First Order Partial Differential Equations

1 1 First Order Partial Differential Equations The profound study of nature is the most fertile source of mathematical discover-ies. - Joseph Fourier (1768-1830) begin our study of Partial Differential equationswithfirstorder Partial Differential doing so, we need to define a (see the appendix on Differential Equations ) that ann-th orderordinary Differential equation is an equation for an unknown functiony(x)n-th Order ordinary Differential equationthat expresses a relationship between the unknown function and its firstnderivatives. One could write this generally asF(y(n)(x),y(n 1)(x), .. ,y (x),y(x),x) =0.( )Herey(n)(x)represents thenth derivative ofy(x). Furthermore, and initialvalue problem consists of the Differential equation plus the values of theInitial value 1 derivatives at a particular value of the independent variable, sayx0:y(n 1)(x0) =yn 1,y(n 2)(x0) =yn 2,.. ,y(x0) =y0.( )If conditions are instead provided at more than one value of the indepen-dent variable, then we have a boundary value problem.

2 If the unknown function is a function of several variables, then the deriva-tives are Partial derivatives and the resulting equation is a Partial differen-tial equation. Thus, ifu=u(x,y, ..), a general Partial Differential equationmight take the formF(x,y, .. ,u, u x, u y, .. , 2u x2, ..)=0.( )Since the notation can get cumbersome, there are different ways to writethe Partial derivatives. First Order derivatives could be written as u x,ux, xu, Partial Differential equationsSecond Order Partial derivatives could be written in the forms 2u x2,uxx, xxu,D2xu. 2u x y= 2u y x,uxy, xyu, , we are assuming thatu(x,y, ..)has continuous Partial , according to Clairaut s Theorem (Alexis Claude Clairaut,1713-1765) ,mixed Partial derivatives are the of some of the Partial Differential equation treated in this bookare shown in However, being that the highest Order derivatives inthese equation are of second Order , these are second Order Partial differentialequations.

3 In this chapter we will focus on First Order Partial differentialequations. Examples are given byut+ux= +uux= +uux= 2uy+u= function of two variables, which the above are examples, a generalfirst Order Partial Differential equation foru=u(x,y)is given asF(x,y,u,ux,uy) =0,(x,y) D R2.( )This equation is too general. So, restrictions can be placed on the form,leading to a classification of First Order Equations . A linear First Order partialdifferential equation is of the formLinear First Order Partial (x,y)ux+b(x,y)uy+c(x,y)u=f(x,y).( )Note that all of the coefficients are independent ofuand its derivatives andeach term in linear inu,ux, can relax the conditions on the coefficients a bit. Namely, we could as-sume that the equation is linear only inuxanduy. This gives the quasilinearfirst Order Partial Differential equation in the formQuasilinear First Order Partial (x,y,u)ux+b(x,y,u)uy=f(x,y,u).

4 ( )Note that theu-term was absorbed byf(x,y,u).In between these two forms we have the semilinear First Order partialdifferential equation in the formSemilinear First Order Partial (x,y)ux+b(x,y)uy=f(x,y,u).( )Here the left side of the equation is linear inu,uxanduy. However, the righthand side can be nonlinear the most part, we will introduce the Method of Characteristics forsolving quasilinear Equations . But, let us First consider the simpler case oflinear First Order constant coefficient Partial Differential Order Partial Differential Equations Constant Coefficient EquationsLet s consider the linear First Order constant coefficientpar-tial Differential equationaux+buy+cu=f(x,y),( )fora,b, andcconstants witha2+b2>0. We will consider how such equa-tions might be solved. We do this by considering two cases,b=0 andb6= the First case,b=0, we have the equationaux+cu= can view this as a First Order linear (ordinary) Differential equation withya parameter.

5 Recall that the solution of such Equations can be obtainedusing an integrating factor. [See the discussion after Equation ( ).] Firstrewrite the equation asux+cau= the integrating factor (x) =exp( xcad ) =ecax,the Differential equation can be written as( u)x=fa .Integrating this equation and solving foru(x,y), we have (x)u(x,y) =1a f( ,y) ( )d +g(y)ecaxu(x,y) =1a f( ,y)eca d +g(y)u(x,y) =1a f( ,y)eca( x)d +g(y)e cax.( )Hereg(y)is an arbitrary function the second case,b6=0, we have to solve the equationaux+buy+cu= would help if we could find a transformation which would eliminate oneof the derivative terms reducing this problem to the previous case. That iswhat we will First note thataux+buy= (ai+bj) (uxi+uyj)= (ai+bj) u.( )4 Partial Differential equationsRecall from multivariable calculus that the last term is nothing but a direc-tional derivative ofu(x,y)in the directionai+bj. [Actually, it is propor-tional to the directional derivative ifai+bjis not a unit vector.]

6 ]Therefore, we seek to write the Partial Differential equation as involving aderivative in the directionai+bjbut not in a directional orthogonal to depict a new set of coordinates in which thewdirection isorthogonal toai+ ayai+ : Coordinate systems for trans-formingaux+buy+cu=fintobvz+cv=fusi ng the transformationw=bx ayandz= consider the transformationw=bx ay,z=y.( )We First note that this transformation is invertible,x=1b(w+az),y=z.( )Next we consider how the derivative terms transform. Letu(x,y) =v(w,z). Then, we haveaux+buy=a xv(w,z) +b yv(w,z),=a[ v w w x+ v z z x]+b[ v w w y+ v z z y]=a[bvw+0 vz] +b[ avw+vz]=bvz.( )Therefore, the Partial Differential equation becomesbvz+cv=f(1b(w+az),z).This is now in the same form as in the First case and can be solved using anintegrating the general solution of the equation3ux 2uy+u= , we transform the equation into new ay= 2x 3y,and z= ,ux 2uy=3[ 2vw+0 vz] 2[ 3vw+vz]= 2vz.

7 ( )The new Partial Differential equation for v(w,z)is 2 v z+v=x= 12(w+3z). First Order Partial Differential Equations 5 Rewriting this equation, v z 12v=14(w+3z),we identify the integrating factor (z) =exp[ z12d ]=e this integrating factor, we can solve the Differential equation for v(w,z). z(e z/2v)=14(w+3z)e z/2,e z/2v(w,z) =14 z(w+3 )e /2d = 12(w+6+3z)e z/2+c(w)v(w,z) = 12(w+6+3z) +c(w)ez/2u(x,y) =x 3+c( 2x 3y)ey/2.( ) Equations : The Method of InterpretationWe consider the quasilinear Partial Differential equationintwo independent variables,a(x,y,u)ux+b(x,y,u)uy c(x,y,u) =0.( )Letu=u(x,y)be a solution of this equation. Then,f(x,y,u) =u(x,y) u=0describes the solution surface, or integral surface,Integral recall from multivariable, or vector, calculus that the normal to theintegral surface is given by the gradient function, f= (ux,uy, 1).Now consider the vector of coefficients,v= (a,b,c)and the dot productwith the gradient above:v f=aux+buy is the left hand side of the Partial Differential equation.

8 Therefore, forthe solution surface we havev f=0,orvis perpendicular to f. Since fis normal to the surface,v= (a,b,c)is tangent to the surface. Geometrically,vdefines a direction field, calledthe characteristic field. These are shown in characteristic field. : The normal to the integralsurface, f= (ux,uy, 1), and the tan-gent vector,v= (a,b,c), are Partial Differential seek the forms of the characteristic curvessuch as the oneshown in Recall that one can parametrize space curves,c(t) = (x(t),y(t),u(t)),t [t1,t2].The tangent to the curve is thenv(t) =dc(t)dt=(dxdt,dydt,dudt).However, in the last section we saw thatv(t) = (a,b,c)for the Partial dif-ferential equationa(x,y,u)ux+b(x,y,u)uy c(x,y,u) =0. This gives theparametric form of the characteristic curves asdxdt=a,dydt=b,dudt=c.( )Another form of these Equations is found by relating the differentials,dx,dy,du, to the coefficients in the Differential equation.

9 Sincex=x(t)andy=y(t), we havedydx=dy/dtdx/dt= , we can show thatdudx=ca,dudy= of these relations can be summarized in the formdt=dxa=dyb=duc.( )How do we use these characteristics to solve quasilinear Partial differen-tial Equations ? Consider the next the general solution: ux+uy u= First identify a=1,b=1,and c= relations between the differentialsisdx1=dy1= can pair the differentials in three ways:dydx=1,dudx=u,dudy= two of these relations are independent. We focus on the First First equation gives the characteristic curves in the xy-plane. This equationis easily solved to givey=x+ second equation can be solved to give u= Order Partial Differential Equations 7 The goal is to find the general solution to the Differential equation. Since u=u(x,y),the integration constant is not really a constant, but is constant withrespect to is in fact an arbitrary constant function. In fact, we could view itas a function of c1,the constant of integration in the First equation.

10 Thus, we letc2=G(c1)for G and arbitrary function. Since c1=y x,we can write thegeneral solution of the Differential equation asu(x,y) =G(y x) the advection equation, ut+cux=0,for c a constant, andu=u(x,t),|x|< ,t> characteristic Equations ared =dt1=dxc=du0( )and the parametric Equations are given bydxd =c,dud =0.( )These Equations imply that u= x=ct+ + before, we can write c1as an arbitrary function of , before doingso, let s replace c1with the variable and then we have that =x ct,u(x,t) =f( ) =f(x ct)where f is an arbitrary function. Furthermore, we see that u(x,t) =f(x ct)indicates that the solution is a wave moving in one direction in the shape of theinitial function, f(x).This is known as a traveling wave. A typical traveling waveis shown in (x)f(x ct) : Depiction of a traveling (x,t) =f(x)att=0 travels withoutchanging that since u=u(x,t),we have0=ut+cux= u t+dxdt u x=du(x(t),tdt.)