1 The Inverse Problem

Transcription

1 The Inverse Problem We have learned that a general solution of a first-order differential equation y = f H, yl (..) is a one-parameter family of (either eplicit or implicit) solutions. Finding a general solution of (..) can be difficult. However, the inverse problem is, in principle, always straightforward. The inverse problem is this: What differential equation is satisfied by a given general solution? Solving this inverse problem amounts essentially to these two steps:. Take the differential of each side of the equation that describes the general solution.. Use the original equation to eliminate the parameter. (In fact, precisely the same technique can be used to verify a general solution of a given differential equation.) ü Eample.. What differential equation is satisfied by the one-parameter family of curves + c y + y =? Here s a plot that shows the curves corresponding to c =, -.5,,.5, : ContourPlotAEvaluateATableA + c y + y ã, 8c,,,.5<EE, 8,, <, 8y,, <E These are ellipses when c 0 and hyperbolas when c < 0. To find the differential equation, we will compute the differential of each side of + c y + y = and then eliminate the parameter. We first enter the equation: soln = I + c y + y == M + y + c y ã

2 inverseproblem.nb Net we use Dt to take differentials, replacing the differential of the (constant) parameter with 0: Deqn = Dt@solnD ê. Dt@cD Ø 0 Dt@D + Dt@yD + c y Dt@yD ã 0 (Here Dt[] and Dt[y] represent and y.) Now, we don t want the parameter c to appear in the differential equation. To eliminate c, we just need to solve for c in the original equation and substitute the result into our differential equation: Deqn = Deqn ê. Solve@soln, cd êê First I - - ym Dt@yD Dt@D + Dt@yD + ã 0 y That s essentially it, but let s simplify (assuming y ¹ 0) and make it prettier: Simplify@Deqn, Assumptions Ø 8y ¹ 0<D êê TraditionalForm I + y - M y y ü Eample.. Verify that + c + y - y = 0. is an implicit general solution of I - y + y M + I - y M y = 0. To verify the solution, we ll begin by using Dt to compute the differential of each side and replacing the differential of the parameter with zero: soln := + c + y - y ã 0; deqn = Dt@solnD ê. Dt@cD Ø 0 c Dt@D + Dt@D + Dt@yD - y Dt@yD ã 0 Net we need to eliminate the parameter from the equation: Solve@soln, cd êê First deqn = deqn ê. % :c Ø - y + y I- - y + y M Dt@D Dt@D + + Dt@yD - y Dt@yD ã 0 Now let s simplify and make the result prettier by viewing it in TraditionalForm: Simplify@deqn, Assumptions Ø 8 ¹ 0<D êê TraditionalForm I + y - ym + I - y M y 0 Now let s have a look at the solution curves. Solving the implicit general solution for c shows that the solution curves are the contours of the function F@_, y_d = c ê. FirstüSolve@soln, cd - - y + y

3 inverseproblem.nb So we can plot solution curves with ContourPlot as follows: yd, 8,, <, 8y,, <, Contours Ø 5D, ContourShading Ø False, Eclusions Ø 8 ã 0<D Since ContourPlot also plots equations, essentially the same picture can be made this way as well: ContourPlot@Evaluate@Table@soln, 8c, -5, 5<DD, 8,, <, 8y,, <D When an eplicit general solution is given, we can take a slightly different approach to find the associated differential equation. ü Eample.. What differential equation is satisfied by the one-parameter family y = Let s start by defining a function to represent the general solution. In[6]:= := c - c -? Since this general solution is eplicit, all we need to do is to eliminate the parameter c from the following pair of equations: In[7]:= eqns = 8y ã y' == y '@D< Out[7]= :y ã c -, y ã c - + Hc - L Fortunately Mathematica has a function named Eliminate for doing just that. In[8]:= Out[8]//TraditionalForm= y y + y Hdeqn = Eliminate@eqns, 8c<DL êê TraditionalForm Now, if we like, we can solve for y and display the equation in the form y = f H, yl. In[9]:= y' == Hy' ê. FirstüSolve@deqn, y'dl êê TraditionalForm Out[9]//TraditionalForm= y y + y

4 4 inverseproblem.nb Here are a few of the solution curves: 8c, -5, 5<DD, 8, -4, 4<D The technique we ve described here can be etended to n-parameter families and nth-order differential equations. In the net eample we will find the second-order differential whose general solution is a given two-parameter family. ü Eample..4 What differential equation has the general solution y = c + c? Let s start by defining a function to represent the general solution. := c + c Since this general solution is eplicit, all we need to do is to eliminate the parameters c and c from the following three equations: eqns = 8y ã y' == y '@D, y'' == y ''@D< c :y ã, y ã - c + c, c Hc + c L y ã Hc + c L Fortunately Mathematica has a function named Eliminate for doing just that. Eliminate@eqns, 8c, c <D êê TraditionalForm y y Hy L That s the differential equation we re after. Notice that it is nonlinear. ü Eercises In 7, (a) find the differential equation that has the given general solution, and (b) make a plot that shows the graphs of several of the solutions.. + y - y = c

5 inverseproblem.nb 5. + y - c y =. + c y - y = 4. c + y - y = 5. y = H - cl 6. y = cosh - cl 7. y = tan - H - cl In 8 0, find the second-order differential equation that has the given general solution. 8. y = c + c 9. y = c + c 0. y = - Hc + c L