Strauss PDEs 2e: Section Exercise 2 Page 1 of 12. For problem (1), complete the calculation of the series in case j(t) = 0 and h(t) = e t.

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1 Strauss PDEs e: Section Exercise Page 1 of 1 Exercise For probem (1, compete the cacuation of the series in case j(t = and h(t = e t. Soution With j(t = and h(t = e t, probem (1 on page 147 becomes u t = ku xx < x <, t > u(, t = e t u(, t = u(x, =. The PDE is the heat equation for a homogeneous one-dimensiona rod. The temperature at the eft end rises exponentiay in time, whereas the right end remains at indefinitey. Initiay the entire rod is at. Method 1 - Using Term-by-Term Differentiation For this method to work, the boundary conditions must be made homogeneous. Make the change of variabes v(x, t = u(x, t r(x, t, where r(x, t is some function that satisfies the boundary conditions. r(, t = e t r(, t = A suitabe function is r(x, t = x e t. As a resut, the initia and boundary conditions for v are v(, t = u(, t r(, t = e t e t = v(, t = u(, t r(, t = = v(x, = u(x, r(x, = x = x. Substitute u(x, t = v(x, t + r(x, t into the PDE now to find the one that v satisfies. Distribute k and bring r t to the right side. v t + r t = k(v xx + r xx v t = kv xx + kr xx r t Consequenty, the PDE for v is v t = kv xx + x e t. (1 Since it is inear and inhomogeneous, we choose to appy the method of eigenfunction expansion to sove it. Consider the eigenvaue probem of the differentia operator invoving the spatia variabe x d φ = λφ (

2 Strauss PDEs e: Section Exercise Page of 1 with the same boundary conditions as v. φ( = φ( = Vaues of λ for which the boundary conditions are satisfied are known as the eigenvaues, and the nontrivia soutions associated with them are caed the eigenfunctions. Equation ( is known as the one-dimensiona Hemhotz equation; the eigenfunctions for it are known to be orthogona and form a compete set, which wi prove usefu ater. Determination of Positive Eigenvaues: λ = µ Suppose that λ is positive. Then equation ( becomes d φ = µ φ. Its soution can be written in terms of hyperboic sine and hyperboic cosine. φ(x = C 1 cosh µx + C sinh µx Appy the boundary conditions to determine C 1 and C. φ( = C 1 = φ( = C 1 cosh µ + C sinh µ = Since C 1 =, the second equation reduces to C sinh µ =. Hyperboic sine is not osciatory, so the ony way this equation is satisfied is if C =. The trivia soution is obtained, so there are no positive eigenvaues. Determination of the Zero Eigenvaue: λ = Suppose that λ is zero. Then equation ( becomes d φ =. The genera soution is obtained by integrating both sides with respect to x twice. φ(x = C 3 x + C 4 Appy the boundary conditions to determine C 3 and C 4. φ( = C 4 = φ( = C 3 + C 4 = Since C 4 =, the second equation reduces to C 3 =. The trivia soution is obtained, so zero is not an eigenvaue.

3 Strauss PDEs e: Section Exercise Page 3 of 1 Determination of Negative Eigenvaues: λ = γ Suppose that λ is negative. Then equation ( becomes d φ = γ φ. Its soution can be written in terms of sine and cosine. φ(x = C 5 cos γx + C 6 sin γx Appy the boundary conditions to determine C 5 and C 6. φ( = C 5 = φ( = C 5 cos γ + C 6 sin γ = Since C 5 =, the second equation reduces to C 6 sin γ =. To avoid getting the trivia soution, we insist that C 6. Then sin γ = γ =, n = 1,,... γ n =, n = 1,,.... The eigenfunctions associated with these eigenvaues for λ are φ(x = C 6 sin γx φ n (x = sin x, n = 1,,... and form a compete set, so the unknown function v can be expanded in terms of them. v(x, t = a n (t sin x To determine a n (t, the generaized Fourier coefficients, substitute this formua for v into the PDE it satisfies, equation (1. t a n (t sin x = k x a n (t sin x + x e t Because v satisfies homogeneous boundary conditions and v, v/ x, and v/ t are continuous (reasonabe assumptions for the temperature profie in a homogeneous soid rod, the two series can in fact be differentiated term by term. da n dt sin x = k a n (t d x sin + x e t The operator appied to the eigenfunction is equa to the eigenvaue times the eigenfunction. da n dt sin x = k a n (tλ n sin x + x e t

4 Strauss PDEs e: Section Exercise Page 4 of 1 Bring both series to the eft side and combine them. [ ] dan dt ka n(tλ n sin x = x e t The eft side is essentiay a Fourier sine series expansion of the function on the right side. To sove for the term encosed in square brackets, mutipy both sides by sin mπx/, where m is an integer, [ ] dan dt ka n(tλ n sin x sin mπx and then integrate both sides with respect to x from to. ˆ [ ] dan dt ka n(tλ n sin x sin mπx ˆ = Bring the constants in front of the integras. [ ] ˆ dan dt ka n(tλ n sin x sin mπx = et = x e t sin mπx ˆ x e t sin mπx (x sin mπx Because the sine functions are orthogona, the integra on the eft side is zero if n m. As a resut, every term in the infinite series vanishes except for one: n = m. Evauate the integras. [ ] ˆ dan dt ka n(tλ n sin x = et ˆ [ ] dan dt ka n(tλ n ( = et Mutipy both sides by / and simpify both sides. da n dt kλ na n = et (x sin x With the hep of the method of eigenfunction expansion, the PDE for v has been reduced to a first-order inhomogeneous ODE, which can be soved with an integrating factor I. [ˆ t ] I = exp ( kλ n ds = e kλnt Mutipy both sides of the previous equation by I. da n dt e kλnt kλ n a n e kλnt = et e kλnt The eft side can be written as d/dt(ia n by the product rue. Combine the exponentias on the right side. d dt (e kλnt a n = e (kλn 1t (3

5 Strauss PDEs e: Section Exercise Page 5 of 1 Integrate both sides with respect to t. Evauate the remaining integra. e kλnt a n = The genera soution for a n is thus e kλnt a n = a n (t = ˆ t ( e (kλn 1s ds + C 7 (kλ n 1 e (kλn 1t + C 7 (kλ n 1 et + C 7 e kλnt. (4 In order to determine C 7, use the initia condition for v in combination with the eigenfunction expansion. v(x, = a n ( sin x = x To sove for a n (, mutipy both sides by sin(mπx/ a n ( sin x sin mπx = x sin mπx and then integrate both sides with respect to x from to. ˆ a n ( sin x Bring the constants in front of the integras. sin mπx = ˆ x sin mπx a n ( ˆ sin x sin mπx = 1 ˆ (x sin mπx The orthogonaity of the sine functions resuts in the disappearance of every term in the infinite series except the n = m term. Evauate the integras. Sove for a n (. a n ( ˆ sin x = 1 ˆ a n ( = 1 ( (x sin x a n ( = Now set t = in equation (4 and use this initia condition to find C 7. a n ( = (kλ n 1 + C 7 =

6 Strauss PDEs e: Section Exercise Page 6 of 1 Soving for it gives So then a n (t = (kλ n 1 et = (kλ n 1 kλ n C 7 = (kλ n 1. kλ n (kλ n 1 ekλnt ( e t kλ n e kλnt. Substitute λ n = γ n = (/ here. = ( [e t + k n π k n π 1 = [e t (kn π + + k n π exp ( k n π ] t exp ( k n π ] t Consequenty, v(x, t = [e t (kn π + + k n π exp ( k n π ] t sin x and, since u(x, t = v(x, t + r(x, t, Therefore, u(x, t = x u(x, t = x e t π e t + [e t (kn π + + k n π exp ( k n π ] t sin x. 1 [e t n(kn π + + k n π exp ( k n π ] t sin x, x.

7 Strauss PDEs e: Section Exercise Page 7 of 1 Figure 1: This figure shows u(x, t at various times for k = 1 and = 1. The curves in red, orange, yeow, green, bue, and purpe correspond to t =, t =.5, t = 1, t = 1.5, t =, and t =.5, respectivey. They are approximate, as ony the first terms in the infinite series have been used. Figure : This is a pot of the two-dimensiona soution surface u(x, t in three-dimensiona xyuspace for k = 1 and = 1. It is approximate, as ony the first terms in the infinite series have been used.

8 Strauss PDEs e: Section Exercise Page 8 of 1 Method - Mr. Strauss s Way The method of eigenfunction expansion wi be appied directy to sove for u as Mr. Strauss does in the textbook, that is, without making the boundary conditions homogeneous. The same eigenvaue probem is considered here as before. d φ = λφ, φ( = φ( = It was shown that the eigenvaues are λ n = (/ and that the eigenfunctions associated with them are φ n (x = sin x, n = 1,,.... The eigenfunctions of the Hemhotz equation are known to form a compete set, so a of the functions in the PDE can be expanded in terms of them. u(x, t = b n (tφ n (x uφ m = b n φ n φ m u t = c n (tφ n (x u t φ m = u x = d n (tφ n (x u x φ m = c n φ n φ m d n φ n φ m ˆ ˆ ˆ uφ n = b n ˆ u t φ n = c n ˆ u x φ n = d n φ n = b n ˆ φ n(x = c n φ n(x = d n It shoud be emphasized that these are generaized Fourier series expansions for u, u/ t, and u/ x, not product soutions that come about from using the method of separation of variabes. Sove the atter equations for the generaized Fourier coefficients. b n (t = c n (t = d n (t = ˆ ˆ ˆ uφ n u t φ n = d ( ˆ dt u x φ n = ( u x φ n }{{} = uφ n = db n dt ˆ u dφ n x = ˆ u x cos x Appy integration by parts once more in order to write d n in terms of b n and the given boundary conditions for u. d n (t = [ u(x, t cos x ˆ ( u sin x ] = [ u(, t cos u(, t + ˆ u sin x ] = [ e t + ( = ( e t + b n = (et b n ˆ u sin x ]

9 Strauss PDEs e: Section Exercise Page 9 of 1 Now that the coefficients are known, substitute the eigenfunction expansions into the PDE. u t = ku xx c n (tφ n (x = k d n (tφ n (x db n dt φ n(x = The summands must be equa as a resut. k (et b n φ n (x db n dt φ n(x = k (et b n φ n (x Divide both sides by φ n (x. db n dt = k (et b n Expand the right side and bring the term with b n to the eft side. db n dt + kn π b n = k e t With the hep of the method of eigenfunction expansion, the PDE has been reduced to a first-order inhomogeneous ODE that can be soved with an integrating factor I. (ˆ t kn π ( kn π I = exp ds = exp t Mutipy both sides by I. ( db n kn dt exp π t + kn π ( kn π b n exp t The eft side can be written as d/dt(i b n by the product rue. = k ( kn e t π exp t [ ( d kn π ] exp dt t b n = k [( kn π ] exp + 1 t Integrate both sides with respect to t. ( kn π ˆ t [( k kn π ] exp t b n = exp + 1 s ds + C 8 Evauate the integra. ( kn π exp t b n = k 1 kn π Divide both sides by I to sove for b n. b n (t = k kn π + et + C 8 exp [( kn + 1 exp π ] + 1 t + C 8 ( kn π t

10 Strauss PDEs e: Section Exercise Page 1 of 1 In order to determine C 8, use the initia condition for u in combination with the eigenfunction expansion. u(x, = b n (φ n (x = b n ( = Set t = in the previous equation and use the initia condition. So then Therefore, u(x, t = b n ( = k kn π + + C 8 = C 8 = k kn π + k b n (t = kn π + et k ( kn π + exp kn π t [ k = kn π + e t exp ( kn π ] t. [ k kn π + e t exp ( kn π ] t sin x, < x. Figure 3: This figure shows u(x, t at various times for k = 1 and = 1. The curves in red, orange, yeow, green, bue, and purpe correspond to t =, t =.5, t = 1, t = 1.5, t =, and t =.5, respectivey. They are approximate, as ony the first terms in the infinite series have been used. Comparing Figure 3 with Figure 1, we concude that appying the method of eigenfunction expansion directy as Mr. Strauss does in the textbook resuts in a series soution that not ony converges very sowy but aso does not satisfy the inhomogeneous boundary condition at x =. These issues stem from the fact that φ( and u(, t are not equa; thus, they can be avoided by making the boundary conditions homogeneous first. This wi be done next in Method 3.

11 Strauss PDEs e: Section Exercise Page 11 of 1 Method 3 - Without Using Term-by-Term Differentiation As in Method 1 the boundary conditions wi be made homogeneous by making the substitution v(x, t = u(x, t r(x, t, where r(x, t is some function that satisfies the boundary conditions. If we choose r(x, t = x e t, then the initia boundary vaue probem that v was shown to satisfy is v t = kv xx + x e t, < x <, t > v(, t = v(, t = v(x, = x. The same eigenvaue probem is considered here. d φ = λφ, φ( = φ( = It was shown that the eigenvaues are λ n = (/ and that the eigenfunctions associated with them are φ n (x = sin x, n = 1,,.... The eigenfunctions of the Hemhotz equation are known to form a compete set, so a of the functions in the PDE can be expanded in terms of them. ˆ ˆ v(x, t = A n (tφ n (x vφ m = A n φ n φ m vφ n = A n φ n = A n v t = B n (tφ n (x v ˆ ˆ t φ v m = B n φ n φ m t φ n = B n φ n = B n x e t = D n (tφ n (x x ˆ ˆ e t φ m = D n φ n φ m et (x φ n = D n φ n = D n v x = E n (tφ n (x v ˆ x φ ˆ v m = E n φ n φ m x φ n = E n φ n = E n It shoud be emphasized that these are generaized Fourier series expansions for the functions, not product soutions that come about from using the method of separation of variabes. Sove the atter equations for the generaized Fourier coefficients. A n (t = B n (t = D n (t = et E n (t = ˆ ˆ ˆ ˆ vφ n v t φ n = d dt ( ˆ (x φ n = et v x φ n = vφ n = da n dt ( ( v x φ n }{{} = = et ˆ v dφ n x = ˆ v x cos x

12 Strauss PDEs e: Section Exercise Page 1 of 1 Appy integration by parts once more in order to write E n in terms of A n and the boundary conditions for v. E n (t = [ v(x, t cos x ˆ ( v sin x ] = [ v(, t cos v(, t + ˆ v sin x ] = = = n π A n [ ( ( A n ˆ v sin x ] Now that the coefficients are known, substitute the eigenfunction expansions into the PDE. v t = kv xx + x e t B n (tφ n (x = k E n (tφ n (x + D n (tφ n (x da n dt φ n(x = k da n dt φ n(x = ( n π ( k n π A n φ n (x + A n et φ n (x The summands must be equa as a resut. da n dt φ n(x = ( k n π A n et φ n (x Divide both sides by φ n (x and bring the term with A n to the eft side. da n dt + k n π A n = et ( et φ n (x This is the same ODE that a n satisfies in equation (3 since λ n = (/, so A n has the same genera soution. Aso, the initia condition for v is the same, so the particuar soution is the same. A n (t = [e t (kn π + + k n π exp ( k n π ] t Therefore, v(x, t = and, because u(x, t = v(x, t + r(x, t, u(x, t = x e t π [e t (kn π + + k n π exp ( k n π ] t sin x 1 [e t n(kn π + + k n π exp ( k n π ] t sin x, x.