Lecture 1 - Introduction and Basic Facts about PDEs

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1 * Introdution to PDEs, Fll 004 Prof. Gigliol Stffilni Leture 1 - Introdution nd Bsi Fts bout PDEs The Content of the Course Definition of Prtil Differentil Eqution (PDE) Liner PDEs VVVVVVVVVVVVVVVVVVVV + homomgeneous non-homogeneous (no foring term) (with foring term) VVVVVVVVVVVVVVVVVVVVV + with vrible nd onstnt oeffiients prboli exmple TTTTTTTTTTTTTTTTTTTTTT hyperboli exmple ellipti exmple (diffusion eqution) (wve eqution) (Lple eqution) (het eqution) u t = κu xx, κ > 0 u tt = u xx u xx = 0 TTTTTTTTTTTTTTTTTTTTT ) nd non-homogeneous se In studying these exmples of PDEs we will lern how to impose onditions to mke the problem well-posed, we will introdue fundmentl mthemtil onepts like distributions, Fourier Trnsform, nd Fourier Series. These tools re by now lssil, but still hevily used in the study of more omplex PDEs, in prtiulr, the nonliner ones. Wht is prtil differentil eqution? This is n eqution involving funtion u(x 1,..., x n ) of n vribles nd its prtil derivtives up to order m: F (u, u x1,..., u xn,..., u xi1 x i,..., u xi1 x i...x i m ) = 0, i j {1,..., n} This funtionl defines the eqution by involving u nd its prtil derivtives. In this se m is known s the order of the eqution. The vribles x 1,..., x n re independent nd the vrible u(x 1,..., x n ) is dependent. u is the unknown for the prtil differentil eqution. My nottion for prtil derivtive is u x = x u u xy = x y u = xy y 1

2 Exmple: u xy + os(x)u x + u y = 5 (1) is prtil differentil eqution of order sine u xy, whih is double prtil derivtive, ppers. If in the eqution the unknown funtion u nd its derivtives pper multiplied only by onstnts, like in u x + 4u xy = u + x () then the eqution is sid to be n eqution with onstnt oeffiients. If there re lso other known funtions involved s ftors like in (1) (i.e. os x) the eqution is sid to be of vrible oeffiients. Liner nd nonliner equtions One n write PDE s n opertor L ting on u: (g(x 1,..., x n ) will be bbrevited g(x).) L = xy + }{{} Lu = g(x 1,..., x n ) (3) The opertor L represents wht one hs to do in order to obtin the eqution. For exmple, in (1): Definition: A PDE is liner if opertor tke derivtives opertor multiplition by os x { os }} x { + ( y ) L(u + bv) = Lu + blv for b, R. (This is linked to the definition of vetor spes.) Definition: We sy tht PDE of the form (3) is homogeneous if g = 0. So S = {u L(u) = 0} is vetor spe. Solutions to inhomogeneous liner eqution re those of (3) with g = 0. Suppose we find u 0 suh tht L(u 0 ) = g(x). Then S = {u+u 0 L(u) = 0}. In ft, if v = u+u 0 when L(u) = 0, then L(v) = L(u + u 0 ) = L(u) + L(u 0 ) = 0 + g(x) so {u + u 0 L(u) = 0} S, nd on the other hnd, if w S L(w u 0 ) = L(w) L(u 0 ) = g(x) g(x) = 0 so u = w u 0 {u L(u) = 0} S {u + u 0 L(u) = 0} Remrk: If L(u) = 0 nd L is liner, then the liner ombintion of n solutions i u i is still solution: n n L( i u i ) = i L(u i ) = 0 (4) i=1 i=1 (Clerly this is not true if the eqution is liner but not homogeneous.) (4) goes under the nme of superposition. n i=1

3 Exmples: In (1), when L = xy + os x + ( y ), L is not liner beuse tking the squre is not liner proedure, tht is, (u + v) = u + v for u, v = 0. In (), where L = x + 4 xy 1, L is liner opertor, so () is liner nd nonhomogeneous. Some well known equtions 1. u x + u y = 0 (trnsport). u xx + u yy = 0 (Lple s eqution) 3. u t + u xxx + uu x = 0 (KdV eqution) 4. u t iu xx = 0 (Srhödinger eqution) Remrks: 1. is liner homogeneous 1st order. is liner homogeneous nd order 3. is nonliner homogeneous 3rd order 4. is liner homogeneous nd order How do we solve PDE? There is no generl rule tht works to solve ll PDEs, though lerly liner homogeneous PDEs re esier to solve thn non-homogeneous or non-liner PDEs. One bsi ide to keep in mind is how to solve ODEs: For exmple: u x u = 0 Find u(x, y). By ignoring y, we get tht u(x, y) = C(y)e x is solution for ny funtion C(y). 3

4 Solution of 1st order liner homogeneous equtions in R Consider the simple homogeneous eqution with onstnt oeffiients: u x + bu y = 0 (5) (, b) u = 0 The ltter eqution sys tht the diretionl derivtive of u in the diretion of v = (, b) is zero. This mens tht the funtion u(x, y) remins onstnt on lines in the diretion of (, b). The equtions of suh lines re (x x 0, y y 0 ) (b, ) = 0 {bx y = R} nd re lled the hrteristi lines for (5). If we think of line s funtion y(x), then d y = b ( = 0). dx Now if u does not hnge long these lines, then u(x, y) = f() bx y= u(x, y) = f(bx y) If one wnts more preise desription of f then some onditions must be speified. Now let s onsider the se of vrible oeffiients. For exmple, yu x + xu y = 0 (6) As bove, this mens tht u is onstnt long urves tht hve tngent vetors (y, x). So, 4

5 similrly to the line, if the urve is represented by y(x), then d x y = dx y dy y = x ( dx d 1 ) dx y = x 1 x = 1 y + C y x = re the hrteristi urves in this se. Thus u(x, y) = f(y x ). If we impose the ondition u(0, y) = e y, then f(y ) = e y f(t) = e t, nd thus u(x, y) = e (y x ). Finlly, n be solved s long s n be solved s n ODE. Now we n solve u x + bu y = in full generlity: First, find speil solution, use u 0 (x, y) = αx. Then Consider the eqution (x, y)u x + b(x, y)u y = 0 (7) dy b(x, y) = (8) dx (x, y) α + b 0 = α = u 0 (x, y) = x u(x, y) = f(bx y) + x u x + bu y + u = 0 (9) Find the set of solutions ssuming = 0: One solution is u = 0. Otherwise there will be t 5

6 lest some point t whih u = 0. At suh points we hve: u x + b u y + = 0 u u v log u v x = ux, v y = u v x + bv y = Thus from the previous: v(x, y) = f(bx y) x ( ) u(x, y) = exp f(bx y) x x = ef(bx y) /e. uy u 6