Copyright c 2008 Kevin Long

Transcription

1 Lecture Separation of variables In the previous lecture we found that integrating both sides of the equation led to an integral dv = 3 8v (.1) dt 8 v(t) dt whose value we couldn t compute because v(t) was an unknown function of time. To avoid this problem, we pulled a rabbit out of the hat: dividing both sides by 3 + 8v and then doing a u-substitution u = v(t) led to du 3 + 8u = dt. The key point is that after the u-substitution the integral on the LHS now involves a fully-determined function of u integrated over u, rather than an undetermined function of t integrated over t, so it can be computed (up to an as-yet undetermined constant of integration). After doing both integrals, we got an algebraic equation that we solved for v(t). The constant of integration was determined by imposing initial conditions. The final step, as always, was to check the solution..1 Separable equations Why did this trick work? More precisely, what was the specific feature of the problem Copyright c 008 Kevin Long dv = 3 8v dt that let us reduce the problem to integration after a simple transformation? And more importantly, can we identify this same feature in other problems? If so, this transformation goes from being an isolated one-off trick to a special case of a general method. A trick to solve a single problem is cute but not particularly useful; a general method able to solve many problems is something worth talking about. To see what s going on let s try the same trick on the equation dv dt = t v. We ll quickly see this won t work; the reason why will help us see what feature of.1 let us transform that problem successfully. Divide both sides by t v, giving 1 dv t v dt dt = dt. 1

2 Make the u-substitution u = v again on the LHS, giving du t u = dt. The LHS integral can t be computed. Why not? Because it involves t as well as u. Because u = v(t) there s a relationship between t and u through the function v, but because we don t know v(t) we can t determine how t depends on u. We re back to square one; we can t do the integral until we ve solved the problem, and we can t solve the problem without doing the integral. But the failure of the trick on this problem 1 highlights why it did work for the first problem. It worked for.1 because the form of the equation was such that when we did the transformation, t disappeared entirely from the LHS, leaving an integral entirely in terms of u. If there s any mixture of t and u on either side of the equation, we re stuck with an integral we can t compute. Expressions such as the sum t u can t be disentangled in that way. However, with any product such as ut, or te u, or t (u + 1), the stuff involving u and the stuff involving t can be disentangled by division. For this division trick to work,then, the equation has to have a particular structure: given an equation dv = f(v, t), dt I have to be able to factor the RHS f(v, t) into a product g(t)/h(v), one factor of which involves only t, the other of which involves only v. Let s introduce a term for such problems: separable equations, and set up a formal definition. Definition.1 A first-order ODE for y(x) of the form is called a separable equation..1.1 Examples dx = g(x) h(y) The equation y = y sin x is separable, because we can identify g(x) = sin x and h(y) = x 1. The equation y = y sin x + is not separable. There is no way to factor y sin x + into a product of g(x) and 1/h(y). The equation y = y + xy is not separable. There is no way to factor y + xy into a product of g(x) and 1/h(y). Copyright c 008 Kevin Long. A procedure for solving separable equations In principle, separable equations can always be solved by integration. In practice, there might be trouble actually computing a given integral, but the important thing is that a separable equation can always be rewritten in terms of integrations. Here s how. To solve the ODE dx = g(x) h(y), multiply both sides by h(y) and integrate with respect to x: h(y(x)) dx dx = g(x) dx. 1 We will be able to solve this problem by another method

3 Make the substitution u = y(x), in which case du = dx dx so that h(u) du = g(x) dx. Define H(u) h(u) du and G(x) g(x) dx. Note that we ve defined u = y(x), then we can write H(u) = H(y(x)). Doing the integrals and including a constant of integration gives us H(y(x)) = G(x) + C. This is an algebraic equation for y(x) in terms of x. In principle (there s that phrase again I ll say more later about the difference between in principle and in practice for such problems), you can usually solve that algebraic equation for y(x) by inverting the function H...1 Some shortcuts There are a few simplifications we can make. Notice that we re always doing the substitution u = y, and that substitution always has the effect of replacing dx dx with du. Notice also that at the end we ll put y back in for u in H(u). Effectively, u is a dummy variable that we put in and take out again; we might as well skip it and just work with y, writing the integral h(u) du as h(y). Notice also that the procedure looks like we re multiplying both sides by dx. Of course dx is not a fraction and in general we can t treat it as one, but as a mnemnonic device we can think of dx = g(x) h(y) (.) = g(x) dx h(y) (.3) h(y) = g(x) dx (.) h(y) = g(x) dx. (.5) You get the same result as we got doing the transformation rigorously with a u-substitution. Feel free to use this mnemnonic device everyone uses it, and it s perfectly OK for you to use it on a test but always remember that it s a notational replacement for the actual u-substitution, and that multiplying by differentials is, in general, not mathematically valid. Keep in mind that a perfectly legitimate test question would be to ask you to justify the trick in terms of substitutions and transformations... Example Solve Copyright c 008 Kevin Long dx = ( 1 + y ) x with initial conditions y() = 1. This is separable, with h(y) = 1 1+y and g(x) = x. Using the shortcuts above, 1 + y = x dx (.6) 1 + y = x dx (.7) arctan y = x 3 + C. (.8)

4 Solve for y by taking tangents of both sides, ) y = tan ( x + C. (.9) That s the solution to the ODE. To solve the IVP we use the initial conditions, plugging x = and y() = 1 into the equation to find Therefore Check 1 = tan ( + C) (.10) C = arctan(1) (.11) C = arctan(1) + (.1) y(x) = tan = π +. (.13) ( x + π ) +. (.1) In the checking step, we ll use the trig identity sec u = 1 + tan u and the differentiation rule d du (tan u) = sec u. Here goes. LHS = ( dx = x sec x + π ) + (.15) = x (1 + tan ( x + π )) + (.16) = x ( 1 + y ) (.17) = RHS (.18) Copyright c 008 Kevin Long Box.1 I should mention that I made a simple sign error in working this problem, and caught it in the checking stage.

5 Box. A COMMON MISTAKE: Many students will set this problem up correctly, do the integrals correctly, then mess up the algebra. Here s an exam solution by an anonymous Texas Aggie: 1 + y = x dx (.19) 1 + y = x dx (.0) arctan y = x + C (.1) y = tan x + tan C. (.) But tan C is just a constant, so we can just write it as C, giving us Now, apply the initial conditions: so y = tan x + C. (.3) 1 = tan + C (.) C = 1 + tan (.5) y(x) = tan x tan. Check the solution to show that it is incorrect (the check gets messy, so it might help to use Mathematica). Where did the Aggie go wrong? (Hint: It is a serious conceptual error, not a simple arithmetic or algebra error).3 Using Mathematica to assist separation of variables Copyright c 008 Kevin Long 5

6 Separation of variables with Mathematica Solve the equation x1 dx y 1 with initial conditions y()=1. This is separable with gx x and hy. 1y Start by defining Mathematica functions for g and h. Note the underscores; they matter (in a way that will be explained later). In[1]:= gx x Out[1]= In[5]:= Out[5]= x hy 1 1 y^ 1 y 1 Define variables for use when we set the initial conditions. In[3]:= x0 Out[3]= In[35]:= y0 1 Out[35]= 1 Doing the integrals In[3]:= Out[3]= In[6]:= Out[6]= Gx Integrategx, x x Hy Integratehy, y tan 1 y Solving the algebraic equation Copyright c 008 Kevin Long Solve for y as a function of x. Note the addition of the constant of integration. In[17]:= Out[17]= algsolnx, C SolveHy Gx C, y y tan 1 C x Use the result of the algebraic solution to define a function (a one-parameter family depending on C, actually) representing the solution of the ODE. In[18]:= Out[18]= odesolnx, C y. algsolnx1 tan 1 C x

7 sepvar.nb Solving for C Plug the initial conditions into the ODE solution and solve for C. In[36]:= csoln SolveodeSolnx0, C y0, C Out[36]= Solve::ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. C Let s return to the issue of the scary warning message later. For now, forge boldly ahead, substituting C=+ into the ODE solution. In[37]:= Out[37]= ivpsolnx odesolnx. csoln1 tan 1 x Here s what the solution looks like In[38]:= Out[38]= PlotivpSolnx, x,, The solution blows up just below x=3. (You should calculate exactly where!) I ve continued the graph after that point to highlight the singularity, but the behavior of the solution to the IVP is undefined once it "blows up." Checking the solution Copyright c 008 Kevin Long You can use Mathematica to simplify checking the solution. Plug in, and let Mathematica do the calculus and algebra: In[39]:= Out[39]= In[0]:= Out[0]= LHS DivpSolnx, x x sec 1 x RHS x 1 ivpsolnx^ x 1 tan x 1

8 sepvar.nb 3 In[1]:= RHS LHS Out[1]= x 1 tan x 1 1 x sec x If you remember your trig identities, you ll see that these are equal. When looking at complicated expressions, the FullSimplify[ ] function is amazingly useful... In[]:= Out[]= FullSimplifyRHS LHS True...so the IVP solution is indeed a solution of the ODE. Now check that the IVP solution satisfies the IC: In[3]:= Out[3]= ivpsolnx0 y0 True...so our solution passes both the ODE check and the IC check. All done! Well, not quite. About that warning message OK, why the scary warning message??? It s because the algebraic equation being solved is In[]:= Out[]= odesolnx0, C y0 tan 1 C 1 which we can clean up a little with our friend FullSimplify[ ] In[5]:= FullSimplifyodeSolnx0, C y0 Out[5]= tan C 1 0 which has multiple solutions because the tangent function is periodic (with period ).The solution C value, but there are infinitely many other solutions obtained by adding to any integer multiple of. Taking into account the multiple solutions to the algebraic equation, the IVP solution is actually In[6]:= Out[6]= Tanx^ Pi n Pi tan x n is the principal where n is an integer. However, the tangent function is periodic with period, so the addition of an n phase shift does not matter. Copyright c 008 Kevin Long Thus the scary warning about inverse function was a false alarm. This will usually be the case for problems of this sort, because we ll be inverting a function to get C, and then putting the value of C right back into the function. You do things like this all the time in hand calculations, but don t always think about what s happening. Here s the plot of the solution where we ve used C 13 solution with the principal value of C. instead of C. As expected, it is identical to the

9 sepvar.nb In[7]:= PlotTanx^ Pi 13 Pi, x,, 6 Out[7]= OK, now we re all done. Copyright c 008 Kevin Long