Introduction to Partial Differential Equations

Introduction to Partial Diﬀerential Equations

1 HeatConductioninaSolidBody

1.1 Heat Transfer in One Spatial Direction Consider a homogeneous rod:

length of rod is L lateral surface of rod is insulated rod has constant thermal conductivity k (k>0) rod has constant speciﬁcheatc (c>0) rod has constant density ρ (mass/volume) rod has congruent cross sections perpendicular to its lateral axis, the x-axis; each cross section has area A

H1. Heat ﬂows only parallel to the lateral axis of the rod.

Consequently, the temperature must be the same at each point in the cross section at x. Let: u = u (x, t) be the temperature at cross section x at time t. (0 x L, t 0) ≤ ≤ ≥ q = q (x, t) be the heat flux (the time rate of heat transfer per unit area) across the cross section at x at time t. (Heat flux is a vector quantity whose direction is given by the sign of q (x, t) . Heat flows from left to right when q (x, t) is positive and flows from right to left when q (x, t) is negative.) e = e (x, t) be the specific internal energy (heat energy per unit mass) at cross section x at time t. f = f (x, t) , assumed known, be the time rate per unit volume at which heat is generated by processes within the rod. (When f>0 there is a heat source. When f<0 there is a heat sink.)

1 H2. Thermal Energy is Conserved: Thetimerateofchangeofthermal energy in any region V in a heat conducting solid is equal to the net rate at which heat ﬂows into V across its boundary plus the net rate at which heat is produced by sources and/or sinks in V.

Apply H2 to the region V of the rod with a x b to obtain: ≤ ≤ b b etρA dx = ( qx + f) Adx (1) − Za Za for all a

b b r (x) dx = s (x) dx for all a

The Lemma and (1) imply that

etρ = qx + f (2) − Now, we need to relate the three unknowns e, q, and u through properties characteristic of heat ﬂow.

H3. (Fourier’s Law of Heat Conduction) Heat flows from regions of higher temperature to regions of lower temperature and the rate of flow is greater the more rapidly temperature changes with respect to distance. Therefore we assume that the rate of heat flow is proportional to the temperature gradient: q (x, t)= kux with k>0. − The thermal conductivity k is defined by this relation.

Use of Fourier’s law in (2) give

etρ = kuxx + f (3)

H4. For many materials and over wide temperature ranges the speciﬁcinternal energy is a linear function of the temperature:

e = cu + d (c>0,dconstants)

The specificheatc is defined by this relation. The units of specificheat are [energy]/[mass][deg].

2 From H4 et = cut. So (3) can be expressed as

ρcut = kuxx + f, the inhomogeneous heat equation (in one spatial dimension). Let k 1 α = (thermal diﬀusivity) and F = f. scρ cρ Then the inhomogeneous heat equation can be expressed as

2 ut = α uxx + F. (IHE)

If there are no sources/sinks present, then F =0and

2 ut = α uxx (HE) which is usually called the heat equation (AKA the diﬀusionequationinone spatial dimension).

1.2 Heat Transfer in Two and Three Spatial Directions Consider a homogeneous solid heat-conducting body: (Mathematically such a body is modeled as a bounded region D in xyz-space.)

solid has constant thermal conductivity k (k>0) solid has constant speciﬁcheatc (c>0) solid has constant density ρ (mass/volume)

In this context, u = u (x, y, z, t) , q = q (x, y, z, t) ,e= e (x, y, z, t) ,f = f (x, y, z, t) .

As before assume:

H2. Thermal energy is conserved H3. Fourier’s Law holds: q = k u − ∇ H4. Speciﬁc internal energy is a linear function of the temperature

3 The heat flux vector q is defined by the following property: Let S be any reasonable surface in the heat conducting medium and let n by a continuous unit normal vector field on S. (In mathematical lingo, the surfaced is orientable.) Then the heat per second flowing across S towardthesideofS specified by n is q dS = q n dS. · · ZZS ZZS

A straightforward adjustment of the reasoning we used in two spatial dimension leads to the inhomogeneous heat equation

2 ut = α (uxx + uyy + uzz)+F (IHE) in three spacial dimensions, and, if there are no sources/sinks present to the heat equation 2 ut = α (uxx + uyy + uzz) (HE) in three spacial dimensions. Here α and F are deﬁned as in the one spatial dimension case.

The derivation referred to above is most easily carried out using the (Gauss) Divergence Theorem: If D is a region in space with bounding surface S oriented by its outer normal n, then under quite general assumptions (see a book on vector analysis or multivariable calculus or Kellog’s Potential Theory)

div F dV = F dS · ZZZD ZZS for any smooth vector field F defined on D S. If the thermal situation is such that there is∪ not heat flow in say the z-direction, then the 3-D heat equations reduce to their 2-D versions in which all functions are independent of z : 2 ut = α (uxx + uyy)+F (IHE) in two spacial dimensions, and, if there are no sources/sinks present to the heat equation 2 ut = α (uxx + uyy) . (HE)

4 1.3 Initial and Boundary Conditions for Heat Conduction Problems and Separation of Variables The foregoing heat equations do not, by themselves, determine the temperature within the solid. In addition, we need to know the initial temperature in the solid and the manner in which heat moves across the boundary of the solid. For example, consider heat ﬂow in the homogeneous rod with no sources or sinks, whose initial temperature proﬁle is given by f (x) , and whose ends are kept at temperature zero. Intuition suggests that the temperature u = u (x, t) in the rod is determined by the initial boundary value problem:

2 ut α uxx =0 for 0 0, (HE) u (−x, 0) = f (x) for 0 x L, (IC) ⎧ u (0,t)=0,u(L, t)=0 for t ≤ 0. ≤ (BC) ⎨ ≥ ⎩

Let’s try to solve this problem.

Grand strategy: (HE) and (BC) are linear and homogeneous in u. By the superposition principle, (finite) linear combinations of solutions to (HE) and (BC) are again solutions to (HE) and (BC). Maybe we can adjust the coefficients in the linear combination so that the sum also satisfies (IC).

Key Step 1: Guess some solutions to the (HE). What is a reasonable educated guess?

Key Step 2: Reﬁne the guess in Step 1 so that the solutions in Step 1 also satisfy (BC).

Key Step 3: Superpose the solutions in Step 2 so that the resulting linear combination will satisfy (IC).

If the grand strategy works we will have found a solution of the given initial boundary value problem for the rod.

Steps 1 and 2: There are no sources or sinks in our problem and the ends of the rod are kept at temperature zero. It is reasonable to expect that the temperature u (x, t) 0 as t . Since solutions to DEs often involve exponential → →∞

5 functions, we look for solutions to the heat equation that decay exponentially in time — say λt u (x, t)=e− X (x) λ>0. For convenience replace λ by α2λ to get

α2λt u (x, t)=e− X (x) λ>0.

α2λt Our hope is to ﬁnd X (x) so that u (x, t)=e− X (x) satisﬁes (HE) and (BC). Let’s see if there is such an X... .

Summary: We found an inﬁnite number of functions

2 2 2 2 nπx u (x, t)=e α n π t/L sin n =1, 2, 3, ... (ss) n − L that satisfy (are solutions of) (HE) and (BC).

Along the way we reviewed or learned the following:

z2 zn ez =1+z + + + + for all complex z, 2! ··· n! ··· z3 z5 z7 sin z = z + + for all complex z, − 3! 5! − 7! − ··· z2 z4 z6 cos z =1 + + for all complex z. − 2! 4! − 6! − ··· Thus, setting z = iθ with θ real in the ﬁrst series and doing some algebra leads to the basic Euler identity

eiθ =cosθ + i sin θ.

Each of the following four equations, also called Euler identities, implies the other three:

iθ iθ e =cosθ + i sin θ, e− =cosθ i sin θ, iθ iθ iθ −iθ e + e− e e− cos θ = , sin θ = − . 2 2i Note that eiθ =1for any real θ. Theta is used here because¯ it often¯ is a polar angle. The Euler identities are also true with θ is complex in¯ which¯ case it is usually replace by z. For the record and later use: If z = x + iy, with x and y real, then

ez = exeiy = ex (cos y + i sin y)=ex cos y + iex sin y,

6 which expresses exp (z) in terms of its real and imaginary parts. Easy conse- quences are:

ez =0for any real or complex number z. 6 ez =1 z =2πin for n an integer. ⇐⇒ cos z =0 z = nπ/2 for n an odd integer. ⇐⇒ sin z =0 z = nπ for n an integer. ⇐⇒ The last two results conﬁrm that the real zeros of the cosine and sine functions are the only zeros of these functions even when they are extended to the complex domain. When are sin z and cos z equal to 1 or 1? − Lemma 2 If y = y (x) is a complex-valued solution to

a (x) y00 + b (x) y0 + c (x) y = f (x) where all coeﬃcients and the right member are real-valued, then the real and imaginary parts of y are real-valued solution of the diﬀerential equation.

The lemma and its proof hold for higher order linear diﬀerential equations.

Pause: Are there other solutions to (HE) and (BC) that decay in time to zero at some other rate? Let’s see. We seek separated solutions to the heat equation and boundary conditions of the form

u (x, t)=T (t) X (x) and ﬁnd ... that the only separated solutions to the (HE) and (BC) are the ones we have already found:

2 2 2 2 nπx u (x, t)=e α n π t/L sin n =1, 2, 3, ... (ss) n − L The use of separated solutions to solve a problem involving PDEs is called the method of separation of variables.

Steps 3: What about (IC)? Can we use the separated solutions in (ss) to ﬁnd a solution to (HE), (BC), and (IC)? Let’s see what is involved ... .

1.4 Standard Notation and Terminology The Laplacian of a function u of one, two, or three variables is deﬁned by

∆u = uxx in 1-variable,

∆u = uxx + uyy in 2-variables,

∆u = uxx + uyy + uzz in 3-variables.

7 With this notation the inhomogeneous heat equation can be expressed as

2 ut = α ∆u + F, (IHE) where F is a function of the space variables and of time, and the heat equation is 2 ut = α ∆u. (HE) IfthesourcetermF is independent of time (or independent of time as t ) and if the boundary conditions are independent of the time (or independent→ of∞ time as t ), then it is natural to expect that the temperature u tends to a steady state→∞(a time-independent state) as t . So in three spatial dimensions →∞ we expect that u (x, y, z, t) u (x, y, z) and ut 0 as t , where u (x, y, z) is the steady state temperature→ distribution. Passing→ to the→∞ limit as t , we see that the steady state temperature distribution will satisfy →∞ F ∆u = , (Poisson’s equation) −α2 when sources/sinks are present and

∆u =0 (Laplace’s equation). when they are not. Solutions to Laplace’s equation are called harmonic functions. Most physical situations in which a steady-state regime emerges are model by either Poisson’s or Laplace’s equation.

1.5 Linear Operators Let L be a operator. ForusthismeansthatL is a function whose domain (set of allowed inputs) consists of real or complex-valued functions from a function space —callit . For us will be a set of continuous or sufficiently differentiable functions so thatF all expressionsF in sight make sense. The key property of a function space is that if u and v are functions in the space so are αu + βv for any choice of scalars α and β. (In mathematical lingo, is a vector space over the real or complex numbers.) Consequently, any finiteF linear combination of functions in will be a function in . The range (set of all outputs) of an operator lie inF another function spaceF or in the set of real or complex numbers. An operator L is linear ( a linear operator) if

L (αu + βv)=αL (u)+βL(v) for all scalars scalars α and β and all functions u and v in . All other operators are called nonlinear. F

8 2 Example L (u)=ut α ∆u is a linear operator (a linear differential operator) from the space of− functions (on some reasonable space-time domain) that are once continuously differentiable in time and twice continuously differentiable in space into the space of continuous functions (on that same domain). Example B (u)=u (0) is a linear operator on the space of continuous functions on [0,L] into the real numbers. Example I (u)=u (x, 0) is a linear operator on the space of continuous (or differentiable or twice differentiable) functions on [0,L] [0, ) into the continuous (or differentiable or twice differentiable) functions× ∞ on [0,L] . An operator equation of the form L (u)=f (or B (u)=g or whatever) may express an ODE, a PDE, a boundary condition, or an initial condition. The equation L (u)=f (or B (u)=g or whatever) is called homogeneous if f (or g or the right member of whatever) is (identically) zero. Otherwise, the equation is called inhomogeneous. The equation L (u)=f (or B (u)=g or whatever) is called linear if L(or B or whatever) is a linear operator.

Superposition Principle If u1 and u2 are solutions of any linear homogeneous equation, L (u)=0, then c1u1 + c2u2 is also a solution of L (u)=0. Consequently, any ﬁnite linear combination of solutions of a linear homogeneous equation is a solution of the equation. Example The initial boundary value problem for the rod without sources/sinks and with its ends held at temperature zero can be expressed as

2 ut α uxx =0 for 0 0, (HE) u (−x, 0) = f (x) for 0 x L, (IC) . ⎧ u (0,t)=0,u(L, t)=0 for t ≤ 0. ≤ (BC) ⎨ ≥ Let ⎩ 2 L (u)=ut α uxx, − I (u)=u (x, 0) ,

B1 (u)=u (0,t) ,B2 (u)=u (L, t) . All three of these operators are linear and the initial boundary value problem for the rod is L (u)=0 for 0 0, (HE) I (u)=f (x) for 0 x L, (IC) ⎧ ≤ ≤ B1 (u)=0,B2 (u)=0 for t 0. (BC) ⎨ ≥ Therefore,⎩ the heat equation is a linear homogeneous equation. The initial condition is linear and inhomogeneous. Each boundary condition is a linear and homogeneous. By the superposition principle, linear combinations of solutions to (HE) and (BC) will again be solutions to these equations.

9 1.6 Concluding Remarks We still have much to learn about the effective use of separation of variables to solve initial boundary value problems. Stay tuned. When we succeed in finding separated solutions to a PDE, a constant, often denoted λ is introduced. It is called a separation constant. The values of the separation constant that lead to nontrivial solutions of the PDE are eigenvalues of a related eigenvalue problem ( a boundary value problem for an ODE that contains λ as a parameter). The class raised two important questions that I want to remark on now. First, you asked what is the advantage to deriving model equations by the integral approach we used as opposed to the more common differential formulation. The answer is two fold. Generally speaking the integral formulation better sorts out what physical assumptions are really essential in order to ar- rive at the model equations. (I’ll comment on this further when we derive the wave equation in the next section.) Second, and this especially pertains to wave phenomena where shocks and other types of discontinuities really need to be modeled, the state variables may not be differentiable up to the order required to derive the equations of state by differential methods. In such cases, the relevant equations of state have integral formulations. (You can’t use Lemma 1 to reduce them to differential equations because the hypotheses of the lemma are not satisfied.) Second, you noted that in deriving the differential equations satisfied by T and X, we implicitly assumed that T and X were never zero. This implicit assumption is always made when separating variables; however, the separated solutions are often zero for certain values of their independent variable. So the following observation is relevant and similar observations apply in any other context in which separation of variables is used. In our heat conducting rod, we actually determined T and X as solutions of

2 T˙ = α λT and X00 = λX. − − When the separated equations are written this way there is no problem with T and/or X taking on the value zero. Furthermore, if T and X are solutions of the foregoing equations, then u = TX is a solution of the heat equation. Just check it: u = TX will satisfy the heat equation if and only if

2 TX˙ = α TX00 and, using the ODEs satisﬁed by T and X,

2 2 2 TX˙ = α λT X = α T ( λX)=α TX00. − − So the heat equation is satis¡ ﬁed¢ and all is well.