Suggested Solution to Assignment 4

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1 MATH 40 (015-16) partia diferentia equations Suggested Soution to Assignment 4 Exercise 41 The soution to this probem satisfies the foowing PDE u t = ku xx, (0 < x <, 0 < t < ) u(0, t) = u(, t) = 0, u(x, 0) = 1 Foowing the process in Page 85 of the textbook, we have and the initia condition impies 1 = A n e nπ ( A n sin nπx )kt sin nπx, By the assumption, we have A n = 4 nπ for odd n and A n = 0 for even ones Then 4 Let T (t)x(x), we have Hence, Since 0 < r < πc/, we get k=1 4 (k 1)π e ( T + rt c T (k 1)π ) kt sin = X X = λ (k 1)πx λ n = ( nπ ), X(x) = sin nπx, n = 1,, T n (t) = [A n cos( n t/) + B n sin( n t/)]e rt/, n = 1,,, where n = r (nπc/) reative to the equation Therefore, λ + rλ + ( nπc ) = 0 [A n cos( n t/) + B n sin( n t/)]e rt/ sin nπx 5 Let T (t)x(x), we have T + rt c = X T X = λ Hence, λ n = ( nπ ), X(x) = sin nπx, n = 1,, When n = 1, since πc/ < r < 4πc/, T 1 (t) = A 1 e λ+ 1 t + B 1 e λ 1 t, 1

2 MATH 40 (015-16) partia diferentia equations where λ ± 1 = r ± r ( πc ) When n, are the roots of the equation λ + rλ + ( πc ) = 0 T n (t) = [A n cos( n t/) + B n sin( n t/)]e rt/, n = 1,,, where n = r (nπc/) reative to the equation λ + rλ + ( nπc ) = 0 Therefore, [A 1 e λ+ 1 t + B 1 e λ 1 t ] sin πx 6 Let T (t)x(x), we have The initia condition impies Therefore, + [A n cos( n t/) + B n sin( n t/)]e rt/ sin nπx n= tt T T = X X = λ, λ n = n, X(x) = sin nx, n = 1,, tt T = λt, T (0) = 0 ct sin x, are soutions So uniqueness is fase for this equation! Exercise 4 1 Let T (t)x(x), we have The initia condition impies for any constantc, T kt = X X = λ X = λx, X(0) = X () = 0 So by soving the above DE, the eigenvaues are [ (n + 1 )π ], the eigenfunctions are X n (x) = sin (n + 1 )πx for n = 0, 1,,, and the soution is n=0 e [ (n+ 1 )π ] kt sin (n + 1 )πx (a) This can be proved as above Here we give another proof Since X (0) = 0, the we can use even expansion, this is, X( x) = X(x) for x 0, then X satisfies X = λx, X( ) = X() = 0 Hence, λ n = [(n + 1 )π] /, X n (x) = cos[(n + 1 )πx/], n = 0, 1,, (b) Having known the eigenvaues, it is easy to get the soution [ A n cos (n + 1 )πct + B n sin (n + 1 )πct n=0 ] cos (n + 1 )πx

3 MATH 40 (015-16) partia diferentia equations 3 We just show how to sove the eigenvaue probem under the periodic boundary conditions; As before, et T (t)x(x), T kt = X X = λ Soving T = λkt gives T = Ae λkt The genera soutions of X + λx = 0 are X = Ce γx + De γx, where et λ is a compex number and γ is either one of the two roots of λ; the other one is γ The boundary conditions yied Hence e γ = 1 and then Therefore, the concentration is Exercise 43 Ce γ + De γ = Ce γ + De γ, γ(ce γ De γ ) = γ(ce γ De γ ) γ = ±nπi/, λ = γ = (nπ/), n = 0, 1,, { 1 X n (x) = A 0 n = 0 A n cos nπx + B n sin nπx, T = e (nπ/) kt n = 1,, 1 A 0 + n=0 ( A n cos nπx + B n sin nπx ) e (nπ/)kt 1 Firsty, et s ook for the positive eigenvaues λ = β > 0 As usua, the genera soution of the ODE is The boundary conditions impy X(x) = C cos βx + D sin βx C = 0, Dβ cos(β) + ad sin(β) = 0 Hence, tan(β) = β a The graph is omitted Seconddy, et s ook for the zero eigenvaue, ie, X(x) = Ax + B, by the boundary conditions, a + 1 = 0 Hence, λ = 0 is an eigenvaue if and ony if a + 1 = 0 Thirdy, et s ook for the negative eigenvaues λ = γ < 0 As usua, the soution of the ODE is Then the boundary condtions impy X(x) = C cosh(γx) + D sinh(γx) C = 0, Dγ cosh(γ) + ad sinh(γ) = 0 Hence, tanh(γ) = γ a The graph is omitted (a) If λ = 0, then X(x) = Ax + B The boundary conditions impy A a 0 B = 0, A + a (A + B) = 0 These two equaities are equivaent to a 0 + a = a 0 a Hence, λ = 0 is an eigenvaue if and ony if a 0 + a = a 0 a 3

4 MATH 40 (015-16) partia diferentia equations (b) By (a), we have X(x) = B(a 0 x + 1), here B is constant 3 If λ = γ < 0, we have Hence, and the boundary conditions impy Therefore, the eigenvaues satisfy and the corresponding eigenfunctions are where C is a constant X(x) = C cosh γx + D sinh γx X (x) = Cγ sinh γx + Dγ cosh γx, Dγ a 0 C = 0, Cγ sinh γ + Dγ cosh γ + a [C cosh γ + D sinh γ] = 0 tanh γ = (a 0 + a )γ γ + a 0 a, X(x) = C cosh γx + a 0 C sinh γx, γ 4 It is easiy known that the rationa curve y = (a 0+a )γ γ +a 0 has a singe maximum at γ = a a 0 a and is monotone in the two intervas (0, a 0 a ) and ( a 0 a, ) Furthermore, max y(γ) = a 0 + a γ [0, ) 1, im a 0 a y(γ) = 0, for γ y (0) = a 0 + a a 0 a Note that tanh γ is monotone in [0, ), tanh γ < 1 when γ [0, ), im tanh γ = 1, and (tanh γ γ) γ=0 = > a 0 + a a 0 a Therefore, the rationa curve y = (a 0+a )γ γ +a 0 a there are two negative eigenvaue and the curve y = tanh γ intersect at two points, that is, 5 When λ = β > 0, β satisfies (10), ie tan β = (a 0 + a )β β a 0 a Since y = tan β is monotonicay increasing when β ((n 1 )π/, (n + 1 )π/) (n = 0, 1,, ) and im tan β =, β (n 1 )π/ im tan β =, β (n+ 1 )π/ whie y = (a 0+a )β β a 0 a is negative, monotonicay increasing when β ( a 0 a, ) and (a 0 + a )β im β β = 0, a 0 a the two curves intersects at infinite many points, that is, there are an infinite many number of positive eigenvaues The graph is simiiar to the Figure 1 in Section 43 in the textbook but y = (a 0+a )β β a 0 is a positive first and then negative now 4

5 MATH 40 (015-16) partia diferentia equations 6 (a) If a > 0, the case turns out to be case 1 in Section 43 and thus there are no negative eigenvaues; if a = 0, the case turns out to be the Neumann boundary condition probem and thus there are no negative eigenvaues, either; if / a < 0, we have (tanh γ) γ=0 = a 0 + a =, using the same way as Exercise 434 a 0 a a above, we concude that there is ony one negative eigenvaue; if a <, we have (tanh γ) γ=0 = > a 0 + a = and thus there are two negative eigenvaues a 0 a a (b) Exercise 43 impies that λ = 0 is an eigenvaue if and ony if a 0 + a = a 0 a, ie, a = 0 or a = 7 Under the condition a 0 = a = a, the eigenvaue satisfies λ = β, tan β = aβ β a Hence, when a and nπ impies < β n < (n+1)π, im a aβ { β n (a) is negative and tends to 0 So Figure 1 in Section 43 β a } (n + 1)π = 0 9 (a) If λ = 0, then X(x) = ax + b for some constants a and b Then the boundary conditions impy a + b = 0 Therefore, X 0 (x) = ax a for some nonzero constant a (b) If λ = β, then X(x) = A cos βx + B sin βx Then the boundary conditions impy Since A, B can not both be 0, we have β = tan β (c) omit (d) If λ = γ, then X(x) = Ae γx + Be γx and A + Bβ = 0, A cos β + B sin β = 0 A + B + Aγ Bγ = 0, Ae γ + Be γ = 0 ( ) 1 + γ 1 γ Then we find the coefficent matrix is aways nonsinguar(since e γ > 1+γ 1 γ when γ > 0, e γ e γ verify by yoursef!), then a = b = 0 So we concude that there is not any negative eigenvaue 10 Let X(x)T (t), by the summary on Page 97, we can have (C n cos β n ct+d n sin β n ct)(cos β n x+ a 0 sin β n x)+(c 0 cosh γct+d 0 sinh γct)(cosh γx+ a 0 sinh γx), β n γ where γ is determined by the intersection point of tanh γ = (a 0+a )γ γ +a 0 a, and the intitia conditions are φ(x) = ψ(x) = C n (cos β n x + a 0 sin β n x) + C 0 (cosh γx + a 0 sinh γx), β n γ D n β n c(cos β n x + a 0 sin β n x) + D 0 γc(cosh γx + a 0 sinh γx) β n γ 5

6 MATH 40 (015-16) partia diferentia equations 11 (a) By the wave equation, de dt = = 0 0 [ 1 c u tu tt + u x u xt ]dx [u t u xx + u x u xt ] = (u t u x ) = u t(, t)u x (, t) u t (0, t)u x (0, t) 0 The Dirichet boundary conditions u(, t) = u(0, t) = 0 impy u t (, t) = u x (, t) = 0 Hence, de dt 0 (b) Same as above Omit here (c) By the computation in (a) and the Robin boundary conditions, we can get that de R dt = u t u x 0 + a u t (, t)u(, t) + a 0 u t (0, t)u x (0, t) 0 1 (a) Let λ = 0, we have v(x) = Ax + B Since v(x) = Ax + B sitisfy the boundary conditions for any A and B, λ = 0 is a doube eigenvaue (b) Let λ = β > 0 and suppose β > 0, we have v(x) = C cos βx + D sin βx Then boundary conditions impy C cos β + D sin β C Dβ = Cβ sin β + Dβ cos β = Therefore, eigenvaues λ > 0 satisfies the equation λ = β, sin β( sin β + β) = (1 cos β) (c) Let γ = 1 λ, then γ is a root of the foowing equation γ sin γ cos γ = sin γ (d) By (c), we have sin γ = 0 or γ = tan γ So the positive eigenvaues are 4n π and 4γ n/ where γ n = tan γ n (nπ π, nπ π ) for n = 1,, The graph is omitted here (e) By (a) and (d), for λ = 0, the eigenfuntions are 1 and x; for λ = 4n π, n = 1,,, the eigenfunctions are cos( nπx ); for λ = 4γ n, where γ n = tan γ n (nπ π, nπ 1 π), n = 1,,, the eigenfunctions are γ n cos γ nx sin γ nx (f) From above, we have A + Bx + + C n e 4γ D n e 4n π n kt cos nπx kt [γ n cos γ nx sin γ nx ] (g) By (f), we have im t A + Bx since im t e λkt = 0 6

7 MATH 40 (015-16) partia diferentia equations 15 Let λ = β, then X(x) = A cos βρ 1x X(x) = C cos βρ x + B sin βρ 1x, 0 < x < a; + D sin βρ x, a < x < Hence, the boundary condtions impy A βρ 1 A cos βρ 1a A = 0; + B sin βρ 1a C cos βρ = C cos βρ a + D sin βρ = 0; + D sin βρ a ; sin βρ 1a + B βρ 1 cos βρ 1a = C βρ sin βρ a + D βρ 1 cos βρ a Hence, when the eigenvaue is positive, ie λ = β > 0, β satisfies ρ 1 cot βρ 1a + ρ cot βρ ( a) = 0 Let λ = 0, then the boundary conditions impy { Ax 0 < a < ; X(x) = B(x ) a < x < Since X(x) shoud be differentiabe at x = a, such A and B can not exist except A = B = 0 Let λ = γ < 0, then X(x) = A cosh βρ 1x X(x) = C cosh βρ x + B sinh βρ 1x, 0 < x < a; + D sinh βρ x, a < x < Hence, the boundary condtions impy A βρ 1 A cosh βρ 1a A = 0; + B sinh βρ 1a C cosh βρ = C cosh βρ a + D sinh βρ = 0; + D sinh βρ a ; sinh βρ 1a + B βρ 1 cosh βρ 1a = C βρ sinh βρ a + D βρ 1 cosh βρ a Hence, when the eigenvaue is negative, ie λ = β > 0, β satisfies ρ 1 coth βρ 1a + ρ coth βρ ( a) = 0 However, since the eft handside is aways positive Therefore, there is no negative eigenvaues 7

8 MATH 40 (015-16) partia diferentia equations 16 Let λ = β 4 > 0 where β > 0, and X(x) = A cosh βx + B sinh βx + C cos βx + D sin βx By the boundary conditions β n = nπ, λ n = ( nπ )4, X n (x) = sin nπx, n = 1,, The detais are as the foowing exercise 17 Let λ = β 4 > 0 where β > 0, and X(x) = A cosh βx + B sinh βx + C cos βx + D sin βx Hence by the boundary conditions, which simpifies to A + C = 0, B + D = 0, A cosh β + B sinh β + C cos β + D sin β = 0, A sinh β + B cosh β C sin β + D cos β = 0, A(cosh β cos β) + B(sinh β sin β) = 0, A(sinh β + sin β) + B(cosh β cos β) = 0 Since eigenfunctions are nontrivia, the determinant of the matrix shoud be zero, that is, and the corresponding eigenfunction is (cosh β cos β) (sinh β sin β) = 0, cosh β cos β = 1 X(x) = (sinh β sin β)(cosh βx cos βx) (cosh β cos β)(sinh βx sin βx) Probem 10 X(x)T (t) = T (t) a T (t) = X(4) (x) X(x) = λ = X (4) λx = 0 and T + λa T = 0 = λ 0 X = 0 X(4) X = 0 X 0 = λ = X 0 0 X If λ = 0, thenx 0 = X(x) = ax + b = X 0 since X(0) = X() = 0 = λ > 0 Letλ = β 4, β > 0, then T (t) = A cos(β at) + B sin(β at) X(x) = Ce βx + De βx + E cos(βx) + F sin(βx) u(0, t) = u xx (0, t) = u(, t) = u xx (, t) = 0 = X(0) = X() = X (0) = X () = 0 = E = 0, F sin(β) = 0, C = D = 0 = sin(β) = 0 = β n = nπ, X n (x) = sin(β n ), (n = 1,, 3, ) are distinct soutions = (A n cos(β nat) + B n sin(β nat)) sin(β n ) where A n, B n are determined by φ(x) = u(x, 0) = ψ(x) = u t (x, 0) = A n sin(β n ) βnab n sin(β n ) 8

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