Differential Equations, Constant Coefficients

Constant Coefficients

When the coefficients of a linear homogeneous equation are constants, the equation can be solved using exponentials. Write the differential equation this way.

a₀y + a₁y′ + a₂y′′ + … (up to a_n × the n^th order derivative) = 0

Let p(x) be the corresponding polynomial a₀+a₁x+a₂x²+…a_nxⁿ. Let r be a root of p(x), and let f = E^rx. Plug f into our differential equation and get p(r)×E^rx. Since p(r) = 0, f is a solution.

Now assume r has multiplicity m. In other words, p(x) has m factors of x-r. Let's try to build a new solution via f = q(x)×e^rx. If q = 1 then f is the function described above, and we're done.

Show that f′ = (rq+q′)E^rx, and f′′ = (r²q+2rq′+q′′)E^rx. In general, the "next" derivative is found by taking the factor on the left of E^rx, call it w, and replacing it with rw+w′. Thus the third derivative is (r³q+3r²q′+3rq′′+q′′′)E^rx. The expression mirrors the binomial theorem.

When we search for a solution, the factor on the right, E^rx, is always nonzero. Divide it out; it simply goes away. All that remains is the first factor, the one that looks like a binomial expansion of q and r.

As the derivatives march along, f′ f′′ f′′′ etc, pull out the first term in each expansion. These are multiplied by a₁ a₂ a₃ etc. This gives p(r)×q(x), which is known to be 0. Thus the first term in the expansion of the derivatives goes away.

Multiply the second term in each expansion by its coefficient and get something like this.

a₁q′ + 2a₂rq′ + 3a₃r²q′ + 4a₄r³q′ + … (up to a_n)

This is the same as p′(x), evaluated at r, and multiplied by q′(x).

Invoke the theory of formal derivatives. If p has multiple roots at r, then p′(r) = 0. This entire sum drops out.

Now take the third term in the expansion of each derivative. Premultiply by the coefficients and get this.

a₂q′′ + 3a₃rq′′ + 6a₄r²q′′ + 10a₅r³q′′ + …

Confused about the "extra" coefficients? They come from the binomial theorem: 2 choose 2, 3 choose 2, 4 choose 2, 5 choose 2, and so on. In other words, k×(k-1)/2. Multiply the above expression through by 2 to get this.

2×1×a₂q′′ + 3×2×a₃rq′′ + 4×3×a₄r²q′′ + 5×4×a₅r³q′′ + …

This is equal to p′′(x), evaluated at r, times q′′(x). If the multiplicity of r is at least 3, i.e. p has at least 3 roots of r, then p′′(r) = 0, and this entire expression goes away.

Collect the fourth term from each derivative, multiply by the respective coefficients, then multiply by 6. The result is p′′′(x), evaluated at r, times q′′′(x). If the multiplicity m is at least 4, p′′′(r) = 0, and this expression goes away. This continues, until we build the m^th expression, whence the m^th derivative of p, evaluated at r, is no longer 0.

If f is a solution, the remaining expressions must drop to 0. The best way to do that is to have the m^th derivative of q disappear. In other words, the m^th derivative of q = 0, and q is any polynomial of degree less than m.

By the previous theorem, the solutions form a vector space. We have identified a subspace of solutions, corresponding to the root r. The subspace is q(x)×E^rx, where q is the set of polynomials of degree < m. There is a convenient basis for this subspace:

E^rx, xE^rx, x²E^rx, x³E^rx, … x^m-1E^rx

Apply this result to each root r in p(x). The basis functions are all linearly independent, (we'll prove this below), thus building a solution space of dimension n. Again, refer to the previous theorem; the solution space has dimension n. We have found all the solutions.

A linear differential equation with constant coefficients can be solved by finding the roots and multiplicities of the corresponding polynomial.

Linear Independence

The functions that form our basis are pure exponentials E^rx, (including E⁰), and exponentials with polynomial modifiers (e.g. x³E^rx).

Suppose a linear combination c₁ produces the zero function, and suppose all the functions in c₁ are pure exponentials. Evaluate E^rj, for each root r, and for each integer j from 0 to n-1. This builds an n×n matrix that is vandermonde, and nonsingular. Therefore these functions are linearly independent.

Suppose there is one exponential with several modifiers. Since E^rx is everywhere nonzero, divide through by E^rx to find a linear combination of powers of x that is equal to 0. A polynomial cannot be the zero function, hence we have a contradiction.

At this point our linear combination has more than one exponential, and at least one of these exponentials is modified by x^j. Select a linear combination that has the smallest modifier x^m, and the fewest number of exponentials carrying this modifier.

If our linear combination of functions is 0, the same is true of its derivative. Remember that x^jE^rx becomes rx^jE^rx + jx^j-1E^rx. Polynomials drop in degree, pure exponentials are multiplied by r, and modified exponentials experience mitosis.

Suppose x^m is the highest multiplicity. If the only representative is a power of x, it has become x^m-1, and we have a linear combination with lesser multiplicity, which is a contradiction.

There is at least one exponential associated with x^m; call it x^mE^rx. Let c₁ be the original linear combination of functions, and let c₂ be the derivative. Both contain the basis function x^mE^rx. Let c₃ = c₁ - c₂/r. Notice that the term x^mE^rx drops out. Thus c₃ has fewer terms modified by x^m, or it has no terms based on x^m. Either way we have a contradiction, since m is minimal, and the number of terms with x^m is minimal. However, we still need to prove c₃ is nontrivial.

Remember that c₁ is based on more than one exponential. consider any other exponential in c₁, such as E^sx. Assume s ≠ 0. This is multiplied by s in c₂, and 1-s/r in c₃. In other words, the term persists in c₃, giving a nontrivial linear combination. If s = 0, the term is simply x^j. The highest power x^j drops out of c₂, hence it remains in c₃.

In all cases, c₃ is a nontrivial linear combination that yields 0, and violates our selection criteria. The pathological linear combination c₁ cannot exist, and these functions are linearly independent. The existence of n different solutions, drawn from distinct basis functions, builds an n dimensional solution space, and completely solves the differential equation with constant coefficients.