How do we know $\sin$ and $\cos$ are the only solutions to $y'' = -y$?

Question

According to Wikipedia, one way of defining the sine and cosine functions is as the solutions to the differential equation $y'' = -y$.

How do we know that sin and cos (and linear combinations of them, to include $y=e^{ix}$) are the only solutions to this equation?

It seems like a clever combination of inverse polynomials or exponentials could do the same.

Answer 1

One way to show uniqueness is the following:

$\textbf{Theorem:}$ A second order differential equation of the form $y''+P(x)y'+Q(x)y=0$ has at most two linearly independent solutions.

$\text{Sketch of Proof: }$ 1) Define the Wronskian determinant as $\det W_3$ where:$$W_3=\begin{bmatrix}y_1 &y_2 &y_3\\y_1' &y_2' &y_3'\\y_1'' &y_2'' &y_3''\end{bmatrix}$$ Here the $y_i$'s solve the aforementioned equation.

2. Show that

$$\exists ~c_i:\sum_{i=1}^3c_iy_i=0\iff \det W_3=0$$

3. Since $y_i''=-P(x)y_i'-Q(x)y_i$, line 3 is a linear combination of lines 1 and 2. Thus the matrix does not have full column rank and it's determinant is zero, and hence the 3 functions have to be linearly dependent.

You can use this theorem as follows: If you find two linearly independent solutions to a linear second order ODE, then every solution to that ODE can be expressed as a linear combination of those two. In your case, you know $\sin x, \cos x$ are solutions and you need to look no further, since they are linearly independent (calculate $W_2$ and it is non-zero).

For the particular case considered, one can also solve the equation using quadrature and obtain a general solution that looks like $y(x)=C_1\sin x+C_2 \cos x$.

EDIT: Proof of Lemma 2

To answer @blue 's comment completely, it is true that the proof of that statement is not JUST linear algebra, but in my opinion, it is still pretty easy to prove with no other machinery to be used other than the fact that a function that has zero derivative is constant. Here is a way that can be used to show both directions:

Specifically, let us assume that $\det W_3=0$ and also that all the $2\times 2$ sub-Wronskians between any two functions of the set $\{y_1,y_2,y_3\}$ are non-zero (if any of them were, then there would exist non-zero $c_i$'s that render the functions linearly dependent hence making the Wronskian vanish trivially).In this case, with

$T_{ij}=y'_iy_j-y'_jy_i=y_j^2\left(\frac{y_i}{y_j}\right)'\neq 0$

it can be easily shown that, after some algebraic simplifications: $$\left(\frac{T_{31}}{T_{21}}\right)'=y_1\det W_3$$

which yields that $T_{31}=AT_{21}$ for some non-zero constant $A$. This in turn can be manipulated, using the identity above, into the form

$$\left(\frac{y_3}{y_1}\right)'-A\left(\frac{y_2}{y_1}\right)'=0\iff y_3-Ay_2-By_1=0$$

and hence there exist non-zero constants that render the functions linearly independent.

Note: We had to assume in the above proof that all sub-Wronskians are non-vanishing. This is the reason why proofs involving Wronskians are usually inductive in the dimension of the linear vector space. If a sub-Wronskian is zero, it is trivial to see that the determinant of the Wronskian will be zero if one has proven it for all dimensions smaller than that. The only non-trivial case can be shown as above for any dimension $n$.

Answer 2

Without the need for other technology, it follows from the existence and uniqueness theorem for solutions of differential equations (Cauchy-Lipschitz, or Picard-Lindelöf, according to sources) in $\mathbb{R}^n$. Write the system in the linearized form $$ \begin{cases} y'=z\\ z' = -y \end{cases} $$ (this is clearly equivalent to $y''=-y$, after introducing an auxiliary variable).

Let $y(x)$ (together with $z(x)=y'(x)$) be any solution of this system of differential equations with initial values $y(0)=y_0,z(0)=z_0$.

Then, you can find $c_1,c_2$ such that the function $\hat{y}(x) = c_1 \sin x + c_2 \cos x$ solves the differential equation and has the same initial conditions $\hat{y}(0)=y_0, \hat{y}'(0) = z_0$. Hence, by the uniqueness theorem, $y(x)=\hat{y}(x)$, which is the result you need.

Answer 3

Consider the initial value problem $y'' = -y$, with $y(0) = 0$ and $y'(0) = 0$. Multiplying both sides by $2y'$, $$2y'' y' = -2y y' \implies ((y')^2)' = -(y^2)' \implies (y')^2 + y^2 = C$$ for some constant $C$. Applying the initial values yields $$C = (y'(0))^2 + (y(0))^2 = 0.$$ Thus, $$0 = (y')^2 + y^2 \ge y^2 \ge 0 \implies y = 0.$$ Hence, the initial value problem has only the unique solution $y = 0$.

Now, fix $a, b \in \Bbb{R}$ and consider two solutions $y_1$ and $y_2$ of the initial value problem $y'' = -y$ with $y(0) = a$ and $y'(0) = b$. Let $y = y_1 - y_2$. Then, $$y'' = (y_1 - y_2)'' = y_1'' - y_2'' = -y_1 + y_2 = -y.$$ Further, $$y(0) = y_1(0) - y_2(0) = a - a = 0$$ and similarly $y'(0) = 0$. Hence, $y$ satisfies the original initial value problem, thus $y = 0$ by the previous result, hence $y_1 = y_2$.

That is, we can choose $a$ and $b$ as we like, and we obtain at most one solution given the initial values. In particular, we can define $\sin$ with $a = 0$ and $b = 1$ and $\cos$ with $a = 1$ and $b = 0$.

Note that this doesn't guarantee a solution to the differential equation, but it's a way to get uniqueness.

Answer 4

Starting from $y'' = -y$, we can add $y$ to form

$$y'' + y = 0$$

This is a homogeneous second order linear differential equation which we can simplify by writing the characteristic polynomial as

$$r^2 + 1 = 0$$

or

$$r = \pm i$$

which are distinct roots. The general solution can be written as

$$y(x) = c_1\sin(x) + c_2\cos(x)$$

This tells us that $y_1 = \sin(x)$ is a solution and so is any constant multiple of it, such as $c_1\sin(x)$. Similarly, $y_2 = \cos(x)$ is another solution (and so is any function of the form $c_2\cos(x)$). So, by the principle of superposition, we can add $c_1\sin(x)$ and $c_2\cos(x)$ to form the general solution.

As any linear combination of $c_1\sin(x)$ + $c_2\cos(x)$ works, it is clear that $\sin(x)$ and $\cos(x)$ cannot be the only solutions.

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Principle of Superposition: If $y_1$ and $y_2$ are any two solutions of the homogeneous equation $y′′ + p(x)y′ + q(x)y = 0$. Then any function of the form $y = c_1 y_1 + c_2 y_2$ is also a solution of the equation, for any pair of constants $c_1$ and $c_2$.

Remark: However, while the general solution of $y′′ + p(x)y′ + q(x)y = 0$ will always be in the form of $c_1 y_1 + c_2 y_2$, where $y_1$ and $y_2$ are some solutions of the equation, the converse is not always true. Not every pair of solutions $y_1$ and $y_2$ could be used to give a general solution in the form $y = c_1 y_1 + c_2 y_2$.

You claim that $\sin(x)$ and $\cos(x)$ are the only solutions. Instead, you should focus on linear combinations of these:

$$ y_1 = \cos(x)$$ $$ y_2 = \sin(x)$$ $$ y_3 = \sin(x) + \cos(x)$$ $$ y_4 = 2\sin(x) + 3\cos(x)$$ $$ y_5 = \sin(x) + i\cos(x)$$ $$ y_6 = 10\sin(x) - 11i\cos(x)$$

among many others where $y_i = c_1\sin(x) + c_2\cos(x)$ for constant $c_1,c_2$.

Answer 5

Suppose $y=y(t)$ is a solution. We can solve this by subtitutions. Let the parameter of the following functions ($x,y,r,\theta$, to be introduced later) be $t$. Consider the substitution $x=y'$. Then this becomes a system of equations: $$x'=-y,\quad y'=x$$ By another substitution $x=r\cos\theta,\ y=r\sin\theta$ we get $$r'=0,\quad \theta'=1$$ (I don't have a pen now so maybe this could be $-1$ instead) Solve this you get $$r=c_1,\ \theta=t+c_2$$ and the original solution is $y=r\sin\theta=c_1\sin(t+c_2)$ which is a linear combination of $\sin$ and $\cos$

Answer 6

If you're comfortable with exponentials of linear maps/matrices, then this answer might be helpful.

Theorem: Let $A \in M_{n \times n}(\Bbb{R})$ be a given $n \times n$ matrix, and consider the differential equation $\xi' = A \xi$. Then, every solution $f: \Bbb{R} \to \Bbb{R}^n$ of this ODE is of the form \begin{align} f(t) = \exp(tA) \cdot \eta \end{align} for some $\eta \in \Bbb{R}^n$. (the $\cdot$ that appears here is matrix multiplication)

There are a couple of facts you need to know. The first is that for any $t \in \Bbb{R}$, the matrix $\exp(tA)$ is invertible and its inverse is $\exp(-tA)$. So, to prove this theorem, let $f: \Bbb{R} \to \Bbb{R}^n$ be any solution to the above ODE. Define a new function $g: \Bbb{R} \to \Bbb{R}^n$ by \begin{align} g(t) = \exp(-tA) \cdot f(t) \end{align} Now, you need to know a bit about multivariable calculus and the "generalised product rule", and you also need to know how to differentiate matrix exponentials. The results are pretty much the same as in single variable calculus, but their proofs require a bit more care. We have that for every $t \in \Bbb{R}$, \begin{align} g'(t) &= \left(\exp(-tA) \cdot (-A) \right) \cdot f(t) + \exp(-tA) \cdot f'(t) \\ &= - \exp(-tA) \cdot A \cdot f(t) + \exp(-tA) \cdot \left(A \cdot f(t) \right) \\ &= 0 \tag{$\ddot{\smile}$} \end{align} In the first line I used the product rule and the rule for differentiating matrix exponentials. In the second line, I used the fact that $f' = A \cdot f$ (by assumption).

Now, $(\ddot{\smile})$ says that the derivative of $g$ is always $0$. Hence, (by a corollary of the mean-value inequality) there is a vector $\eta \in \Bbb{R}^n$ such that for all $t \in \Bbb{R}$, \begin{align} g(t) = \eta \tag{$*$} \end{align} In other words, we have shown that $g$ is a constant function. Multiplying both sides of $(*)$ by $\exp(tA)$ immediately gives us that for all $t$, $f(t) = \exp(tA) \cdot \eta$, which is what we wanted to prove.

If this is unfamiliar, I suggest, you consider the special case $n=1$, so that there are no matrices, and everything is just multiplication by real numbers.

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To apply this general result to your specific question, we take $n=2$, and consider the ODE $\xi' = A \xi$, where $A$ is the $2 \times 2$ matrix \begin{align} A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \end{align} If you write out the equation $\xi' = A \xi$ in components, we get \begin{align} \begin{cases} \xi_1' &= \xi_2 \\ \xi_2' &= -\xi_1 \end{cases} \end{align} Differentiating the first again, and substituting into the second gives us $\xi_1'' = -\xi_1$, which is exactly what you asked (you just wrote $y'' = -y$ instead). It is sometimes helpful to consider a second order ODE as 2 first-order ODE's as I have done.

By what I proved above, we know that every solution to this ODE is of the form $\exp(tA) \cdot \eta$, for some $\eta \in \Bbb{R}^2$. You can verify that the matrix exponential in this case is given by \begin{align} \exp \begin{pmatrix} 0 & t \\ -t & 0 \end{pmatrix} &= \begin{pmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{pmatrix} \end{align}

Hence, the multiplying out the solution gives \begin{align} \begin{pmatrix} \xi_1(t) \\ \xi_2(t) \end{pmatrix} &= \begin{pmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{pmatrix} \cdot \begin{pmatrix} \eta_1 \\ \eta_2 \end{pmatrix} \\ &= \begin{pmatrix} \eta_1 \cos(t) + \eta_2 \sin(t) \\ -\eta_1 \sin(t) + \eta_2 \cos(t) \end{pmatrix} \end{align}

Hence, the solution to $y'' = -y$ (which I apologise, but in my notation is $\xi_1'' = -\xi_1$) is given by a linear combination of sines and cosines: \begin{align} \xi_1(t) = \eta_1 \cos(t) + \eta_2 \sin(t), \end{align} for some $\eta_1,\eta_2 \in \Bbb{R}$.

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To learn about matrix exponentials and their properties, and how to compute them, I suggest you take a look at Hirsch and Smale's book "Differential Equations, Dynamical Systems and Linear Algebra".

Answer 7

Intuitively, we can tell that $sin(x)$ and $cos(x)$ and all their multiples and linear combinations would be part of the solution since differentiating them twice gives us the same thing. E.g -> $$(sin(x))''=-sin(x)$$ and $$(cos(x))''=-cos(x)$$ But you suggest that functions of the form $y = \frac{1}{g(x)}$ can be possible, where $g(x)$ is a polynomial of degree $>=2$, so that when it is differentiated twice, the derivative is not $0$. Here's why this is not a possible function: $$y'=-\frac{g(x)'}{g(x)^2}$$ $$y''=\frac{2}{g(x)^3}-\frac{g(x)''}{g(x)^2}$$ Now $y''=-y$ suggests $$-\frac{1}{g(x)}=\frac{2}{g(x)^3}-\frac{g(x)''}{g(x)^2}$$ multiplying by $g(x)^2$, $$-g(x)=\frac{2}{g(x)}-g(x)''$$ Substituting $y=1/g(x)$, $$-g(x)=2y-g(x)''$$, which means, $$y=\frac{g(x)''-g(x)}{2}\neq\frac{1}{g(x)}$$ Thus, this rules out all functions of the form $y=1/g(x)$.

Answer 8

Assume a solution for $y$ will be proportional to $e^{mx}$ for a constant $m$. This means $$\frac{d^2}{dx^2} (e^{mx}) + e^{mx} = 0 \to m^2 e^{mx} + e^{mx} = 0$$

The only solutions to this are $m = \pm i$. This means that $$y = c_1 e^{ix} + c_2e^{-ix} = c_1(\cos(x)+i\sin(x)) + c_2(\cos(x)-i\sin(x))$$

This equals $$y = (c_1+c_2)\cos(x) + i(c_1-c_2)\sin(x)$$

Renaming the variables finds $$y = d_1\cos(x)+d_2\sin(x)$$

Answer 9

Note that $y=0$ is trivial solution. Also, $sinx$ and $cosx$ both satisfy the homogeneous ODE $y''+y=0$ and $\frac{sinx}{cosx}\ne$ constant hence both are linearly independent solutions of given ODE of second order. So the solution space (a vector space in itself of dimension $2$) must have basis $B=${$sinx,cosx$}. Claiming another $y*$ as a solution of given ODE will eventually lead to $y*=c_1cosx+c_2sinx$ where $c_1$ and $c_2$ are arbitrary constants.