1 Quasi-Linear Partial Differential Equations

1 1 Quasi-Linear Partial Differential EquationsDefinition th order Partial differential equationis an equationinvolving the firstnpartial derivatives ofu,F(x,y,..,u, u x, u y,.., nu xn,..) = first-order two variablesx,yis an equation of typea(x,y) u x+b(x,y) u y=c(x,y)u(x,y).We will be able to solve Equations of this form; in fact of a slightly moregeneral form, so calledquasi-linear:a(x,y,u) u x+b(x,y,u) u y=c(x,y,u).2 SolutionDefine a curve in thex,y,uspace as followsr(t) = (x(t),y(t),z(t)) satisfyingdrdt= (a,b,c)with initial conditionr(0) on the surfacez=u(x,y).

2 From Picard s theorem, assuming thata,bandcare well-behaved functionsofx,yandu, we know that there is a unique we can compare this curve with the curveu=u(x(t),y(t)).dudt= u xdxdt+ u ydydt=a u x+b u y=c=dzdtand since both curves start at the same point, we must haveu= course, this only gives a single curve on the surface, but if we are specifieda whole line of initial points of the typeu(xs,ys) =usthen we can solve theequation r= (a,b,c) with initial conditionsx(0) =xs,y(0) =ysandu(0) =usto give a collection of lines all lying on the surface we getu=u(t,s).

3 If we can now eliminate the variablestandsusing the equationsx=x(t,s)andy=y(t,s), then we get our required solutionu=u(x,y). Examplex u x y u y=x y1given thatu(0,s) = : Define the following curves x=xx(0) = 0 y= yy(0) =s u=x yu(0) = 0 Solving the first two Equations is straightforward to givex(t) =ety(t) =se can now substitute into the third equation to get u=et se tu(0) = ,u(t,s) =et+se t (1 +s)=x+y 1 required solution is thereforeu(x,y) =x+y 1 Second MethodThe method described above is straightforward, but it may prove difficult inpractice to change variables froms,tback tox,y.

4 A second method makes thiseasier by automatically suppressing thetvariable to start object is to find two independent integrable identities that, when inte-grated, do not involvet(but may, and usually do, involve the initial conditionvariables). Once these two relations are found, it is then usually straightforwardto eliminatesfrom the two to leave a relation involvingu, Exampley u x+x u y+ 2xy= 0withu(s,2s) = : Set up the differential Equations onx(t),y(t) andu(t): x=yx(0) =s y=xy(0) = 2s u= 2xyu(0) = integrable relations are the following.

5 X x y y= 0 u+ 2x x= 0,2which integrate tox2/2 y2/2 =Aandu+x2=B, where the constantsAandBare determined from the initial conditions to getx2 y2= 3s2u+x2= can now eliminatesfrom the two relations to getu+x2= (x2 y2) (x,y) =y2 General SolutionThe second method is particularly useful in finding the general solution to aquasi-linear , that is, one for which the initial conditions are not we follow the same steps as before, we again end up with two integratedrelations that have two undetermined constantsA(s) andB(s).

6 Howeverscanstill be eliminated from the two Equations in the sense that if the relations aregiven asR(x,y,u) =A(s) andT(x,y,u) =B(s) thenR(x,y,u) =A(B 1(T(x,y,u))) =F(T(x,y,u))for a general Example(3y 2u) u x+ (u 3x) u y= 2x yDefine the curves satisfying the Equations x= 3y 2u y=u 3x u= 2x can find two integrable relations, x+ 2 y+ 3 u= 0,x x+y y+u u= 0,which integrate tox+ 2y+ 3u=A(s),x2/2 +y2/2 +u2/2 =B(s).3 Eliminatings, we getx2+y2+u2=F(x+ 2y+ 3u)which is the general solution of the differential initial conditions are now specified, we can find whatFis from them asfollows.

7 Suppose we specify thatu(s,s) = 0. Therefore substitutingx=s,y=sandu= 0 in the general solution we get2s2=F(3s).Hence, if we write 3s=x+ 2y+ 3u, we get the specific solutionx2+y2+u2= 2s2= 2(x+ 2y+ 3u3) the terms we finally get9u2+ 12(x+ 2y)u (7x2 8xy+y2) = 0to giveu(x,y) = 23(x+ 2y) 13 11x2+ 8xy+ ExercisesFind the general solution, and then solve using the given data, for the u x+ u y+u/2 = 0u(s,0) = 12.(y+u) u x (x+u) u y+ (x y) = 0u(s,2s) = (y+u2) u x+y(x2 u2) u y+u(x2+y) = 0u(0,s) = y( u x u y) = 2u(s,0) = s25.

8 (3x u) u x+ (3y u) u y=x+yu(s,2s) = 04 Tutorials the solutionu(t,x) of the following Equations :1. tu 2xu= 0u(t,0) = 0u(t,1) = 0u(0,x) = 6 sin(9 x)2. Same as above but withu(0,x) ={1 0<x<1/22 1/2<x< tu 2xu= 0 xu(t,0) = 0 xu(t,1) = 0u(0,x) =f(x)Show that the solution isu(t,x) = n=1 Ane ntcos(n x)wheref(x) = n=1 Ancos(n x).Find formulas forAnin terms off(x).4. tu 2xu+u= 0u(t,0) = 0 =u(t,1)u(0,x) = 15. 2xu+ 2yu= 0u(0,y) =y,u(x,0) =u(1,y) =u(x,2) = 06. 2tu 2xu= 0u(t,0) =u(t,1) xu(t,0) = xu(t,1) tu(0,x) = 0u(0,x) ={2x0<x<1/31 x1/3<x<15}}