Math Problem Statement

Solve the differential equation du/dx + kux^n =0, using a power series expansion u(x) = \sum_{m=0}^{\infty}{a_m x^m}, i.e. find the coefficients a_m. This method sometimes known as the method of Frobenius.

Solution

To solve the differential equation

dudx+kuxn=0\frac{du}{dx} + k u x^n = 0

using a power series expansion, u(x)=m=0amxmu(x) = \sum_{m=0}^{\infty} a_m x^m, we apply the method of Frobenius. Here's the step-by-step approach:

Step 1: Express u(x)u(x) and its derivative as power series

Let u(x)=m=0amxmu(x) = \sum_{m=0}^{\infty} a_m x^m.

Now, differentiate u(x)u(x) term by term:

dudx=m=1mamxm1.\frac{du}{dx} = \sum_{m=1}^{\infty} m a_m x^{m-1}.

Step 2: Substitute into the differential equation

The original equation is:

dudx+kuxn=0.\frac{du}{dx} + k u x^n = 0.

Substitute u(x)=m=0amxmu(x) = \sum_{m=0}^{\infty} a_m x^m and dudx=m=1mamxm1\frac{du}{dx} = \sum_{m=1}^{\infty} m a_m x^{m-1} into this equation:

m=1mamxm1+k(m=0amxm)xn=0.\sum_{m=1}^{\infty} m a_m x^{m-1} + k \left( \sum_{m=0}^{\infty} a_m x^m \right) x^n = 0.

Step 3: Adjust the powers of xx

Now, expand the second term:

k(m=0amxm)xn=km=0amxm+n.k \left( \sum_{m=0}^{\infty} a_m x^m \right) x^n = k \sum_{m=0}^{\infty} a_m x^{m+n}.

This gives the full equation:

m=1mamxm1+km=0amxm+n=0.\sum_{m=1}^{\infty} m a_m x^{m-1} + k \sum_{m=0}^{\infty} a_m x^{m+n} = 0.

Step 4: Align powers of xx

We now need to match powers of xx. Rewrite the sums so that both sums have the same power of xx. In the first sum, let’s shift the index by setting m1=jm-1 = j, or m=j+1m = j+1:

j=0(j+1)aj+1xj+km=0amxm+n=0.\sum_{j=0}^{\infty} (j+1) a_{j+1} x^j + k \sum_{m=0}^{\infty} a_m x^{m+n} = 0.

This can be written as:

j=0(j+1)aj+1xj+km=0amxm+n=0.\sum_{j=0}^{\infty} (j+1) a_{j+1} x^j + k \sum_{m=0}^{\infty} a_m x^{m+n} = 0.

Step 5: Combine the sums and solve for the coefficients

To simplify, we combine terms with the same power of xx. The equation becomes:

j=0[(j+1)aj+1+kajn]xj=0.\sum_{j=0}^{\infty} \left[ (j+1) a_{j+1} + k a_{j-n} \right] x^j = 0.

For this sum to be zero for all xjx^j, the coefficients must satisfy:

(j+1)aj+1+kajn=0.(j+1) a_{j+1} + k a_{j-n} = 0.

Step 6: Find the recurrence relation for ama_m

This gives a recurrence relation for the coefficients:

aj+1=kj+1ajn.a_{j+1} = -\frac{k}{j+1} a_{j-n}.

Thus, we can express each ama_{m} in terms of the previous coefficients, depending on kk and nn.

Summary of the Solution

The recurrence relation for the coefficients ama_m is:

am+1=km+1amn.a_{m+1} = -\frac{k}{m+1} a_{m-n}.

This provides the general form for the coefficients, allowing you to recursively calculate ama_m from initial conditions or boundary values.

Would you like further details or clarifications on specific steps?

Related Questions:

  1. How can you choose initial conditions for the series solution?
  2. How does the value of nn influence the solution?
  3. Can you provide an example using specific values of kk and nn?
  4. What happens if n=0n = 0 (i.e., when the differential equation is linear)?
  5. How would the series solution behave for large values of xx?

Tip:

When solving differential equations using power series, pay close attention to the recurrence relations as they define how each coefficient depends on the previous terms.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Power Series
Method of Frobenius
Recurrence Relations

Formulas

u(x) = \sum_{m=0}^{\infty} a_m x^m (Power Series Expansion)
\frac{du}{dx} = \sum_{m=1}^{\infty} m a_m x^{m-1} (Derivative of Power Series)
Recurrence Relation: a_{j+1} = -\frac{k}{j+1} a_{j-n}

Theorems

Method of Frobenius

Suitable Grade Level

Undergraduate Level (Mathematics or Engineering Students)

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