2 (x2 + y 2 ) 1 ) y = 2(x 2 + y 2 ) 1 (2x 2 + 2y 2 ) (x 2 + y 2 ) 2 = 0.

Transcription

1 u xx + u yy = (x 1 (x + y ) 1 ) x + (y 1 (x + y ) 1 ) y = (x + y ) 1 (x + y ) (x + y ) =. 4. The statement of this problem is somewhat unclear in whether they mean (u xx ) + (u yy ) = (the more likely one) or (u ) xx + (u ) yy =, so either interpretation would be considered correct. With the first interpretation it is obvious that all function in the stated form satisfy that u xx = u yy =. With the second there would need to be additional constraints on a, b, c, d for it to work. 5. The general solution is u = xf (t) + G(t), hence one can let u = t + x(1 t ). 6. u tt = (g(x + ct) + g(x ct)) t = c(g (x + ct) g (x ct)), u xx = c 1 (g(x + ct) g(x ct)) x = c 1 (g (x + ct) g (x ct)). 7. (e at sin bx) t = ae at sin bx, (e at sin bx) xx = b e at sin bx, hence a = kb. 8. (u x ) t = 1 3u x, hence u x = 1 3 +e 3t f(x) for some arbitrary function f, hence u(x, t) = x 3 +e 3t F (x)+ G(t) for arbitrary function F (which is the anti-derivative of f) and G. 1. To sketch wave profile, pick some k, A, D or c, sketch u(x, t) for different values of t, and if u is complex-valued you can sketch either the real or imaginary part. Dispersion relations: a) ω = idk. b) ω = ±ck. c) ω = k 3. d) ω = k. e) ω = ck. 14. Dispersion relation is ω = ( 1 + δk k 4 )i hence this is diffusive. When δ = k + 1/k the solution has growth rate. When k + 1/k > δ the solution decays From equation (1.7) in the text we have d b dt a uadx = Aφ a Aφ b. Differentiate with respect to b (or use some other argument, for example as in the textbook), we have Au t = A x φ Aφ x, hence u t +φ x = A φ/a. 3. By chain rule, u x = u ξ, u t = cu ξ + u τ, hence the equation (1.1) becomes u τ = λu, hence the general solution is u = e λτ F (ξ) = e λt F (x ct). 4. u t +cu x = λu. If w = ue λt, u = we λt hence u t +cu x = w t e λt λwe λt +cw x e λt, λu = λwe λt, hence w t + cw x =. 5. By method of characteristics u t + xtu x = has characteristics x = Ce t /, hence the general solution is u = F (xe t / ). Together with the initial value condition we know that F = f hence u = f(xe t / ). The general solution of u t + xu x = e t is u = e t + F (xe t ), so with the initial condition, the solution should be u = e t + f(xe t ) 1. 6(b). The characteristics are x = Ct, and the general solution is u = e t F (x/t). Use the initial condition we get F = e f, hence u = e (t 1) f(x/t). 7. The general solution is u = e λt F (x ct). The{ initial-boundary condition tells us that F (x) = for x > and e λt x > F ( ct) = g(t) for t >, hence F (x) = e λx/c g(x/c) x. 1

2 1. By the method of characteristics, u(x, t) = F (x ct)e (αt u)/β. Set t = we have F (x) = f(x)e f/β hence u(x, t) = f(x ct)e (αt u+f(x ct))/β. 14. Characteristics are x = Ce ut hence u = F (xe ut ). Together with the initial condition we get u = xe ut. A solution does not exist for all t. For example, there doesn t exist any u at point x = t = 1 because s < e s for all s R d dt l u dx = l uu tdx = l kuu xxdx = kuu x l l k(u x) dx, hence l u dx l u dx for t. 3. Let w = u g + (x/l)(h g), then w(, t) = w(l, t) =, u t = ku xx will imply w t = kw xx g + (x/l)(h g ). 4. The steady state satisfy = ku xx hu and u() = u(1) = 1, hence u = e(h/k)1/ (x 1/) +e (h/k)1/ (1/ x) e (h/k)1/ / +e (h/k)1/ /. 5. u t = w t e αx βt βwe αx βt = w t e αx βt βu, u x = w x e αx βt + αu, u xx = w xx e αx βt + αw x e αx βt + αw x e αx βt + α u, hence = u t Du xx + cu x + λu = (w t Dw xx )e αx βt + (c Dα)w x e αx βt + (λ β Dα + cα)u, so when α = c/(d) and β = λ Dα + cα = λ + c /(4D), = w t Dw xx. 6. The steady state doesn t depend on the initial condition. It is u = 1 k x(1 x). 1. The flux is Du x + u /. Replace u = ψ x we have ψ xt = Dψ xxx + ψ x ψ xx. Integrate along x we have ψ t = Dψ xx + (ψ x ) / + F (t). Replace ψ t with ψ t + t F (s)ds we can get rid of F. Now let ψ = D ln v we get Dv t /v = D (v xx v (v x ) )/v + D (v x ) /v, hence v t = Dv xx For u t = Du xx cu x, the time independent solution satisfies = Du xx cu x. So the solution is u = C 1 +C e cx/d. For u t = Du xx cu x +ru, the time independent case reduces to = Du xx cu x +ru, the characteristic polynomial is Dλ cλ+r = whose roots are r = c± c 4Dr D. Hence, when c = 4Dr the general solution is u = (C 1 + C x)e cx D, when c > 4Dr the general solution is u = C 1 e xc+x c 4Dr D + C e xc x c 4Dr D, when c < 4Dr the general solution is u = C 1 e xc D cos( x 4Dr c D ) + C e xc D sin( x 4Dr c D ). 9. u = ax + b then u xx =. u = a ln r + b then u xx + u yy = a( x r ) x + a( y r ) y = a( r x +r y r 4 ) =. a u = ρ+b then u xx + u yy + u zz = a(( x ρ ) 3 x + ( y ρ ) 3 y + ( z ρ ) 3 z ) =. 1. (a) d b dt a πrudr = πa( Du r a ) πb( Du r b ). Differentiate on b we get bu t b = Dbu rr b + Du r b, hence u t = Du rr + D r u r = D 1 r (ru r) r. b a 4πr udr = 4πa ( Du r a ) 4πb ( Du r b ). Differentiate on b then you get the differential equation. (b) d dt

3 You can do it however you want, for example, in the 3rd equation on page 51, add a term b a ρ gdx to the right. 3. Verification is by chain rule. Sketch u = 1 ( 1 1+(x t) (x+t) ). 4. The initial condition is u n (x, ) = sin nπx l, (u n ) t (x, ) =. The frequency is cn l, they decrease as l increases and as c (tension) increases. 5. d dt E = l (ρ u t u tt + τ u x u tx )dx = τ 1 (u tu xx + u x u tx )dx = τ u t u x l =. 9. I x + CV t + GV =, so I xx + CV xt + GV x =. Substitute V x = LI t + RI, we get that I satisfy the telegraph equation. The fact that V satisfy telegraph equation follows analogously. When R = G = the speed of wave is (LC) 1/ div(gradu) = div((u x, u y, u x )) = u xx + u yy + u zz. 7 Quiz 1: u t + (x + 1)u x = 1, u(x, ) = sin x. Characteristics are x = Ce t 1. Hence u = t + F ((x + 1)e t ), hence F (x) = sin(x 1) and u = t + sin((x + 1)e 1 1) This is divergence theorem. The heat generated in Ω equals the heat flowing out at the boundary. 4. Let φ = (φ 1, φ, φ 3 ), then div(wφ) = (wφ 1 ) x + (wφ ) y + (wφ 3 ) z = (w x φ 1 + w y φ + w z φ 3 ) + w((φ 1 ) x + (φ ) y +(φ 3 ) z ) = φ gradw+wdivφ. Let φ = gradu then Green s identity follows from this and the divergence theorem. 5. λ = Ω u udv Ω u dv = Ω gradu dv Ω u dv <, 6. Let w = u + v where v is at the boundary, then Ω gradw dv = Ω gradu dv + Ω gradv dv + Ω gradu gradvdv. By 4 and the assumption, the last term is, hence Ω gradw dv Ω gradu dv. 7. Use cρu t = divφ Maximum are at r =, θ = π/4, 5π/4, minimum are at r =, θ = 3π/4, 7π/4. 3

4 . u = (x + y )/4 a /4. 4. The solution is spherical symmetric because the function and the boundary conditions are both spherical symmetric, i.e. u = u(ρ). Hence u = 1 reduces to u ρρ + ρ u ρ = 1, hence (ρ u ρ ) ρ = ρ, hence ρ u ρ = 1 3 ρ3 + C 1, u = 1 3 ρ + C1 ρ, hence u = 1 6 ρ C1 ρ + C. Apply the boundary condition one gets u = 1 6 ρ + b3 3ρ 1 6 a b3 3a. 5. u = Aatan(x) + B, solve for constants A and B using the boundary condition. 6. u = A log r + B. u = 1 log log r. 8. Use chain rule. 9. curle = implies that such a potential exists. V = divgradv = dive = This is a parabolic equation. u = F (kx t)+(x+kt)g(kx t), or you can write it in other equivalent ways.. Let p = x + t, q = t, then u x = u p, u xx = 4u pp, u t = u p + u q, u xt = u pp + u pq, hence the equation becomes u p = 4u qp, hence u = F (x + t)e t/4 + G(t). 3. It is hyperbolic. Under the change of variable, by chain rule, u x = 4 x u τ, u xx = 16 x u ττ 4 x u τ, u xt = 4 x u τξ + 4 x u ττ, so = xu xx + 4u xt = 4 x u τ 16u τξ, hence u = e ξ/4 f(τ) + g(ξ). 4. Use chain rule and product rule. 5. Elliptic. Find the eigenvalues of matrix [ Parabolic. The general solution calculation is similar to 3 above. 7. a) Elliptic when xy > 1 and hyperbolic when xy < 1. b) Elliptic. ]. 4

5 Midterm 1 1. Solve the following initial or initial/boundary value problems: (1) u t = xu x, u(x, ) = x. Here u is a function of x and t. (5 points) () u t + u x = sin x, u(x, ) = for x, u(, t) = t for t. Here u is a function of x and t. (15 points) Answer: (1) General solution is u = F (xe t ), hence u = x e t. () General solution is u = cos x+f (x t), so u = cos x (x t)+1 when x t, and u = cos x+cos(x t) when x t.. (1) Find the general solution of u tt = u tx. (15 points) () Find the solution of the initial value problem: u tt = u tx, u(x, ) =, u t (x, ) = x. (1 points) Answer: (1) u t = f(x + t), so u = F (x + t) + G(x) where F and G are arbitrary functions. () F (x) + G(x) =, F (x) = x, so u = 1 (x + t) x. 3. Consider the 1 dimensional advection-diffusion equation: u t = u x + u xx. (1) Use change of coordinate of the form p = x Ct, q = t to reduce it to the 1 dimensional heat equation. (13 points) () Recall that the solution of initial value problem of 1-dimensional heat equation: v t = v xx when t >, v(x, ) = f(x) can be given by the Poisson integral representation: v(x, t) = f(y)g(x y, t)dy, where G(x, t) = 1 4πt e x 4t. Can you write down the analogous formula for the following initial value problem: u t = u x + u xx when t >, u(x, ) = f(x)? (1 points) (3) Consider the following problem with periodic boundary condition: u t = u x + u xx when < x < 1, u(, t) = u(1, t), u x (, t) = u x (1, t). Show that I(t) = 1 u (x, t)dx is a non-increasing function by calculating d dti. (7 points) Answer: (1) u t = Cu p + u q, u x = u p, u xx = u pp, hence when C = 1, u q = u pp. () u(x, t) = f(y)g(x + t y, t)dy. (3) d dt I = 1 uu tdx = 1 uu x + uu xx dx = u 1 + uu x 1 1 (u x) dx. 4. Consider the equation u xx + u yy = x + y on R \(, ). Find all radial symmetric solutions (In other words, all solutions of the form u(x, y) = g( x + y )). You may want to use the fact that the Laplace operator in polar coordinate (r, θ) is = u rr + 1 r u r + 1 r u θθ. (5 points) Answer: u rr + u r /r = r, so (ru r ) r = r 3, ru r = A r4, u r = A/r r3, and u(r) = B + A log r r4. 5

6 11.1. u = R φ(y)g(x y, t)dy R φ(y)g(x y, t) dy M G(x y)dy = M, where G is the heat kernel. R 3. u(x, t) = R φ(y)g(x x y, t)dy = u G(x y, t)dy = u G(s, t)ds = u ( G(s, t)ds + x G(s, t)ds). We know G(s, t)ds = 1/, x G(s, t)ds = x /sqrtt G(s, 1)dt which converges to as t, hence lim t u(x, t) = u / = 1/. 1. x+ct 3. The solution of the latter Cauchy problem is u(x, t) = 1 c φ(s)ds, and the solution of the first x ct Cauchy problem is the partial derivative of the solution of the latter Cauchy problem in t direction which by fundamental theorem of calculus is 1 (φ(x ct) + φ(x + ct)). 6

7 13 Quiz u tt = 4u xx + e x+t, u t (x, ) =, u(x, ) = sin x. Solution: By Dahamel s principle, u tt = 1 (sin(x t) + sin(x + t)) + t 1 x+t s 4 x t+s er+s drds = 1 (sin(x t)+sin(x+t))+ 1 t 4 ex+t s e x t+3s ds = 1 (sin(x t)+sin(x+t))+ 1 4 (ex+t e x+t ) 1 1 (ex+t e x t ). 7

8 u 1 u = ( 1 (f 1 (x ct)+f 1 (x+ct))+ 1 x+ct c x ct g1 (s)ds) ( 1 (f (x ct)+f (x+ct))+ 1 x+ct c x ct g (s)ds) 1 ((f 1 f )(x ct) + (f 1 f )(x + ct)) + 1 x+ct c x ct (g1 g )(s)ds = δ 1 + δ T. It shows that this Cauchy problem is stable and well posed Do odd extension of the initial condition, one gets u(x, t) = 1 4kπt e (x y) 4kt e (x+y) 4kt dy By Duhamel s principle, u(x, t) = t 1 x+c(t τ) c x c(t τ) sin sdsdτ = 1 t c cos(x+c(t τ)) cos(x c(t τ))dτ = 1 c (sin(x + ct) + sin(x ct) sin(x)) L( t f(τ)dτ) = e st ( t f(τ)dτ)dt = ( τ e st dt)f(τ)dτ = 1 s e sτ f(τ)dτ = L(f) s. 8. Let v = Lu in the t direction, we have sv u = v xx, v x (, s) = v(, s), because we want bounded solution, v = u s(1+ s) e sx + u s, hence u = u L 1 1 ( s(1+ s) e sx ) + u. Remark: for those who know complex analysis, we can evaluate L 1 1 ( s(1+ s) e sx ) using the inverse formula on pp. 17 of the textbook. The answer is 1 πi e st 1 ( s(1+i s) e six 1 s(1 i s) e six )ds = 1 π sin( six)+ s cos( six) s(1+s)e ds. st Fu = e x +iξx dx = 1+ξ. So, F 1 ( 1 (1+ξ ) ) = 1 4 u u = 1 4 e y x y dy = 1 4 ( x e x + e x ). 15. Do Fourier transform in the x direction, let v = Fu, we have v t = Ds v + ( cisv, so v(s, t) = (Fφ)(s)e ( Ds +cis)t, and u(x, t) = F 1 ((Fφ)(s)e ( Ds +cis)t ) = φ F 1 (e ( Ds +cis)t 1 ) = φ 4πDt e (x ct) /(4Dt) ). 8

9 19 Solution of other exercise problems..6 u(x, t) = 1 (e x ct + e x+ct ) + 1 c (sin(x + ct) sin(x ct)) I don t see a.3.4 in my textbook?.5.3. u(x, t) = t w(x, t τ; τ)dτ = t f(x ct + cτ, τ)dτ Use.5.3, u(x, t) = t (x t+τ)e τ dτ = (x t)(1 e t ) te t + e t = x t+ xe t e t Do Laplace transform in t direction, let v = L(u), then s v = c v xx gs 1, with boundary condition v(x, s) as x, v(, s) =. So v(x, s) = g s 3 (e sx/c 1), u = g (x /c tx/c) when x < tc and g t when x > tc. Note that when you use the table on page 114, all functions are for t < or s < Do Laplace transform in t direction, v = Lu, sv = v yy, so v(x, y, s) = C(x, s)e sy. Use the other boundary conditions one gets C(, s) = 1/s, sc(x, s) + C x (x, s) =, so C(x, s) = e xs /s. The result can now be obtained from Table Use the hint, write v as in Example.18, then integrate the differential form vdy. You can also use Fourier transform directly to get a solution but without the arbitrary constant C (a) ω = k 3. (b) Let v = F u, suppose the initial condition is u(x, ) = f(x), then v t +is 3 v =, so v(s, t) = F (f)e is3t, u = f K(x, t) where K(x, t) = 1 t isx π ds = 1 R e is3 π t 1/3 Ai( x ). t 1/3 9

10 Midterm 1. (1) Find the inverse Fourier transform of cos(x)e x. (6 points) () Find the Laplace transform of cos(x)e x. (6 points) (3) Find the convolution ( between e x and e x. (6 points) Solution: (1) 1 1 π (eix + e ix )e x e isx dx + ) 1 (eix + e ix )e x e isx dx () cos xe x xs dx = 1+s (1+s) +1. (3) ex y e y dy = e x+1/4 π.. Consider the initial-boundary value problem on the region x >, t >. (1) Reduce it to a problem of the form by adding a function to u.(1 points) u t = u xx, u(x, ) = sin x, u x (, t) = 1 v t = v xx + f(x, t), v(x, ) = g(x), v x (, t) = = 1 π () Find the solution of the original initial-boundary value problem about u. ( points) Solution: (1) Let v = u sin x, then v t = v xx sin x, v(x, ) =, v x (, t) =. ( 1+(s+1) + 1+(s 1) ). () u(x, t) = sin x t sin y(g(x + y, t τ) + G(x y, t τ))dydτ, where G(x, t) = 1 4π e x /(4t). 3. Consider the following problem: on the region < x < 1, t >. u tt = u xx 4u, u(, t) = u(1, t) =, u(x, ) = f(x) (1) For any integer n, find a solution of u tt = u xx 4u, u(, t) = u(1, t) = of the form u = φ(t) sin(nπx). (1 points) () Find the solution of the original problem for f(x) = sin(πx) sin(3πx). (1 points) Solution: (1) u = (A cos( n π + 4t) + B cos( n π + 4t)) sin(nπx). () u = cos( π + 4t) sin(πx) cos( 9π + 4t) sin(3πx). 4. Find the bounded solution of the following problem: u tt = u xx, u t (x, ) = u(x, ) =, u x (, t) = u(, t) + sin t on the region x >, t >. You may want to use the Laplace transform or the general solution of 1-d wave equations. (16 points) Solution 1: u = F (t x) + G(t + x), u t (x, ) = u(x, ) = implies that we can set F = on (, ] and G =. The boundary condition is saying that F = F sin t so F (s) = s sin rer s dr, 1

11 u = { t x sin re r t+x dr t > x t x. Solution : Laplace transform in t direction, v = Lu, then s v = v xx, v(x, s) = C(s)e xs, sc = C + L(sin t), so v = L(sin t) 1 1+s e xs, u = (sin t e t+x )H(t x). 5. Find the bounded solution of u xx + u yy = u, u(x, ) = f(x) on the region y >. You may want to use the Fourier or Laplace transform. (14 points) Solution 1: Do Fourier transform in the x direction, let v = F (u), then s v + v yy = v, u(x, y) = f F 1 (e s +1y ). Solution : Do Laplace transform in the y direction, let v = L(u), then v xx + s v sf g = v, where g(x) = u y (x, ). Because v should decay as s, v = F 1 1 ( s ξ 1 (sf (f) + F (g))). Furthermore, boundedness implies that there shouldn t be a pole when s ξ 1 =, which is only possible when F (g) = ξ + 1F (f), hence u = L 1 (f F 1 1 ( )). s+ ξ +1 11

12 Quiz 3 { sin(πx) < x < 1 Find the Laplace transform of f(x) =. x > 1 Solution: Lf = 1 sin(πx)e sx dx = i 1 Quiz 4 u tt + u xx = u t, u x (, t) = u(1, t) =, u(x, ) = f(x). 1 (e sx+iπx + e sx iπx )dx = i ( e s+iπ 1 s iπ Solution: Eigenfunctions are cos(π(n+1/)x), so the result is 1/)x)e (1/ π (1/+n) +1/4)t. n= ( 1 + e s iπ 1 s+iπ ). f(s) cos(π(n+1/)s)ds) cos(π(n+ 1

13 a) b) d) are straightforward. c) a n = π π f(x) sin xdx. e) u(x, t) = n π ncπ f(x) sin xdx sin nct sin nt. 4. u(x, t) = n ( 1 f(s) sin(nπs)ds)e t (cos(t n π /4 1)+(n π /4 1) 1/ sin(t n π /4 1)) sin(nπx). 5. (a) u =. (b) u(x, t) = n l ( l f(s) sin(nπs/l)ds)e kn π /l +ht sin(nπx/l) When λ <, y(x) = C 1 e x λ +C e x λ, so the boundary condition is C 1 +C + λc 1 λc =, 3(e λ C 1 + e λ C ) + ( λe λ C 1 λe λ C ) =, so there is a non-zero solution iff λ is the solution of e 4t 8t = t 4t. By taking derivatives and intermediate value theorem there is a unique such t in (, 1/). is not an eigenvalue. There are infinite positive eigenvalues by Sturm Liouville theory. 8. The boundary condition is not symmetric e.g. y 1 (a) = y 1 (b) = y (a) = y (b) = y (a) = y (b) = 1, y 1(a) = y 1(b) =. It is not self adjoint either. For u, v satisfying the boundary conditions, b a u v = v u vu b a + b a uv, and v u vu b a can not be guaranteed to be If u = X(x)T (t), the first pde can be separated into (x X ) /(x X) = T /T = λ. If one let u = X(x)Y (y) in the second PDE, it becomes X Y + (x + xy + y )XY =, and is not separable. 4. If y is an eigenfunction with eigenvalue λ, λ = π 1 y(x y ) dx π = π 1 x y dx 1 y dx π 1 y dx. The differential equation is the Cauchy-Euler equation, hence eigenfunctions are x 1/ sin( 4λ n 1 log x), and λ n = n π (logπ) In θ direction the eigenfunctions are Θ n (θ) = sin((n + 1)θ), with corresponding eigenvalues (n + 1). So u(r, θ) = 4 ( ) π/ π n= f(s) sin((n + 1)s)ds (r/r) n+1 sin((n + 1)θ). 11. There is a typo and the h should be the same as g. This is the divergence theorem. 13

14 7 Final exam 1. Find the general solution of the following PDE: (1) u x = e x u t. (15 points) () u tt = 6u xx + 1. (1 points) Answer: (1) u = F (t + e x ). () u = t + F (x + 6t) + G(x 6t).. Solve the Laplace equation u = u xx + u yy = on the unit disc D with Dirichlet boundary condition u D (x, y) = x 3, here D is the boundary of the unit disc, which is the unit circle {(x, y) : x + y = 1}. (15 points) Answer: u(1, θ) = cos 3 θ = cos θ(cos θ+1)/ = cos 3θ/4+3 cos θ/4, so u(r, θ) = r 3 cos 3θ/4+3r cos θ/4 = x 3 /4 + 3x(1 y )/4. 3. Solve the following initial value problem: u t = u xx + u x + g(x), u(x, ) = f(x), on the region t >. (15 points) Answer: u = 1 4πt R (x+t s) f(s)e 4t ds + t 1 4πτ R (x+τ s) g(s)e 4τ dsdτ. 4. Find bounded solution for the following initial-boundary value problem: u tt + u xx u x u =, u(, t) = u(1, t) =, u(x, ) = f(x), on the region < x < 1, t >. (15 points) Answer: u = n=1 1 f(s)e s/ sin(nπs)dse x/ sin(nπx)e n π +3/4t. 5. Find the bounded solution for the following initial-boundary value problem: u t = u xx, u(, t) = u(π, t) =, u t (x, ) = sin(3x), on the region < x < π, t >. (15 points) Answer: u(x, t) = 1 9 sin(3x)e 9t. 6. Find the solution for the following initial-boundary value problem: u tt = u xx, u(x, ) { = u t (x, ) =, u x (, t) = f(t), on the region x >, t >. (15 points) x t f( s/ )ds x < t Answer: u = x t. 14