1. Review: Eigenvalue problem for a symmetric matrix

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1 1. Review: Eigenvlue problem for symmetric mtrix 1.1. Inner products For vectors u,v R n we define the inner product by u,v := nd the norm by u := u,u. We cll two vectors u,v orthogonl if u,v =. Recll the Cuchy-Schwrz inequlity: For ny u,v R n u,v u v. (1) n i=1 u i v i For mtrix A R n n we hve Au,v = u,a v where A denotes the trnspose mtrix. If A = A we cll the mtrix symmetric. Lemm 1.1. For symmetric A R n n we hve mx Au,u = mx Au (2) u =1 u =1 Proof. Let M 1 denote the vlue of the mximum on the left, let M 2 denote the vlue of the mximum on the right. By Cuchy-Schwrz (1) we hve for ny w Aw,w Aw w M 2 w 2, (3) hence M 1 M 2. Next we show M 2 M 1 : For n rbitrry u with u = 1 we let λ := Au, f := λu, g := Au nd consider X := A f,g + Ag, f = λ Au,Au + λ A 2 u,u = 2λ Au 2 = 2λ 3. }{{} Au,Au On the other hnd we see by multiplying out nd using Aw,w M 1 w 2 tht X = 1 2 A( f + g), f + g 1 2 A( f g), f g M 1 2 f + g, f + g + M ( 1 2 f g, f g = M 1 f 2 + g 2) = M 1 2λ 2, hence we hve X = 2λ 3 M 1 2λ 2. Therefore λ = Au M 1 nd hence M 2 M 1 (since u with u = 1 is rbitrry). Now consider n invrint subspce V of A. This mens tht u V implies Au V. Then we obtin by the sme rgument mx u V u =1. Au,u = mx Au (4) u V u =1. 1

2 1.2. Eigenvlue problem Let A be rel n n mtrix. A number λ C is clled n eigenvlue of the mtrix if there exists nonzero vector v C n such tht Av = λv. (5) All eigenvlues re rel: Multiplying (5) from the left by v (br denotes complex conjugte) gives v Av = λv v. Since A = A we hve v Av = (Av) v = λv v yielding λ = λ. Eigenvectors for different eigenvlues re orthogonl: λ j v ( j),v (k) = Av ( j),v (k) = The centrl result is tht we hve full number of eigenvectors: Theorem 1.2. Assume tht A R n n is symmetric. Then 1. There re n rel eigenvlues λ 1,...,λ n v ( j),av (k) = λ k v ( j),v (k), hence (λ j λ k ) 2. There re corresponding rel eigenvectors v (1),...,v (n) which re orthogonl on ech other. 3. Any vector u R n cn be written s liner combintion of eigenvectors: n u,v u = c j v ( j) ( j), c j = v ( j),v ( j) v ( j),v (k) = The key ide of the proof is to mximize Au,u over ll u with u = 1. The solution of this problem yields n eigenvector: Lemm 1.3. Assume tht A R n n is symmetric nd V is n invrint subspce, i.e., v V implies Av V. 1. There exists v V with v = 1 such tht Av,v is mximl: 2. Actully v is n eigenvector: There is λ R with Av,v = M = mx Au, u u V u =1. Av = λv, λ = M Proof. Here we re mximizing continuous function over compct set, so there exists such v with v = 1. Let λ := Av, v. By (4) we hve Av λ nd hence Av λv 2 = Av 2 2λ Av,v +λ 2 v 2 λ 2 2λ 2 + λ 2 =. }{{}}{{} λ 1 Now we cn give the proof of Theorem 1.2: First mximize Au,u over ll u R n with u = 1. By Lemm 1.3 this yields vector v (1) nd n eigenvlue λ 1. Define the spce V 1 s ll vectors which re orthogonl on v (1) : V 1 := {u R n u,v (1) = }. This is n invrint subspce: If u V 1, then Au stisfies Au,v (1) = u,av (1) = λ 1 u,v (1) u V = 1, 2

3 hence Au V 1. Now we mximize Au,u over ll u V 1 with u = 1. By Lemm 1.3 this yields vector v (2) nd n eigenvlue λ 2. Define the spce V 2 s ll vectors which re orthogonl on v (1) nd v (2) : V 2 := {u R n u,v ( j) = for j = 1,2}. This is n invrint subspce: If u V 2, then Au stisfies for j = 1,2 Au,v ( j) = u,av ( j) = λ j u,v ( j) u V = 2, hence Au V 2. By continuing in the sme wy we obtin eigenvlues λ 1,...,λ n with λ 1 λ 2 λ n nd eigenvectors v (1),...,v (n) which re by construction orthogonl on ech other. 2. Review: Liner System of ODEs 2.1. Initil vlue problem We consider mtrix of size n n where the entries cn depend on t: Let A(t) be n n n mtrix with entries i j (t) for i, j = 1...n. We re given n initil vector u () with entries u () i, i = 1...n nd right hnd side function f (t) with entries f i (t), i = 1...n. We wnt to find function u(t) with entries u i (t), i = 1...n such tht If the function f (t) is zero we cll the problem homogeneous. u (t) + A(t) u(t) = f (t) for t (6) u() = u () (7) If the functions i j (t) nd f i (t) re continuous for t, then one cn show tht the initil vlue problem (6), (7) hs unique solution for ll t Homogeneous ODE with constnt coefficients, symmetric mtrix We now consider specil cse of (6), (7): f (t) = (homogeneous ODE) the mtrix A is constnt the mtrix A is symmetric: A = A. Then we cn solve the IVP in the following wy: 1. Find specil solutions of the form u i (t) = v i g(t): We wnt to find solutions of the form u(t) = vg(t) with nonzero vector v nd function g(t). Plugging this into the ODE gives vg (t) + A vg(t) = A v = g (t) g(t) v (if we ssume g(t) ). Since the left hnd side does not depend on t nd v is not the zero vector this eqution cn only hold if g (t)/g(t) is equl to constnt λ. 3

4 Therefore we hve to solve n eigenvlue problem: Find λ nd vector v (not the zero vector) such tht A v = λ v Solving this eigenvlue problem gives rel eigenvlues λ j nd eigenvectors v ( j) for j = 1,...,n which re orthogonl on ech other: v ( j), v (k) = if j k. (8) For ech eigenvector v ( j) we hve now specil solution of the PDE: Since g (t) g(t) = λ j we hve g(t) = c j e λ jt nd our specil solution is c j ve λ jt for j = 1,2,3, Write the solution u(t) of the initil vlue problem s liner combintion of the specil solutions: We need to find coefficients c 1,...,c n such tht u = c j v ( j) e λ jt (9) We find the coefficients by setting t = : Then the initil condition gives u () = c j v ( j) (1) We now tke the inner product of this eqution with the vector v (k) : Then the orthogonlity (8) gives u (), v (k) = c k v (k), v (k) or u (), v (k) c k = v (k), v (k) (11) 2.3. Fundmentl System S(t) Let u (1) (t) denote the solution of the homogeneous ODE with initil condition u (1) () = [1,,...,],..., Let u (n) (t) denote the solution with initil condition u (n) () = [1,,...,]. We define the n n mtrix [ ] S(t) := u (1) (t),..., u (n) (t). Now the solution of the homogeneous ODE with initil condition u() = u () is obtined s liner combintion of u (1) (t),..., u (n) (t): We hve u(t) = u (1) (t) u () u (n) (t) u () n, i.e., u(t) = S(t) u () 2.4. Duhmel s principle Review: Derivtive of integrl We consider function F(t) defined s n integrl (t) F(t) = f (s,t)ds. (t) Then the fundmentl theorem of clculus nd the chin rule imply tht we hve for the derivtive F (t) = (t) (t) f t (s,t)ds + f (b(t),t) b (t) f ((t),t) (t). (12) 4

5 Generl cse Duhmel s principle sttes: If we cn solve the homogeneous initil vlue problem u (t) + A(t) u(t) = for t s (13) u(s) = u () (14) (here the initil condition is given t s insted of ), then we cn obtin formul for the solution of the inhomogeneous initil vlue problem (6), (7) s follows: 1. Define the function u hom (t) for t s the solution of the homogeneous initil vlue problem with initil condition u hom () = u () 2. Define for ny s > the function z (s) (t) for t s s the solution of the homogeneous initil vlue problem with initil condition z (s) (s) = f (s) 3. Now the solution u(x,t) of the inhomogeneous initil vlue problem (6), (7) is given by t u(t) = u hom (t) + z (s) (t)ds (15) s= The proof is esy: For the prticulr solution u prt (t) := t s= z (s)(t)ds we hve obviously u prt () =. Using (12) we obtin t u prt(t) = z (s) (t)ds + z (t)(t) s= }{{} f (t) Since z (s) (t) + A z (s)(t) = we obtin u prt(t) + A u prt (t) = f (t) Cse of ODE with constnt coefficients If we hve constnt coefficient mtrix A nd fundmentl system S(t) (which is n n n mtrix) we cn use Duhmel s principle s follows: 1. The homogeneous solution is obtined s 2. The solution z (s) (t) is obtined s u hom (t) = S(t) u () z (s) (t) = S(t s) f (s) 3. The solution u(x,t) of the inhomogeneous problem is therefore t u(t) = S(t) u () + S(t s) f (s)ds s= 5

6 3. Het Eqution 3.1. Derivtion s conservtion lw We consider long thin br of some mteril. Let u(x,t) denote the temperture t time t t position x long the br. The temperture corresponds to het energy stored in the br. Let r > denote the energy needed to increse the temperture of br of length 1 by 1 degree. This differs for different mterils ( specific het ). If the mteril (or the cross section of the br) is not constnt in x we hve function r(x) >. Then the totl energy stored in the br between x = nd x = b t time t is E b (t) = r(x)u(x, t)dx (16) Becuse of the conservtion of energy this cn only chnge in time if het flows cross the endpoints nd b. Let F(x,t) denote the het flux t the point x t the time t, i.e., the mount of energy which flows cross x (from left to right) per time unit. Therefore we must hve E b(t) = F(,t) F(b,t) (17) since F(,t) is the energy per time coming in t the left endpoint, nd F(b,t) is the energy per time coming out t the right endpoint. We cn now rewrite the left hnd side nd the right hnd side of this eqution s follows: r(x) t u(x,t)dx = x F(x,t)dx This eqution hs to hold for ll,b long the br, hence we must hve [ ] r(x) t u(x,t) + x F(x,t) dx =. r(x) t u(x,t) + x F(x,t) =. (18) The het flux F(x,t) should be proportionl to the chnge in temperture in x ( Fourier s lw of cooling ). The energy should flow from the hotter to the colder region. Therefore we should hve with constnt κ > ( het conductivity ) tht F(x,t) = κ x u(x,t). (19) If the mteril is not constnt in x we my hve function κ(x) > for the het conductivity. Then we obtin from (18), (19) the PDE r(x) t u(x,t) x [κ(x) x u(x,t)] =. This is the most generl form of the het eqution. It is liner homogeneous PDE. If we dditionlly hve het sources which give dditionl energy f (x,t) per spce nd time unit t position x nd time t we obtin the inhomogeneous PDE r(x) t u(x,t) x [κ(x) x u(x,t)] = f (x,t) We will minly consider the cse where the mteril properties re constnt in x: Then the functions r nd κ re constnt, nd we hve with k := κ/r tht u t (x,t) ku xx (x,t) = f (x,t) Initil vlue problem for infinite br For n infinite br we wnt to find function u(x,t) with x R nd t >. We prescribe the initil temperture u (x) t time zero. We obtin the following initil vlue problem: Find function u(x,t) for x R nd t such tht r(x)u t (x,t) = x [κ(x)u x (x,t)] for x R, t > (2) u(x,) = u (x) for x R (21) 6

7 Initil vlue problem for br of finite length It is more relistic to consider br of finite length where x b. In this cse we wnt to find function u(x,t) with x [,b] nd t >. We hve to specify wht hppens t the boundry points x = nd x = b. There re three resonble choices for x = : Dirichlet condition (D): We prescribe the temperture t x = for ll times: u(,t) = g 1 (t) for t > If the left end of the br touches lrge pool of ice wter, we will hve (pproximtely) u(,t) = (in degree Celsius). Note tht in this cse the het flux cross x = is in generl nonzero, so the totl energy in the br is not conserved. Neumnn condition (N): We prescribe the het flux t x = for ll times: κ()u x (,t) = g 1 (t) for t > If the left end of the br touches perfect insultor we will hve κ()u x (,t) =. Note tht in this cse no energy crosses the boundry t x =. Robin condition (R): Assume tht we hve thin imperfect insultor with temperture q(t) on the other side of the insultor. Then it is resonble to ssume tht the het flux to the left is proportionl to the temperture difference on both sides of the insultor: With constnt α 1 > we hve κ()u x (,t) = α [u(,t) q(t)] or κ()u x (,t) + α 1 u(,t) = g 1 (t) We lso hve three choices for the boundry condition t x = b: Dirichlet condition (D): u(b,t) = g 2 (t) Neumnn condition (N): κ(b)u x (b,t) = g 2 (t) Robin condition (R) with constnt α 2 > : κ(b)u x (b,t) α 2 u(b,t) = g 2 (t) We cn hve ny of the 9 combintions (DD), (DN), (DR),..., (RR) for the boundry conditions t the left nd right endpoints. We obtin the following initil vlue problem: Find function u(x,t) for x R nd t such tht for t > we hve the PDE r(x)u t (x,t) = x [κ(x)u x (x,t)] for x (,b) nd boundry conditions of the types (D), (N), (R) t x = nd x = b. For t = we hve the initil condition u(x,) = u (x) for x (,b). We prescribe the initil temperture u (x) t time zero. Remrk 3.1. If we hve homogeneous Neumnn conditions t both endpoints, i.e., u x (,t) =, u x (b,t) = for t > then the het flux t both endpoints is zero nd (17) shows tht the totl energy E b (t) must be constnt: r(x)u(x, t)dt = r(x)u (x)dx. In ll other cses we my hve het flow cross one of the boundries nd therefore the totl energy in [,b] my chnge. 7

8 3.2. Mximum Principle We now consider the het eqution u t = ku xx with constnt k > Mximum principle for intervl Theorem 3.2. Assume u(x,t) is clssicl solution of the het eqution for x [,b], t [,T ]. This mens tht u is continuous for x [,b], t [,t], nd the functions u t,u xx re continuous for x (,b), t (,T ] nd stisfy u t (x,t) = ku xx (x,t) for x (,b), t (,T ]. Then we hve mx x [,b] t [,T ] { } u(x,t) = mx mx u(x,), mx u(,t), mx u(b,t). x [,b] t [,T ] t [,T ] Mximum principle for rel line Theorem 3.3. Assume u(x,t) is bounded clssicl solution of the het eqution for x R, t [,T ]. This mens tht u is continuous for x R, t [, T ] nd there exists C such tht u(x,t) C x R, t [,T ]. Furthermore the functions u t,u xx re continuous for x (,b), t (,T ] nd stisfy u t (x,t) = ku xx (x,t) for x (,b), t (,T ]. Then we hve mx x [,b] t [,T ] u(x,t) = mx x [,b] u(x,) Consequences Note tht we lso hve minimum principles with min insted of mx. (Just pply the mximum principle to u(x,t)). Uniqueness of solution for IVP on the rel line: Assume we hve two bounded clssicl solutions u(x, t) nd ũ(x, t) which solve the sme initil vlue problem on the rel line. Then the difference v(x,t) = u(x,t) ũ(x,t) solves the initil vlue problem with zero initil dt. By Theorem 3.3 we therefore must hve v(x,t) =. Uniqueness of solution for IBVP on n intervl with Dirichlet conditions: Assume we hve two clssicl solutions u(x,t) nd ũ(x,t) which solve the sme IBVP for n intervl x [,b]. Then the difference v(x,t) = u(x,t) ũ(x,t) solves the IBVP with zero initil dt nd zero boundry dt. By Theorem 3.3 we therefore must hve v(x,t) =. Montonicity: Assume we hve two initil functions u (x) nd ũ (x) on the rel line with Then we must hve u (x) ũ (x) for ll x R. u(x,t) ũ(x,t) for ll x R, t for the corresponding solutions of the IVP for the het eqution. (Just pply the mximum principle to v(x, t) = ũ(x, t) u(x,t).) We obtin similr monotonicity property for the IBVP on n intervl with Dirichlet conditions. 8

9 Well-posedness : This mens tht smll chnges in the given dt only cuse smll chnges in the solution. Consider the IBVP on n intervl with Dirichlet conditions with given functions u (x), g 1 (t), g 2 (t) which yields solution u(x,t). Consider slightly different given functions ũ (x), g 1 (t), g 2 (t) with u (x) ũ (x) ε, g 1 (t) g 1 (t) ε, g 2 (t) g 2 (t) ε. We clim tht we then must hve for the corresponding solution ũ(x,t) tht u(x,t) ũ(x,t) ε for ll x [,b], t. We cn prove this by considering u, (x) := u (x) ε, g 1, (t) := g 1 (t) ε, g 2, (t) := g 2 (t) ε. By the monotonicity we must hve for the corresponding solution u (x,t) tht u(x,t) ε u (x,t) ũ(x,t) yielding u(x,t) ũ(x,t) ε. By considering u,+ (x) := u (x) + ε etc. we obtin ũ(x,t) u(x,t) ε Het eqution on the rel line Initil vlue problem for homogeneous cse We wnt to find function u(x,t) for x R nd t such tht u t (x,t) = ku xx (x,t) for x R, t > (22) u(x,) = u (x) for x R (23) For clssicl solution the function u(x,t) must be continuous for x R, t. Moreover, the derivtives u t, u xx must be continuous functions for x R nd t > Invrince properties Assume tht u(x,t) nd ũ(x,t) re solutions of the het eqution (22) for x R, t >. Then we cn obtin other solutions v(x,t) of the het eqution s follows: 1. Linerity of the problem: For ny c 1,c 2 R is lso solution of the het eqution (22). 2. Trnsltion invrince in time: For ny c is lso solution of the het eqution (22). 3. Trnsltion invrince in spce: For ny c R is lso solution of the het eqution (22). 4. Diltion invrince: For ny > is lso solution of the het eqution (22). v(x,t) := c 1 u(x,t) + c 2 ũ(x,t) v(x,t) := u(x,t + c) v(x,t) := u(x c,t) v(x,t) := u( 1/2 x,t) Note tht the properties (1.) nd (2.) lso hold in the more generl problem with functions r(x), κ(x), both for the problem on the infinite br, nd for the problem on the finite br. Note tht the properties (3.), (4.) only hold in the cse of constnt r,κ nd for n infinite br. 9

10 Solution U(x, t) where initil condition is step function We will first consider the het eqution u t = u xx with k = 1. We first consider the initil condition given by the Heviside function (unit step function): { for x < u (x) = (24) 1 for x > (we cn define e.g. u () = 1 2, though this does not relly mtter). We wnt to find solution U(x,t) of the het eqution (22) such tht lim U(x,t) = t { for x < 1 for x > We ssume tht we hve solution U(x,t) with this initil condition. We will nlyze its properties which will led us to formul for U(x, t). Let g(x) := U(x,1) denote the solution t time t = 1. We wnt to show tht the solution t ny time t > is diltion of this function g. We use the diltion invrince of the het eqution: for ny > the function Ũ(x,t) := U( 1/2 x,t) lso stisfies Ũ t = Ũ xx. Note tht the function Ũ(x,t) stisfies the sme initil condition (25) s U(x,t). Therefore the difference V (x,t) =U(x,t) Ũ(x,t) stisfies the het eqution with initil condition zero. Therefore we should hve V (x,t) =. [The function V (x,t) is continuous for t >, but we don t know this for x =, t =. So we cnnot directly pply our mximum principle. But one cn use slightly more generl version to show V (x,t) =.] Therefore we should hve U(x,t) = U( 1/2 x,t) for ny >. Choosing = t 1 gives (25) U(x,t) = U(t 1/2 x,1) = g(t 1/2 x). (26) We cn now find the function g by plugging U(x,t) = g(t 1/2 x) into the PDE U t = U xx nd obtin Let g(t) := g (t), then we obtin the first order liner ODE g (t 1/2 x)( 1 2 )t 3/2 x = g (t 1/2 x)t (t 1/2 x)g (t 1/2 x) = g (t 1/2 x) 1 2 sg (s) = g (s) g (s) + s 2 g(s) = which we cn esily solve by using the integrting fctor µ(s) = e s2 /4 yielding the generl solution g(s) = C 1 e s2 /4 s g(s) = C 1 e p2 /4 d p +C 2 with C 1,C 2 R. We now use the initil condition (25) for U(x,t) = g(t 1/2 x): For x > we must hve U(x,t) 1 s t, i.e., g(s) 1 s s. For x < we must hve U(x,t) s t, i.e., g(s) s s. 1

11 Recll tht nd hence with the chnge of vribles p/2 = z, d p = 2dz e z2 dz = π Therefore we obtin e p2 /4 d p = 2 e z2 dz = π. (27) lim g(s) = C! 1 π +C2 = 1 s lim g(s) = C! 1 π +C2 = s yielding 2C 2 = 1 nd 2C 1 π = 1, i.e., g(s) = s 1 π e p2 /4 d p = s/2 2 π e z2 dz = erf(s/2) with the error function erf(x) := 2 x π e x2 dx (which is vilble in Mtlb nd most progrmming lnguges). Therefore we obtin for the solution of our initil vlue problem U(x,t) = g(t 1/2 x), with g(s) = erf(s/2) This function stisfies the het eqution U t (x,t) = U xx (x,t) for x R, t >. For t it stisfies the initil condition for x < lim U(x,t) = 1 t 2 for x = 1 for x > Fundmentl solution We define the fundmentl solution S(x,t) by tking the prtil derivtive with respect to x of U(x,t): Recll tht g (s) = g(s) = 1 2 π e s2 /4 yielding S(x,t) := x U(x,t) = x g(t 1/2 x) = t 1/2 g (t 1/2 x). S(x,t) = 1 ) ( 2 πt exp x2 4t The function U(x,t) stisfies for t > the het eqution U t (x,t) = U xx (x,t). By tking the prtil derivtive of this with respect to x we obtin tht S t (x,t) = S xx (x,t), so S(x,t) lso stisfies the het eqution. The initil condition of U(x,t) is the step function u (x) in (24). Hence the initil condition of S(x,t) should be the derivtive u (x). The derivtive of the step function does not exist s clssicl function. However, in the sense of generlized functions (.k.. distributions) the derivtive of the step function is the delt function: Formlly, we hve S(x,) = δ(x). The delt function hs n infinitely shrp pek t x =, so tht its integrl is 1. Recll tht the delt function formlly stisfies for ny continuous function f tht f (x)δ(x)dx = f (). x= 11

12 Therefore we hve for continuous function u (x) u (x) = u (y)δ(x y)dy y= i.e., we express u formlly s n infinite liner combintion of shifted delt functions. Anlogously we now wnt to write the solution of the het eqution using the shifted fundmentl solutions S(x y,t) multiplied by u (y). We clim tht the solution of the initil vlue problem with u(x,) = u (x) is given by u(x,t) = S(x y,t)u (y)dy y= We now hve to prove tht this formul relly gives solution for the initil vlue problem. We ssume tht the function u (x) is continuous nd bounded for x R. First we hve to show tht the integrl exists s n improper integrl for t >. This follows from the fct tht u (x) is bounded nd the fst decy of S(x,t) for z nd z. Note tht lso the functions t S(x,t) nd z S(z,t) decy fst for for z nd z, hence lso the integrls t S(x y,t)u (y)dy, x 2 S(x y,t)u (y)dy y= exist for t > s improper integrls. One cn see tht these integrls must be equl to u t (x,t) nd u xx (x,t), respectively. Since S(x,t) stisfies the het eqution we hve nd hence u t (x,t) = u xx (x,t). y= t S(x y,t) = 2 x S(x y,t) Note tht for ny t > with z = t 1/2 x, dz = t 1/2 dx S(x,t)dx = 1 ) 2 exp ( x2 dx = 1 ( ) πt x= 4t 2 exp z2 dz = 1 π z= 4 by (27). This mkes sense since the integrl of the delt function is 1, i.e., initilly we hve totl energy 1. Since the totl energy on the rel line is preserved we must hve tht the integrl over the solution is 1 t ll times. We lso hve q(r) := 1 R ( ) 2 exp z2 dz 1 s R. (28) π z= R 4 1 r ) 2 exp ( x2 dx = q(t 1/2 r) (29) πt x= r 4t With the chnge of vribles w = x y, dw = dy we hve u(x,t) = S(x y,t)u (y)dy = S(w,t)u (x w)dw y= w= Consider now fixed point x. We wnt to show tht u(x,t) u (x ) s t. We split the integrl over R into two prts ( t 1/4,t 1/4 ) nd R \ ( t 1/4,t 1/4 ): u(x,t) = I 1 + I 2 t 1/4 I 1 := S(w,t)u (x w)dw, I 2 := S(w,t)u (x w)dw t 1/4 R\( t 1/4,t 1/4 ) As u (x) is bounded there exists C with u (x) C. We then hve using (29) [ ] I 2 C S(w,t)dw = C 1 q(t 1/4 ) R\( t 1/4,t 1/4 ) 12

13 nd we hve for t I 2 C(1 1) =. For I 1 we cn find upper nd lower bounds by using the mximum nd minimum of u (x): ( ) ( ) min u (x) q(t 1/4 ) I 1 mx u (x) q(t 1/4 ) x t 1/4 x x +t 1/4 x t 1/4 x x +t 1/4 For t we hve tht the min nd mx must converge to u (x ) by the continuity of u, nd we hve tht q(t 1/4 ) 1, hence I 1 u (x ) s t. Therefore we hve for ech fixed x R tht lim u(x,t) = u (x). t Problem with sources (inhomogeneous cse) For the inhomogeneous het eqution u t (x,t) u xx (x,t) = f (x,t) x R, t > with het sources f (x,t) we cn now pply the Duhmel principle: 1. We find the function u hom (x,t) by solving the homogeneous het eqution with initil condition t t = u hom (x,) = u (x) yielding u hom (x,t) = S(x y,t)u (y)dy. y= 2. For s > we find the function z (s) (x,t) by solving the homogeneous het eqution with initil condition t t = s z (s) (x,s) = f (x,s) yielding z (s) (x,t) = S(x y,t s) f (y,s)dy. y= 3. Now the solution u(x,t) of the inhomogeneous initil vlue problem (6), (7) is given by t u(x,t) = u hom (x,t) + z (s) (x,t)ds s= t u(x,t) = S(x y,t)u (y)dy + S(x y,t s) f (y,s)dyds (3) y= s= y= 3.4. Het eqution on n intervl Formultion of the problem We consider the het eqution for x in n intervl [,b]: Tht mens t > we wnt u t (x,t) = ku xx (x,t) for x (,b) 13

14 nd boundry conditions t x = nd x = b. In the cse (D) of Dirichlet conditions t both endpoints this mens tht for t > u(,t) =, u(b,t) =. In the cse (N) of Neumnn conditions t both endpoints this mens tht for t > u t (,t) =, u t (b,t) =. We cn lso hve cse (DN) where we hve Dirichlet condition t the left endpoint nd Neumnn condition t the right endpoint: For t > : u(,t) =, u t (b,t) =. We could lso hve cse (ND) where the conditions re the other wy round. We cn lso hve Robin boundry conditions u () + α 1 u() = or u (b) αu(b) = with α 1,α 2 >. In ny of these cses we require tht u(x,t) stisfies n initil condition t t = : u(,t) = u (x) for x (,b) with given function u (x). We sy u(x,t) is clssicl solution if the function u(x,t) is continuous for x [,b], t, nd the functions u t (x,t), u xx (x,t) re continuous for x (,b), t > Inner product nd symmetry We consider the het eqution on n intervl [,b] with e.g. Dirichlet condition u = for x = nd Neumnn condition u x = for x = b (ll other cses work in the sme wy). We define the inner product of two functions u,v on the intervl (,b) s u,v := u(x)v(x)dx Assume tht the functions u,w hve two continuous derivtives on [,b] nd stisfy the boundry conditions u() = u (b) = (31) v() = v (b) = (32) Let us write Au for u. We clim tht the opertor A with these boundry conditions is symmetric in the following sense: This follows by using integrtion by prts twice: u (x)v(x)dx = [ u (x)v(x) ] b }{{} Au,v = u,av + u (x)v (x)dx = [ u(x)v (x) ] b }{{} The terms t x =,b vnish becuse of the boundry conditions (31), (32). u(x)v (x)dx 14

15 3.6. Recipe for solution: Seprtion of Vribles We now consider the initil boundry vlue problem (IBVP) for the het eqution: Find function u(x,t) such tht with the initil condition u t (x,t) u xx (x,t) = x (,b), t > u(,t) =, u (b,t) = t > (33) u(x,) = u (x) x (,b) for given function u (x). Note tht with the opertor Au := u xx we cn write the PDE formlly s with boundry condition (33) nd the initil condition t u + Au = u(,) = u. This is reminiscent of the ODE initil vlue problem (6), (7). Therefore we wnt to use similr strtegy to solve the problem: 1. Find specil solutions of the form where v(x) is not the zero function: Plugging this into the PDE gives where g stisfies the pproprite boundry conditions: E.g. in cse (DN) we hve u(x, t) = v(x)g(t) (34) v(x)g (t) = kv (x)g(t) (35) v() =, v (b) =. Assuming g(t) we cn write (35) s kv (x) = g (t) g(t) v(x). The left hnd side does not depend on t. Since v(x) is not the zero function this cn only work if g (t) is constnt g(t) λ. Therefore we hve to solve n eigenvlue problem: Find λ nd function v(x) (not the zero function) such tht (in cse (DN)) v (x) = λv(x) x (,b) v() =, v (b) = Solving this eigenvlue problem gives rel eigenvlues λ 1 λ 2 λ 3 with λ j s j, nd eigenfunctions v j (x) which re orthogonl on ech other: v j,v k = if j k. (37) (36) For ech eigenfunction v j (x) we hve now specil solution of the PDE: Since g (t) g(t) = λ j we hve g(t) = c j e λ jt nd our specil solution is c j v j (x)e λ jt for j = 1,2,3,

16 2. Write the solution u(x,t) of the initil vlue problem s liner combintion of the specil solutions: We need to find coefficients c 1,c 2,c 3,... such tht u(x,t) = c j v j (x)e λ jt (38) We find the coefficients by setting t = : Then the initil condition gives u (x) = c j v j (x) (39) We now tke the inner product of this eqution with the function v k (x): Then the orthogonlity (37) gives u,v k = c k v k,v k or c k = u,v k (4) v k,v k Summry of Recipe: We need to do the following steps: Find ll eigenvlues λ j nd eigenfunctions v j (x) of the eigenvlue problem (36). Find the coefficients c k by computing the integrls in (4). Then u(x,t) is given by the series (42). We clim tht this recipe lwys gives the solution of our initil vlue problem. Exmple 1: We wnt to solve the het eqution u t = u xx for x [,1]. We impose impose Dirichlet conditions t both endpoints (cse (D)) u(,t) =, u(1,t) = nd the initil condition u(x,) = u (x) = 1 x (,1). 1. Solve the eigenvlue problem: Find λ R nd function v(x) (not zero function) such tht v (x) = λv(x), v() =, v(1) =. For given λ the ODE v (x) λv(x) = hs the following generl solution: If λ = the generl solution is v(x) = C 1 1 +C 2 x Now v() = gives C 1 =, nd then v(1) = gives v(1) =. Hence the only solution is v(x) =, nd λ = is not n eigenvlue. If λ > the generl solution is v(x) = C 1 sin(λ 1/2 x) +C 2 cos(λ 1/2 x). Now v() = gives C 2 =. Then v(1) = gives C 1 sin(λ 1/2 ) =. For solution v(x) which is not the zero function we need sin(λ 1/2 ) =. We know tht sinx = if nd only if x = jπ for some integer j. Hence we need λ 1/2 = jπ λ = j 2 π 2 j = 1,2,3,... (recll tht we wnt to find λ > ). Therefore we hve found the eigenvlues λ j nd eigenfunctions v j (x): λ j = j 2 π 2, v j (x) = sin( jπx) j = 1,2,3,... 16

17 2. Find the coefficients c k : We hve to compute c k = u,v k v k,v k for k = 1,2,3,... We first compute the denomintor using sin 2 xdx = 1 2 x 1 4 sin(2x) nd z = kπx: 1 v k,v k = sin 2 (kπx)dx = 1 [ 1 kπ 2 z 1 4 sin(2z)] kπ z= = 1 2. We then compute the numertor using cos(kπ) = ( 1) k for integer k: Hence we obtin Therefore the solution of the initil vlue problem is 1 u,v k = sin(kπx)dx = 1 kπ [ cos(kπx)]1 x= = 1 ( 1 ( 1) j ) kπ u(x,t) = c k = { 4 kπ k=1,3,5,... for k odd for k even 4 π 2 t kπ sin(kπx)e k2 Exmple 2: We wnt to solve the het eqution u t = u xx for x [,1]. We impose Neumnn conditions t both endpoints (cse (N)) u(,t) =, u(1,t) = nd the initil condition u(x,) = u (x) = x x (,1). 1. Solve the eigenvlue problem: Find λ R nd function v(x) (not zero function) such tht v (x) = λv(x), v () =, v (1) =. For given λ the ODE v (x) λv(x) = hs the following generl solution: If λ = the generl solution is v(x) = C 1 1 +C 2 x, v (x) = C 2 Now v () = gives C 2 =, nd then v (1) = is lso stisfied. Hence we cn obtin solution (other thn v = ) s v(x) = 1. Therefore λ = is n eigenvlue, nd the corresponding eigenfunction is v (x) = 1. If λ > the generl solution is v(x) = C 1 sin(λ 1/2 x) +C 2 cos(λ 1/2 x), v (x) = C 1 λ 1/2 cos(λ 1/2 x) C 2 λ 1/2 sin(λ 1/2 x) Now v () = gives C 1 =. Then v (1) = gives C 2 λ 1/2 sin(λ 1/2 ) =. For solution v(x) which is not the zero function we need sin(λ 1/2 ) =. We know tht sinx = if nd only if x = jπ for some integer j. Hence we need λ 1/2 = jπ λ = j 2 π 2 j = 1,2,3,... (recll tht we wnt to find λ > ). Therefore we hve found the eigenvlues λ j nd eigenfunctions v j (x): λ j = j 2 π 2, v j (x) = sin( jπx) j = 1,2,3,... 17

18 2. Find the coefficients c k : We hve to compute For k = we obtin c k = u,v k v k,v k for k =,1,2,... c = u,v 1 v,v = xdx 1 1 1dx = 2 1 = 1 2. For k = 1,2,3,... we hve for the denomintor using cos 2 xdx = 1 2 x sin(2x) nd z = kπx: 1 v k,v k = cos 2 (kπx)dx = 1 [ 1 kπ 2 z sin(2z)] kπ z= = 1 2. We then compute the numertor using the ntiderivtive cosxdx = cosx + xsinx: Then the chnge of vribles z = kπx gives { 1 kπ z dz u,v k = x cos(kπx)dx = cos(z) x= z= kπ kπ = 1 k 2 [cosz + zsinz]kπ π2 z= = 1 ( ) k 2 π 2 ( 1) k for k even 1 = for k odd Hence we obtin Therefore the solution of the initil vlue problem is u(x,t) = for k = c k = 4 for k odd k 2 π 2 otherwise k=1,3,5,... 4 π 2 t k 2 π 2 cos(kπx)e k2 2 k 2 π Convergence of the series? So fr we hve forml recipe: Find ll eigenvlues λ j nd eigenfunctions v j (x) of the eigenvlue problem (36). Find the coefficients c k by computing the integrls in (4). Then u(x,t) is given by the series (42). Assume we re ble to find ll eigenvlues nd eigenfunctions, nd cn compute ll coefficients c k. Then the following questions remin: 1. Does the series for u (x) u (x) = converge? In which sense does it converge? Is the limit of the series relly u (x)? 2. For t > : does the series u(x,t) = converge? In which sense does it converge? c j v j (x)e λ jt c j v j (x) (41) x (,b), t > (42) 3. Does the function u(x, t) defined by the series (42) stisfy the het eqution? This sems resonble, since by construction ech of the functions v j (x)e λ jt re solutions of the het eqution. Therefore for finite N the function N c jv j (x)e λ jt stisfies the het eqution. But for N one needs to prove this. 4. Does the function u(x,t) stisfy the initil condition? I.e., if we let t tend to zero, will we hve lim u(x,t) = u (x) t For finite series N ( ) we cn just let t in ech term. But for n infinite series this is more difficult. 18

19 3.8. Review: Uniform convergence We consider sequence f j (x), j = 1,2,3,... of continuous functions for x [,b]. We would like to know whether the limit function f (x) = lim f j (x) (43) j exists nd is lso continuous function. Exmple: Let f j (x) = tn 1 ( jx) on the intervl [ 1,1]. Then for ech x [ 1,1] the limit (43) exists: lim j π 2 for x < tn( jx) = for x = π 2 for x > Obviously, the limit function f (x) is not continuous. Here the functions f j (x) converge pointwise, i.e., for ech fixed x the sequence f j (x) is convergent. A stronger notion of convergence of functions is uniform convergence: We sy the functions f j (x) converge uniformly to the function f (x) on the intervl [,b] if mx f j (x) f (x) s j x [,b] In this cse the limit function must lso be continuous: Theorem 3.4. Assume tht the functions f j (x) re continuous for x [,b]. Assume tht for ech x [,b] the limit f (x) := lim j f j (x) exists, nd tht Then the function f (x) is continuous. mx f j (x) f (x) x [,b] s j Now we pply this theorem to the cse of series: f (x) = Corollry 3.5. Assume tht the functions g j (x) re continuous for x [,b]. Let α j := mx g j (x) If g j (x) (44) x [,b] α j < then f n = n g j converges uniformly to f, nd the limit function f (x) is continuous on [,b]. Proof. For ech x [,b] the limit (44) must exist since g j (x) converges. We then hve for ny x [,b] f (x) f n (x) = j=n+1 g j (x) g j (x) j=n+1 α j j=n+1 Since α j converges the expression on the right hnd side must converge to zero s n. Therefore the sequence f j stisfies the ssumption of Theorem

20 3.9. Convergence for the exmples We wnt to see whether in the exmples the series (39) for u (x) nd the series (38) for u(x,t) (t > ) converge uniformly. Exmple 1: For u (x) = 1 we hve obtined the series u (x) = k=1,3,5,... 4 kπ sin(kπx) One cn prove tht the series converges to 1 for ech fixed x (,1). For x = we see tht the series gives. Therefore the series cnnot converge uniformly (if it did, it would converge to continuous function. Note tht we cnnot expect to find clssicl solution of the het eqution for exmple 1: A clssicl solution must be continuous for x [,b] nd t. In this exmple we wnt u(x,) = 1 for ll x, nd u(,t) = for ll t. Therefore it is impossible for the function u(x,t) to be continuous t (x,t) = (,). For u(x,t) with t > we hve the series u(x,t) = k=1,3,5,... 4 π 2t kπ e k2 sin(kπx) Note tht 4 π 2t kπ e k2 sin(kπx) 4 π 2t kπ e k2 =: α k Note tht α k decys exponentilly fst if t >. Therefore α k < k=1 nd by Corollry 3.5 the series for u(x,t) converges uniformly. Now we consider u x (x,t): Tking the prtil derivtive with respect to x of ech term of the series gives the series ũ(x,t) := k=1,3,5,... 4 π 2t kπ e k2 k cos(kπx), α k = 4 π 2t kπ e k2 k Since k=1 α k < the series converges uniformly, nd we cn lso conclude tht u x (x,t) = ũ(x,t). The sme rgument lso works for u xx, u t nd in fct for ny prtil derivtive. In prticulr we obtin tht u t (x,t) u xx (x,t) = Exmple 2: Here we obtined for u (x) = x the series k=1,3,5,... u (x) = π 2t kπ e k2 sin(kπx) [ k 2 + k 2] = k=1,3,5,... 4 k 2 π 2 cos(kπx) Note tht 4 k 2 π 2 cos(kπx) 4 k 2 π 2 =: β k where k=1 β k = 4 π 2 k=1 1 k 2 <. Hence the series for u (x) converges uniformly. For u(x,t) with t > we hve the series u(x,t) = k=1,3,5,... 4 π 2t k 2 π 2 cos(kπx)e k2. 2

21 Here 4 π 2t k 2 π 2 cos(kπx)e k2 cos(kπx) 4 π 2t k 2 π 2 e k2 =: α k Note tht α k decys exponentilly fst for t >. Therefore we obtin s in Exmple 1 tht the series for u(x,t) converges uniformly. The sme holds for ll prtil derivtives. As in Exmple 1 it follows tht u(x,t) stisfies the het eqution. But there is one difference: Note tht α k depends on t, but for ny t > we hve the bound α k = 4 k 2 π 2 e k2 π 2t 4 k 2 π 2 = β k where k=1 β k <. Therefore one is llowed to perform the limit for t term by term: [ ] lim u(x,t) = t lim c k v k (x)e λ kt = t k= k= c k v k (x) = u (x). Actully, one cn see tht this limit is uniform in x, so one obtins tht the function u(x,t) is ctully continuous for x [,b] nd t. Therefore u(x,t) is clssicl solution of the het eqution. A. Appendix: Hilbert spce, orthogonl systems nd eigenvlue problems A.1. Hilbert spce We consider rel vector spce H with the following properties 1. There is n inner product u,v stisfying the following: For ny u,v,w H, α,β R αu + βv,w = α u,v + β v,w, u,v = v,u, u,u > for u We define the norm u := u,u 1/2. 2. With respect to this norm the spce H is complete: every Cuchy sequence converges. Such spce is clled Hilbert spce. Recll tht we cll sequence u 1,u 2,u 3,... of elements of H Cuchy sequence if for ny ε > there exists N such tht M,N N : u M u N < ε. Exmple 1: H = R n, u,v = n u jv j Exmple 2: H is the set of ll sequences (u 1,u 2,u 3,...) with u2 j <. Here u,v = u jv j Exmple 3: H = L 2 (I), the spce of ll functions u on n intervl I = (,b) such tht u(x)2 dx is bounded. Remrk: (i) One needs to use the Lebesgue integrl for the proper definition. (ii) Functions u nd v which differ only on set of mesure zero stisfy u v = nd re considered identicl. 21

22 A.2. Orthogonl vectors We sy vectors v 1,v 2,v 3,... form n orthogonl system if v j nd v j,v k = for j k We cn define the normlized vectors w j := v j / v j. These vectors form n orthonorml system, i.e., w j,w k = {1 if j = k if j k Assume tht the vectors w 1,...,w N form n orthonorml system. We wnt to pproximte given vector u H by vector of the form N ũ = c j w j It is esy to see tht the error ũ u becomes miniml if ũ u is orthogonl on ll the vectors w 1,...,w n, i.e., ũ u,wj = j = 1,...,N (45) As ũ,w j = c j this condition implies Note tht we hve becuse of (45) c j = u,w j. u 2 = (u ũ) + ũ,(u ũ) + ũ = u ũ 2 + ũ 2 As ũ 2 = N c2 j we hve u 2 = u ũ 2 + N c 2 j. (46) Now we consider n infinite orthonorml system w 1,w 2,w 3,... For given vector u H we define the coefficients c j := u,wj nd By (46) we hve ũ N := N c j w j. N c 2 j u 2 nd therefore we obtin for N c 2 j u 2 This is clled Bessel inequlity. Since the series S := c2 j converges, the sequence S N := N c2 j sequence. Then the sequence ũ N must lso form Cuchy sequence since for M N must form Cuchy ũ M ũ N 2 = M c 2 j = S M S N. j=n+1 As the spce H is complete the sequence ũ N must hve limit ũ H. We then hve u 2 = u ũ 2 + ũ 2 = u ũ 2 + c 2 j. If we hve for ll u H tht c 2 j = u 2 22

23 ( Prsevl identity ) we cll the system w 1,w 2,w 3,... complete orthonorml system. In this cse we hve for ll u H u ũ N s N. Exmple 1: For the spce H = L 2 [,1] of squre integrble functions on the intervl [,1] the functions form complete orthonorml system. w j (x) = 2 1/2 sin(π jx), j = 1,2,3,... Exmple 2: For the spce H = L 2 [,1] of squre integrble functions on the intervl [,1] the functions w j (x) = 2 1/2 cos(π jx), j = 1,2,3,... form n orthonorml system, but it is not complete: E.g. for u(x) = 1 we hve u = 1, but we hve 1 c k = u,w k = cos(πkx)dx = nd therefore ũ(x) =, c2 j =. The Bessel inequlity is stisfied, but we do not hve the Prsevl identity. We cnnot represent ll squre integrble functions s series ũ = c jw j. Another wy to see this is the following: All the functions w j (x) hve integrl zero: 1 w j(x)dx =. Therefore we lso must hve 1 ũ(x)dx =. Actully, one cn show tht one cn obtin ny function u L 2 [,1] with integrl zero s series u = c jw j. If we dd to our orthonorml system the function 1 the new system is complete orthonorml system for ll of L 2 [,1]. 1, 2 1/2 cos(πx), 2 1/2 cos(2πx), 2 1/2 cos(3πx),... A.3. Liner opertors Let T : H H be liner mpping. We cll this mpping bounded if the set S = {Tu u H, u 1} is bounded, i.e., C = sup v S v is finite. We cll mpping compct if the set S is compct, i.e., every sequence in S hs convergent subsequence. We cll liner mpping T self-djoint if u,v H : Tu,v = u,t v A.4. The eigenvlue problem We sy µ R is n eigenvlue of T if there exists nonzero v H such tht In this cse v is clled the eigenvector for the eigenvlue µ. T v = µv Theorem A.1. Assume tht H is Hilbert spce nd T : H H is liner compct self-djoint opertor. Assume tht is not n eigenvlue of T. Then 1. There re infinitely mny eigenvlues µ 1, µ 2, µ 3,... They re rel nd stisfy µ 1 µ 2 µ 3, µ k s k 23

24 2. There re corresponding eigenvectors v j which re orthogonl on ech other: v j,v k = for j k 3. The eigenvectors form complete orthogonl system: every u H cn be written s n infinite liner combintion of eigenvectors: Let c j = u,v j / v j,v j, then u = c j v j This mens tht the u N c jv j s N. I don t give the proof of this theorem. But it is similr to the proof of Theorem 1.2: We first mximize Tu,u over ll u H with u = 1 yielding vector v 1. One cn gin show tht this function must be n eigenvector with eigenvlue µ 1. Then one considers the subspce V 1 = {u H u,v 1 = } of ll vectors orthogonl to v 1. Now we mximize Tu,u over ll u V 1 with u = 1 yielding vector v 2. One obtins tht this function must be n eigenvector with eigenvlue µ 2 where µ 2 µ 1. In this wy one obtins ll eigenvectors v 1,v 2,v 3,.... A.5. Appliction to the ODE eigenvlue problem Let us consider e.g. the Dirichlet cse: Find λ nd function v such tht v (x) = λv(x) x (,b) (47) v() =, v(b) = We first consider the BVP with given right hnd side function f (x): Find u(x) such tht u (x) = f (x) x (,b) u() =, u(b) = In this cse it is esy to see tht we cn obtin the unique solution by tking the ntiderivtive twice nd djust the constnts to stisfy the boundry conditions. We denote the opertor which mps the function f to the function u by T : We hve u = T f If u stisfies u = f nd the boundry conditions, w stisfies w = g nd the boundry conditions we hve u,w = u, w f,t g = T f,g Therefore the opertor T is self-djoint. One cn lso show tht the opertor cn be defined for ny f L 2 nd gives function u = T f L 2. Furthermore one cn show tht the opertor T is compct. Therefore Theorem A.1 shows tht the opertor T hs infinitely mny eigenvlues µ j with eigenfunctions v j such tht T v j = µ j v j Hence µ j v j stisfies µ j v j = v j. Therefore v j is n eigenfunction of our BVP (47) with λ j := µ 1 j : v j = λ j v j The eigenvlues µ j of the opertor T stisfy lim j µ j =. Hence we obtin tht the eigenvlues λ j stisfy lim j λ j =. 24

25 A.6. The generl cse with coefficient functions r(x), κ(x) Recll from 3.1 tht for the cse of vrible mteril properties long the br we obtin the het eqution in the form r(x)u t (x,t) x [κ(x)u x (x,t)] = x (,b) (48) with functions r(x) (specific het) nd κ(x) (het conductivity) where r(x) >, κ(x) > for x [, b]. Let us consider the cse of the Dirichlet problem (the other cses work in the sme wy) with the boundry conditions We wnt to find solution u(x,t) which stisfies n initil condition u(,t) =, u(b,t) = t >. (49) Then the seprtion of vribles method leds to the following eigenvlue problem: Find λ nd function v(x) (not the zero function) such tht u(x,) = u (x) x (,b). (5) d dx [ κ(x)v (x) ] = λr(x)v(x) x (,b) v() =, v(b) = (51) We now define the inner product u,v in different wy: If A denotes the opertor u,v := Au := r(x) 1 d dx u(x)v(x)r(x)dx. [ κ(x)v (x) ] we clim tht the opertor is self-djoint for functions u, v which re zero t x =,b: Au,v = d dx = [ κ(x)u(x)v (x) ] b }{{} [ κ(x)u (x) ] v(x)dx = [ κ(x)u (x)v(x) ] b + }{{} [ κ(x)v (x) ] dx = u,av u(x) d dx Therefore we cn use the stndrd recipe to solve the IBVP (48), (49), (5): Find ll eigenvlues λ j nd eigenfunctions v j (x) of the eigenvlue problem (51). Find the coefficients c k using Then u(x,t) is given by the series c k = u,v k b v k,v k = u (x)v k (x)r(x)dx v k(x) 2 r(x)dx u(x,t) = k=1 c k v k (x)e λ kt Remrk A.2. The sme method lso works for the PDE with n dditionl term q(x)u(x,t) u (x)κ(x)v (x)dx r(x)u t (x,t) x [κ(x)u x (x,t)] + q(x)u(x,t) = x (,b) where the function q stisfies q(x). In this cse we hve to solve the eigenvlue problem d [ κ(x)v (x) ] + q(x)u(x) = λr(x)v(x) x (,b) dx v() =, v(b) =. 25

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