---------------------------------------------------------------------------- Mon Dec 6 06:53:54 MST 2010 today's office hrs, normally 1-2 will be held on Wed. 1-2pm notice that problem 10.2.6 was removed from this week's assignment (it will be discussed in class on Tuesday; involves Neumann functions, the linearly indep. second solution of Bessel's equation that is singular at the origin but is necessary if two-point BC must be satisfied, as is the case for annular geometry) ---------------------------------------------------------------------------- Sat Nov 6 22:58:11 MDT 2010 the other condition is that the solution must be finite at r=0 (that does away with all the inverse powers and the log term) > Problem 6.1.5: > > Is it possible to solve for the constants in this > problem with only one BC? Or is there a special BC > (possibly at the origin) that I am not seeing, > other than > > u(a) = 0 ? > ---------------------------------------------------------------------------- Wed, 03 Nov 2010 20:16:10 problem 6.1.3 Sec. 10.5 states: u_zz + (1/z) u_z + u = 0 has solution J0(z) our equation is u_rr + (1/r) u_r - k^2 u = 0 change variables z = a*r and use chain rule to convert the z-equation to the r-equation, and detrmine for what value of the parameter "a" you get the exact same form. The solution wiull be J0(a*r). You do not need to know anything about Bessel functions to do this problem, just about change of variables, once you know what form you are shooting for. ---------------------------------------------------------------------------- Wed Oct 6 16:22:00 MDT 2010 A: the equilibrium solution does not depend on time; it still satisfies the diffusion eq. and the BC, but since v_t=0, we have v_xx = 0, v(0) = 0, v(1) = 1 Q1: > I have found the solution > for the problem with BC u(x,0)=u(x,1)=0 to be the sum(A exp(-(n*pi)^2 t) > sin(n*pi*x)). u(0,t)=u(1,t)=0 ******* A: yes, but A should be A_n and you need to find it. The full solution is U(x,t) = u(x,t) + v(x) with U(x,0) = phi(t), the given IC. After you find v(x) you need to figure out what to use for u(x,0) and compute the A_n. Q2: > On problem 5.1.8 the problem would have two parts > due to the boundary conditions being inhomogeneous. I have found the solution > for the problem with BC u(x,0)=u(x,1)=0 to be the sum(A exp(-(n*pi)^2 t) > sin(n*pi*x)). Is this on the right track? If so how do I find the other part > of the solution and which part is considered the equilibrium solution? ---------------------------------------------------------------------------- Wed Oct 6 01:43:02 MDT 2010 A: use the formula for computing A0 = (2/l) Int_0^l f(x) dx (the constant associated with the sine series in computing x^3 is equal to zero; just set x=0 in the formulas related to x^3 expansion to show that.) When you have a constant (like in the x^2 case) when you integrate you get x, and you do need to plug in the sine series for it. Q: > I'm working on the first problem from the homework, and it mostly makes > sense. However, i don't know how to handle the constant of integration that > come out of integrating the cosine series of x^2 to find the sine series of > x^3. Likewise, when integrating x^3 to get the cosine series of x^4 another > constant of integration comes out, plus there's the previous constant > integrated, giving Cx (do we then plug in the sine series of x into Cx?). I > assume, but don't know, that the new constant obtained from integrating the > second time is the first term (A_o/2) of the cosine series of x^4. If this > is the case i still don't know how to find its value. ---------------------------------------------------------------------------- Wed Sep 29 22:58:05 MDT 2010 A: the solution is a sum of elementary solutions made up of arbitrary multiples of each eigenfunction. As the relationship between r and n determines the type of the solution in time (exponential, ocillatory) that means for different values of n you rewrite the solution appropriately to reflect the (r,n) dependence Q: >problem 4.1-2 > All i'm seeing is that r will determine whether the time dependent part has > complex exponents. But i don't know how to use this information to finish > the problem. I guess i'm having a hard time making the jump from > interpretting the solution for different values of r and getting different > solutions for different values of r. ---------------------------------------------------------------------------- Wed Sep 29 16:02:58 MDT 2010 Q: hint for 4.3.17: you want l real; any real number has 4 fourth roots: +/- (abs(l))^ (1/4) and +/- i(abs(l))^(1/4) where i is the imaginary unit The solutions of the equation X""+l X = 0 are sin, cos and sinh, cosh of (abs(l))^(1/4)x ---------------------------------------------------------------------------- Wed Sep 22 18:02:41 MDT 2010 Q: > I had a questions for the about the homework due tomorrow. Problem 2.3 #2 > asks if M(T) increases or decreases for different circumstances. Is a one >word answer adequate or do we need to prove why? A: a logical justification is required, e.g. a drawing depicting possible situations that shows why it is so. Hint: what can you say about the world record in 100 meter dash? Is it possible that it will be higher or lower next year compared to what it is now? ---------------------------------------------------------------------------- Wed Sep 15 21:47:32 MDT 2010 Notice the change in the hint I posted for problems 2.1.(10-11). Things work out OK either way, but this form puts the coefficient where it belongs and allows its identification with the wave speed, dx/dt = c hint: for 2.1(10,11) follow the discussion of the wave equation with and without forcing; factor the operator as L = l1*l2, then work with L*u = f => l1*l2*u = f => { l2*u = v , l1*v = f} For the forced problem, consider changing variables: write as l1 = p_t + c1*p_x , l2 = p_t + c2*p_x (p_x = operator of partial differentiation by x, etc) and change coordinates to: r = x - c1*t , s = x - c2*t This means there are two wave speeds, c1 and c2. ---------------------------------------------------------------------------- Wed Sep 8 21:47:34 MDT 2010 > I'm in your PDE class and I'm having a very difficult time with first order > problems of the form g(x,y)U_x + h(x,y)U_y + f(u,x,y)=0. > After solving dy/dx=h/g I am at a loss. In looking over your solution to U_x > + U_y = 1 I don't understand how the substitution of t works and what it's > place in the solution is. Let me write (pu/px) to stand for (partial u/partial x). The idea is that there is a curve in the (x,y) plane along which the PDE becomes an ODE. A curve can be represented in various ways: (1) x=X(y), (2) y=Y(x) or (3) (x,y) = (X(t),Y(t)) where t is a parameter. When you have one of these, then you can write (chain rule) (1') du/dx = (pu/px) + (dy/dx)(pu/py) (2') du/dy = (dx/dy)(pu/px) + (pu/py) (3') du/dt = (dx/dt)(pu/px) + (dy/dt)(pu/py) > > I'm now working on > U_x + 2xU_y + U y = 0. > and I got to the solution for U_x + 2xU_y = 0 as did the student who's > question you posted on the course website but I don't know where to go from > here. > Why is the first equation du/dx + y u = 0 and what do I do with the c in y = > 2x^2 +c when I substitute? (actually y = x^2 + c) This is like case (1-1') above. c is a constant and you treat it as such when you integrate the ODE. You will get another constant. When you write the final solution you substitute back c = y-x^2 and replace the other constant by an arbitrary function F(y-x^2). To finish you must set x=0 and choose F so your solution equals f(y). ---------------------------------------------------------------------------- Q: I am having alittle problems with problem 1 on homework 2. > I solved the following without the yu > u_x + 2 x u_y + y u = 0 > I got u(x,y) = f(y-x^2) by > dy/dx = 2x and solving to get y = x^2 + C....Solved for C to get U(x,y). > Is this the right path to take to solve the PDE with the yu included? A: yes, to get to this point you had to solve du/dx = 0 with dy/dx = 2x What changes is the first equation becomes du/dx + y u = 0; you need to solve the other equation to find y in terms of x before you can proceed. ---------------------------------------------------------------------------- Sun Aug 29 15:22:18 MDT 2010 Q: problem number 9 from section 1.2... (du/dx)+(du/dy)=1. The homogeneous case is fairly straightforward. but when i approach this problem i don't know how to deal with the 1 on the right hand side of the equation. I get as far as finding that dx/dt=1, dy/dt=1 and du/dt=1... A: solve: along x = t+x0, y=t+y0; u = t + u0. Eliminate t, show that u is an arbitrary function of x-y (since it is constant when x-x0 = t = y-y0 => x-y = x0-y0 = const.). ---------------------------------------------------------------------------- Thu Aug 26 17:13:57 MDT 2010 Q: > I want to know that how do we show that a function > (c1+c2·(sinx)^2+c3·(cosx)^2) can form a vector space and why its dimension > is 2. A: there are three functions here, 1, cosx^2, sinx^2. To show that they are independent compute the wronskian and show it is not 0: W(f, g, h) = det| f g h | | f' g' h'| | f" g" h"| You then need to use the fact that the span of a set of vectors forms a vector space. The number of independent vectors in this set gives the dimension. The span of, say, three vectors f, g, h is the set of all vectors of the form V = { x | x = c1*f + c2*g + c3*h, c1, c2, c3 real no.s} The sum of any x, y in V is also an element of V. ---------------------------------------------------------------------------- Q: > What is the definition of a vector space? set V elements f,g,h,... numbers x,y,z,... (real, complex,...) operations: + (f + g is defined) * x*f is defined if for every f, g in V: f+g is also in V for every number x, f in V: x*f is also in V Then V is a vector space. (This is a "sloppy" definition, but it is enough to do the problems; I will give the precise definition in class) ---------------------------------------------------------------------------- Q: Sec 1.2 #5 For this problem: When your equation is x*u_x+y*u_y=0 does that imply dy/dy=y/x? A: not quite; I would introduce a new variable, say t,: du/dt = 0 along lines dx/dt = x, dy/dt = y ---------------------------------------------------------------------------- Q: Sec 1.2 #10 I'm not certain how to approach this problem. Could you give me some hints? The equation is u_x+u_y+u=exp(x+2*y) with u(x,0)=0 find t: so that PDE becomes (1) du/dt + u = exp(x+2*y) along lines on which dx/dt = ...., dy/dt = .... Once you find what these lines are (i.e. once you get x(t), y(t), substitute in equation (1) and solve. -----------------------------------------------------------------------------