Separation of Variables - University of Arizona

1 Separation of VariablesA typical starting point to study differential equations is to guess solutions of a certain we will deal with linear PDEs, the superposition principle will allow us to form new solu-tions from linear combinations of our guesses, in many cases solving the entire problem. To beginwith, we will consider functions of two variablesu(v1,v2)(for exampleu(x,y)oru(r, )), wherethe domain is very particular: it must be of the form(v1,v2) [a,b] [c,d]. It will also be nec-essary to have homogeneous boundary conditions on opposite boundariesv1=aandv1=b(oralternativelyv2= candv2=d).

2 The Separation principleSuppose we have a problem with variablesx1,x2,..,xn, and letf1(x1),f2(x2),..,fn(xn)be func-tions of one variable each. If it happens thatf1(x1) +f2(x2) +..+fn(xn) = 0,for all(x1,x2,..,xn) (1)then each term in (1) is a constant. This is easy to show: just take partial derivatives of the lefthand expression with respect to eachxi. This givesf i(xi) = 0sofi(xi) = i; each iare calledseparation constants. The point of Separation of Variables is to get to equation (1) to begin with,which can be done for a good number of homogeneous linear The wave equationAs a first example, consider the wave equation with boundary and initial conditionsutt=c2uxx, u(0,t) = 0 =u(L,t), u(x,0) = (x), ut(x,0) = (x).

3 (2)We attempt an educated guess: find solutions of the formu(x,t) =X(x)T(t)which satisfy ev-erything except the inhomogeneous initial conditions. These will be calledseparated solutions . Ofcourse, not every solution will be found this way, but we have a trick up our sleeve: the superpo-sition principle guarantees that linear combinations of separated solutions will also satisfy boththe equation and the homogeneous boundary conditions. The proper choice of linear combinationwill allow for the initial conditions to be (x,t) =X(x)T(t)into the equation in (2) givesXT =c2TX.

4 We can separate thex- andt- dependence by dividing to giveT c2T=X the Separation principle, it follows that the left and right hand sides are equal to a constant,which we will call . Similarly, insertingu(x,t) =X(x)T(t)into the boundary conditions in (2)means thatX(0)andX(L)must be zero. We therefore get two ODEs: a boundary value problemforXX + X= 0, X(0) = 0 =X(L),(3)and an unconstrained equation forT,T +c2 T= 0.(4)1 Notice immediately that the problem forXis actually an eigenvalue problem which was solvedpreviously. We have a countable number of solutionsXn= sin(n xL), n=(n L)2, n= 1,2,3.

5 Equation (4) has two linearly independent solutions :sin(c t)andcos(c t). Setting = nforeachn, we find that separated solutions have the formsin(cn t/L) sin(n x/L),cos(cn t/L) sin(n x/L), n= 1,2,3,..We might hope that a linear combination of separated solutions solves the whole problem,u(x,t) = n=1[Ancos(cn t/L) +Bnsin(cn t/L)] sin(n xL).(5) using the initial conditions in (2) means that (x) = n=1 Ansin(n xL), (x) = n=1(n cL)Bnsin(n xL).These are simply Fourier sine series, and determining the coefficients is just a matter of takinginner products of both sides with the orthogonal eigenfunctionsXn= sin(n x/L).

6 This givesAn= ,Xn Xn,Xn , Bn=(Ln c) ,Xn Xn,Xn ,(6)where , is the usualL2inner this point, the result (5) may seem anti-climatic. It s difficult to imagine exactly what asuperposition of an infinite number of oscillations might even look like. It turns out that we areoften more interested in the individual components of the solution, the separated solutions whosespatial and temporal structure is easy to understand. The eigenfunctions that make up the spatialdependence are often calledmodes(or normal modes) whose shape defines the underlying mode has its own frequency of oscillation, n=n cL, n= 1,2,3.

7 In many problems, in fact, the set of frequencies{ n}is much more interesting than the completesolution (5)! These frequencies form the basis for the description of many physical phenomenon,including the production of sound waves and atomic The diffusion and Laplace equationsThe preceding strategy can be immediately adapted to other linear equations with the same do-main and boundary conditions such as the diffusion equationut=Duxx, u(0,t) = 0 =u(L,t), u(x,0) = (x),(7)and Laplace s equationuxx+uyy= 0, u(0,y) = 0 =u(L,y), u(x,0) =h(x), u(x,H) =g(x),(8)2 For the diffusion equation (7), again setu=X(x)T(t)and separate the Variables to giveT T=X X=.

8 Remarkably, we obtain exactly the same eigenvalue problem forXand as before (3). The differ-ence is in the equation forT, which readsT = D T,which has one linearly independent solutionT= exp( D t). The separated solutions are there-foreexp( D(n /L)2t) sin(n x/L), n= 1,2,3,..In this case, the modesXndecay in time rather than oscillate. Note that we had predicted exactlythis when discussing conserved and dissipated quantities for the wave and diffusion linear combination of separated solutions isu(x,t) = n=1 Anexp( D(n /L)2t) sin(n xL).(9)Invoking the initial condition, it follows that (x) = n=1 Ansin(n xL).

9 Which means the coefficients are the same as in (6).For the Laplace equation (8), Separation of variablesu=X(x)Y(y)leads to Y /Y=X /X= . We again get the the same eigenvalue problem (3) forXand an equation forYof the formY = is always positive, there are two linearly independent solutionsY= exp( y)andY=exp( y). The separated solutions are thereforeexp(n y/L) sin(n x/L),exp( n y/L) sin(n x/L), n= 1,2,3,..A superposition of these isu(x,y) = n=1[Anexp(n y/L) +Bnexp( n y/L)] sin(n xL).(10)We now try to satisfy the inhomogeneous boundary conditions in (8).

10 Settingy= 0andy=Hgivesh(x) = n=1(An+Bn) sin(n xL), g(x) = n=1[Anexp(n H/L) +Bnexp( n H/L)] sin(n xL).Each represents a Fourier sine series, so upon taking inner products with the eigenfunctionsXn=sin(n x/L), one getsAn+Bn= h,Xn Xn,Xn , Anexp(n H/L) +Bnexp( n H/L) = g,Xn Xn,Xn 3 For each value ofn, this represents a system of two equations for the two unknownsAn,Bn, whichcan be solved in that we could have used different linearly independent solutions forY = Y, such asY= cosh( y)andY= sinh( y), so that the solution readsu(x,y) = n=1[Cncosh(n y/L) +Dnsinh(n y/L)] sin(n xL).