Partial Differential Equations

Transcription

1 Partial Differential Equations Spring Exam 3 Review Solutions Exercise. We utilize the general solution to the Dirichlet problem in rectangle given in the textbook on page 68. In the notation used there and on page 67 we have a = b = f x = 5x f x = 5 x g y = 5y and g y = 5 y. It follows that the coefficients in the series solution are A n = B n = C n = D n = 5x sin nπx dx = n+ nπ 5 x sin nπx dx = 4 n + n π n 3 π 3 5y sin nπy dy = 4 n + n+ n π n 3 π 3 5 y sin nπy dy = nπ. The complete solution is then given by ux y = A n sin nπx nπ y sinh + nπ x + C n sinh sin nπy + with A n B n C n and D n as above. B n sin nπx D n x sinh nπy sin nπy Exercise. This is simply an exercise in using the chain rule. If z = e x/ then on applying the chain rule we have dy dx = dy dz dz dx = dy dz e x/. Another application of the chain rule with the product rule then yields d y dx = d dy dx dz e x/ = d dy e x/ + dy dx dz dz e x/ = d y dz dz dx e x/ + dy dz e x/ = d y dz e x + dy dz e x/. Since e x/ = z/ this implies that d y dx = z d y 4 dz + z dy 4 dz.

2 Substituting this into the differential equation y + e x y = and again using the fact that e x/ = z/ we get z d y 4 dz + z dy 4 dz + z 4 y =. Multiplying through by 4 finally gives us which is Bessel s equation of order. z d y dz + z dy dz + z y = Exercise 3. The central idea in both parts of this exercise is to use equation 6 of section 4.8 namely J p+ = p x J px J p x to express the given Bessel function in terms of those of lower order. a. Equation with p = 3/ tells us that J 5/ x = 3 x J 3/x J / x. Our goal is to get formulas for J / x and J 3/ x and plug them in to. According to Example in section 4.7 J / x = sin x. 3 πx We now need a similar formula for J 3/ x. For this we turn to identity 4 of section 4.8. If we divide by x and set p = / we find that J 3/ x = J /x + x J /x = πx cos x + x πx sin x + x πx sin x = πx cos x + sin x x πx where we have used the expression 3 for J / x. If we plug this and 3 into we obtain J 5/ x = 3 x πx cos x + 3 x πx sin x πx sin x 3 = πx x sin x 3x cos x which is what we were asked to show. b. This is simply a repeated use of. Starting with p = 4 we get J 5 x = 8 x J 4x J 3 x. 4

3 Taking p = 3 gives J 4 x = 6 x J 3x J x and plugging this into 4 and simplifying we find that 48 J 5 x = x J 3 x 8 x J x. If we repeat this process using p = and then p = in we eventually find that J 5 x = 7 x J x 4 x + x 9 J x 3 x. [Remark: There is actually a nice way to make the procedure just described much more efficient i.e. amenable to hand computation. The idea is to write the relationship in matrix form as If we let then this becomes It then follows that That is Jp+ x J p x A p = = p/x p/x Jp x J p x Jp x v p = J p x v p+ = A p v p.. v 5 = A 4 v 4 = A 4 A 3 v 3 = A 4 A 3 A v = A 4 A 3 A A v. J5 x J 4 x = A 4 A 3 A A J x J x Computing the matrix product A 4 A 3 A A a relatively straightforward procedure this immediately expresses J 5 x and J 4 x too in terms of J x and J x. This same idea can be used to quickly express any given Bessel function of the first kind in terms of Bessel functions of lower order.]. Exercise 4. As with Exercise we simply refer to the general solution to the vibrating membrane problem as given on pages and 3 in our textbook. The statement of the problem implies that the initial shape of the membrane is fr θ = i.e. the membrane is initially flat which according to the integral formulae on page immediately tells us that a mn = b mn = for all m and n. Moreover using the given radius and initial velocity the integral factor appearing in the formula for a n is π 5 r r 3 cos 3θJ λ n rr dθ dr = π dr cos 3θ dθ = }{{}

4 we have simply written the r-integrand as since it is clearly immaterial to this computation. This implies that a n = for all n. Likewise the integral factor appearing in b mn separates as π dr cos 3θ sin mθ dθ = }{{} which implies that b mn = for all m and n the fact that the θ integral vanishes can be observed in one of several ways: by direct antidifferentiation; by observing that the integrand is an odd π-periodic function and we are integrating over a complete period; or by utilizing previously established orthogonality results. We now turn to the computation of the a mn coefficients. integral factor we find that it separates to become 5 r r 4 J m λ mn r dr cos 3θ cos mθ dθ. }{{} } {{ } A π B Again focusing only on the The integral B is equal to unless m = 3 as above there are a few ways to verify this the simplest probably being orthogonality in which case it is equal to π. This means a mn = unless m = 3. Setting m = 3 we can quickly evaluate the integral A by taking advantage of the useful identity on page of the textbook. It gives 5 r r 4 J 3 λ 3n r dr = 5 r r 4 J 3 λ 3n r dr = 57 J α3n 5 α 3n. Putting these computations together we find that taking a quick glance at equation 8 on page 3 a 3n = 5πcα 3n J 4 α 3n AB = 4 56 J 5 α 3n cα 3 3nJ 4 α 3n. This can be further simplified although this isn t strictly necessary using identity with p = 4 which tells us that J 5 α 3n = 8 α 3n J 4 α 3n J 3 α 3n = 8J 4α 3n α 3n since J 3 α 3n =. Using this in our previous expression and simplifying slightly we end up with a 3n = 5 cα 4 3nJ 4 α 3n.

5 Finally we conclude that the shape of the membrane at time t is given by ur θ t = = a 3nJ 3 λ 3n r cos 3θ sin cλ 3n t 5 cos 3θ c α3nj 4 4 α 3n J α3n 3 5 r cα3n t sin 5 the elasticity constant c was unspecified so it must appear in our solution as a parameter. Exercise 5. This one s extra credit. No hints! Exercise 6. a. The given differential equation is not in S-L form. However if we multiply it by x / it becomes which is the same as x 3/ y 3x x / y + nn + x / y = x 3/ y + nn + x / y =. This is in S-L form with px = x 3/ qx = and rx = x / with parameter λ = nn +. The given problem is singular because: The boundary conditions are not of the correct form see equation of section 6.. The function px is not positive for x since p = p =. The function rx is not positive for x since r = r =. b. Suppose that y and y are eigenfunctions of the problem in question with distinct eigenvalues. In class we showed that y and y are guaranteed to be orthogonal on the interval [ ] with respect to the weight function rx = x / provided that px y xy x y xy x =. 5 Because p = p = as noted above this is immediate thereby verifying orthogonality. Exercise 7. Separating variables we assume that ux t = XxT t and plug into the given PDE yielding X T xxt =. Dividing by xxt this becomes X xx T T =

6 or X xx = T T = λ for some constant λ since the first two terms depend on different independent variables. Reorganizing slightly we obtain the two ODEs X + λxx = 6 T + λt =. 7 Because we seek nontrivial solutions the given boundary conditions also tell us that X = X =. The equation for X is in S-L form with px = qx = and rx = x. The conditions X = and X = are regular see equation of section 6. but the fact that r = means that the boundary value problem for X is actually singular. Nonetheless the fact that the boundary conditions are regular implies the eigenfunctions for distinct eigenvalues are orthogonal on [ ] with respect to the weight rx = x. Moreover one can show directly that the eigenvalues form an increasing sequence λ < λ < λ 3 < that tends to infinity that up to a scalar multiple the eigenfunction X n x associated with λ n is unique and that the Fourier convergence theorem holds for these eigenfunctions. Taking this as given equation 7 tells us that T t = T n t = e λnt so that the normal modes of the original boundary value problem are and the general solution is given by The initial condition gives u n x t = X n xe λnt ux t = c n X n xe λnt. fx = ux = c n X n x with according to the Fourier convergence result quoted above c n = fxx n xx dx. Xnxx dx According to my notes I originally intended to use the interval x. This would have kept the problem regular since then p and r would have been positive throughout the interval. This is automatically true of any regular S-L problem and a brief sketch of the derivation of these facts in this particular singular case is given below.

7 [Remark: Changing variables specifically by first substituting y = x / X followed by setting s = x 3/ /3 one can transform the equation X + λxx = into the parametric Bessel equation of order /3. Imposing the boundary conditions one can then show that nonzero solutions are possible only for λ > and after back substitution one can show that in this case the eigenvalues and eigenfunctions are λ n = 9α /3n 4 X n x = x / J /3 α/3n x 3/ for n = Aside from the completeness statement i.e. that Fourier convergence holds this proves the claims made above.]