Preface

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This secton is devoted to a special class of variable coefficients linear differential equations. This remarkable class was discovered by Leonhard Euler who showed that these differential equations could be solved explicitely via elementary functions.

Euler Equations

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An Euler equation (also known as the Euler-Cauchy equation, or equidimensional equation) is a linear homogeneous ordinary differential equation with variable coefficients of the following form:

where the coefficients \( a_0 , a_1 , \ldots , a_n \) are constants and the driving function f(x) is given. Since the leading term in the Euler equation is a multiple of xⁿ, the origin should be excluded from consideration. The Euler equation is usually considered on semi-infinite intervals either \( (0, \infty ) \quad\mbox{or}\quad (-\infty , 0) . \) Let be the linear differential operator corresponding to the Euler equation. We start with finding functions annihilated by the Euler operator; in other words, we are going to determine the fundamental set of solutions for the homogeneous equation \( L\left[ x, \texttt{D} \right] y = 0 \) on the positive semi-axis \( (0, \infty ) . \)

There are known two methods to solve the homogeneous Euler equation. One is to try a power function \( y = x^m , \) which leads to Here \( \displaystyle m^{\underline{k}} = m(m-1) \cdots (m-k+1) \) is k-th falling factorial of number m. Applying the above Euler operator to a trial solution x^m, we obtain Canceling common multiple x^m yields an auxiliary algebraic equation where P(m) is an auxiliary polynomial of degree n (in accordance to the degree of the Euler operator). If m is a root of the above algebraic equation, then \( y = x^m \) is a solution of the n-th order Euler homogeneous equation. We postpone analyzing the fundamental set of solutions, which depends on whether the roots of the auxiliary algebraic equation are real or complex and whether some of them have multiplicities others than 1.

Because of the particularly simple equidimensional structure, the Euler equation can be replaced with an equivalent constant coefficients differential equation, which can then be solved explicitly. Using change of independent variable \( x = e^t , \quad t = \ln x ,\) it follows that

and so on. So we see that every derivative with respect to x, upon transformation to a new variable t, has a multiple reciprocal to x; therefore, all these power multiples are canceled out and we obtain a constant coefficient differential equation. We summarize our observation in the table:

Solutions of P(m) = 0	Fundamental set of solutions
Simple roots m1, m2	\( y_1 = x^{m_1} , \ y_2 = x^{m_2} , \ldots \)
Double roots \( m_1 = m_2 \)	\( y_1 = x^{m_1} , \ y_2 = x^{m_1} \ln x \)
Triple roots \( m_1 = m_2 = m_3 \)	\( y_1 = x^{m_1} , \ y_2 = x^{m_1} \ln x , \ y_3 = x^{m_1} \left( \ln x \right)^2 \)
Complex roots \( m_1 = \overline{m_2} = \alpha + \beta {\bf j} \)	\( y_1 = x^{m_1} \cos \left( \beta\,\ln x \right) , \ y_2 = x^{m_1} \sin \left( \beta\,\ln x \right) \)

To make our exposition transparent, we knock out second order Euler equations and then consider the third order equations.

 

Consider second order homogeneous Euler equation

where a, b, and c are some given real numbers. For a trial solution \( y= x^m , \) we have Plugging into the Euler equation, we get Thus, \( y= x^m \) is a solution of the Euler equation if and only if m is a root of the auxiliary polynomial which simplifies to If P(m) has two distinct roots, either real or complex, we know how to solve the equation. To recall briefly, if P(m) has two distinct real roots m₁ and m₂, then \( x^{m_1} \quad\mbox{and}\quad x^{m_2} \) are two linearly independent solutions of the Euler equation; hence, the general solution becomes for some arbitrary constants C₁ and C₂.

If P(m) has two distinct complex roots \( m_1 \quad\mbox{and}\quad m_2 , \) they must be complex conjugates, so we can write \( m_1 = \alpha + \beta {\bf j} \) and \( m_2 = \alpha - \beta {\bf j} , \) where α and β are real numbers and j is a unit vector in positive vertical direction on the complex plane, so \( {\bf j}^2 =-1 . \)

It's perhaps not so clear what a complex power of x means, but since x is assumed to be positive, we have \( x = e^{\ln x} = e^t . \) If the complex power makes any kind of sense, we ought to have

Thus, we define To justify this definition, we need to check Verifications of these equations we leave to Mathematica. We can also check that our two complex solutions \( y_1 = x^{\alpha + {\bf j}\,\beta} \) and \( y_2 = x^{\alpha - {\bf j}\,\beta} \) are linearly independent by computing their Wronskian Therefore, we conclude that in the case of complex roots the general complex solution of the Euler equation is where A₁ and A₂ are arbitrary complex numbers. Each complex number is an ordered pair of two real numbers, so the above complex solution contains four arbitrary real constants. Since we are after a real-valued solution, we have to exclude two of them and impose additional constraint: \( A_2 = \overline{A_1} , \) so they should be complex conjugate. In this case, we reduce our general form of solution to the form for some real constants C₁ and C₂.

Finally, we consider the case when the auxiliary polynomial P(m) has one double root: \( P(m) = a\left( m- m_1 \right)^2 . \) Then we have the only one solution expressed through a power function: \( y_1 = x^{m_1} . \) So we we need to find a second solution that is independent of y₁. This can be done by the method of reduction of order. To apply this method we look for a solution of the Bernoulli form \( y = u(x)\,y_1 , \) where y₁ is the solution we already know. Before plugging this in, we need to rewrite the differential equation in the right form by expanding the auxiliary polynomial: \( P(m) = a \left( m- m_1 \right)^2 = a\, m^2 - 2a\, m_1 m + a\,m_1^2 . \) Equating it to \( P(m) = a\,m^2 + \left( b-a \right) m + c , \) we find

Finally, there is the case where P(m) has only one real root r of multiplicity two. In this case, we get one solution \( y_1 = x^r \) of the Euler equation and we need to find a second solution that is independent of y₁. This can be done by the method of reduction of order. To apply this method we look for a solution of the form \( y = u(x)\,y_1 = u(x)\,x^r .\) Before plugging this in, we need to rewrite the differential equation in the right form.

We need to express coefficients in terms of the double root:

Thus, we must have Plugging these values into the Euler equation, we can rewrite it as Differentiating our trial solution u x^r, we obtain Now plug these derivatives into equation \( x^2 y'' + x\left( 1 - 2r \right) y' + r^2 y =0 \) to obtain Thus, we get the equation where dependent variable is missing: If we set \( v= u' , \) we get a separable equation for v: Integration yields where C is a constant of integration. Therefore, we have We’re only looking for one u that works, so we can choose the value of C, say C = 1. This gives us Again, we can set the constant of integration to 0 because we’re only looking for one u that works. Plugging into our trial solution, \( y = u(x)\,y_1 = u(x)\,x^r ,\) we conclude that the second solution of the Euler equation in the case of a double root r is \( y = x^r \ln x , \quad x>0. \) It’s pretty clear that these are independent solutions, so the general solution of the Euler equation in the case of a double root r is Example: Homogeneous Euler equation (distinct real roots) Example: Homogeneous Euler equation (one double root) Example: Initial value problem for Euler equation (with complex roots) Example: Reducing the nonhomogeneous Euler equation to constant coefficient case Example: variation of parameters for the nonhomogeneous Euler equation Example: building the Euler equation

 

 

For arbitrary real numbers a, b, c, and d, the auxiliary polynomial

corresponding to the third order Euler equation

has always one real null. Two other nulls could be either complex conjugate pair (recall that we consider only equations with real coefficients) or two real numbers. Depending on these cases, the general solution is expressed as a linear combination through the corresponding fundamental set of solutions. However, the pattern is clear. Therefore, we conclude this section with two illustrative examples.

Example: Third order homogeneous Euler equation (one triple root) Example: variation of parameters for the third order nonhomogeneous Euler equation (three distinct roots)

 

 

Homework Assignments

 

References

 

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