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I'm doing a multiple choice exercise and I'm stuck at a question, where there are two possible answers (may be both or none correct).

Consider the differential equation

$$y''+y'+y=0$$

for $y(x)$. Which of the following statements are true?

  • The set of complex solutions forms a two-dimensional complex vector space.

  • The set of real solutions forms a two-dimensional real vector space.

I'm really not comfortable with vector spaces and the differentiation of real and complex vector space makes me even more unconfident. I solved the differential equation with the ansatz $y(x) = e^{\lambda x}$ and got to $\lambda^2 + \lambda + 1 = 0$, an equation which does not possess any real solutions. However, when I proceeded, I figured that I could rewrite the exponential function (with complex exponent) with the help of the sine and the cosine, so in the end I got a general solution with no complex coefficients.

What kind of vector space is this?

Huy
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    Hint: How many parameters do you need to describe the solutions in each case? Beware, in the complex case, we're talking about complex parameters and in the real case, we're talking about real parameters. – Raskolnikov Dec 10 '10 at 19:05
  • I don't quite get it. The solutions of the differential equation should be $y(x) = e^{-\frac{x}{2}} \cdot (A \cdot \sin{(\frac{\sqrt{3}x}{2})+B \cdot \cos{(\frac{\sqrt{3}x}{2}})})$. So I will always have two constants? – Huy Dec 10 '10 at 19:19
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    @Huy: Those constants are the parameters. Your real solutions are all linear combinations of $e^{-x/2}\sin(\sqrt{3}x/2)$ and $e^{-x/2}\cos(\sqrt{3}x/2)$, so the dimension is at most $2$ (over $\mathbb{R}$); since the two are actually linearly independent... In the complex case, you have a similar situation but with your solutions being (complex)-linear combinations of $e^{(-1+\sqrt{3}i)x/2}$ and $e^{(-1-\sqrt{3}i)x/2}$. – Arturo Magidin Dec 10 '10 at 19:25
  • So in both cases I have a vector space with dimension (at the most) 2? – Huy Dec 10 '10 at 19:44
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    @Huy: Certainly dimension at most 2, because you can span them with two vectors. The question is whether the vectors are linearly independent (in the first case, over $\mathbb{R}$; in the second, over $\mathbb{C}$). If they are linearly independent, then the dimension is exactly $2$; if they are linearly dependent, then the dimension would be strictly less than $2$ because you can generate it with fewer than $2$ vectors (but it won't be zero, because there are nontrivial solutions). – Arturo Magidin Dec 10 '10 at 19:47
  • Ok, $e^{(-1+\sqrt{3}i) x/2}$ and $e^{(-1-\sqrt{3}i)x/2)}$ are linearly independent, right? – Huy Dec 10 '10 at 20:10
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    @Huy: Surely you can figure it out and be confident in your answer... If $ae^{(-1+\sqrt{3}i)x/2} + be^{(-1-\sqrt{3}i)x/2} = 0$ (the zero function), must $a=b=0$? If so, yes, they are linearly independent. If not, then that gives you a dependence. – Arturo Magidin Dec 10 '10 at 20:15
  • No, they don't have to be 0. However, there are also (infinitely many?) zeroes for A = B = 1 for $e^{-x/2} \cdot (\sin + \cos)$... – Huy Dec 10 '10 at 20:18
  • Or wait. Do you actually mean it has to be the zero function? That is to say I can't choose x? If so, yes, a must equal b must equal 0... – Huy Dec 10 '10 at 20:22
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    @Huy: Of course it has to be the zero function (notice I even said so). We are in a vector space of functions, so the zero vector is the zero function. We are checking linear independence of the functions. That means that for every $x$ you must get $0$ as the result, because we are assuming that the expression you have equals the zero vector/zero function. – Arturo Magidin Dec 10 '10 at 20:23
  • So definitely $A = B = 0$, so they are linearly independent. Thanks for helping me out. – Huy Dec 10 '10 at 20:26
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    An important point that: the zero vector $\neq$ zero number. That's why I like the use of the notation $\vec{0}$ in this context. It makes things immediately clearer. – Raskolnikov Dec 10 '10 at 20:33
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    @Huy: Great! Glad you worked through it. You can post an answer to your own question and accept it after a suitable waiting period, to avoid having unanswered questions. – Arturo Magidin Dec 10 '10 at 20:47
  • @Huy: I see you already accepted an answer, so I'm posting this as a comment. There is a theorem that says that if you have a linear differential equation of order k, where all the coefficients are continuous, then the space of the solutions is a vector space of dimension k, if the equation is homogeneous, or an affine space of dimension k, if the equation isn't homogeneous. I thought this might help somehow ;) – Andy Dec 11 '10 at 13:55

2 Answers2

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Both statements are true.

Huy
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As Huy says, both statements are true.

Well, if you are a hammer, all the problems seem nails to you, but since this is a linear differential equation, Laplace transform again gives us the general solution in both, real and complex, cases.

$$ {\cal L}[y''] + {\cal L}[y'] + {\cal L}[y] = 0 \ \Longleftrightarrow \ (s^2 F(s) -sy(0) -y'(0) ) + (sF(s) -y(0)) + F(s) = 0 \ , $$

where $F(s) = {\cal L}[y]$. Now, put $a= y(0)$ and $b= y'(0)$ and solve this equation for $F(s)$:

$$ F(s) = \dfrac{as + (a+b)}{s^2+s+1} \ . $$

Since the polynomial $s^2+s+1$ has no real roots, the distinction between the real and complex cases arrives at this point.

In the complex case, you write

$$ s^2 + s + 1 = (s-\alpha)(s-\beta) \ , $$

where $\alpha = \dfrac{-1+i\sqrt{3}}{2}$ and $\beta = \dfrac{-1-i\sqrt{3}}{2}$ are the roots of your characteristic polynomial. Hence the solution of your differential equation is

$$ y(t) = A{\cal L}^{-1} \left[\dfrac{1}{s-\alpha} \right] + B {\cal L}^{-1} \left[\dfrac{1}{s-\beta} \right] = A e^{\alpha t} + B e^{\beta t} \ , $$

where $A$ and $B$ are complex numbers, depending on $a , b$, which again I let you the pleasure to compute :-) .

Anyway, the fact is that the solution of your differential equation is the $\mathbb{C}$-linear span of two linearly independent complex functions

$$ e^{\alpha t} \qquad \text{and} \qquad e^{\beta t} \ . $$

Thus a $\mathbb{C}$-vector space of dimension two.

In the real case, you write

$$ s^2 + s + 1 = \left(s + \frac{1}{2} \right)^2 + \frac{3}{4} $$

and this time, the solution set of your differential equation is

$$ y(t) = A {\cal L}^{-1} \left[ \frac{s + \frac{1}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} } \right] + B {\cal L}^{-1} \left[ \frac{ \frac{\sqrt{3}}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} } \right] = A e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2}t\right) + B e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2}t\right) \ , $$

where $A$ and $B$ are real numbers, depending on $a,b$, which... Anyway again: the solution set is the $\mathbb{R}$-linear span of two linearly independent real functions

$$ e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2} t \right) \qquad \text{and} \qquad e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2} t \right) \ . $$

Thus, a real vector space of dimension two.

Alternatively, you could read what Wikipedia says about linear differential equations.

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Agustí Roig
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