Lecture notes. May 2, 2017

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1 Lecture notes May, 7 Preface These are ecture notes for MATH 47: Partia differentia equations with the soe purpose of providing reading materia for topics covered in the ectures of the cass. Severa exampes used here are reproduced from Partia ifferentia Equations, an Introduction: Water Strauss. Notation efaut notation uness mentioned. Independent variabes: x, y, z Spatia variabes, t time variabe Functions: u, f, g, h, φ, ψ u x = u x, u y = u y Lec. What is a PE? efinition.. PE Consider an unknonw function ux, y,... of severa independent variabes x, y,.... A partia differentia equation is a reation/identity/equation which reates the independent variabes, the unknown function u and partia derivatives of the function u. efinition.. Order of a PE The order of a PE is the highest derivative that appears in the equation. The most genera first order partia differentia equation of two independent variabes can be expressed as Simiary, the most genera second order differentia equation can be expressed as F x, y, u, u x, u y =. F x, y, u, u x, u y, u xx, u yy, u xy =. Exampe.. Here are a few of exampes of first order partia differentia equations: u t + u x =, Transport equation. 3 u t + uu x =, Burger s equation mode probem for shock waves 4 Exampe.. Here are a few exampes of second order differentia equations u t u x = sin t + x. 5 u xx + u yy + u zz = Lapace s equation Governing equations for eectrostatics, potentia fows, u t = u xx + u yy, Heat equation/iffusion equation. 7 u tt = u xx, Vibrating string. 8 u t = iu xx, Schrödinger s equation. 9 u t = u xx + u 4, Heat equation with radiation. xu xx + yu yy = x + y. Remark.. The primary difference between a partia differentia equation and an ordinary differentia equation is that there is ony one independent variabe in an ordinary differentia equation,. the unknown function ut is a function of the singe independent variabe t, where as in a partia differentia equation, the unknown function is a function of more than one independent variabe

2 . Properties of partia differentia equations In any partia differentia equation, can aternativey be denoted in operator notation as For exampe, the equation for a vibrating string 8 can be written in operator form as Burger s equation 4 can be written in operator form as Equation in operator form is given by.. Linearity L[u] = fx, y.... L [u] = u tt u xx, f x, t =. 3 L [u] = u t + uu x, f x, t =. 4 L 3 [u] = xu xx + yu yy, f 3 x, y = x + y. 5 efinition.3. Linearity of a PE Given a partia differentia equation in operator form L[u] = fx, y,..., the equation is inear if the operator L is inear i.e. the operator L satisfies L[u + v] = L[u] + L[v], L[cu] = cl[u], 6 for any function u, v and constant c. The operator L corresponding to a inear partia differentia equation is caed a inear operator. Equations 8 and are inear since the corresponding operators L and L 3 satisfy the conditions for inearity: L [u + v] = u + v tt u + v xx = u tt u xx + v tt v xx = L [u] + L [v]. L [cu] = cu tt cu xx = cu tt cu xx = cl [u]. L 3 [u + v] = xu + v xx + yu + v yy = xu xx + yu yy + xv xx + yv yy = L 3 [u] + L 3 [v]. However, equation 4 is non-inear, since L 3 [cu] = xcu xx + ycu yy = cxu xx + cyu yy = cl 3 [u]. L [u + v] = u + v x + u + vu + v y = u x + u + vu y + v x + u + vv y u x + uu y + v x + vv y = L [u] + L [v]. 7.. Homogeneity efinition.4. Homogeneous partia differentia equation Given a inear partia differentia equation L[u] = fx, y,..., the equation is homogeneous if f = and inhomogeneous otherwise. Thus, the equation of a vibrating string 8 is inear since f = and equation is inhomogeneous since f 3 = x + y Exercise.. Cassify equations in the notes as inear or non-inear. If inear, futher cassify them as homogeneous or inhomogeneous...3 Superposition principe One important characteristic of inear differentia equations is the superposition principe. Suppose L is a inear operator. Then if u and u are soutions to L[u] =, then v = c u + c u aso satisfies L[v] =. This foows from inearity of the operator: L[v] = L[c u + c u ] = L[c u ] + L[c u ] = c L[u ] + c L[u ] =. Simiary if u h satisfies L[u h ] = and u p satisfies L[u p ] = f, then v = u p + cu h aso satisfies L[v] = f Exercise: prove this. For a majority of the cass, we wi be discussing soutions to inear partia differentia equations with constant coefficients.

3 .3 Soutions to differentia equations A function u is soution to the partia differentia equation L[ux, y] = fx, y x, y, 8 if the above equation hods for a x, y. In OE and, we know that an nth order inear differentia equation has an n parameter famiy of soutions. For exampe, the second order differentia equation u + 4u =, has a two parameter famiy of soutions u = c sin x + c cos x, where c, c are constants determined based on either the initia vaues or the boundary vaues of u. Let us ook at the nature of soutions to partia differentia equations. Consider the foowing partia differentia equation for the unknown function ux, y u xx u =, x, y R. 9 It is straight forward to verify that ux, y = f ye x + f ye x, where the functions f y and f y are arbitrary functions. Let us ook at another exampe. The wave equation in different coordinates can be rewritten as Integrating the above equation in x we get Integrating the equation in y, we get u xy =, x, y R. u y = fy. ux, y = y fsds + f x. Reabeing the y fsds = f y, the above soution can be written as ux, y = f y + f x. Even in this case, we see that the soution contains two arbitrary functions f y and f x. Thus in PEs, we have arbitrary functions which describe the genera soution as opposed to arbitrary constants for OEs. However, just as in the case of OEs, these aribtrary functions are fixed by a combination of initia and/or boundary vaues of the unknown function u. Lec In inear OEs, a combination of initia and boundary conditions were used to pick out particuar soutions from a given famiy of soutions. These conditions often had physica interpretation. For exampe, the dynamics of a spring mass system is governed by the differentia equation x + k x = The above differentia equation has a two parameter soutions given by By prescribing the initia position, and inita veocity of the partice xt = c cos kt + c sin kt 3 x = x and v = x = v, we get the particuar soution xt = x cos kt + v sin kt. 4 k Aternativey, the governing equation for the potentia in a wire of ength is given by d dx εdu = fx, < x < 5 dx where εx is the dieectric constant of the wire and fx is the given charge density. In this setup, a particuar soution is obtained by prescribing the potentia boundary conditions at both ends of the wire u = u, u = u. Simiary, to make a probem we-posed in PEs, auxiiary conditions in the form of initia and boundary conditions need to be specified. In this section, we discuss the types of boundary conditions that typicay appear in second order PEs. To discuss this, we consider three sampe PEs: 3

4 Wave equation in the exterior of an object The heat equation in the interior of a domain where u is the temperature of the object. u tt = c u xx + u yy + u zz x, y, z R \, t < t < 6 u t = κ u, x, y < t <, 7 The equation for eectrostatics in the composite domain = shown beow: Here u represents the potentia associated with the Eectric fied. u xx + u yy = x, y, 8 u xx + u yy = x, y. 9 Let S denote the surface of the object and n denote the outward norma to the boundary S.. Initia conditions efinition.. An initia condition specifies the unknown function at a particuar time t. Typicay, if the equation has n time derivatives, then n initia conditions need to be prescribed. For exampe, in the wave equation above, two initia conditions corresponding to the position of the partice and its initia veocity need to be prescribed. ux, y, z, t = φx, y, z, u t x, y, z, t = ψx, y, z. By the same reasoning, for the heat equation, we just need to prescribe the initia temperature of the object. Boundary conditions ux, y, t = = T x, y. Just ike in OEs, we can aso specify boundary conditions. Boundary conditions for PEs are usuay specified associated to the spatia variabes. As the name suggests, the boundary conditions are specified on the boundary S of the object. Two of the commony speficied boundary conditions are: a irichet conditions, b Neumann conditions. efinition.. A irichet boundary condition is when the unknown function in the PE is specified on the boundary of the object. For exampe, in the wave equation, the position of the boundary coud be specified at a times in order for the probem to be we posed ux, y, z, t = gx, y, z, t, x, y, z S t < t <. This corresponds to a irichet boundary condition. Simiary, for the heat equation, the object coud be surrounded by a heat bath which maintains the temperature of the boundary at a fixed vaue: ux, y, t = hx, y, t x, y S < t <. And in eectrostatics, the potentia on the boundary coud be specified. If the materia is grounded then the potentia on the boundary is uniformy ux, y = x, y S. efinition.3. A Neumann boundary condition corresponds to a prescription of the norma derivative of the unknown function u u n = u n = u n x S. For the heat equation, the Neumann boundary condition corresponds to heat fux through the boundary. Thus, if the object is insuated on the boundary, the boundary condition for the heat equation woud be u n x, y, t = x, y S, < t <. For the heat equation, either the irichet OR the Neumann conditions need to be specified. Aternativey, the Robin boundary condition can aso be specified for the heat equation. efinition.4. The robin boundary condition corresponds to a inear combination of the irichet and the Neumann boundary conditions: αux, y, t + βu n x, y, t = fx, y, t x, y S < t <. 4

5 .3 Transmission conditions At materia interfaces, ike in the eectrostatics exampe; the vaue of the potentia or the eectric fied is usuay unknown. However, what is physicay known in such situations is both the potentia and the eectric fied are continuous in the whoe domain =. This resuts in the foowing boundary conditions at the interface S = im x x ux = im x x x S x x We wi use the foowing notation to represent the above equation [[u]] S =, where [[φ]] Γ denotes the jump in φ across the interface Γ. Simiary, the continuity of the eectric fied impies [[εu n ]] S = where εx = ε for x and εx = ε for x are the known materia properties. Such boundary conditions are often referred to as transmission boundary conditions or jump conditions.4 Radiation conditions Finay, for unbounded domains ike the wave equation described above, the behaviour of the soution as x aso needs to be specified. For exampe, the Sommerfed radiation condition given by u im r r u =, t imposes that the waves be radiating outward to. 3 Lec 3 In the previous section, we mentioned that sufficient number of boundary conditions are required in order to make a PE we posed. Here we formay define we-posedness. efinition 3.. We-posedness. A PE defined on a domain with the prescribed set of boundary conditions these may be a combination of the types of boundary conditions described above or even competey different boundary conditions is we-posed if it satisfies the foowing conditions: Existence: There exists at east one soution ux, t satisfying a the conditions both the PE in the voume and the boundary conditions Uniqueness: There exists at most one soution Stabiity or continuous dependence on data: The unique soution ux, t depends continuousy on the data of the probem, i.e., sma perturbations in the data ead to sma perturbations in the soution. Why do we care about we-posedness? At the end of the day, the partia differentia equations we write down are modes for describing physica phenomenon that we observe" which quantifiaby agrees with the measurements. Moreover, these measurements have errors owing to the resoution of the devices used to obtain them. Simiary, the boundary data avaiabe for the PE has simiar measurement errors. Thus, since we cannot distinguish data or measurements which are sighty perturbed, it is desirabe that the corresponding soutions and hence the computed measurements from the soution of the PEs aso ony be sighty perturbed. Let us ook at an exampe. Consider the wave equation in frequency domain: with the boundary conditions and the radiation conditions at u xx + u yy + ω ux, y = fx, y x, y R \, ux, y = hx, y x, y, u im r r.5 r iku =. The data for this probem consists of two functions fx, y R \ and hx, y defined on the boundary. The above probem, i.e. the PE aong with the boundary condition and the radiation condition at is we-posed. We wi show 5

6 through the course of this cass that the probem indeed has a unique soution for a given fx, y and hx, y under reasonabe assumptions on their smoothness, i.e. assuming f,h have > continuous derivatives. Let us iustrate, what we mean by stabiity. Assume ux, y and vx, y satisfy the foowing PEs with the boundary conditions and the radiation conditions at with the boundary conditions and the radiation conditions at Then if f is cose to f in the sense that u xx + u yy + ω ux, y = f x, y x, y R \, ux, y = h x, y x, y, u im r r.5 r iku =. v xx + v yy + ω vx, y = f x, y x, y R \, f f L R \ = vx, y = h x, y x, y, v im r r.5 r ikv =. max f x, y f x, y < ε, x,y R \ h h L = max x, y h x, y < ε, x,y and h is cose to h then the soutions u and v are aso cose to each other u v L R \ = max ux, y vx, y C f f L R \ + h h L Cε x,y R \ where C is a constant independent of f, f, h, h and hence u and v. Unfortunatey, not a probems arising in physics are we-conditioned. For exampe, consider the same probem above with fx, y =. In some appications in medica imaging the goa is to detect the boundary of an obstace. So the domain and its boundary are unknown. u xx + u yy + ω u = x, y R \, ux, y = hx, y x, y, u im r r.5 r iku =. However, what is avaiabe is the measurement of the soution u,m R cos θ j, R sinθ j for a coection of anges θ j, severa boundary conditions h x, y, and severa frequencies ωm. Given these measurements, the goa is to determine the boundary of. This probem is known to be i-posed. Severa domains that are not cose" resut in exteremy cose measurements u,m R cos θ j, R sin θ j. Whie the forward probem" of computing the soution ux, y given the domain and the boundary conditions hx, y is we-posed, the inverse" probem of determining given the measurements u,m R cos θ j, R sin θ j, boundary condition h is i-posed. 3. Waves in We now turn our attention to soving PEs. We start off with the wave equation in -dimensiona space over a of R. The governing equation is u tt c u xx = x R t >. efining the equation over the whoe rea ine aows us to skip the compexity of handing boundary condition at the same time aows us to buid some insight into the nature of soutions to the wave equation. In mathematics, it is natura to construct simpe anaytic soutions to a given partia differentia equation and then use these as buiding bocks for more compicated probems. And so we proceed. We wi construct the soution using two methods. First, being the method of characteristics. The PE can be decomposed into a system of two first order differentia equations. u tt c u xx = t c x t + c x u =. 6

7 Setting Pugging vx, t in the equation above, we get t + c x = u t + cu x = vx, t, v t cv x = t c x v = t c x t + c x u =. Thus, we get the foowing system of first order partia differentia equations for ux, t and vx, t. v t cv x =, u t + cu x = vx, t. We can sove the first equation for vx, t, using the method of characteristics discussed before. v t cv x = vx, t c, =, i.e. the directiona derivative of vx, t in the direction c, is thus vx, t is constant aong the direction c, in the x, t pane. Thus the genera soution to v t cv x =, is given by vx, t = fx + ct, Exercise 3.. Verify that this indeed is a soution to v t cv x = for any differentiabe f for an arbitrary differentiabe function f. Pugging it into the equation for ux, t, we get u t + cu x = fx + ct, We construct a soution to this PE in two stages. First, we compute a particuar soution u p x, t which satisfies u p t + cu p x = fx + ct, to which we add a genera soution u h x, t to the homogeneous probem The soution ux, t is then given by u h t + cu h x =. ux, t = u h x, t + u p x, t. Proceeding as above, it is straightforward to construct the homogeneous soution to the probem u h t + cu h x =. In this case, the function u h x, t is constant on ines in the direction c, in the x, t pane. Thus, a genera soution to the above PE is given by u h x, t = gx ct. For the particuar soution u p, we make an ansatz of the form u p x, t = hx + ct. The reason for making such an ansatz is that the differentia operator t + c x maps functions of the form hx + ct to themsef. Pugging it into the differentia equation, we get t + c x u p = t + c x hx + ct = t x + ct h x + ct + x x + ct h x + ct = ch x + ct = fx + ct. Thus, if h satisfies the ode ch s = fs, then u p x + ct = hx + ct is a particuar soution to the PE u p t + cu p x = fx + ct. Combining a of this, we see that the genera soution to the wave equation on the ine is for arbitrary twice differentiabe functions h and g. ux, t = hx + ct + gx ct, Exercise 3.. Formay verify that this is indeed a soution to the PE. 7

8 We see that the soution has two waves, a wave going to the eft at speed c corresponding to hx + ct, and a wave going to the right at speed c corresponding to gx ct. It is easy to see that by setting hs to be a bump function centered at the origin. At time t =, hx + ct = hx corresponds to a bump centered at the origin, at time t =, the soution hx + ct = hx + c is a bump centered at c. Let s rederive the same soution using a different technique. From the discussion above, we notice that the directions x + ct and x ct pay an important roe in the soution to the differentia equation. So et us change coordinates to ξ = x + ct and η = x ct. Using the chain rue x = ξ ξ x + η η x = ξ + η, and t = ξ ξ t + η dη dt = c ξ c η Then combining these partia derivatives, we get The wave equation in the new coordinates can be written as The soution to the above equation is given by t + c x = c ξ, t c x = c η. t + c x t c x u = c ξ c η u = c u ξη =. u = hξ + gη = hx + ct + gx ct, which is exacty what we had before. The arbitrary functions can be determined if we are given initia conditions for the probem. To be consistent with the text, I reabe the genera soution of the wave equation on the rea ine: to u tt c u xx = x R,, t >, ux, t = fx + ct + gx ct. 3 Now, we wish to determine functions f and g such that ux, t satisfies the initia conditions: ux, = φx, u t x, = ψx. We wish to find functions f, g in terms of φ, ψ. Pugging in t = in equation 3 we get ifferentiating equation 3 with respect to t we get Pugging in t =, we get ux, = fx + gx = φx. t ux, t = cf x + ct cg x ct. t ux, = cf x cg x = ψx. Thus, we need to sove this system of equations to obtain the unknown functions f, g in terms of the given functions φ, ψ. Exercise 3.3. Show, that the soution to the above system of equations is infact given by and fs = φs + c gs = φs c s s ψ + A. ψ + B. Since f + g = φ, we concude that A + B = and the soution to the initia vaue probem is given by ux, t = [φx + ct + φx ct] + c x+ct Exercise 3.4. Verify that this indeed is a soution to the initia vaue probem: x ct u tt c u xx = x R, t >, ux, = φx, u t x, = ψx. ψsds 3 8

9 4 Lec 4 4. Properties of soutions to wave equation in 4.. Finite speed of propogation The soution to the wave equation can be expressed as waves traveing in either direction with speed c ux, t = fx + ct + gx ct = [φx + ct + φx ct] + c x+ct x ct ψsds. This property has a further consequence, Information traves at finite speed in this case at a sspeed c. This is aso referred to as the principe of causaity. Furthermore, from the anaytic form of the soution, it is cear that the initia data at x,, i.e. φx and ψx ; at time t ony infuences the soution between x ct < x < x + ct. This region as a function of t is referred to as the domain of infuence of the point x. Another way to state the same resut is that if φx, ψx = for x > R, then at time t the soution is sti for x > R + ct, i.e. the domain of infucence of the segment x R is the sector x R + ct. A reated concept is that of the domain of dependence. What segment of the intia data does ux, t depend on? This segment is referred to as the domain of dependence. Again, it is cear from the expression of the soution above that the soution depends on intia data supported in [x ct, x + ct], more particuary ψs for s [x ct, x + ct], φx + ct and φx ct. 4.. Conservation of energy Consider an infinite string with density ρ and tension T. The wave equation in this set up is given by ρu tt = T u xx x R t >. In physics, we know that the kinetic energy of an object is given by: KEt = mv = ρ u t x, tdx, since the veocity of the partice at ocation x is the rate of change of dispacement u with respect to time which is u t. The potentia energy of a stretched spring is given by P Et = k x = infty T u xx, tdx, where u x denotes the stretching the string. In the wave equation, the tota energy of the spring is given by Et = KEt + P Et = ρ u t x, tdx + T u xx, tdx, is conserved. It is an empty statement if the energy is not finite to begin with. So we wi make a simpifying assumption to ensure that the initia energy is finite. We wi assume that φ and ψ are supported on x R. Thus, E = ρ ψx dx + T φ xdx <. To proved that the energy is conserved, we wi show that the rate of change of energy is zero. de dt = d dt ρ u t x, t + T d u x, 3 dt = ρ d dt u t + T d dt u x, 33 = ρ = = u t u tt dx + T u t ρu tt dx + T u x u t The above resut aso impies that Et = E for a t. u x u tx 34 T u xx u t, Integration by parts 35 u t ρu tt T u xx dx =, Since ux, t is supported on x R + ct. 36 9

10 4..3 Uniqueness The conservation of energy gives us a straightforward proof of uniqueness of soutions to the wave equation on the ine. Consider two soutions of the wave equation with the same initia data u x, t and u x, t. Both these functions satisfy u tt = c u xx, u x, = φx, u t x, = ψx. u tt = c u xx, u x, = φx, u t x, = ψx. Then, their difference w = u u aso satisfies the wave equation but with zero initia data w tt = u tt u tt = c u xx u xx = c w xx, wx, = u x, u x, = φx φx =. w t x, = u t x, u t x, = ψx ψx =. Thus, w is a soution to the wave equation. However, E = since w t = w x = for a x R for t = initia conditions for w. From energy conservation principe, we concude that Et = for a t, i.e. ρ w t x, t + T w x x, t =. This impies that w t = w x = for a x and a t. Thus, wx, t is a constant and is since it is at t =. Thus, the wave equation cannot have two distinct soutions with the same initia data. At this stage, we ve constructed soutions to show existence, we ve proven uniqueness. Stabiity of soutions to the wave equation on the rea ine wi be on HW. 4. iffusion in The diffusion equation in one dimension is given by u t ku xx = fx, t x R, ux, = φx. where fx, t and φx are known functions, and k > is the diffusion constant. The process of deriving a soution to the above probem is fairy invoved so we wi postponed the discussion to ater. However, in the same spirit as that of the wave equation, we study eementary properties of the soutions of the diffusion equation. To simpify things even further, we are going to restrict our attention the diffusion equation on a ine segment < x <. The PE is then given by u t ku xx = fx, t, < x <, t > 4.. Maximum and minimum principe ux, = φx, u, t = ft, u, t = gt. Soutions to the diffusion/heat equation with no source term, i.e. fx, t = for a x and t satisfy the maximum principe. Let u be a soution to the diffusion equation with no source term, i.e. u t = ku xx < x < < t T. ux, = φx, u, t = ft, u, t = gt Then ux, t achieves its maxium on one of the boundaries, t =, x = or x =, i.e max x t T ux, t = max ux, t = max{max φx, max ft, max gt} 37 {t=} {x=} {x=} The intuitive expanation is the foowing. Suppose x, t is the ocation of the maximum of ux, t. Then u t x, t =, and u x x, t =. Furthermore, u xx x, t. If we coud show that u xx < and not, then we are done since we woud have a contradiction in u t = ku xx as the eft hand side is and the right hand side is not. To convert the above intuition into a proof, we buy ourseves a itte wigge room. Consider vx, t = ux, t + εx for an arbitrary ε >. Then v t kv xx = u t u xx kε x x x = kε = kε < As discussed above, if x, t is a oca maximum of vx, t, then v t x, t = and v xx x, t. Thus, v t x, t v xx x, t, which contradicts v t v xx = kε <. We sti haven t rued out the top id x, T.

11 Exercise 4.. Show that vx, t cannot achieve its maximum at x, T using a simiar argument. Thus, vx, t achieves its maximum on one of the boundaries t =, x = or x =. If M is the maximum of u on these boundaries, then max vx, t = M + ε, and vx, t M + ε, for a x, t T. From the definition of v, ux, t + εx = vx, t M + ε = ux, t M + ε x x [, ] t [, T ] Since, ε in the above equation is arbitrary, we concude that ux, t M in the entire domain and thus the resut. It is very important to note that u satisfies the diffusion equation with no source term, i.e. u t = ku xx, or fx, t =. From the maximum principe, it is straightforward to derive the minimum principe for ux, t by considering ux, t instead of ux, t Exercise 4.. Assuming the maximum principe to be true, prove the minimum principe. 4.. Uniqueness Given the maximum and the minimum principe, it is straightforward to show uniqueness for the heat equation on the ine segment. Consider two distinct soutions u and u to the diffusion equation with the same initia condition φx and the same boundary data ft, gt. Then, by inearity, the difference w = u u satisfies w t kw xx = u t u t ku xx u xx =. wx, = u x, u x, = φx φx =. w, t = u, t u, t = ft ft =. w, t = u, t u, t = gt gt =. The maximum and minimum of w on the boundaries t =,x = and x = are. By the maximum principe, the maximum is achieved on the boundary. Thus wx, t for a x, t in the domain. Simiary, the minimum is aso achieved on the boundary. Thus wx, t for a x, t in the domain. Therefore, wx, t = for a x, t which proves the resut. We prove uniqueness, using an aternate energy principe argument. However, instead of physicay arguing an energy conservation, we take a more mathematica approach this time around. We mutipy the equation w t kw xx by w and integrate for a x at a fixed time t. = w = ww t wkw xx. Integrating between x = to x =, we get = ww t kww xx dx = d dt = w dx = w t kww xx dx, 38 w t kww x + kw x, Integration by parts 39 kw xdx, Since w = for x = and x = 4 Thus, wx, t dx is a positive decreasing function of t. Since, wx, dx =, we concude that wx, t = for a t and hence wx, t = for a x and a t in the domain. 5 Lec 5 In this section, we wi derive a soution for the diffusion or heat equation on the rea ine u t = ku xx < x <, t > 4 ux, = φx 4 As before, we proceed with constructing simpe soutions to the differentia equation and construct a soution to the initia vaue probem by using these simpe soutions as buiiding bocks.

12 5. Properties of soutions to the diffusion equation Before we proceed to construct soutions to the differentia equation, et us study some interesting properties of soutions to the PE.. Spatia and tempora transation invariance of the PE: Spatia invariance: If ux, t is a soution to the PE, then for a fixed y, ux y, t is aso a soution to the PE. Physicay, if ux, t represents a heat source paced at the origin,, then ux y, t represents a heat source paced at y,. Tempora invariance: If ux, t is a soution to the PE, then for a fixed τ, ux, t τ aso is a soution to the PE. In the same spirit as above, ux, t τ woud represent a heat source paced at, τ. Exercise 5.. Verify that the heat equation satisfies the spatia and tempora invariance property.. Linear combinations: Since the heat equation is a inear homogeneous PE, inear combinations of soutions are sti soutions to the differentia equation. Combining the spatia invariance and inear combination property if ux, t is a soution to the PE, then N c j ux y j, t, where c j are constants aso represents a soution to the heat equation. j= 3. Intergra soutions: A imiting argument of the inear combination case woud show that if Sx, t represents a soution to the differentia equation, then vx, t = Sx y, tgydy, aso represents a soution to the differentia equation for a nice enough function gx defined on the rea ine. Here by nice, we mean the set of functions for which the improper integra on the right hand side converges. Negecting the issue of convergence for now, if we were aowed to switch the derivative and the integra we get: v t = t Sx y, tgydy = S t x y, tgydy. Simiary, Then, v xx = x x Sx y, tgydy = S xx x y, tgydy. v t kv xx = S t x y, t ks xx x y, tgydy =, Since Sx y, t satifies heat equation. 4. iation between spatia and tempora variabes. Another common technique for constructing more soutions to the differentia equation is to rescae the spatia and tempora variabes. Suppose ux, t represents a soution to the heat equation, we seek α such that vx, t = uλ α x, λt is aso a soution to the heat equation for any λ. v t x, t = t uλα x, λt = λu t λ α x, λt, v xx x, t = x x uλα x, λt = λ α u xx λ α x, λt. For v to satisfy the heat equation, we require = v t kv xx = λu t kλ α u xx. Thus, if λ α = λ, i.e., α =.5, vx, t = u λx, λt satisfies the heat equation if ux, t satisfies the heat equation, for a λ >. An interesting consequence of the property is the foowing emma. Lemma 5.. if φx is diation invariant, i.e. φ λx = φx for a λ, then the corresponding soution ux, t satisfies u λx, λt = ux, t for a λ >. Proof. Let ux, t be a soution to the initia vaue probem: u t = ku xx, ux, = φx. 43

13 Let λ > and consider the initia vaue probem: v t = kv xx vx, = φ λx. 44 From the diation property, vx, t = u λx, λt aso satisfies the heat equation and has boundary data vx, = u λx, = φ λx. Since, φx = φ λx, the two initia vaue probems 43 and 44 are identica. By uniqueness of soutions to the heat equation, we conude that ux, t = vx, t for a x, t. ux, t = u λx, λt for a λ >. A further interesting consequence of the above emma is the foowing emma which we state without proof. Lemma 5.. If ux, t satisfies the heat equation and u λx, λt = ux, t then ux, t = g c t x for any constant c It is straightforward to show that if ux, t = g c t x, then ux, t satisfies the diation property. This foows by u λx λx, λt = g c = g c x t = ux, t. λt Showing the resut in the other direction is a itte tricky for which we wi essentiay construct a soution to the heat equation and appea to uniqueness of soutions to initia vaue probems of the heat equation. 5. Green s function Based on the properties discussed above, we wi ook for a soution to the diffusion/heat equation: of the form u t = ku xx, ux, = φx, 45 ux, t = where Sx, t = x Qx, t and Qx, t satisfies the heat equation Q t = kq xx, Qx, = Sx y, tφydy, { x < x. To ignore issues of convergence for the time being, we assume that φx is compacty supported, i.e. φx = for a x > R. Why do we choose to represent the soution in this form? We, et us just iustrate that if we find such a soution Q, then ux, t = Q x y, tφy, x satisfies the heat equation 45 By property 3 discussed above, we see that ux, t does satisfy the heat equation. Now a we need to worry about is the initia data of ux, t. For this we sha use one of the favorite theorems of a PE theorist, integration by parts ux, t = Q x y, tφydy = x Pugging in t = in the above equation, we get ux, = = = x x = φx Q x y, tφydy y = Qx y, tφy + = Qx y, tφ ydy Qx y, φ ydy Qx y, φ y + φ y + x x φ ydy Qx y, tφ ydy Since φ is compacty supported Qx y, φ ydy 3

14 Thus, given such a particuar soution to the heat equation, we can sove any initia vaue probem for the heat equation. The initia data for Qx, t i.e. Qx, = for x > and for x <, satisfies the diation property. Thus, using property 5 discussed above, we may ook for a soution of the form x Qx, t = g. 4kt Thus, we have reduced the task of finding a function of two variabes to a function of one variabe and that wi end up converting the PE to an OE. Foowing the derivation in the text in section.4, we see that if Qx, t satisfies the heat equation, then g satisfies the ordinary differentia equation: g p + pg =. Using your favorite method to sove the above OE, we concude that and finay g p = c e p, Qx, t = + π x 4kt e p dp, Sx, t = πkt e x 4kt t >, ux, t = 4πkt e x y 4kt φydy. Sx, t defined above, is what is refered to as the Green s function or the fundamenta soution to the heat equation in one dimension, or aternativey as the source function or the propogator for the diffusion equation. At time t, Sx, t is a Gaussian with variance 4kt. Thus, for sma t the buk of Sx, t is concentrated around the origin and for arge t, Sx, t is essentiay fat. Furthermore Sx, tdx = t >. Thus, the Green s function for the heat equation is essentiay an averaging operator. The soution at time t and ocation x is the weighted average of the initia data φx. For sma t, the weights are concentrated in a sma vicinity of x where as for arge t, the weights are eveny spread out. The physica interpretation is as foows: suppose you heat up a section of a conducting rod to a higher temperature. Then say φx = for δ < x < δ and otherwise. Then as time progresses, the heat wi essentiay spread out to the whoe rod ti the temperature is uniform. Thus, we see that even non smooth initia data smoothens out due to the averaging phenomenon. Infact as t increases, the smoothness of the function ux, t as a function of x increases. 6 Lec 6 6. Refections for the heat equation We now consider a simpe prototype of soutions to heat equation with boundary wherein the soutions can be constructed using eementary techniques. The methods discussed here are in fact anaytica toos used to deveop more efficient numerica sovers for a arge cass of probems. Consider the heat equation on haf of the rea ine: u t = ku xx < x <, PE, 46 ux, = φx, < x <, Initia conditions, 47 u, t =, < t <, Boundary conditions. 48 What goes wrong if we try to proceed as before. The soution of the heat equation on the whoe rea ine is given by x y ux, t = exp φydy. 4πkt 4kt Since φx in now defined on, as opposed to, earier, one might be tempted to ook for a soution to the initia-boundary vaue probem of the form x y u x, t = Sx y, tφydy = exp φydy. 4πkt 4kt 4

15 For the above representation, the PE is ceary satisfied. Proceeding as in Section 5, the initia condition ux, = φx is aso satisfied. However, the boundary condition, y u, t = exp φydy, 4πkt 4kt is not satisfied. Thus, we need to add in a soution, which sti satisfies the PE, does not ater the initia conditions but annihiiates the boundary condition generated by u x, t. One technique commony used to achieve this task, particuar for simpe boundaries such as haf panes or circes, is caed the method of refections. The idea is simpe, the physica domain of the probem is < x <, < t <. We wi pace auxiiary heat sources with strategicay chosen strengths so as to annihiiate the soution u x, t on the boundary, t. The kerne Sx, t is symmetric in x about the origin, i.e., it is an even function of x. The effect of the heat source paced at x, to the soution at, t is identica to the effect of the heat source paced at x, to the soution, t Sx, t = S x, t. Thus, a simpe way to annihiiate the soution u x, t, woud be to add in the soution due to an initia condition φx at x,, i.e., x y x + y ux, t = Sx y, tφydy + Sx + y, t φydy = exp exp φydy. 4πkt 4kt 4kt 49 Another way to interpret the resut above is the foowing: Consider the odd extension of the function φx given by φx x >, φ odd x = φ x x <, x =. Consider the initia vaue probem u t = ku xx, < x < < t <, 5 ux, = φ odd x, < x <. 5 The soution vx, t to the above initia vaue probem is exacty the soution given in equation 49. Let us verify this. vx, t = = = Sx y, tφ odd ydy, Sx y, tφ ydy + Sx + y, tφydy + Sx y, tφydy, Sx y, tφydy. We now caim that ux, t soves the initia-boundary vaue probem given by equations 46,47, and 48. As discussed above, ux, t aso satisfies the initia vaue probem given by equations 5, and 5. Thus, ux, t satisfies the PE, and ux, = φ odd x. In particuar, ux, = φx for x >. Finay, 7 Lec 7 u, t = S y, tφy 7. Refections for the wave equation Sy, tφy = Since S is even. We now turn our attention to a simiar initia-boundary vaue probem for the wave equation defined on the haf ine given by: v tt = c v xx, < x <, < t < PE, 5 vx, = φx, v t x, = ψx, < x <, Initia conditions, 53 v, t =, < t <, Boundary conditions. 54 For the heat equation defined on the haf ine, we saw that using an odd-extension for the initia condition annihiiates the soution vx, t on the ine x =. Proceeding in a simiar fashion, we consider the foowing initia vaue probem: v tt = c v xx, < x <, < t <, vx, = φ odd x, v t x, = ψ odd x, < t <. 5

16 where φ odd and ψ odd are the odd extensions of the functions φ and ψ respectivey. The soution to this initia vaue probem is given by vx, t = [φ oddx + ct + φ odd x ct] + c x+ct x ct ψ odd sds. 55 Equation 55 is infact aso a soution to our initia boundary vaue probem given by equations 5, 53, and 54. vx, t satisfies the PE and initia conditions given by 5, and 53, since it aso satisfies the auxiiary initia vaue probem with initia conditions φ odd and ψ odd respectivey. Furthermore, v, t = [φ oddct + φ odd ct] + c since φ odd ct = φ odd ct and a fsds = for any odd function f. a 7. uhame s principe ct ct ψ odd sds =, We now turn our attention to soving the inhomogeneous initia vaue probems for both the heat and the wave equations. Before, we proceed to obtaining a soution to the inhomogeneous PEs, we turn our attention to uhame s principe. Lemma 7.. uhame s principe. Consider the foowing constant coefficient nth order ordinary differentia equation: with initia data L[y]t = y n + a n y n t a yt = ft. 56 y =, y =... y n =. Suppose wt is a soution to the homogeneous probem with initia data w =, w =,... w n =, w n =. The choice of the initia data for w wi be carified in the proof. Then the soution to the inhomogeneous probem with given boundary conditions is yt = Proof. Taking a derivative of y in the above expression, we get y t = d dt t wt sfsds = wt tft + Using an inductive argument, it is easy to show that y j t = t t wt sfsds. Exercise 7.. Fi in the detais for the inductive argument above. dw t sfsds = dt From Equation 57, it is cear that yt satisfies the initia conditions: Taking the derivative w n, we get y n t = d dt Finay, t t w n t sfsds = w n t tft+ L[y]t = y n + a n y n t +... a yt = ft + = ft + t t t dw t sfsds, Since w =. dt w j t sfsds, j =,,... n. 57 y =, y =... y n =. w n t sfsds + L[w]t sfsds t t w n t sfsds = ft+ t a n w n t sfsds +... = ft L[w] = as w satisfies the homogeneous OE. w n t sfsds, Since w n =. t a wt sfsds 6

17 What does the above emma te us? The response for a forcing function fs supported on a infinitesimay sma time interva [, ds] is roughy given by yt wtfdsds < t < ds i.e., a soution of the homogeneous probem with initia data for an impuse w =, w =,... w n =, w n =. Furthermore, given a soution to the homogeneous probem with impuse initia conditions, the soution of the inhomogeneous probem is given by a convoution of the forcing function and the soution wt. Another way to interpret the inhomogeneous soution obtained using uhame s principe is the foowing: The soution to the inhomogeneous OE 56 given by yt = t wt sfs ds = t P s [f]tds, where P s [f]t is the operator which given a function f defined on < t < on output returns the soution to the initia vaue probem L[y] =, y j s = j =,,... n, y n s = fs, i.e., we propogate the OE with initia data fs for time t s. Since, in our case, the ode is a constant coefficient inear OE, by uniqueness P s [f]t = wt sfs. This abstract principe naturay extends to soutions of inhomogeneous constant coefficient PEs as we. 7.3 Inhomogeneous heat equation Consider the inhomogeneous heat equation given by: u t ku xx = fx, t, < x <, < t <. ux, = φx. Using inearity of the PE, we spit the task of computing the soution into two parts: ux, t = u h x, t + u p x, t, where the soution to the homogeneous probem u h x, t is responsibe for the initia data and satisfies: and the particuar soution u p t satisfies t u h k xx u h =, u h x, = φx, t u p k xx u p = fx, t, u p x, =. We have aready derived the soution of the homogeneous probem to be u h x, t = Sx y, tφydy. Using uhame s principe then, the particuar soution to the inhomogeneous probem is given by u p x, t = where P s [f]x, t is the soution of the initia vaue probem: with initia conditions t P s [f]x, tds, u t = ku xx < x <, < t, ux, s = fx, s, i.e., we propogate the soution to the homogeneous heat equation with initia data fy, s for time t s to obtain the soution P s [f]x, t. The soution to the above initia vaue probem is given by: P s [f]x, t = Sx y, t sfy, sdy 7

18 Thus, u px, t = t P s [f]x, tds = t Sx y, t sfy, sdyds = t 4πkt s Let us formay verify that u p is indeed a soution to the inhomogeneous probem. exp x y 4kt s fy, sdyds. t u p x, t = t t = = t t Sx y, t sfy, sdyds, t Sx y, t sfy, sdyds + im Sx y, t sfy, sdy s t k xx Sx y, t sfy, sdyds + im Sx y, εfy, t εdy ε = k xx u p t + im Sx y, εfy, t εdy ε = k xx u p t + fx, t Since Sx, t satisfies heat equation The ast equation foows from the observation that the initia data corresponding to the soution of the heat equation is φx. im t Sx y, tφydy, 7.4 Inhomogeneous wave equation We now turn our attention to the inhomogeneous wave equation given by: u tt c u xx = fx, t, < x <, < t. ux, = φx, u t x, = ψx. Using inearity of the PE, we spit the task of computing the soution into two parts: ux, t = u h x, t + u p x, t, where the soution to the homogeneous probem u h x, t is responsibe for the initia data and satisfies: and the particuar soution u p t satisfies tt u h c xx u h =, u h x, = φx, t u h x, = ψx, tt u p c xx u p = fx, t, u p x, =, t u p x, =. We have aready derived the soution of the homogeneous probem to be u h x, t = [φx + ct + φx ct] + c x+ct x ct ψsds. Using uhame s principe then, the particuar soution to the inhomogeneous probem is given by u p x, t = where P s [f]x, t is the soution of the initia vaue probem: with initia conditions t P s [f]x, tds, u tt = c u xx < x <, < t, ux, s =, u t x, s = fx, s, i.e., we propogate the soution to the homogeneous wave equation with initia veocity fy, s for time t s to obtain the soution P s [f]x, t. The soution to the above initia vaue probem is given by: P s [f]x, t = c x+ct s x ct s 8 fy, sdy

19 Thus, u px, t = t P s [f]x, tds = c t x+ct s x ct s fy, sdy ds. Let us now verify that u p x, t indeed satisfies the inhomogeneous equation with the initia conditions. It is straight-forward to check that u p x, = for a x. ifferentiating under the integra sign can be tricky in this setup. Let Then And now we are in business. t u p x, t = c Gx, t, s = u p x, t = c t u p x, t = c x+ct s t Gx, t, s = t fy, sdy = [ + t x ct s From the expression above, we readiy see that Taking one more derivative, we get tt u p x, t = t [ x+ct s x ct s t [ Gx, t, t + t fy, sdy. Gx, t, sds. t ] t Gx, t, s ] c fx + ct s, s c fx ct s, s ds = = t c fx + ct s, s c fx ct s, s ds t u p x, =. ] fx + ct s, s + fx ct s, s ds [ fx + ct t, t + fx ct t, t + t fx + ct s, s = x fx + ct s, s t x + ct s = x fx + ct s, s c t fx ct s, s = x fx ct s, s t x ct s = x fx ct s, s c tt u p x, t = fx, t + c t Exercise 7.. Proceed as above to show that xx u p x, t = c t t fx + ct s, s fx ct s, s ds t fx + ct s, s fx ct s, s ds. fx + ct s, s + fx ct s, s ds ] t fx + ct s, s + fx ct s, s ds Combining these resuts, we concude that u pp satisfies the inhomogeneous wave equation with forcing function fx, t and the initia conditions We-posedness We have aready shown existence and uniqueness for the inhomogeneous wave equation. T he proof for uniqueness is same as that for the homogeneous probem, and existence foows from the formua we derived above. The ast ingredient to show that the inhomogeneous probem is we-posed is stabiity. For this purpose, we consider the wave equation on the finite time interva t [, T ]. We wi show stabiity in the L or the supremem norm, i.e f L R [,T ] = sup x, t [,T ] fx, t, φ L R = sup φx. x, The statement of stabiity of soutions to the inhomogeneous wave equation woud be ux, t L R [,T ] C T fx, t L R [,T ] + C T φx L R + C 3 T ψx L R, where the constants C, C, and C 3 are independent of the functions f, φ, ψ, and u and ony possiby depend on T. The soution to the inhomogeneous initia vaue wave equation is given by u h x, t = [φx + ct + φx ct] + c x+ct x ct 9 ψsds + c t x+ct s x ct s fy, sdyds.

20 A straight forward estimate shows that the above resut hods with constants C T = T, C T =, C 3 T = T. The reason the above resut impies stabiity is the foowing. Let u and u be soutions to the inhomogeneous intia vaue wave equation with forcing functions f, and f, and initia vaues φ x, ψ x and φ x, ψ x respectivey. Suppose the functions f and f are cose" in the distance defined above, i.e. f f L R [,T ]. Simiary, suppose that the initia vaues are cose in the sense φ φ L R, ψ ψ L R. Then, the above resut tes us that the maximum difference between u and u is aso sma. 7.5 Proofs of convergence for the diffusion equation The proof discussed above is just a forma proof of ux, t satisfying the inhomogeneous heat equation. In fact, the proof that im ε Sx y, εφydy = φx, i.e., the soution to the homogeneous initia vaue probem for the heat equation given by ux, t = Sx y, tφy, achieves the right initia vaue was aso forma and not rigorous. Furthermore, even the fact that ux, t above soves the heat equation was aso forma. In this section, we discuss the toos necessary to make both of these proofs rigorous, i.e. that ux, t defined above does indeed sove the heat equation and satisfies the right initia conditions. Simiar arguments can be used to make the proofs for the inhomogeneous case rigorous and wi be discussed in practice probem set 3. We first note that the soution ux, t is expressed as a convoution of S, t and φ, ux, t = Sx y, tφydy = Lemma 7.. If φx is a bounded continuous function, then ux, t = Sx y, tφydy, Sz, tφx z, dz. satisifes the heat equation for x R and t >. Furthermore, ux, t is a smooth function, i.e. it has continuous partia derivatives of a orders for a t >. Proof. We know that Sx y, t satisfies the heat equation for a x R and t >. Taking the time derivative of ux, t, we get x ux, t = x Sx y, tφydy. We can switch the order of differentiation and integration as ong as the foowing integras converge absoutey Sx y, tdyφy and x Sx y, tφydy. The first part is easy to show Sx y, tφy dy max φy y R Sx y, tdy = max y R φy The first inequaity foows from triange inequaity for integras and using the positivity of the kerne Sx y, t and the second one foows from the fact that Sx y, tdy = for a x R and t >. The second part is a sighty more tedious R cacuation, but proceeds in a simiar fashion x Sx y, tφy dy = x y x y exp φy 4πkt kt 4kt dy 58 = c p p exp φx p x y kt Change of variabe: p = 59 t 4 kt max φy c p exp p 6 y R t 4 which converges.

21 Exercise 7.3. Show that the integras converge for a n. Thus x Sx y, tφydy = p n exp p 4, x Sx y, tφydy. and the integra on the right converges uniformy. Simiary, taking higher orer derivatives in x or p resuts in integrands of the form max φy ct p n exp p y R 4, where ct <. Thus ux, t has partia derivatives of a orders and the derivatives commute with the integra. Furthermore, since Sx y, t soves the heat equation for t >, we concude that ux, t aso soves the heat equation for t > since, t ux, t k xx ux, t = t Sx y, tφydy k xx Sx y, tφydy = t k xx Sx y, tφydy = Lemma 7.3. If φ is continuous at x and max y R φy <, then Proof. Since Sx y, t =, consider φx im t Sx y, tφydy = φx. Sx y, tφydy Sx y, tφx φydy, Using the change of variabes in the proof above, the above integra can be rewritten as exp p φx φx p ktdp 4π 4 We wish to show that im t exp 4π p φx φx p ktdp = 4 The intuition to show the above resut is the foowing, as t, φx p 4kt gets coser and coser to φx except for reay arge vaues of p, owing to the continuity of φ at x. However, for arge vaues of p, exp p 4 ends up driving the integra to. Foowing this intuition, we break p R into two regimes, p kt δ where δ is such that φx φy ε/ for x y δ and the second regime being p δ kt exp p φx φx p ktdp exp p φx φx p kt dp+ 6 4π 4 4π 4 p δ/ kt 4π p kt δ p kt δ p δ/ kt exp p 4 exp p φx φx p kt dp 6 4 ε + p δ/ exp kt p φx φx p kt dp 4 63 ε/ + max φy y R p δ/ exp p dp 64 kt 4 ε/ + ε/ 65 Given an ε, we have aready picked δ >. Now given this δ, we can choose t sma enough so that max φy exp p dp = C exp p dp ε/. y R 4 4 This foows from the fact that the integra converges. exp p >N p dp, 4

22 8 Separation of variabes We now turn our attention to soving PEs on finite intervas. 8. Wave equation Consider the initia vaue probem for wave equation with homogeneous dirichet conditions on the finite interva To simpify probem, we ook for soutions of specia form u tt = c u xx, < x < 66 u, t = = u, t 67 ux, = φx, u t x, = ψx. 68 ux, t = XxT t. 69 The procedure of computing such specia soutions where the soution is expressed as a product of functions of each independent variabe is caed separation of variabes. Ceary, not a functions of the form woud satisfy a wave equation. Pugging in the representation for ux, t in equation 74 into the PE, we get ividing by c XxT t, we get XxT t = c X xt t. 7 T t c T t = X x Xx. Since the equation on the eft is soey a function of t and the equation on the right is soey a function of x, the ony way for that to be so is if both were constants, i.e. T t c T t = λ X x Xx = λ. The experts in OEs in this room know that there are three cases: λ = β, λ = β, λ =. We eave it as an exercise beow to show that λ = β and λ = resut in trivia soutions i.e. ux, t = for a x, t. Let us focus on λ = β. The soution for Xx is given by Xx = C cos βx + sin βx. The corresponding soution for T t is given by T t = A n cos βct + B n sin βct. On imposing the boundary conditions u, t = = u, t, we get XT t = = XT t, Since T t, we concude that Xx must satisfy the boundary conditions X = and X =. On imposing the boundary conditions, we get X = C + = = C =. X = sin β =. To obtain a non-trivia soution, we must have, thus β must satisfy sin β =, i.e β = nπ n N. Thus, nπx X n x = sin, which are exacty the eigenfunctions for the operator L with homogeneous dirichet boundary conditions given by L[X] = X x, X = = X. The correpsonding soution to the wave equation is given by nπct nπct u n x, t = A n cos + B n sin sin nπx