Partial differential equations (ACM30220)

Transcription

1 (ACM3. A pot on a stove has a handle of length that can be modelled as a rod with diffusion constant D. The equation for the temperature in the rod is u t Du xx < x <. ( Solve Eq. ( subject to the following mixed inhomogeneous boundary conditions: and subject to the initial condition u(x t > u u x (x t > u(xt. Using only the first (n term in the resulting Fourier series solution find an estimate for the time at which the temperature of the end point x is 99% of its final (t value. eave the answer in terms of D and. Hints: Break up the solution: u(xt u hom (xt + u PI (x. Try separation of variables and a sine-and-cosine series for the homogeneous part. Be very careful with the boundary conditions! You will need to find an orthogonality relation for sine functions. Use sin θ sin ϕ [cos(θ ϕ cos(θ + ϕ]. We need to break up the solution into two parts: u u hom (xt + u PI (x where t u hom Du hom

2 with boundary conditions u hom (x x u hom (x. and with boundary conditions t u PI Du PI u PI ( u (du PI /dx x. Homogeneous part: inearity constant coefficients: attempt separation of variables: u(xt X(xT(t. Substitute into the PDE: X(x dt dt X DT(td dx. Divide out by X(xT(t: dt T dt D d X X dx. But now the HS is a function of t alone and the RHS is a function of x alone. Hence HS RHS Const. : λd. Also substitute the trial solution into the BCs and the ICs: Initial condition: u(xt X(xT( f(x < x < Boundary condition: T(tX( T(tX (. Focussing on the X(x-equations we have: X Equation in the bulk < x < : λ < x < X X ( X (. X + λx ( We will focus on λ > and justify this choice at the end. We have X A sin( λx + B cos( λx with BCs X( B. Hence X A sin λx only. At x X( A λ cos λ.

3 For A we require cos λ hence ( λ n + π n. Thus λ λ n π ( n + n. Now substitute λ n [n + (/]π / back into the T(t-equation: Solving gives or dt T dt λd λ nd. T(t T(e λndt T(t T(e n π Dt/. Now recall the ansatz: u(xt X(xT(t. Thus we have a solution ( (n X(xT(t T(A n sin + e (n+/ π Dt/. Calling T(B n : C n this is ( (n X n (xt n (t C n sin + e (n+/ π Dt/ n. The label n is just a label on the solution. However each n produces a different solution linearly independent of all the others (NOTE THE STARTING-VAUE OF THE INDEX. We can add all of these solutions together to obtain a general solution of the PDE: u hom (xt X n (xt n (t n n ( (n C n sin + e (n+/ π Dt/. Particular integral We solve d u PI dx with u PI ( u and (du PI /dx. The solution is u PI (x u. 3

4 Initial conditions We are almost there. However we still need to take care of the initial condition or u(xt u hom (xt + u PI (x ( (n u(xt C n sin + + u n ( (n C n sin + + u n Next note that the functions are orthogonal on []: I nm π π n ( (n C n sin + u : g(x. { sin ( (n + π Thus consider the IC again: ( (n sin + } n sin ( (m + dx sin((n + /y sin((m + /ydy cos[(n my]dy π π (/δ mn. u hom (xt n g(x. Multiply both sides by cos(m/ and integrate: π ( (n f(x sin + dx π n C n δ mn n C m (/. π cos[(n + m + y]dy ( (n C n sin + C m sin ( (n + sin ( (m + dx Hence C n ( (n g(x sin + dx. 4

5 In other words In conclusion u(xt u 4u π C n u u π u π u π 4u π n π sin ( (n + sin ( (n + dx d sin (( n + y dy cos (( n + y n + π n + πdt/ e (n+/ n + ( sin ( (n +. Note: We have fixed λ >. The other choices do not satisfy the boundary conditions. For example λ gives X hom Ax + B. However this is the particular integral not the homogeneous solution! Also λ < gives X hom A sinh(µx + B cosh µx µ λ. The boundary condition X hom gives B. Thus we are left with and X hom A sinh(µx X hom Aµ cosh(µx. However cosh( is never zero so X hom ( can not be satisfied for non-zero A-values. Finally an approximation of the solution at π including only the first-order mode of oscillation is u(x t u 4u Dt/4 e π. π 5

6 Set u(x t c.99u : 4u π e π Dt c/4.99u u u π e π Dt c/4 4 π e π Dt c/4.u. e π Dt c/4.π 4( π Dt c.π log 4 4 ( π Dt c 4 log 4.π [ t c 4 D π log ( 4.π t c.9643 ( /D. ]. Diffusion-like problems again: Solve the problem with initial and boundary conditions u t u x α u < x < π t > u(t u u(πt u u(xt f(x where α u and u are real constants and [ ] } f(x sin( {u 3x + u u cosh(πα cosh(αx + sinh(αx. sinh(πα et : xx α. We are to solve u t u. Because the BCs are non-zero we propose the solution u(xt u Hom (xt + u E (x where u Hom (xt solves the equation with zero boundary conditions: t u Hom u Hom u Hom (x u Hom (x π 6

7 and where u E (x is an equilibrium solution that satisfies the equation with the finite boundary conditions: t u E u E u E (x u u E (x π u. et s solve the equilibrium part first. We have t u E hence we must solve xx u E (x α u E (x with solution u E (x A cosh(αx + B sinh(αx. Choose A and B to satisfy the BCs: Hence A u B u u cosh(πα α. sinh(πα [ ] u u cosh(πα u E (x u cosh(αx + sinh(αx. sinh(πα Now consider the homogeneous problem: t u Hom xx u Hom α u Hom. We attempt separation of variables by writing Substituting into the PDE gives or u Hom (xt e λt φ(x. λφ(x φ (x α φ(x φ (x + ( λ α φ(x. We look for a periodic solution (sines and cosines. We therefore take that is α < λ. The solution is thus k λ α > φ(x A sin kx + B cos kx. The boundary conditions are φ( hence B. Also φ(π hence sin(πk. Thus k is an integer or a half-integer k n/ and sin(πk sin(πn. This fixes the eigenvalue λ: k 4 n ( λ α n. λ λ n 4 n + α n. 7

8 Thus But u Hom (xt u Hom (xt n n C n sin ( n x C n e (n /4+α t sin ( n x. u(xt u E (x { [ ] } u u cosh(πα f(x u cosh(αx + sinh(αx sinh(πα sin ( 3 x. Hence C 3 and C n otherwise. Thus Putting it all together u hom (xt e (9/4+α t sin ( 3 x. u(xt u Hom (xt + u E (x [ ] u u cosh(πα u E (x u cosh(αx + sinh(αx sinh(πα u hom (xt e (9/4+α t sin ( 3 x. 8

9 3. The wave equation and the hyperbolic property: et u 3 c t ij A ij u ( x i x j be a hyperbolic problem defined on the domain (xt R 3 ( with initial data u(xt f(x u t (xt g(x. Here x (xyz : (x x x 3 and A ij is a constant matrix. By suitable coordinate transformations show that Eq. ( can be rewritten as u c t 3 i u λ i yi where λ i > are the eigenvalues of A ij. Show that by a further linear transformation this problem can be reduced to the familiar wave equation. Start with a real symmetric matrix A R n n. By the spectral theorem the eigenvalues are real and the eigenvectors are orthonormal and span R n : Ax i λ i x i x i x i x j δ ij. α i x i α i R x R n Consider the matrix R whose columns are the eigenvectors of A: R x x n. Consider now the matrix product x R T AR. A x n x. x n λ λ n x x n λ x λ n x n 9

10 Note also that R is orthogonal since ( R T R ij δ ij. One can also show that RR T I too such that R T R RR T I (a matrix whose transpose is also its inverse is called orthogonal. Now here comes the trick: introduce a linear change of variables such that R ji y i x j Note the order here R ji!!. Also to save typing/writing/chalk we are going to use the Einstein summation convention unless indication to the contrary is given a sum over repeated indices is implied. Now R ki y i R ki y i R ki y i R ki R ji x j R ji (R T ik x j δ jk x j R ki y i x k. Hence R ki y i x k R lj y j x l. Now consider A kl x k x l R ki A kl R lj y i y j (R T i ka kl R lj y i y j ( R T AR ij ij i y i y j λ i δ ij y i y j λ i. yi Hence Equation ( can be re-written in the y-variables as u c t i λ i u y i

11 Finally introduce X i y i / λ i. Note that this is a sensible (linear transformation to make as λ i is real. Thus Equation ( in the X-coordinates can be written as u c t i u X i i.e. where c u t Xu + + X Xn is the aplacian in n dimensions. Thus the familiar wave equation is recovered. 4. Consider Burgers equation u t + uu x < x < subject to the initial data u(xt f(x. Suppose that f ( < on some finite interval I R. Prove the following statements: The inversion procedure [x x + f(x t x η(xt] fails; Characteristic lines cross in finite time; The slope u x tends to infinity in finite time. This is bookwork. The characteristic solution is given by u Const. f(x along dx dt f(x subject tox( x. The inversion fails in finite time: We need to invert the characteristic curve x x + f(x t to write x η(xt. Consider two curves: y (x x; y (x x + f(x t in x y space. A sufficient condition for a solution to the problem y y to exist is if dy dx > since then the curve y (x is increasing and it crosses the line y x precisely once (Case Fig.. However if f(x is decreasing on some interval I then the curves

12 (a Case (b Case Figure : Question 4 y and y can intersect in more than one place which means that there is NOT a one-to-one relationship between x and x (Case Fig.. This suggests that the method of characteristics will fail in general if The failure happens at the critical time min x dy dx min x ( + f (x t <. t c min x f (x. 5. Consider Burger s equation u t + uu x < x < subject to the initial data u(xt e x. Using the results of Q. ( compute the minimum breaking time. Obtain the characteristic solution and sketch the space-time diagram. Hint: Try to draw as accurate a sketch as possible. Use the data in Tab. if necessary. et u(x t f(x. We have the characteristic solution u Const. f(x along In other words dx dt f(x x( x. u e x along dx dt e x.

13 t Solving for trajectories gives and or 3 3 x Write down some of these curves: We plot these curves in Fig.. Figure : Question 5 x x + te x t (x x e x t mx + c m e x c x e x. t 4.487x t.783x t.6487x t x t.6487x.844 t.783x.783 t 4.487x To find the minimum breaking time we consider the half-line where f( is decreasing (x >. On this interval f (x e x. The minimum breaking time from Question is t cmin min x > f (x min ( +e +x. x > 3

14 x e x x e x Table : 6. By interchanging the order of integration in I y x e xy sin(ydxdy show that Hint: sin(y y π. d ds tan (s + s. [6 marks] We have But I I [ dy sin(y dy sin(y dy sin(y. y ] dxe xy [ ] x y e xy x [ ] dx e xy sin(ydy (. 4

15 Now consider J e as sin(sds cos(se as a a e as cos(sds [ a sin(se as e as cos(sds a e as sin(sds a J J + a. Substitution into Eq. (* gives I dx + x d dx tan (xdx tan ( tan ( π. ] ( ae as sin(sds 5

16 7. Using Question 6 prove that the family of functions converges to the Dirac delta function. Hints: δ (x / e ikx dk (3 π / As a first step integrate Eq. (3 and re-express δ (x in terms of a sine function. You may assume that the integral exists and is bounded for n. sin(ss n ds [4 marks] We have δ (x π π π e ikx k/ ix k / e ix/ e ix/ ix e ix/ e ix/ i sin(x/ sin x/ x/ π sinc(x/. Verify Property : sin(x/ sinc(x/dx dx π x/ π sin(s ds s sin(s ds π s sin(s ds π s 6

17 where the last line follows by Question 6. Verify property 3 (not a part of the tutorial: et f(x have a Taylor series at x a. Then δ (x af(xdx δ (yf(y + ady dyδ (y p f(adyδ (y + f(a + I( f (p (a y p p! p f (p (a p! δ (yy p dy where I( π p p p4 p4 f (p (a p! f (p (a p! p4 π f (p (a p! f (p (a p! f (p (a p! δ (yy p dy sin(y/ y p dy y/ sin(y/ y p dy π y/ ( p π ( p J(p sin(s s p ds s and where J(p is a bounded integral. The general term in I( is Hence lim I( and lim and Property 3 is proved. f (p (a p! sin(ss p ds p 4 ( p J(p p. δ (x af(xdx f(a + lim I( f(a 7