Lecture Notes on PDEs, part I: The heat equation and the ...

1 Lecture Notes on PDEs, part I:The heat equation and the eigenfunction methodFall 2018 Contents1 Motivating example: Heat conduction in a metal bar22 The heat equation : Initial and boundary conditions .. Linearity and homogeneous PDEs .. More on superposition ..93 Orthogonal functions Linear algebra analogy: using orthogonal bases inRn.. What is the right space of functions? .. Some examples; importance of the interval ..144 Solving the eigenvalue problem .. Eigenfunctions and orthogonal bases .. Example: Neumann boundary conditions ..195 The eigenfunction method to solve The method (for the heat equation ).

2 Dirichlet boundary conditions .. Neumann boundary conditions .. An example .. Another example: finite number of modes .. efficient approach: ..306 Separation of A first example .. The method .. Use in solving unfamiliar problems .. : ..3517 Appendix: additional Some context: PDEs from conservation laws .. Deriving the heat equation ..37 Notation:Hereafter, subscripts will denote derivatives with respect to a variable. Forinstance,ut(x,t) = t(u(x,t)), uxx= 2u x2, differential equation (PDE) for a function of more than one variable is a anequation involving a function of two or more variables and its partial Motivating example: Heat conduction in a metal barA metal bar with lengthL= is initially heated to a temperature ofu0(x).

3 The temper-ature distribution in the bar isu(x,t). At the ends, it is exposed to air; the temperatureoutside is constant, so we require thatu= 0 at the endpoints of the time, we expect the heat to diffuse or be lost to the environment until the temperatureof the bar is in equilibrium with the air (u 0).Physicist Joseph Fourier, around 1800, studied this problem and in doing so drew attentionto a novel technique that has since become one of the cornerstones of applied approach outlined below hints at some of the deep structure we will uncover in theremainder of the temperature is modeled by theheat equation (see subsection for a derivation) u t= 2u x2, t >0 andx (0, ).

4 Since the temperature is fixed at both ends, we haveu(0,t) = 0, u( ,t) = 0 for , the initial heat distribution ist= 0 isu(x,0) =f(x)2wheref(x) is some positive function that is zero at 0 and . The temperature should de-crease as heat leaks out of the bar through the ends; eventually it all dissipates. The solutionu(x,t) should predict summary, our goal is to find a functionu(x,t) defined on [0, ] satisfying u t= 2u x2t >0 andx (0, ),(1a)u(0,t) =u( ,t) = 0 fort 0(1b)u(x,0) =u0(x).(1c)Our objective here is just to find a solution to the first two parts, (1a) and (1b) and worryabout the initial condition , let us guess a solution of the formu=e t (x).

5 (2)Substituting into the PDE (1a), we find that (x) = (x).Now substitute into the boundary conditions (1b) (note thate tcancels out here) to get (0) = 0, ( ) = convenience set = follows that (2), our guess foru, satisfies the PDE (1a) andthe boundary conditions (1b) if the functiong(x) solves theboundary value problem (x) + 2 (x) = 0, (0) = 0, ( ) = 0.(3)This problem is not an initial value problem (conditions are imposed at both ends), but itis a constant-coefficient ODE, so we can still solve it explicitly. The general solution is =c1sin( x) +c2cos( x).Imposing the condition (0) = 0 we find that =c1sin( x).The second condition, (1) = 0, requires thatsin( ) = solutions exist whenever is a non-zero integer.

6 We have now found an infinitesequence of solutions to (3): n(x) = sin(nx), n= 1,2,3, 3 Observe that (3) is a linear, homogeneous problem. In particular, 1, 2are solutions to (3) = c1 +c2 2is a solution.(4)This means that for any constantan,the functionane n2t n(x)(5)is a solution to the heat conduction problem with initial datau0(x) =ansin(nx).Now the crucial question:what happens when the initial data is not a sine? No singlesolution of the form (5) will work. Fourier s breakthrough was the realization that, using thesuperposition principle (12), the solution could be written as aninfinitelinear combinationof all the solutions of the form (5):u(x,t) = n=1ane n2t n(x).

7 Thenu(x,t) solves the original problem (10) if the coefficientsansatisfyu0(x) = n=1an n(x).(6)This idea is a generalization of what you know from linear algebra (representing vectors interms of a basis) but with basisfunctions{sin(nx) :n= 1,2,3, }.In fact, this set of functions has the rather remarkableorthogonalityproperty 0 m(x) n(x)dx= 0sin(mx) sin(nx)dx= 0, m6=n.(7)To solve for the coefficientam, we can multiply (6) by sin(mx) and integrate: 0u0(x) sin(mx)dx= 0 n=1ansin(mx) sin(nx) move the integral inside the sum (it is not trivial to show this is allowed!). By theproperty (7), only one of the terms in the sum will be non-zero: 0u0(x) sin(mx)dx= 0 n=1ansin(mx) sin(nx)dx= n=1an 0sin(mx) sin(nx)dx=( n=1,n6=man 0)+am 0sin(mx) sin(mx)dx=am 0sin2(mx) , the infinite sum has been reduced to a simple equation foram:am= 0u0(x) sin(mx)dx 0sin2(mx)dx.

8 (8)This process works for allm, so the solution to the heat conduction problem (5) with arbitraryinitial conditionu0(x) isu(x,t) = n=1ane n2tsin(nx)with the coefficients given by the formula (8). Of course, all of the manipulations here areformal and unjustified - it is far from clear whether the series converges, or if it is valid toswap integrals and sums, and so on (Fourier did not know this either when first applyingthe method; it took several decades to settle the issue).2 The heat equation : preliminariesLet [a,b] be a bounded interval. Here we consider the PDEut=uxx, x (a,b), t >0.(9)foru(x,t). This is theheat equationin the interval [a,b].

9 Remark (adding a coefficient):More generally, we could considerut=kuxxwherek >0 is a diffusion coefficient . However, since the constant can be scaled out bydefining a rescaled time =t/kto getu =uxxthere is no loss of generality in studying the structure of (10). Note that it isessential thatthe coefficient is positive; a negativekwill produce drastically different Initial and boundary conditionsAninitial boundary value problem(IBVP) for the heat equation consists of the PDEitself plus three other conditions specified atx=a,x=bandt= 0. As a simple example: u t= 2u x2t >0 andx (a,b),(10a)u(a,t) = 0 andu(b,t) = 0fort >0(10b)u(x,0) =f(x).(10c)5 There are three components:The PDE: equation (10a) is the PDE (sometimes just the equation ), which thThe besolution must satisfy in the entire domain (x (a,b) andt >0 here).

10 Boundary conditions (BCs):Equations (10b) are theboundary conditions, imposedat the boundary of the domain (but not the boundary intatt= 0). Each boundary condi-tion is some condition onuevaluated at the conditions (ICs): equation (10c) is theinitial condition, which specifies theinitial values ofu(at the initial timet= 0).The initial boundary value problem (10a)-(10c) has a unique solution provided some tech- nical conditions hold on the boundary can think of the boundary of the solution domain to have three sides:{x=a},{x=b}and{t= 0},with the last side left open (the solution fills this in ast ). The initialcondition is really a boundary condition att= (important BCs):There are three basic types of boundary conditions.