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CORSO ESTIVO DI MATEMATICA Differential Equations Of CORSO ESTIVO DI MATEMATICA Differential Equations of Mathematical Physics G. Sweers http://go.to/sweers or http://fa.its.tudelft.nl/ sweers ∼ Perugia, July 28 - August 29, 2003 It is better to have failed and tried, To kick the groom and kiss the bride, Than not to try and stand aside, Sparing the coal as well as the guide. John O’Mill ii Contents 1Frommodelstodifferential equations 1 1.1Laundryonaline........................... 1 1.1.1 Alinearmodel........................ 1 1.1.2 Anonlinearmodel...................... 3 1.1.3 Comparingbothmodels................... 4 1.2Flowthroughareaandmore2d................... 5 1.3Problemsinvolvingtime....................... 11 1.3.1 Waveequation........................ 11 1.3.2 Heatequation......................... 12 1.4 Differentialequationsfromcalculusofvariations......... 15 1.5 Mathematical solutions are not always physically relevant . 19 2 Spaces, Traces and Imbeddings 23 2.1Functionspaces............................ 23 2.1.1 Hölderspaces......................... 23 2.1.2 Sobolevspaces........................ 24 2.2Restrictingandextending...................... 29 2.3Traces................................. 34 1,p 2.4 Zero trace and W0 (Ω) ....................... 36 2.5Gagliardo,Nirenberg,SobolevandMorrey............. 38 3 Some new and old solution methods I 43 3.1Directmethodsinthecalculusofvariations............ 43 3.2 Solutions in flavours......................... 48 3.3PreliminariesforCauchy-Kowalevski................ 52 3.3.1 Ordinary differentialequations............... 52 3.3.2 Partial differentialequations................ 53 3.3.3 TheCauchyproblem..................... 55 3.4CharacteristicsI ........................... 56 3.4.1 Firstorderp.d.e........................ 57 3.4.2 Classification of second order p.d.e. in two dimensions . 58 3.5CharacteristicsII........................... 61 3.5.1 Secondorderp.d.e.revisited................ 61 3.5.2 Semilinearsecondorderinhigherdimensions....... 63 3.5.3 Higherorderp.d.e....................... 65 iii 4 Some old and new solution methods II 67 4.1Cauchy-Kowalevski.......................... 67 4.1.1 Statementofthetheorem.................. 67 4.1.2 Reductiontostandardform................. 68 4.2Asolutionforthe1dheatequation................. 71 4.2.1 The heat kernel on R .................... 71 4.2.2 Onboundedintervalsbyanorthonormalset....... 72 4.3AsolutionfortheLaplaceequationonasquare.......... 76 4.4Solutionsforthe1dwaveequation................. 78 4.4.1 Onunboundedintervals................... 78 4.4.2 On a bounded interval.................... 79 4.5AfundamentalsolutionfortheLaplaceoperator......... 81 5 Some classics for a unique solution 85 5.1Energymethods........................... 85 5.2Maximumprinciples......................... 89 5.3Proofofthestrongmaximumprinciple............... 93 5.4Alexandrov’smaximumprinciple.................. 95 5.5Thewaveequation..........................100 5.5.1 3spacedimensions......................100 5.5.2 2spacedimensions......................105 iv Week 1 From models to differential equations 1.1 Laundry on a line 1.1.1 A linear model Consider a rope tied between two positions on the same height and hang a sock on that line at location x0. x0 XX XXX ¡ XXX ¡ XXX ¡ XX F2 ¡µ XXX ¡ XX F XXyXX 1 ¡β α XXXXXX¡¡ Then the balance of forces gives the following two equations: F1 cos α = F2 cos β = c, F1 sin α + F2 sin β = mg. We assume that the positive constant c does not depend on the weight hanging on the line but is a given fixed quantity. The g is the gravitation constant and m the mass of the sock. Eliminating Fi we find mg tan α +tanβ = . c We call u the deviation of the horizontal measured upwards so in the present situation u will be negative. Fixing the ends at (0, 0) and (1, 0) we find two straight lines connecting (0, 0) through (x0,u(x0)) to (1, 0) . Moreover u0(x)= tan β for x>x0 and u0(x)= tan α for x<x0, so − + u0(x−)= tan α and u0(x )=tanβ. 0 − 0 1 This leads to + mg u0(x ) u0(x−)= , (1.1) 0 − 0 c and the function u can be written as follows: mg c x (1 x0) for x x0, u (x)= − mg − ≤ (1.2) x0 (1 x) for x>x0. ½ − c − Hanging multiple socks on that line, say at position xi asockofweightmi with i =1...35, we find the function u by superposition. Fixing x (1 s) for x s, G (x, s)= (1.3) s (1 −x) for x>s,≤ ½ − we’ll get to 35 mig u (x)= G (x, xi) . − c i=1 X Indeed,ineachpointxi we find + x=xi + mig ∂ mig u0(xi ) u0(xi−)= G (x, xi) = . − − c ∂x x=x c · ¸ i− In the next step we will not only hang point-masses on the line but even a blanket. This gives a (continuously) distributed force down on this line. We will approximate this by dividing the line into n units of width x and consider 1 4 1 the force, say distributed with density ρ(x), between xi x and xi + x to − 2 4 2 4 be located at xi. We could even allow some upwards pulling force and replace mig/c by ρ(xi) x to find − 4 n u (x)= ρ(xi) xG(x, xi) . 4 i=1 X Letting n and this sum, a Riemann-sum, approximates an integral so that we obtain→∞ 1 u(x)= G (x, s) ρ(s) ds. (1.4) Z0 We might also consider the balance of forces for the discretized problem and see that formula (1.1) gives 1 1 u0(xi + x) u0(xi x)= ρ(xi) x. 2 4 − − 2 4 − 4 By Taylor 1 1 2 u0(xi + x)=u0(x)+ xu00(xi)+ ( x) , 2 4 2 4 O 4 1 1 2 u0(xi x)=u0(x) xu00(xi)+ ( x) , − 2 4 − 2 4 O 4 and 2 xu00(xi)+ ( x) = ρ(xi) x. 4 O 4 − 4 After dividing by x and taking limits this results in 4 u00(x)=ρ(x). (1.5) − 2 Exercise 1 Show that (1.4) solves (1.5) and satisfies the boundary conditions u(0) = u(1) = 0. Definition 1.1.1 The function in (1.3) is called a Green function for the boundary value problem u00(x)=ρ(x), u−(0) = u(1) = 0. ½ 1.1.2 A nonlinear model Consider again a rope tied between two positions on the same height and again hang a sock on that line at location x0. Now we assume that the tension is constant throughout the rope. x0 XX XXX ¡ XXX ¡ XXX ¡ XX F2 ¡µ XXX ¡ XX F XXyXX 1 ¡β α XXXXXX¡¡ To balance the forces we assume some sidewards effect of the wind. We find F1 = F2 = c, F1 sin α + F2 sin β = mg. Now one has to use that + u0 x u0 x− sin β = 0 and sin α = − 0 + 2 2 1+ ¡u0 x¢0 1+ u¡0 x¢0− q q and the replacement of (1.1)¡ is¡ ¢¢ ¡ ¡ ¢¢ + u x u x− mg 0 0 0 0 = . + 2 − 2 c 1+ ¡u0 x¢0 1+ ¡u0 x¢0− q q A formula as in (1.2) still¡ holds¡ ¢¢ but the superposition¡ ¡ ¢¢ argument will fail since the problem is nonlinear. So no Green formula as before. Nevertheless we may derive a differential equation as before. Proceeding as before the Taylor-expansion yields ∂ u (x) 0 = ρ(x). (1.6) −∂x 2 1+(u0 (x)) q 3 1.1.3 Comparing both models The first derivation results in a linear equation which directly gives a solution formula. It is not that this formula is so pleasant but at least we find that whenever we can give some meaning to this formula (for example if ρ L1) there exists a solution: ∈ Lemma 1.1.2 Afunctionu C2 [0, 1] solves ∈ u00(x)=f(x) for 0 <x<1, u−(0) = u(1) = 0, ½ if and only if 1 u(x)= G (x, s) f(s)ds Z0 with x (1 s) for 0 x s 1, G (x, s)= s (1 −x) for 0 ≤ s<x≤ ≤ 1. ½ − ≤ ≤ The second derivation results in a non-linear equation. We no longer see immediately if there exists a solution. Exercise 2 Suppose that we are considering equation (1.6) with a uniformly g distributed force, say c ρ(x)=M. If possible compute the solution, that is, the function that satisfies (1.6) and u(0) = u(1) = 0.ForwhichM does a solution 1 1 exist? Hint: use that u is symmetric around 2 and hence that u0( 2 )=0. 0.2 0.4 0.6 0.8 1 -0.1 -0.2 -0.3 -0.4 -0.5 Here are some graphs of the solutions from the last exercise. Remark 1.1.3 If we don’t have a separate weight hanging on the line but are considering a heavy rope that bends due to its own weight we find: 2 u00(x)=c 1+(u (x)) for 0 <x<1, − 0 ( u(0) = u(1)q = 0; 4 1.2 Flow through area and more 2d Consider the flow through some domain Ω in R2 according to the velocity field v =(v1,v2) . Definition 1.2.1 A domain Ω is definedasanopenandconnectedset.The boundary of the domain is called ∂Ω and its closure Ω¯ = Ω ∂Ω. ∪ ¢¸ ¶¶7 ¡¡µ ½½> 6 ¢¸ ¶¶7 ¡¡µ ½½> ©©* (x, y) ¶¶7 ¡¡µ ½½> ©©* y 4 r ¶¶7 ¡¡µ ½½> ©©* ¡¡µ ½½> ©©* ? ¾ x - 4 If we single out a small rectangle with sides of length x and y with (x, y) in the middle we find 4 4 flowing out: • 1 1 y+ 2 y x+ 2 x 4 1 4 1 Out = ρ.v1 x + 2 x, s ds + ρ.v2 s, y + 2 y ds, y 1 y 4 x 1 x 4 Z − 2 4 Z − 2 4 ¡ ¢ ¡ ¢ flowing in: • 1 1 y+ 2 y x+ 2 x 4 1 4 1 In = ρ.v1 x 2 x, s ds + ρ.v2 s, y 2 y ds. y 1 y − 4 x 1 x − 4 Z − 2 4 Z − 2 4 ¡ ¢ ¡ ¢ Scaling with the size of the rectangle and assuming that v is sufficiently differentiable we obtain y+ 1 y 1 1 Out In ρ 2 4 v1 x + x, s v1 x x, s lim − = lim 2 4 − − 2 4 ds + y 0 x y y 0 y y 1 y x 4x↓0 4 4 4x↓0 4 Z − 2 4 ¡ ¢4 ¡ ¢ 4 ↓ 4 ↓ x+ 1 x 1 1 ρ 2 4 v2 s, y + y v2 s, y y + lim 2 4 − − 2 4 ds y 0 x x 1 x y 4x↓0 4 Z − 2 4 ¡ ¢4 ¡ ¢ 4 ↓ y+ 1 y x+ 1 x ρ 2 4 ∂v ρ 2 4 ∂v =lim 1 (x, s) ds + lim 2 (s, y) ds y 0 y y 1 y ∂x x 0 x x 1 x ∂y 4 ↓ 4 Z − 2 4 4 ↓ 4 Z − 2 4 ∂v (x, y) ∂v (x, y) = ρ 1 + ρ 2 .
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