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Elementary Partial Di erential Equations - BYU Math

Elementary Partial Differential Equations William V. differential Equations (PDEs) is one of the oldest subjects in math-ematical analysis. Its development extends back to Euler s work in the 1700s,together with Brooks Taylor and arising in the study of PDEs have motivated many of the prin-ciple developments in classical and modern analysis. For example, harmonicanalysis (Fourier), complex analysis (Cauchy, Riemann), theory of integralequations (Fredholm, Hilbert), Hilbert and Banach space theory, fixed pointtheorems (Schauder), theory of distributions (L. Schwartz) and many present the theory of PDEs is one of the most active fields of researchin modern mathematics. Each monthMathematical Reviewscontains manypages of reviews of publications on PDE s. As another example ProfessorC. Miranda published a monograph on PDEs of Elliptic Type in 1954. Itcontained a bibliography of more than 600 research papers published between1924-1953.

Elementary Partial Di erential Equations William V. Smith Introduction. Partial di erential equations (PDEs) is one of the oldest subjects in math-ematical analysis. Its development extends back to Euler’s work in the 1700s, …

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Transcription of Elementary Partial Di erential Equations - BYU Math

1 Elementary Partial Differential Equations William V. differential Equations (PDEs) is one of the oldest subjects in math-ematical analysis. Its development extends back to Euler s work in the 1700s,together with Brooks Taylor and arising in the study of PDEs have motivated many of the prin-ciple developments in classical and modern analysis. For example, harmonicanalysis (Fourier), complex analysis (Cauchy, Riemann), theory of integralequations (Fredholm, Hilbert), Hilbert and Banach space theory, fixed pointtheorems (Schauder), theory of distributions (L. Schwartz) and many present the theory of PDEs is one of the most active fields of researchin modern mathematics. Each monthMathematical Reviewscontains manypages of reviews of publications on PDE s. As another example ProfessorC. Miranda published a monograph on PDEs of Elliptic Type in 1954. Itcontained a bibliography of more than 600 research papers published between1924-1953.

2 In the revised edition published in 1968, Miranda estimated that Copyrightc 2011, William V. Smith. All rights bring the bibliography up to date more than 1600 items would have to beadded. The process has continued to the present time, it is impossible to present in a single course a com-plete survey of what is known as PDEs and the properties of their advanced monographs exist and in many cases their contents for this courseA study of classical theories for some of the simplest PDEs. We shalluse as a source, V. Smirnov,A Course in Higher Mathematics, vols. II andIV, and C. H. Wilcox, Notes on PDEs. 1 The modern functional analytictheories of PDEs must wait for further 1. Heat Conduction in a Slab. p. 3 Chapter 2. Wave Propagation on a Taut String. p. 13 Chapter 3. Steady Temperature in a Circular Cylinder. - p. 32 Chapter 4. Basic Concepts in the Theory of Heat Conduction.

3 P. 48 Chapter 5. The Cauchy Problem for the Heat Equation in 1 Space Dimen-sion. p. 77 Chapter 6. Steady Temperature in a Finite Cylinder. Vibration of a DrumHead p. 98 Appendices: Classification of PDEs. Bessel functions and Sturm-Liovilleproblems. Fourier series. p. by 1. Heat Conduction in a Classical PDEs2 v t=k v=k( 2v x2+ 2v y2+ 2v z2)Heat or Diffusion Equation 2v t2=c2 vWave Equation v= (x,y,z)Poisson s Equation v= 0 Laplace s Equation 2 See appendix I p. 117 for other =v(x,t). Equations : v t=k 2v x2for 0< x < l,t >0v(x,0) =g(x)for 0 x l(initial condition)v(0,t) =voandv(l,t) =v1,fort 0(boundary conditions)The Classical Questions:Existence of a Solution?Uniqueness of Solution?Continuous Dependence on the Data?The Steady State (x) = limt v(x,t) should satisfy 2vs x2= 0,0< x < l,andvs(0) =vo,vs(l) =v14It follows thatvs(x) =vo(l xl) +v1(xl)The Reduced (x,t) =vs(x) +u(x,t) v t= u t, 2v x2= 2u x2 Thus,(1) u t=k 2u x2for 0< x < l,t >0(2)u(x,0) =g(x) vs(x) f(x)for0 x l(3)u(0,t) = 0 andu(l,t) = 0 fort 0 Separation of Variables.

4 Look for functionsu(x,t) =X(x)T(t)6 0which satisfy (1) and (3) (but not necessarily (2)).5XT =kX T(1)X (x)X(x)=T (t)kT(t)= const. = (3)X(0)T(t) = 0 andX(l)T(t) = 0 fort 0 These conditions are satisfied ifT (t) +k T(t) = 0 fort >0andX (x) + X(x) = 0 for 0< x < lX(0) = 0 andX(l) = 0 These are problems involving constant coefficient linear ordinary differ-ential Equations and are therefore explicitly solvable:X(x) =Xn(x) = sin( nxl),n = 1, 2, 3, .. = n= ( nl)2T(t) =Tn(t) =e k( nl)2tandu(x,t) =un(x,t) = sin(n xl)e k(n l) solve (1), (2), (3) try(4)u(x,t) = n=1cnsin(n xl)e k(n l)2twherec1,c2, to be (x,0) =f(x) = n=1cnsin(n xl)(Fourier Sine Series)(5)cn=2l l0f(x) sin(n xl)dx,n = 1, 2, 3, ..(5) follows from (4) by the orthogonality formal solutionis defined by (4), (5). Convergence theory for Fourier series may now Theorem for Fourier Sine Series. Assume that(a)f(x) C[0,l](b)f(0) =f(l) = 0(c)f (x) is sectionally continuous on [0,l].

5 L0sin(n xl) sin(k xl) = 0 whenn6=k. See page function is sectionally continuous if it is possible to divide its domain into a finitenumber of sections on which the function is continuous (or has removable discontinuities),7 Then the Fourier sine series coefficients (5) satisfy(6) n=1|cn|< moreoverf(x) = n=1cnsin(n xl)for 0 x land the convergence is uniform and absolute on 0 x Notation: ={(x,t) : 0< x < l,t >0}; = closure of inR2={(x,t) : 0 x l,t 0}.Classical Solution. A functionuis a classical solution of the heatconduction problem in a slab if and only ifu C( ), u/ t C( ), 2u/ x2 C( ) and (1), (2) and (3) all Theorem. Iff(x) satisfies (a), (b), (c) then the formal so-lution (4), (5) converges uniformly on and defines a classical (6) shows that (4) converges uniformly on and henceu C( )bounded and has left and right-hand limits at each point in its domain.

6 Naturally, allbounded continuous functions are also sectionally continuous, the converse of course, isnot V. Churchill,Fourier Series and Boundary Value Problems, 2nd ed. (2) and (3) hold (for a detailed argument that this is so, see pages .To prove that u t, 2u x2 C( ) and (1) holds note that|cnsinn xle k(n l)2t| |cn|e k(n l)2tofor allx Randt the series n=1n2cnsinn xle k(n l)2tconverges uniformly for allx Randt follows that u t, u x, 2u x2 C( ) can all be calculated by termwise differentiation and are continuous forallx R,t >0. Finally, (1) holds because each term in (4) is a solution ofthe heat equation. +={(x,t) :x R,t >0}( is as above). T= {(x,t) :t < T} T= closure of T= {(x,t) :t T} = boundary of = 6 The Weierstrass M-Test may be used. See page T= TMaximum Principle. Letu C( ) satisfy u/ t C( ), 2u/ x2 C( ) and the heat equation in . Then for allT >0,max Tu(x,t) = max Tu(x,t)Proof.)

7 (By contradiction.) Assume the conclusion is false, , max Tu(>max Tu) occurs at (xo,to) T T. Define the functionv(x,t) =u(x,t) (t to), >0 Then for all such ,v(xo,to) =u(xo,to) = max Tu >max TuLet maxvoccur at (x1,t1) T. Thenv(xo,to) = max Tu >max Tu+ to= max Tvprovided is sufficiently small. Thus maxvmust occur at some (x1,t1) T T. It follows that 2v x2(x1,t1) 0, v t(x1,t1) 010whence 2u x2(x1,t1) = 2v x2(x1,t1) 0but u t(x1,t1) = v t(x1,t1) + >0which contradicts (1) (the heat equation). QEDU niqueness Theorem. The BV problem (1), (2), (3) can have onlyone classical Letu1andu2be any two classical solutions with the same initialvaluesf(x). Thenu(x,t) =u1(x,t) u2(x,t) is a classical solution withf(x) 0. Thusmax Tu= max Tu= (x,t) 0 (x,t) TSimilarly, u(x,t) is a classical solution withf(x) 0. Following thesame reasoning, u(x,t) 0. It follows thatu(x,t) 0. QED7 is a logical symbol which simply means for all.

8 11 The maximum principle implies that iff(x) satisfies (a), (b), (c) andu(x,t) is the corresponding classical solution thenmax |u(x,t)| max0 x l|f(x)|This impliesContinuous Dependence on the Data. Let{fn(x)}be a sequenceof functions satisfying (a), (b), (c). Let{un(x,t)}be the corresponding so-lutions of (1), (2), (3). Suppose further thatfn(x) 0 uniformly asn on 0 x l. Thenun(x,t) 0 whenn , uniformly in .12 Chapter 2. Wave Propagation on a Taut small amplitude vibrations of a taut string, moving in a plane, aregoverned by the wave equation 2u t2=c2 2u x2,c >0 Interpretation[See chalkboard illustration]Note that the change of variable =ctreduces the wave equation to(1) 2u 2= 2u x2 The integration8of (1) can be based onLemma 1 Let be a domain in the (x, )-plane that is intersected by eachlinex = const. in an interval (possibly empty) and let (a1,a2),(b1,b2)be the smallest intervals such that o={(x, ) :a1< x < a2andb1< x+ < b2}[See chalkboard illustration]8By integration we mean finding the solution of the thatu C2( ) and (1) holds for all (x, ).

9 Then there existfunctionsf( ),g( ) such that(a)f C2(a1,a2),g C2(b1,b2)(b)u(x, ) =f(x ) +g(x+ )in Remark 1 This is d Alembert s solution of (1).Remark 2fandgare unique up to constant new coordinates =x , =x+ and letu(x, ) =v( , )Then (chain rule for Partial derivatives) u x= v + v , u = v + v 2u x2= 2v 2+ 2 2v + v 2, 2u 2= 2u 2 2 2v + 2v 29 That is, the pair (f,g) is equivalent to the pair (f+C,g C) whereCis any 2u x2 2u 2= 4 2v = 0 in ={( , ) : (x, ) }Also,v C2( ). Thus for each o (b1,b2), ( v ) = 0 on a non-empty -interval ( is connected) and hence v( , o) =G( o) C1(b1,b2).Repeating this with o (a1,a2) givesv( , ) =f( ) +g( ) on .whereg( ) = G( )d C2(b1,b2)and hencef=v g C2(a1,a2). QEDC orollary. Under the hypotheses of Lemma 1uhas an extensionu C2( o) which satisfies (1) in Propagation on a Long the (x, ) - plane considerthe domains ={(x, ) : >0,a < x < b}, ={(x, ) : >0, +a < x < b }15[See chalkboard illustration]Letu C2( ) describe a motion of the string.

10 By the lemma the valuesofuin oare independent of what happens at the ends of the stringand f,g C2(a,b)10such thatu(x, ) =f(x ) +g(x+ )in Moreover,fandgcan be determined by the initial valuesu(x,0) =uo(x)and u(x,0) =u1(x), a < x < bIndeed,u(x,0) =f(x) +g(x) =uo(x),f (x) +g (x) =u o(x) u(x,0) = f (x) +g (x) =u1(x)Thus2f (x) =u o(x) u1(x),2g (x) =u o(x) +u1(x)2f(x) =uo(x) xau1( )d +C,2g(x) =uo(x) + xau1( )d +C 10 is a logical symbol meaning there exists. 16 Hence2(f(x) +g(x)) = 2uo(x) +C+C = 2uo(x)whenceC+C = 0 orC = CThusu(x, ) =12uo(x ) 12 x au1( )d +C+12uo(x+ ) +12 x+ au1( )d Cor(2)u(x, ) =12{uo(x ) +uo(x+ )}+ x+ x u1( )d ,(x, ) .This leads us to theInitial Value Problem for the Wave the idealized case of an infinitely long string,a= ,b= , (2)gives the solution of the boundary value problem17(3) 2u 2= 2u x2for < x < , >0(4)u(x,0) =uo(x)and u(x,0) =u1(x) for < x < +={(x, ) : < x < , >0}.


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