Existence and Uniqueness for First-Order Linear ODEs

calculus

A self-contained proof using only the Fundamental Theorem of Calculus.

Here’s an undergraduate-friendly proof of the existence and uniqueness of solutions to general first order homogeneous ODEs.

TheoremExistence and Uniqueness

Let I⊂RI\subset \RR be an interval (possibly all of R\RR) and PP be a continuous function on II. Then for every a∈Ia\in I, b∈Rb\in\RR the differential equation y′+P(x)y=0y^\prime + P(x)y=0 has a unique solution f ⁣:I→Rf\colon I\to \RR with f(a)=bf(a)=b, given by

f(x)=bexp⁡(∫[a,x]P)f(x)=\frac{b}{\exp\left(\int_{[a,x]}P\right)}

First, we verify existence: one way to guess the solution is to divide the equation through by yy, giving

y′y=−P\frac{y^\prime}{y}=-P

The left hand side of this is the result of differentiating log⁡y\log y with respect to xx, so we may write

()′=−P   ⟹   log⁡(y)=−∫P\left(\right)^\prime = -P\,\implies\, \log(y)=-\int P Exponentiating both sides gives the proposed solution, y=exp⁡(−∫P)y=\exp\left(-\int P\right)

Proof

Let a∈Ia\in I and b∈Rb\in\RR be arbitrary. As PP is continuous on II it is integrable, and thus for any x∈Ix\in I it is integrable on the subinterval [a,x][a,x]. Define the function A ⁣:I→RA\colon I\to\RR by A(x)=∫[a,x]PA(x)=\int_{[a,x]}P By the fundamental theorem of calculus, AA is differentiable and A′(x)=(∫[a,x]P)′=P(x)A^\prime(x)=\left(\int_{[a,x]}P\right)^\prime = P(x) Finally, define the function f ⁣:I→Rf\colon I\to \RR as f(x)=bexp⁡(−A(x))f(x)=b\exp(-A(x)). Note that A(a)=∫{a}P=0A(a)=\int_{\{a\}}P=0 so f(a)=bexp⁡(−A(a))=bexp⁡(0)=b.f(a)=b\exp(-A(a))=b\exp(0)=b.

We show ff satisfies the differential equation by direct computation. Since ff is a composition of differentiable functions, its differentiable and we proceed via the Chain Rule:

f′(x)=b(exp⁡(−A(x)))′=bexp⁡′(−A(x))(−A(x))′=bexp⁡(−A(x))(−P(x))=−P(x)f(x)\begin{align*} f^\prime(x)&=b\left(\exp(-A(x))\right)^\prime\\ &=b\exp^\prime(-A(x))(-A(x))^\prime\\ &=b\exp(-A(x))(-P(x))\\ &=-P(x)f(x) \end{align*}

Thus, f′(x)+P(x)f(x)=0f^\prime(x)+P(x)f(x)=0 for all xx, so ff is indeed a solution to the differential equation y′+Py=0y^\prime+Py=0, with f(a)=bf(a)=b.

Now that we know y′+P(x)y=0y^\prime + P(x)y=0 has solutions, we show that they are uniquely determined by initial condition.

Proof

Let gg be an arbitrary solution to y′+Py=0y^\prime + Py =0 with g(a)=bg(a)=b. We wish to show g(x)=bexp⁡(−A(x))g(x)= b\exp(-A(x)) for A(x)=∫[a,x]PA(x)=\int_{[a,x]}P as above. The rather striking idea is to consider the function h(x)=g(x)exp⁡(A(x))h(x)=g(x)\exp(A(x)) and attempt to show that hh is constant. Computing its derivative, we see

h′=(gexp⁡(A))′=g′exp⁡(A)+gexp⁡′(A)A′=g′exp⁡(A)+gexp⁡(A)P=(g′+Pg)exp⁡(A)=0\begin{align*} h^\prime&=\left(g\exp(A)\right)^\prime\\ &=g^\prime\exp(A)+g\exp^\prime(A)A^\prime\\ &=g^\prime\exp(A)+g\exp(A)P\\ &=(g^\prime+Pg)\exp(A)\\ &=0 \end{align*}

Where the last equality to zero follows as g′+Pg=0g^\prime + Pg=0 is exactly the assumption that gg solves our differential equation. Thus, we see h′(x)=0h^\prime(x)=0 for all xx,so hh is constant. Evaluating at x=ax=a we see h(a)=g(a)exp⁡(A(a))=g(a)exp⁡(0)=g(a)=bh(a)=g(a)\exp(A(a))=g(a)\exp(0)=g(a)=b Where A(a)=0A(a)=0 as it is the integral over [a,a]={a}[a,a]=\{a\} as we saw previously. Thus, h(x)h(x) is a constant function with h(a)=bh(a)=b, so in fact h(x)=bh(x)=b for all x∈Rx\in\RR. But solving this for gg gives b=g(x)exp⁡(A(x))  ⟹  g(x)=bexp⁡(−A(x))b=g(x)\exp(A(x))\implies g(x)=b\exp(-A(x)) and so gg is exactly the solution we already knew about. Thus, this is the only solution passing through (a,b)(a,b).

Non-Homogeneous Equations

The general first order linear equation is

y′+Py=Qy^\prime + P y = Q

Where P,QP,Q are continuous functions of xx. To ‘guess’ a solution to this is a bit more involved, but still possible. We’ll start by looking at the special case P(x)=xP(x)=x. Here the left hand side is just the expanded product rule of (xy)′(xy)^\prime, so we have

(xy)′=Q   ⟹   xy=∫[a,x]Q   ⟹   y=∫[a,x]Qx(xy)^\prime = Q \,\implies\, xy = \int_{[a,x]} Q\,\implies\, y=\frac{\int_{[a,x]} Q}{x}

In general of course we are not so lucky: y′+Pyy^\prime + Py is not the result of a product rule. But we can change this! Letting ff be some arbitrary function, let’s multiply through the whole equation to yield

fy′+fPy=fQf y^\prime + fP y = f Q

Can we find a particular function ff for which the left side is a product rule? If so, (fy)′=fy′+fPy(fy)^\prime = f y^\prime +fPy. This implies f′=fPf^\prime = fP, which is exactly our earlier homogeneous case: we know the solution to this is

f(x)=exp⁡(∫P)f(x)=\exp\left(\int P\right)

With this choice of ff, we can rewrite our equation as

(fy)′=fQ(fy)^\prime = fQ

which can be solved directly by antidifferentiation:

fy=∫fQ   ⟹   y=∫fQffy = \int fQ\,\implies\, y = \frac{ \int fQ}{f}

Plugging in our known candidate for ff gives our proposed solution:

y(x)=∫exp⁡(∫P)Qexp⁡(∫P)y(x)=\frac{\int \exp\left(\int P\right)Q}{\exp\left(\int P\right)}

TheoremExistence and Uniqueness

Let I⊂RI\subset \RR be an interval (possibly all of R\RR) and P,QP,Q be continuous functions on II. Then for every a∈Ia\in I, b∈Rb\in\RR the differential equation y′+P(x)y=Q(x)y^\prime + P(x)y=Q(x) has a unique solution f ⁣:I→Rf\colon I\to \RR with f(a)=bf(a)=b given by f(x)=b+∫[a,x][Qexp⁡(∫[a,x]P)]exp⁡(∫[a,x]P)f(x)=\frac{b+\int_{[a,x]}\left[Q\exp\left(\int_{[a,x]}P\right)\right]}{\exp\left(\int_{[a,x]} P\right)}

Exercise: Prove this! Existence = calculation with derivatives.

Uniqueness: if f,gf,g are both solutions, what do we know about h=f−gh=f-g? It solves the homogeneous problem h′+Ph=0h^\prime + Ph =0 with h(a)=0h(a)=0. But this has just one solution (we’ve proven uniqueness) and its the zero solution! So their difference is zero.

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