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2 Introduction to perturbation methods 2.1 What are perturbation methods? Perturbation methods are methods which rely on there being a dimensionless parameter in the problem that is relatively small: 1. The most common example you may have seen before is that of high-Reynolds number uid me- chanics, in which a viscous boundary layer is found close to a solid surface. Note that in this case the standard physical parameter Re is large: our small parameter is Re 2.2 A real research example This comes from my own research . I will not present the equations or the working here: but the problem in question is the stability of a polymer extru- sion ow. The parameter varied is wavelength: and for both very long waves (wavenumber 1) and very short waves ( 1) the system is much simplied. The long-wave case, in particular, gives very good insight into the physics of the problem. If we look at the plot of growth rate of the instability against wavenumber (inverse wavelength): we can see good agreement between the perturbation method solutions (the dotted lines) and the numerical calculations (solid curve): this kind of agree- ment gives condence in the numerics in the middle region where perturbation methods can’t help. 3 Regular perturbation expansions We’re all familiar with the principle of the Taylor expansion: for an analytic function ), we can expand close to a point as: ) = ) + "f ) + 00 ) + For general functions ) there are many ways this expansion can fail, including lack of convergence of the series, or simply an inability of the series to capture the behaviour of the function; but the paradigm of the expansion in which a H J Wilson & J M Rallison. J. Non-Newtonian Fluid Mech. 72 , 237{251, (1997) 10

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small change to makes a small change to ) is a powerful one, and the basis of regular perturbation expansions. The basic principle and practice of the regular perturbation expansion is: 1. Set = 0 and solve the resulting system (solution for deniteness) 2. Perturb the system by allowing to be nonzero (but small in some sense). 3. Formulate the solution to the new, perturbed system as a series "f 4. Expand the governing equations as a series in , collecting terms with equal powers of ; solve them in turn as far as the solution is required. 3.1 Example dierential equation Suppose we are trying to solve the following dierential equation in 0: "f ) = 0 (0) = 2 (1) Ignore the fact that we could have solved this equation directly! We’ll use it as a model for more complex examples. We look rst at = 0: ) = 0 (0) = 2 ) = 2 Now we follow our system and set = 2 "f ) + ) + ) + where in order to satisfy the initial condition (0) = 2, we will have (0) = (0) = (0) = = 0. Substituting into (1) gives "f ) + ) + +2 "f ) + ) + "e ) = and we can collect powers of + 2 = 0 ) + = 0 ) + ) = 0 ) + ) = 0 The order (or 1) equation is satised automatically. Now we simply solve at each order, applying the boundary conditions as we go along. 11

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Order terms. ) + ) = 4 ) = and the boundary condition (0) = 0 gives = 4: ) = 4( Order terms. The equation becomes ) + ) = 4 ) = ) + ) = 16 with solution ) = 8( ) + and the boundary condition (0) = 0 gives = 8: ) = 8( Order terms. The equation is ) + ) = 0 which becomes ) + ) = 48( The solution to this equation is ) = 16( + 3 ) + Applying the boundary condition (0) = 0 gives = 16 so ) = 16( + 3 The solution we have found is: ) = 2 + 4 ) + 8 + 16 + 3 ) + This is an example of a case where carrying out a perturbation expansion gives us an insight into the full solution. Notice that, for the terms we have calculated, ) = 2 +1 (1 suggesting a guessed full solution ) = =0 +1 (1 = 2 =0 [2 (1 )] (1 Having guessed a solution, of course, verifying it is straightforward: this is indeed the correct solution to the ODE of equation (1). 12

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3.2 Example eigenvalue problem We will nd the rst-order perturbations of the eigenvalues of the dierential equation 00 y "y = 0 in 0 , with boundary conditions (0) = ) = 0. [Exercise: repeat this with the nal term as "y (easy) or "y (harder).] First we look at the case = 0: 00 y = 0 This has possible solutions: < cosh [ ] + sinh [ = 0 Ax > cos [ ] + sin [ The rst two solutions can’t satisfy both boundary conditions. The third must have = 0 to satisfy the condition (0) = 0, and the second boundary condition leaves us with sin [ ] = 0 = ;m = 1 ;::: Now we return to the full problem, posing regular expansions in both and = sin mx "y " Substituting in, we obtain for the dierential equation: sin mx sin mx = 0 "y 00 "m " sin mx sin mx = 0 As we would expect, the order 1 equation is already satised, along with the boundary conditions. Order The ODE at order becomes 00 sin mx sin mx sin mx cos 2 mx We expect a solution of the form sin mx cos mx Cx cos mx cos 2 mx and substituting this form back in to the left hand side gives us Cm sin mx cos 2 mx Em sin mx cos 2 mx 13

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which xes . The solution is sin mx cos mx cos mx cos 2 mx: Now we apply the boundary conditions to determine the eigenvalue: (0) = 0 gives 0 = and then the condition ) = 0 becomes: 0 = 1) 1) which simplies to determine m [( 1) 1] = m even odd Thus the eigenvalues become = 1 16 25 15 3.3 Warning signs As I mentioned earlier, the Taylor series model of function behaviour does not always work. The same is true for model systems: and a regular perturbation expansion will not always capture the behaviour of your system. Here are a few of the possible warning signs that things might be going wrong: One of the powers of produces an insoluble equation By this I don’t mean a dierential equation with no analytic solution: that is just bad luck. Rather I mean an equation of the form + 1 = 0 which cannot be satised by any value of The equation at = 0 doesn’t give the right number of solutions An th order ODE should have solutions. If the equation produced by setting = 0 has less solutions then this method will not give all the possible solutions to the full equation. This happens when the coecient of the highest derivative is zero when = 0. Equally, for a PDE, if the solution you nd at = 0 cannot satisfy all your boundary conditions, then a regular expansion will not be enough. The coecients of can grow without bound In the case of an expansion ) = )+ "f )+ )+ , the series may not be valid for some values of if some or all of the ) become very large. Say, for example, that !1 while ) remains nite. Then "f ) is no longer strictly larger than ) and who knows what even larger terms we may have neglected. . . 14

1 What are perturbation methods Perturbation methods are methods which rely on there being a dimensionless parameter in the problem that is relatively small 1 The most common example you may have seen before is that of highReynolds number 57357uid m ID: 22963

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2 Introduction to perturbation methods 2.1 What are perturbation methods? Perturbation methods are methods which rely on there being a dimensionless parameter in the problem that is relatively small: 1. The most common example you may have seen before is that of high-Reynolds number uid me- chanics, in which a viscous boundary layer is found close to a solid surface. Note that in this case the standard physical parameter Re is large: our small parameter is Re 2.2 A real research example This comes from my own research . I will not present the equations or the working here: but the problem in question is the stability of a polymer extru- sion ow. The parameter varied is wavelength: and for both very long waves (wavenumber 1) and very short waves ( 1) the system is much simplied. The long-wave case, in particular, gives very good insight into the physics of the problem. If we look at the plot of growth rate of the instability against wavenumber (inverse wavelength): we can see good agreement between the perturbation method solutions (the dotted lines) and the numerical calculations (solid curve): this kind of agree- ment gives condence in the numerics in the middle region where perturbation methods can’t help. 3 Regular perturbation expansions We’re all familiar with the principle of the Taylor expansion: for an analytic function ), we can expand close to a point as: ) = ) + "f ) + 00 ) + For general functions ) there are many ways this expansion can fail, including lack of convergence of the series, or simply an inability of the series to capture the behaviour of the function; but the paradigm of the expansion in which a H J Wilson & J M Rallison. J. Non-Newtonian Fluid Mech. 72 , 237{251, (1997) 10

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small change to makes a small change to ) is a powerful one, and the basis of regular perturbation expansions. The basic principle and practice of the regular perturbation expansion is: 1. Set = 0 and solve the resulting system (solution for deniteness) 2. Perturb the system by allowing to be nonzero (but small in some sense). 3. Formulate the solution to the new, perturbed system as a series "f 4. Expand the governing equations as a series in , collecting terms with equal powers of ; solve them in turn as far as the solution is required. 3.1 Example dierential equation Suppose we are trying to solve the following dierential equation in 0: "f ) = 0 (0) = 2 (1) Ignore the fact that we could have solved this equation directly! We’ll use it as a model for more complex examples. We look rst at = 0: ) = 0 (0) = 2 ) = 2 Now we follow our system and set = 2 "f ) + ) + ) + where in order to satisfy the initial condition (0) = 2, we will have (0) = (0) = (0) = = 0. Substituting into (1) gives "f ) + ) + +2 "f ) + ) + "e ) = and we can collect powers of + 2 = 0 ) + = 0 ) + ) = 0 ) + ) = 0 The order (or 1) equation is satised automatically. Now we simply solve at each order, applying the boundary conditions as we go along. 11

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Order terms. ) + ) = 4 ) = and the boundary condition (0) = 0 gives = 4: ) = 4( Order terms. The equation becomes ) + ) = 4 ) = ) + ) = 16 with solution ) = 8( ) + and the boundary condition (0) = 0 gives = 8: ) = 8( Order terms. The equation is ) + ) = 0 which becomes ) + ) = 48( The solution to this equation is ) = 16( + 3 ) + Applying the boundary condition (0) = 0 gives = 16 so ) = 16( + 3 The solution we have found is: ) = 2 + 4 ) + 8 + 16 + 3 ) + This is an example of a case where carrying out a perturbation expansion gives us an insight into the full solution. Notice that, for the terms we have calculated, ) = 2 +1 (1 suggesting a guessed full solution ) = =0 +1 (1 = 2 =0 [2 (1 )] (1 Having guessed a solution, of course, verifying it is straightforward: this is indeed the correct solution to the ODE of equation (1). 12

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3.2 Example eigenvalue problem We will nd the rst-order perturbations of the eigenvalues of the dierential equation 00 y "y = 0 in 0 , with boundary conditions (0) = ) = 0. [Exercise: repeat this with the nal term as "y (easy) or "y (harder).] First we look at the case = 0: 00 y = 0 This has possible solutions: < cosh [ ] + sinh [ = 0 Ax > cos [ ] + sin [ The rst two solutions can’t satisfy both boundary conditions. The third must have = 0 to satisfy the condition (0) = 0, and the second boundary condition leaves us with sin [ ] = 0 = ;m = 1 ;::: Now we return to the full problem, posing regular expansions in both and = sin mx "y " Substituting in, we obtain for the dierential equation: sin mx sin mx = 0 "y 00 "m " sin mx sin mx = 0 As we would expect, the order 1 equation is already satised, along with the boundary conditions. Order The ODE at order becomes 00 sin mx sin mx sin mx cos 2 mx We expect a solution of the form sin mx cos mx Cx cos mx cos 2 mx and substituting this form back in to the left hand side gives us Cm sin mx cos 2 mx Em sin mx cos 2 mx 13

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which xes . The solution is sin mx cos mx cos mx cos 2 mx: Now we apply the boundary conditions to determine the eigenvalue: (0) = 0 gives 0 = and then the condition ) = 0 becomes: 0 = 1) 1) which simplies to determine m [( 1) 1] = m even odd Thus the eigenvalues become = 1 16 25 15 3.3 Warning signs As I mentioned earlier, the Taylor series model of function behaviour does not always work. The same is true for model systems: and a regular perturbation expansion will not always capture the behaviour of your system. Here are a few of the possible warning signs that things might be going wrong: One of the powers of produces an insoluble equation By this I don’t mean a dierential equation with no analytic solution: that is just bad luck. Rather I mean an equation of the form + 1 = 0 which cannot be satised by any value of The equation at = 0 doesn’t give the right number of solutions An th order ODE should have solutions. If the equation produced by setting = 0 has less solutions then this method will not give all the possible solutions to the full equation. This happens when the coecient of the highest derivative is zero when = 0. Equally, for a PDE, if the solution you nd at = 0 cannot satisfy all your boundary conditions, then a regular expansion will not be enough. The coecients of can grow without bound In the case of an expansion ) = )+ "f )+ )+ , the series may not be valid for some values of if some or all of the ) become very large. Say, for example, that !1 while ) remains nite. Then "f ) is no longer strictly larger than ) and who knows what even larger terms we may have neglected. . . 14

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