Applied Mathematics 505b January 22, Today Denitions, survey of applications. Denition A PDE is an equation of the form F x 1 ;x 2 ::::;x n ;~u

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1 Applied Mathematics 505b January 22, Applied Mathematics 505b Partial Dierential Equations January 22, 1998 Text: Sobolev, Partial Dierentail Equations of Mathematical Physics available at bookstore by early next week. Collateral reading list available on my web page rmc/courses/am505 Administration The exam for this course is a portion of the comprehensive exam for the Ph.D. program of Applied Mathematics. counts for 70% of the mark for the course sample exams will be handed out Presentation and participation Everyone will have to give a one-hour lecture on a topic to be assigned This counts for 10% of the mark of the course Assignments There will be three or four assignment sets. to be done on your own count for 20% of the course mark Oce hours Rob Corless WSC 115 x 8794 rmcpineapple.apmaths.uwo.ca Home: Notes will be available on the web, typed by Mrs. Pat Malone, in.tex,.dvi,.ps, and.html formats (maybe.pdf if I get my act together). Outline We will stick fairly closely to Sobolev, emphasizing the pragmatic side. If time permits we will make a brief excursion into solitons. This course covers only essentially linear equations. The main purpose is to acquaint you with a central theme of applied mathematics, and with results that every applied mathematician should know. To do PDEs properly requires a lifetime, not just one term course. But we will do our best.

2 Applied Mathematics 505b January 22, Today Denitions, survey of applications. Denition A PDE is an equation of the form F x 1 ;x 2 ::::;x n ;~u;~u x 1 ;:::~u xn ;:::;~u xi 1 x i 2 :::x im =0 which is supposed to hold everywhere in the (nonempty) interior of some nice domain. Together with the PDE we may have boundary conditions holding on or some portion thereof. PDEs where one variable plays the role of time are also called dynamical systems (loosely). Examples of PDE If the Laplace operator is written u = 2 u x u x :::+ 2 u ; x 2 n then arguably the most important equation in mathematical physics is Laplace's equation, u =0 in : Solutions of this equation are called harmonic functions, or potential functions. The Laplacian operator occurs so often because it is invariant under rigid body motions (change of Cartesian coordinates) and thus occurs very naturally in applications having isotropy. In the case n =2wehave u = u xx + u yy =0: Associate with u a conjugate harmonic function v such that the Cauchy-Riemann equations u x = v y u y =,v x hold: Then f(z) =f(x+iy) =u(x; y)+iv(x; y) is analytic. The wave equation for u(x 1 ;x 2 ;:::;x n ;t), u tt = c 2 u; describes linear vibration of a string, membrane, or solid; propagation of sound waves in tubes (n = 1), waves on the surface of shallow water (n = 2), acoustic or light waves when n =3. The heat equation (diusion equation) u t = ku is satised by the temperature of a body conducting heat, when the density and specic heat are constant.

3 Applied Mathematics 505b January 22, An important class of applications not covered explicitly in Sobolev concerns Stochastic Dierential Equations such as dx dt = b(x)+(x) (1) where is a stochastic perturbation of this otherwise ordinary dierential equation. If we set a ij = ik jk, then the Fokker-Plank equation corresponding to 1is u t = 1 2 dx i;j=1 2 x i x j (a ij u), which gives the probability density function of trajectories of 1. That is, prob fx(t) 2 Bg = B dx i=1 x i (b i u) ; (2) u(t; z) dz : For more information on this application, consult Chaos, Fractals, and Noise, by Lasota and Mackey, Springer-Verlag Other PDEs of interest include Maxwell's equations, the biharmonic equation 2 u =0, and Schrodinger's equation. These are all linear. Nonetheless the theory and results are extremely rich. Nonlinear equations belong to the 20 t h century, really, except perhaps for the following. Navier-Stokes equations D Dt = t + u i x i D nonlinear operator. Dt ~u =,r~p + 1 Re ~u + f ~u = 0 Korteweg-de Vries equation Equation of minimal surface z = u(x; y) We will see more examples later. Some things which are not PDE: u t + cuu x + u xxx (1 + u 2 y)u xx, 2u x u y u xy +(1+u 2 x)u yy =0: delay-dierential equations, for example dierence equations _y(t) =ay(t, 1) y(0) = y 0 y(t) = f(t), 1 <t<0 u ij = 1 4 (u i,1;j + u i+1;j + u i;j,1 + u i;j+1 )

4 Applied Mathematics 505b January 22, functional equations f(xy) =f(x)+f(y) But ODE are really just a simple kind of PDE; simple in that they are nite-dimensional, whereas PDE are innite-dimensional. Derivation of PDE by the calculus of variations. Preliminaries: Green's theorem and \integration by parts" Green's theorem: r fdv ~ = ~f ~nds ~n= unit inward normal. If ~ f = ~ F or = scalar function in then r ~ f =r ~ F + ~ F r so r ~ FdV = ~F rdv + ~f ~nds ~F rdv =, r FdV ~ + F ~nds Now suppose we wish to nd u = u(x; y) that extremizes I(u) = L(x; y; u; u x ;u y )dxdy with u(x; y) = a given function on. The usual heuristic argument goeslike this: assume rst that a smooth extremizing function, call it u (x; y), exists. Then if everything is \nice enough", it must be true that if I() = L(x; y; u + ; u x + x ; u y + y )dxdy for some smooth function (x; y) (any such function, so long as = 0 on so as not to disturb the boundary condition), then necessarily I 0 (0) = d d I() =0 =0: But I 0 () = = by Green 0 s theorem = L u + x L ux + y L uy dxdy L u, x L ux, y L uy! dxdy +! L u, x L ux, y L uy dxdy L ux ;L uy ~nds because 0 on : Because this must be zero for every smooth test function (x; y) with =0on, we must have L u, x L ux, y L uy =0

5 Applied Mathematics 505b January 22, everywhere in. This gives a PDE (the Euler-Lagrange equation) for u. Example I = q 1+u 2 L x+u 2 y dxdy surface area of z = u(x; y).

6 Applied Mathematics 505b January 22, we haven't made an error detectable by these simple checks. Next we suppose u(x; y) onisvery gently curved, so u x, u y, u xx, u yy, u xy are all small on the surface. Then neglecting higher order terms, we get In fact, if we expand u xx + u yy =0: L = 1+u 2 x+u 2 y 1=2= u2 x+ 1 2 u2 y+::: 1 and minimizing 2 u2 x u2 y dv (Dirichlet's integral) would lead to u = 0. (Exercise). This begs the question of how we solve (1 + u 2 y)u xx, 2u x u y u xy +(1+u 2 x)u yy =0 u=f(x; y) on, of course. In practice it's often better to work directly from the variational formulation. For this convex problem, quite a lot can be said. From Variational Calculus with Elementary Convexity by Troutman (Springer, 1983) we nd that we need not assume existence { or, rather, if is convex, existence is guaranteed. Exercises (Some of these questions may be promoted to an assignment, later. For now, do them, but don't hand them in.) 1.1 Solve u=0on=[0;1] [0; 1] subject to u(x; 0) = u(x; 1)=0,u(0;y)=(y, 1 2 ), u(1;y) = 0. [This question just reminds you about separation of variables, and the Dirac function.] What can you say about the validity of your answer? 1.2 Write down and justify (i.e. derive) the Euler-Lagrange equations for the function minimizing (a) I 1 (u) = (b) I 2 (u) = L(x; y; z; u; u x ;u y ;u z )dv L(x; y; u; u x ;u y ;u xx ;u xy ;u yy )dv Assume Green's theorem can be used (i.e. is a Green's domain). 1.3 Solve _y(t) =ay(t, 1) t>0; y(t)=f(t)on,1 <t<0 (Hint: use the Laplace transform.) (and look up the complex inverse formula) 1.4 Solve u t + cu x =0where u(x; 0) = f(x), given. 1.5 Solve u t = u xx on 0 <x<1 ; t>0 u(0;t) = u(1;t)=0 u(x; 0) = (x, 1 2 ): The solution is the elliptic theta function 1 (z; q) with q = e.