trial solution to Fourier's heat problem on a bar is the infinite series u(x,t) = sum(b[n] sin(n Pi x/L) exp(-(n Pi c/L)^2 t), n = 1 ..infinity) where b[1], b[2], b[3], ... are constants to be determined. This form of the trial solution already satisfies the conditions u(0,t)=u(L,t)=0 for t>0. THEOREM. The coefficients in the trial solution will satisfy the condition u(x,0)=f(x) provided b[n] = (2/L) int(f(x) sin(n Pi x/L), x = 0 .. L), n=1,2,3, ... EXAMPLE. A welding rod of length L=Pi units rests in water at 100 degrees Celsius. It is removed, then laterally insulated, with the ends kept in an ice pack at 0 degrees Celsius. Let c=1. Find the temperature u(x,t) using fourier's solution to the heat equation (above). ANSWER: u(x,t) = sum( b[n] sin(nx) exp(-n^2 t), n = 1 .. infinity), b[n]=400/Pi/n for n odd, b[n]=0 for n even. QUESTION. What is the first normal mode? ANSWER: u[1](x,t) = b[1] sin(x) exp(-t), where b[1]=400/Pi. COMPUTER PROBLEM. Create 6 plots of y=u(x,t) for t=0,0.5,1,1.5,2,2.5. Display the maple code that does it. ANSWER: Choose a number N=15 of terms in the series solution to start, then adjust this value to get a detailed plot for each of the 6 values of t. The code: N:=15: u:=(x,t)->sum((400/Pi)*(1/(2*k+1))*sin((2*k+1)*x)*exp(-(2*k+1)^2*t),k=0..N); tVals:=[0,0.5,1,1.5,2,2.5]: Q:=[seq(plot(u(x,tVals[j]),x=0..Pi),j=1..6)]; Q[1];Q[2];Q[3];Q[4];Q[5];Q[6]; plots[display](Q); See also Asmar's maple code for similar problems, in which plots[animate] is used to display a filmstrip, then a single line of code is issued to display the individual frames. STEADY-STATE TEMPERATURE The negative exponentials in the solution to Fourier's bar problem account for the limit condition limit u(x,t) = 0 as t approaches infinity. We say that the steady-state temperature for the Fourier bar problem is u=0. DEF. The steady-state temperature is the solution of the heat equation when u_t=0. This is the solution of u_xx=0, which is the linear function u(x)=Ax+B (by the method of quadrature for ODE). EXAMPLE. Steady-state solution A welding rod of length L has one end at temperature T1 and the other end at temperature T2, with lateral surface insulation. Describe the steady-state solution. ANSWER. u(x)=Ax+B where A=(T2-T1)/L and B=T1. NONZERO BOUNDARY CONDITIONS The heat problem is the nonzero-ends BVP u_tt = c^2 u_xx, u(0,t)=T1, u(L,t)=T2, u(x,0)=f(x) on 00. THEOREM. The solution of the nonzero-ends BVP is u(x,t) = u1(x) + u2(x,t) u1(x)=(T2-T1)x/L+T1 (steady-state solution) u2(x,t)=sum(v[n],n=1..infinity) (series of prod sols) v[n]=b[n] exp(-(c n Pi/L)^2t) sin(n Pi x/L), b[n]=(2/L)int(f1(x)sin(n Pi x/L),x=0..L), f1(x)=f(x)-u1(x) EXAMPLE. Solve the nonzero-ends welding rod problem with u(0,t)=0, u(Pi,t)=100, assuming L=Pi, c=1. ANSWER: b[n]=200/(n Pi), u1(x)=100 x/Pi. QUESTION 1. Does limit u(x,t)=u1(x) at t=infinity? ANS. Yes, but difficult to write a proof. QUESTION 2. Does u(x,0)=100? ANS. Yes. The question reduces to finding the Fourier series expansion of the triangular wave which has shape x -> 100(1-x/Pi) on 0 < x < Pi. APPENDIX. Derivation of the heat equation. The class agreement is to read this derivation by yourself, due to the widely different backgrounds present in the audience (77 students from many different engineering and science programs). 3.6 Heat conduction in bars with varying boundary conditions ==== We study these problems: DIRICHLET CONDITIONS u(0,t)=0, u(L,t)=0 (zero temperature ends) NEUMANN CONDITIONS u_x(0,t)=0, u_x(L,t)=0 (zero heat flux) ROBIN CONDITIONS u(0,t)=0, u_x(L,t)+kappa u(L,t)=0 (kappa = convection coefficient) GENERALIZED FOURIER SERIES A study of series which look formally like Fourier series, but differ in essential ways. EXAMPLE. Bar with insulated ends. This is the Neumann problem u_t = c^2 u_xx u_x(0,t)=0, u_x(L,t)=0 (zero heat flux) u(x,0)=f(x) Solved anew by separation of variables to give an answer with cosines instead of the sines present in the Dirichlet problem. THEOREM. Separation u=X(x)T(t) in the Neumann problem gives the two problems X'' - k X = 0, X'(0)=0, X'(L)=0, X nonzero, T' - k c^2 T = 0, T nonzero. Then up to a constant multiple, X(x)=cos(n Pi x/L), T(t) = exp(-(c n Pi x/L)^2 t) THEOREM. Neumann Problem The solution of the problem [see display above] is given by the formula u(x,t)= a[0] + sum(a[n] v[n],n=1..infinity), where v[n] = cos(n Pi x/L) exp(-(c n Pi/L)^2 t), a[0]=(1/L)int(f(x),x=0..L), a[n]=(2/L)int(f(x) cos(n Pi x/L),x=0..L) EXAMPLE. Bar with one end ice-packed, other with convection loss. This is the Robin problem u_t = c^2 u_xx u(0,t)=0, u_x(L,t)+kappa u(L,t)=0 (radiating end) u(x,0)=f(x) Solved anew by separation of variables to give an answer with sines and exponentials, with a twist, because we don't get a classical Fourier series for f(x). THEOREM. Separation u=X(x)T(t) in the Robin problem gives the two problems X'' - k X = 0, X(0)=0, X'(L)+kappa X(L)=0, X nonzero, T' - k c^2 T = 0, T nonzero. Then up to a constant multiple, X(x)=sin(mu[n] x), T(t) = exp(-c^2 mu[n]^2 t) where mu[n]=nth root of the nonlinear algebraic equation tan( mu L) + mu/kappa = 0 THEOREM. Orthogonality Relations The system of functions {sin(mu[n] x)} for n=1..infinity are orthogonal on [0,L]. This means that for any two functions f, g selected from this system, 1. int(f(x)g(x),x=0..L)=0 for f unequal to g, 2. int(f(x)^2,x=0..L) is positive, THEOREM. Robin Problem The solution of the problem [see display above] is given by the formula u(x,t)= sum(c[n] v[n],n=1..infinity), where v[n] = sin(mu[n] x) exp(-c^2 mu[n]^2 t), c[n]=(1/d[n])int(f(x) sin(mu[n] x),x=0..L), d[n]=int((sin(mu[n] x))^2,x=0..L) EXAMPLE. Bar with one end ice-packed, other with convection loss. This is the Robin problem u_t = u_xx u(0,t)=0, u_x(1,t)+u(1,t)=0 (radiating end) u(x,0)=x(1-x) Solved by appeal to the previous example to give an answer with sines and exponential. We don't get a classical Fourier series for f(x)=x(1-x). The example is largely numerical, because of the difficult integrations required for the generalized Fourier coefficients. ANSWER. u(x,t)=0.2133 exp(-mu[1]^2 t) sin(2.0288 x) + 0.1040 exp(-mu[2]^2 t) sin(4.9132 x) - 0.0220 exp(-mu[3]^2 t) sin(7.9787 x) + ... mu[1]=2.0288, m[2]=4.9132, mu[3]=7.9787, ... Constant mu[n] is found from the equation tan(mu)+mu=0, using this maple code: # Find the eigenvalues of the Sturm-Liouville BVP h:=mu->tan(mu)+mu; plot({tan(x),-x},x=0..20,y=-20..1,discont=true); d:=0.1;guess:=[seq((2*n+1)*Pi/2+d,n=0..10)]; for j from 1 to nops(guess) do mu[j]:=fsolve(h(mu)=0,mu=guess[j]); od; # Find generalized Fourier coefficients f:=x->x*(1-x); for j from 1 to nops(guess) do c[j]:=int(f(x)*sin(mu[j]*x),x=0..1)/int(sin(mu[j]*x)^2,x=0..1); od; DEF. A series of the form f(x) = sum of constants c[n] times functions g[n](x) is called a GENERALIZED FOURIER SERIES provided 1. Function g[n] is defined on [a,b] with real or complex values 2. The system { g[n] } for n=1 to infinity is an orthogonal system on [a,b]. This means that if u and v are two functions in the list { g[n] }, and u is different from v, then 2a. int (u(x)^2,x=a..b) > 0 [u is nonzero]. 2b. int ( u(x)v(x),x=a..b)=0 [u and v are orthogonal]. THEOREM. Generalized Fourier Coefficients In a generalized Fourier series, c[n] = int (f(x)g[n](x),x=a..b)/int(g[n]^2(x),x=a..b)