Hard Problems from Advanced Partial Differential Equations (18.306)

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1 Har Problems from Avance Partal Dfferental Equatons (18.306) Kenny Kamrn June 27, We are gven the PDE 2 Ψ = Ψ xx + Ψ yy = 0. We must fn solutons of the form Ψ = x γ f (ξ), where ξ x/y. We also mpose the conton that Ψ x Ψ y = 0 along the lne x = y. What are the possble values for γ an what are the corresponng functons f (ξ) that solve ths PDE? Soluton: Pluggng nto the PDE, we quckly get that f must fulfll the followng ODE: ξ 2 (ξ 2 + 1)f + 2ξ(ξ 2 + γ)f + γ(γ 1)f = 0 (1) We frst note that ths sn t a stanar Sturm-Luvlle problem snce our unetermne constant γ appears n the f term. We can convert a secon orer ODE to a Sturm- Louvlle problem only when the unetermne constant,.e. the egenvalue, occurs only n the coeffcent of the f term. So we must use our ngenuty to tackle ths problem from a fferent perspectve. We notce that ξ s the cotangent of the angle mae by the pont (x, y) wth the horzontal. Ths prompts us to try a change of varables so as to take avantage of the angular corresponence to ξ: Smlarly, we also get ξ = cot θ = = sn 2 θ ξ (cot θ) θ 2 2 = sn 2 θ sn 2θ + sn 4 θ ξ 2 θ θ 2 Let s go ahea an efne f (ξ) = g(cot θ). We wll nstea solve for g an from there we can get f. Applyng our new varable an our new functon, we have 0 = ξ 2 (ξ 2 + 1)f + 2ξ(ξ 2 + γ)f + γ(γ 1)f = cot 2 θ csc 2 θ(sn 2 θ sn 2θ g + sn 4 θ g ) + 2 cot θ(cot 2 θ + γ)( sn 2 θ) g + γ(γ 1) g 1

2 2 = cos θ g 2γ sn θ cos θ g + γ(γ 1) g = 0. (2) Observng the coeffcents of Equaton 2, we are remne of the entty whch helps us notce the followng: 2 sn θ cos θ = (cos 2 θ) (3) θ (cos 2 γ θ) 2 θ 2 (cos γ θ g) = cos 2 γ [cos γ θ g 2γ cos γ 1 θ sn θ g γ( (γ 1) cos γ 2 θ sn 2 θ + cos γ θ) g] = cos 2 θ g 2γ cos θ sn θ g + [γ(γ 1) sn 2 θ γ cos 2 θ] g = cos 2 θ g 2γ cos θ sn θ g + γ(γ 1) g γ 2 cos 2 θ g. Ths looks an awful lot lke the left-han se of Equaton 2, so let s use ths fact to rewrte the ODE: ) 2 2 (cos 2 γ θ θ 2 (cos γ θ g) + γ cos 2 θ g = cos 2 θ g 2γ sn θ cos θ g + γ(γ 1) g = 0. Ths yels 2 2 θ 2 (cosγ θ g) = γ (cos γ θ g). (4) Now, we rejoce, because lettng φ = cos γ θ g, prouces a secon-orer lnear ODE wth constant coeffcents: φ = γ 2 φ = s the general soluton for g. φ(θ) = A cos(γθ) + B sn(γθ) = g(θ) = (cos γ θ)[a cos(γθ) + B sn(γθ)] Now, let us conser the gven contons. Ψ 0 as y for constant x means that Ψ(x, y ) = x γ f(ξ) ξ 0 = x γ g(arccot(ξ)) ξ 0 = 0. Thus we can euce that g(arccot(0)) = g(π/2) = 0 whenever x = 0. Lookng back to our general soluton, we see that when θ = π/2, the prefactor (cos γ θ) becomes 0 γ whch wll go to nfnty an volate our conton unless γ < 0. Thus we requre γ < 0. But uner ths requrement, g(π/2) automatcally equals 0, meanng that γ < 0 s the only constrant 2

3 neee n the x = 0 case. Now, we must check f we nee to mpose anymore restrctons on γ so as to ensure that our conton hols when x = 0. In ths case, we nee the leang term x γ to cancel n orer to prevent Ψ from gong to along the y-axs. Notce that cos γ θ = cos γ (arccot(x/y)) = ( x 2 + y 2 /x) γ. Thus, we re-wrte Ψ(x, y) = x 2 + y 2 [A cos(γ arccot(x/y)) + B sn(γ arccot(x/y)], an see that when x = 0, Ψ s a constant tmes y γ. Ψ goes to zero when y as long as γ s negatve, thus we nee not a anymore restrctons to γ. So, we have that the only constrant s γ < 0. For the other conton, we start by wrtng t usng Ψ = x γ g(θ). Ψ x = γx γ 1 g + x γ θ x = x γ 1 (γg sn θ cos θg ) Ψ y = x γ 1 g θ y = x γ 1 g cos 2 θ Ψ x Ψ y = x γ 1 (γg (sn θ cos θ + cos 2 θ)g ). Our conton says ths nees to be zero whenever x = y, or, equvalently, θ = π/4, 5π/4. Thus n the θ = π/4 case, we have Pluggng n our g: γg(π/4) g (π/4) = 0 = γg(π/4) = g (π/4). g (π/4) = (γ cos γ 1 θ sn θ(a cos γθ+b sn γθ)+γ cos γ θ( A sn γθ+b cos γθ)) θ=π/4 π π π π π = (γ cos γ )[A cos γ + B sn γ 4 A sn γ + B cos γ ] The conton γg(π/4) = g (π/4) mples: A cos(γπ/4) + B sn(γπ/4) = A cos(γπ/4) + B sn(γπ/4) A sn(γπ/4) + B cos(γπ/4) = B cos(γπ/4) = A sn(γπ/4). Ths yels B = A tan(γπ/4) when γ s not even, an A = 0 wth B arbtrary when γ s even. In the case of θ = 5π/4, we easly reach the analagous result B cos(5γπ/4) = A sn(5γπ/4) 3

4 nto whch we plug n our recently etermne expressons for A an B, yelng tan(γπ/4) = tan(5γπ/4). Ths combne wth the π-perocty of the tangent functon means that γπ/4 = 5γπ/4 + πn for some nteger n. Therefore, we euce the atonal constrant that γ must be an nteger. So, altogether, we have: For γ Z an γ < 0, Ψ = { A x γ cos γ θ [cos(γθ) + tan(γπ/4) sn(γθ)] f γ s not even, B x γ cos γ θ [sn(γθ)] f γ s even. where θ = arccot(ξ) as prevously efne, an the constants A an B are arbtrary. 2. The nonlnear KV equaton u t 6uu x + u xxx = 0 can be shown to have a smlarty soluton of the form u = (3t) 2/3 g(η) for η = x(3t) 1/3. Applyng ths soluton reuces the PDE to the ODE g (η) + (6g(η) η)g (η) 2g(η) = 0. Show that ths ODE reuces to the secon-orer ODE V ηv 2V 3 = 0 upon a proper choce of the constant µ n the varable change g(η) = µ(v (η)/η) V (η) 2 (an gven that V ecays exponentally for large η). How oes the soluton relate to Ary functons for large η? Soluton: We frst plug the gven varable change rectly nto the ODE for g, yelng the followng ODE for V : µv 2V V + 6µ 2 V V 6V V 6µV 2 V µηv 12µV (V ) V 3 V + 2ηV V 2µV + 2V 2 = 0. We try µ = 1. A keen observaton allows us to notce that wth ths selecton for µ, the ODE can be wrtten as 2 η (V ηv 2V 3 ) 2V (V ηv 2V 3 ) = 0. 2 η Lettng F (η) = η (V ηv 2V 3 ), we can rewrte the ODE as whch quckly reuces to F = 2V F η (log F ) = 2V η 4

5 leavng us wth the general soluton F (η) = Ae 2 V (η)η for some constant A. We are gven that V s exponentally ecayng for large η, so n ths regme, the antervatve V η s also an exponentally ecayng functon. An thus as η gets large, the rght se approaches Ae 2 0 = A. So as η, F (η) = (V ηv 2V 3 ) A η = V ηv 2V 3 Aη + B (5) for some constant B. As η, all three terms on the left se of (5) approach 0 snce V s exponentally ecayng. Therefore, A an B must both be 0, an our ODE has succesfully reuce to the secon-orer ODE V ηv 2V 3 = 0 (6) Ths marks the completon of the frst part of ths problem. V ecayng exponentally means that for large η, the V 3 term n Equaton 6 s exponentally smaller than the other two terms. Lkewse we wll neglect the V 3 term. We are left to analyze the ODE V ηv = 0 (7) for η 1. We note that f we coul somehow convert every (/η) nto η an vce versa, then the ODE woul be frst-orer an easly solvable. Recallng that Fourer Transforms have ths feature, we express V n terms of ts transform: Now, we can wrte an V (η) = (1/2π) Vˆ (k) e kη k. V (η) = (1/2π) k 2 Vˆ (k) e kη k kη k= e ˆ e kη ηv (η) = η Vˆ (k) e kη k = Vˆ (k) η e kη k = Vˆ (k) V (k) k. k= As long as the ntegraton contour s chosen to ensure V ˆ (k)e kη + = 0, Equaton 7 can be wrtten n terms of Vˆ as: ( ) ˆ (1/2π) k 2 Vˆ (k) + V (k) e kη k = 0 5 k k

6 By unqueness of the Fourer Transform, ths means Ths frst orer ODE has the soluton ˆV (k) k 2 ˆV (k) = 0. (8) ˆV (k) = Ce k3 /3. (9) Now that we have a canate for Vˆ (k), we can only accept t once we verfy that there s k + an approprate contour whose enponts are at ± such that f(k) = Vˆ (k)e kη = k 0. The functon f(k) = Ce (kη+k3 /3) s analytc k C, so we can eform the ntegraton contour however we wsh wthn C as long as the enponts reman at ±. Let k = re θ. The exponent n f(k) s omnate by ts cubc term for k large, thus for the sake of etectng ecay behavor, we may neglect the kη term. The real part of ths reuce exponent s Re(k 3 /3) = Re(r 3 e 3θ /3) = (r 3 /3) sn(3θ). Decay behavor requres ths to be negatve n the regon through whch the ens of the contour pass. Consequently, the ens must approach ± through regons fulfllng 2πn < 3θ < π + 2πn. Ths 0 < θ < π/3 goo regon can be more smply wrtten as 2π/3 < θ < π 4π/3 < θ < 5π/3. So, our goal s met for any contour that approaches + through the 0 < θ < π/3 regon, through the 2π/3 < θ < π regon, an oes not attan a large magntue n a regon outse the three lste above. For example, the contour k(t) = t + 1+t 2 woul suffce. Therefore, we can fnally say that the Vˆ we receve s nee legtmate. Transformng out, our soluton for V s V (η) = (1/2π) Ce (kη+k3 /3) k (10) = (1/π) C cos(kη + k 3 /3) k (11) 0 = C A(η) (12) Ths soluton, though only one of two nepenent solutons, s the one we esre snce t ecays exponentally as η +. 6