Series solution to nonlinear differential equation

In summary, by truncating the given differential equation at n=12, the form of the solution is derived as \phi = 1+ \lambda x + \sum_{n=3}^{12}a_{n} x^{n/2}. To obtain expressions for all the coefficients, the approach is to equate the coefficients of like powers of x. The substitution x = u2 can be used to simplify the equation.
  • #1
teroenza
195
5

Homework Statement



By truncating the differential equation below at n=12, derive the form of the solution, obtaining expressions for all the ancoefficients in terms of the parameter [itex]\lambda [/itex].

Homework Equations


The ODE is:
[tex] \frac{\mathrm{d^2}\phi }{\mathrm{d} x^2} = \frac{\phi^{3/2}}{x^{1/2}} [/tex]

I am told that the solutions have the form:

[tex] \phi = 1+ \lambda x + \sum_{n=3}^{12}a_{n} x^{3/2} [/tex]

The Attempt at a Solution


The way I have solved series equations before was to equate the coefficients of like powers of x. In this case, however, I have those x underneath the root, and I don't know how to go about doing this. I have the terms, via Mathematica. But I don't know how to get relationships between the coefficients.

Code:
p[x_] := 1 + \[Lambda]*x + Sum[Subscript[a, n]*x^(n/2), {n, 3, 12}];
pdd[x_] :=
  Sum[Subscript[a, n]*(n^2 - 2*n)/4 x^((n - 4)/2), {n, 3, 12}];

LHS = pdd[x] - x^(-1/2)*(p[x])^(3/2)
 
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  • #2
teroenza said:

Homework Statement



By truncating the differential equation below at n=12, derive the form of the solution, obtaining expressions for all the ancoefficients in terms of the parameter [itex]\lambda [/itex].

Homework Equations


The ODE is:
[tex] \frac{\mathrm{d^2}\phi }{\mathrm{d} x^2} = \frac{\phi^{3/2}}{x^{1/2}} [/tex]

I am told that the solutions have the form:

[tex] \phi = 1+ \lambda x + \sum_{n=3}^{12}a_{n} x^{3/2} [/tex]

The Attempt at a Solution


The way I have solved series equations before was to equate the coefficients of like powers of x. In this case, however, I have those x underneath the root, and I don't know how to go about doing this. I have the terms, via Mathematica. But I don't know how to get relationships between the coefficients.

Code:
p[x_] := 1 + \[Lambda]*x + Sum[Subscript[a, n]*x^(n/2), {n, 3, 12}];
pdd[x_] :=
  Sum[Subscript[a, n]*(n^2 - 2*n)/4 x^((n - 4)/2), {n, 3, 12}];

LHS = pdd[x] - x^(-1/2)*(p[x])^(3/2)
Do you mean [tex] \phi = 1+ \lambda x + \sum_{n=2}^{12}a_{n} x^{n} [/tex]?
 
  • #3
No, but I did make an error in the above. It should be:

[tex] \phi = 1+ \lambda x + \sum_{n=3}^{12}a_{n} x^{n/2} [/tex]

But I can't figure out how to edit my original post.
 
  • #4
That suggests to me the substitution x = u2. Haven't tried it, though.
 

What is a series solution to a nonlinear differential equation?

A series solution to a nonlinear differential equation is a method of finding an approximate solution to a differential equation by using a series of terms that are added together. This method is often used when it is difficult or impossible to find an exact solution to the equation.

How is a series solution to a nonlinear differential equation calculated?

A series solution is calculated by first expanding the nonlinear function into a Taylor series, which is a series of terms with increasing powers of the independent variable. Then, the series is substituted into the differential equation and the coefficients of the terms are determined by solving a system of equations. The solution is then given by the sum of the terms in the series.

What are the advantages of using a series solution to a nonlinear differential equation?

One advantage of using a series solution is that it can provide an approximate solution to a differential equation even if an exact solution cannot be found. Additionally, the method can often be used to obtain more accurate solutions compared to other methods, such as numerical approximation.

Are there any limitations to using a series solution to a nonlinear differential equation?

Yes, there are limitations to using a series solution. One limitation is that the method can only be applied to certain types of equations, such as those that are polynomial or can be transformed into a polynomial form. Additionally, the solution obtained may only be valid for a certain range of values for the independent variable.

How accurate is a series solution to a nonlinear differential equation?

The accuracy of a series solution depends on the number of terms used in the series. As more terms are added, the solution becomes more accurate. However, it is important to note that even with an infinite number of terms, the solution is still an approximation and may not be exact.

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