Approximaths

Perturbation Theory in Action

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We would like to use perturbation methods similar to those presented for solving polynomials in the previous posts, but this time to solve differential equations. We will solve the following basic first-order equation with the corresponding initial condition:

\displaystyle y' = y,\; y(0)=1

This equation can be solved exactly very easily. The answer is y(x)=e^x. Now, let us pretend that we don’t know how to solve this equation exactly and use perturbation theory techniques. We write the function y(x) in the form (in a similar way to how we calculated the roots of polynomials):

\displaystyle y_{\epsilon}(x)=\sum_{n=0}^{\infty}a_n(x)\epsilon^n=a_0(x)+a_1(x)\epsilon+a_2(x)\epsilon^2+\ldots

According to the definition above, we have:

\displaystyle y_{\epsilon}(x)=a_0(x)+a_1(x)\epsilon+a_2(x)\epsilon^2+a_3(x)\epsilon^3+\ldots
\displaystyle y'_{\epsilon}(x)=a_0'(x)+a_1'(x)\epsilon+a_2'(x)\epsilon^2+a_3'(x)\epsilon^3+\ldots

Now we can perturb the differential equation above:

\displaystyle y'=\epsilon y

and write (up to terms of degree 3):

\displaystyle a_0'(x)+a_1'(x)\epsilon+a_2'(x)\epsilon^2+a_3'(x)\epsilon^3=\epsilon(a_0(x)+a_1(x)\epsilon+a_2(x)\epsilon^2)
\displaystyle a_0'(x)+a_1'(x)\epsilon+a_2'(x)\epsilon^2+a_3'(x)\epsilon^3-a_0(x)\epsilon-a_1(x)\epsilon^2-a_2(x)\epsilon^3=0
\displaystyle a_0'(x)+(a_1'(x)-a_0(x))\epsilon+(a_2'(x)-a_1(x))\epsilon^2+(a_3'(x)-a_2(x))\epsilon^3=0

Therefore, we have to solve the following system of equations:

\displaystyle a_0'(x)=0
\displaystyle a_1'(x)-a_0(x)=0
\displaystyle a_2'(x)-a_1(x)=0
\displaystyle a_3'(x)-a_2(x)=0

Applying the initial conditions a_0(0)=1 and a_n(0)=0 for n\ge1, we obtain:

\displaystyle a_1'(x)-1=0\implies a_1=x
\displaystyle a_2'(x)-a_1(x)=0\implies a_2=\frac{1}{2}x^2
\displaystyle a_3'(x)-a_2(x)=0\implies a_3=\frac{1}{3!}x^3

Therefore:

\displaystyle y_{\epsilon}(x)=1+x\epsilon+\frac{1}{2}x^2\epsilon^2+\frac{1}{3!}x^3\epsilon^3

and y(x) is approximated as follows (setting \epsilon=1):

\displaystyle y_{\epsilon}(x)=1+x+\frac{1}{2}x^2+\frac{1}{3!}x^3

which corresponds to the first terms of the Taylor series of e^x (the exact solution).

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