Transcription of Chapter 3. Second Order Linear PDEs
1 Chapter 3. Second Order Linear IntroductionThe general class of Second Order Linear PDEs are of the form:a(x,y)uxx+b(x,y)uxy+c(x,y)uyy+d(x,y )ux+e(x,y)uy+f(x,y)u=g(x,y).( )The three PDEs that lie at the cornerstone of applied mathematics are: theheat equation, the wave equation and Laplace s equation, (i)ut=uxx,the heat equation(ii)utt=uxx,the wave equation(iii)uxx+uyy=0,Laplace s equationor, using the same independent variables,xandy(i)uxx uy=0,the heat equation( )(ii)uxx uyy=0,the wave equation( )(iii)uxx+uyy= s equation( )Analogous to characterizing quadratic equationsax2+bxy+cy2+dx+ey+f=0,as either hyperbolic, parabolic or elliptic determined byb2 4ac>0,hyperbolic,b2 4ac=0,parabolic,b2 4ac<0,elliptic,2 Chapter 3.
2 Linear Second Order Equationswe do the same for PDEs. So, for the heat equationa=1,b=0,c=0sob2 4ac=0 and so the heat equation is parabolic. Similarly, the waveequation is hyperbolic and Laplace s equation is elliptic. This leads to anatural question. Is it possible to transform one PDE to another where thenew PDE is simpler? Namely, under a change of variabler=r(x,y),s=s(x,y),can we transform to one of the followingcanonicalforms:urr uss+ ,hyperbolic,( )uss+ ,parabolic,( )urr+uss+ ,elliptic,( )where the term stands for lower Order terms. For example, con-sider the PDE2uxx 2uxy+5uyy=0.( )This equation is elliptic since the ellipticb2 4ac=4 40= 36<0.
3 Ifwe introduce new coordinates,r=2x+y,s=x y,then by a change of variable using the chain ruleuxx=urrr2x+2ursrxsx+usss2x+urrxx+uss xx,uxy=urrrxry+urs(rxsy+rysx) +usssxsy+urrxy+ussxy,uyy=urrr2y+2ursrysy +usss2y+urryy+ussyy,givesuxx=4urr+4urs+u ss,uxy=2urr urs uss,uyy=urr 2urs+ Introduction3 Under ( ), equation ( ) becomesurr+uss=0,which is Laplace s equation (also elliptic). Before we consider transforma-tions for PDEs in general, it is important to determine whether the equa-tion type could change under transformation. Consider the general classof PDEsauxx+buxy+cuyy=0( )wherea,b, andcare functions ofxandyand noting that we have sup-pressed the lower terms as they will not affect the type.
4 Under a change ofvariable(x,y) (r,s)with the change of variable formulas ( ) givesa(urrr2x+2ursrxsx+usss2x+urrxx+ussx x)+b(urrrxry+urs(rxsy+rysx) +usssxsy+urrxy+ussxy)( )+c(uyy+urrr2y+2ursrysy+usss2y+urryy+uss yy)=0 Rearranging ( ), and again neglecting lower Order terms, gives(ar2x+brxry+cr2y)urr+ (2arxsx+b(rxsy+rysx) +2crysy)urs( )+ (as2x+bsxsy+cs2y)uss= +brxry+cr2y,B=2arxsx+b(rxsy+rysx) +2crysy,( )C=as2x+bsxsy+cs2y,gives (again suppressing lower Order terms)Aurr+Burs+Cuss=0,whose type is given byB2 4AC= (b2 4ac)(rxsy rysx)2,4 Chapter 3. Linear Second Order Equationsfrom which we deduce thatb2 4ac>0, B2 4AC>0,b2 4ac=0, B2 4AC=0,b2 4ac<0, B2 4AC<0,giving that the equation type is unchanged under transformation.
5 We nowconsider transformations to canonical form. As there are three types ofcanonical forms, hyperbolic, parabolic and elliptic, we will deal with eachtype Canonical FormsIf we introduce the change of coordinatesr=r(x,y),s=s(x,y),( )the derivatives change according to:First Orderux=urrx+ussx,uy=urry+ussy,( ) Second Orderuxx=urrr2x+2ursrxsx+usss2x+urrxx+us sxx,uxy=urrrxry+urs(rxsy+rysx) +usssxsy+urrxy+ussxy, ( )uyy=urrr2y+2ursrysy+usss2y+urryy+ussyy, If we substitute ( ) and ( ) into the general Linear equation ( ) andre-arrange we obtain(ar2x+brxry+cr2y)urr+(2arxsx+b(rxs y+rysx) +2crysy)urs+ (as2x+bsxsy+cs2y)uss+ ( )Our goal now is to target a given canonical form and solve a set of equa-tions for the new Canonical Parabolic Canonical FormComparing ( ) with the parabolic canonical form ( ) leads to choos-ingar2x+brxry+cr2y=0,( )2arxsx+b(rxsy+rysx) +2crysy=0,( )
6 Since in the parabolic caseb2 4ac=0, then substitutingc=b24awe findboth equations of ( ) are satisfied if2arx+bry=0.( )with the choice ofs(x,y)arbitrary. The following examples +6uxy+9uyy=0.( )Here,a=1,b=6 andc=9 showing thatb2 4ac=0, so the PDE isparabolic. Solvingrx+3ry=0,givesr=f(3x y).As we wish to find new coordinates as to transform the original equationto canonical form, we chooser=3x y,s= Second derivativesuxx=9urr,uxy= 3urr+3urs,uyy=urr 2urs+uss.( )Substituting ( ) into ( ) givesuss=0! Not to be confused with factorial (!).6 Chapter 3. Linear Second Order EquationsSolving givesu=f(r)s+g(r).wherefandgare arbitrary functions.
7 In terms of the original variables,we obtain the solutionu=y f(3x y) +g(3x y).Example 4xyuxy+4y2uyy+xux=0.( )Here,a=x2,b= 4xyandc=4y2showing thatb2 4ac=0, so the PDEis parabolic. Solvingx2rx 2xyry=0,orxrx 2yry=0,givesr=f(x2y).As we wish to find new coordinates, , we choose simpler=x2y,s= first derivatives givesux=2xyur.( )Calculating Second derivativesuxx=4x2y2urr+2yur,( )uxy=2x3yurr+2xyurs+2xur,( )uyy=x4urr+2x2urs+uss.( )Substituting ( ) and ( ) into ( ) gives4y2uss 4x2yur= Canonical Forms7or, in terms of the new variables,rands,uss rs2ur=0.( )An interesting question is whether different choices of the arbitrary func-tionfand the variableswould lead to a different canonical forms.
8 Forexample, suppose we choser=2 lnx+lny,s=lny,we would obtainuss ur us=0,( )a constant coefficient parabolic equation, whereas, choosingr=2 lnx+lny,s=2 lnx,we would obtainuss ur=0,( )the heat Hyperbolic Canonical FormIn Order to obtain the canonical form for the hyperbolic type, uss+ ,( )it is necessary to choosear2x+brxry+cr2y= (as2x+bsxsy+cs2y),2arxsx+b(rxsy+rysx) +2crysy=0.( )The problem is that this system is still a very hard problem to solve (bothPDEs are nonlinear and coupled!). Therefore, we introduce a modified hy-perbolic form that is much easier to work 3. Linear Second Order Modified Hyperbolic FormThe modified hyperbolic canonical form is defined asurs+ ,( )noting thata=0,b=1 andc=0 and thatb2 4ac>0 still!
9 In Order totarget the modified hyperbolic form, it is now necessary to choosear2x+brxry+cr2y=0,( )as2x+bsxsy+cs2y=0.( )If we re-write ( ) and ( ) as followsa(rxry)2+2brxry+c=0,( )a(sxsy)2+2bsxsy+c=0,( )then we can solve equations ( ) and ( ) separately leads to two first Order Linear PDEs forrands. The solutions of thesethen gives rise to the correct canonical variables. The following 5uxy+6uyy=0( )Here,a=1,b= 5 andc=6 showing thatb2 4ac=1>0, so the PDEis hyperbolic. Thus, ( ) becomesr2x 5rxry+6r2y=0,s2x 5sxry+6s2y=0,and factoring gives(rx 2ry)(rx 3ry)=0,(sx 2sy)(sx 3sy)=0, Canonical Forms9from which we chooserx 2ry=0,sx 3sy=0,giving rise to solutionsr=f(2x+y),s=g(3x+y).
10 As we wish to find new co-ordinates as to transform the original equationto canonical form, we chooser=2x+y,s=3x+ Second derivativesuxx=4urr+12urs+9uss,uxy=2urr+ 5urs+3uss,( )uyy=urr+2urs+ ( ) into ( ) givesurs= givesu=f(r) +g(s).wherefandgare arbitrary functions. In terms of the original variables,we obtain the solutionu=f(2x+y) +g(3x+y). (x+y)uxy+yuyy=0.( )Here,a=x,b= (x+y)andc=yshowing thatb2 4ac= (x y)2>0,so the PDE is hyperbolic. Solvingxr2x (x+y)rxry+yr2y=0,10 Chapter 3. Linear Second Order Equationsor, upon factoring(xrx yry)(rx ry)= the same equation, we choose the first factor forrand thesecond forsxrx yry=0,sx sy=0.( )Upon solving ( ), we obtainr=f(xy),s=g(x+y).
