Concatenation of solutions

Question

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1. What the problem is asking (in plain language)

We have a first‑order autonomous‑in‑t ODE

[ u’(t)=\alpha(t)\,g\bigl(u(t)\bigr),\qquad t\in I\subset\mathbb R, ]

where

(\alpha : I_{1}\to\mathbb R) and (g : I_{2}\to\mathbb R) are continuous,
(I_{1},I_{2}) are intervals, and
there is a special value (u_{0}\in I_{2}) with (g(u_{0})=0) (an equilibrium).

Two partial solutions are given:

(\varphi_{1}:(a,b_{1})\to I_{2}) solves the ODE on ((a,b_{1})) and approaches the equilibrium as (t\to b_{1}^{-});
(\varphi_{2}:(b_{2},c)\to I_{2}) solves the ODE on ((b_{2},c)) and approaches the equilibrium as (t\to b_{2}^{+}).

The intervals may overlap at a single point ((b_{1}=b_{2})) or be disjoint ((b_{1}<b_{2})).

We are asked to glue the two solutions together, staying at the equilibrium on the “gap’’ ([b_{1},b_{2}]), and to prove that the resulting function

[ \varphi(t)= \begin{cases} \displaystyle \varphi_{1}(t), & a<t<b_{1},\[4pt] u_{0}, & b_{1}\le t\le b_{2},\[4pt] \displaystyle \varphi_{2}(t), & b_{2}<t<c, \end{cases} ]

is itself a (global) solution of the same ODE on the whole interval ((a,c)).

In other words, we must show that (\varphi) is continuous on ((a,c)) and differentiable at every point, with derivative satisfying

[ \varphi’(t)=\alpha(t)\,g\bigl(\varphi(t)\bigr)\qquad\text{for all }t\in(a,c). ]

The only delicate points are the “junction’’ points (b_{1}) and (b_{2}); everywhere else (\varphi) coincides with a known solution, so the ODE is already satisfied there.

2. Detailed proof

2.1 Extend the two given solutions to the endpoints

Define the extended functions

[ \tilde\varphi_{1}:[a,b_{1}]\to I_{2},\qquad \tilde\varphi_{2}:[b_{2},c]\to I_{2} ]

by

[ \tilde\varphi_{1}(t)= \begin{cases} \varphi_{1}(t), & a<t<b_{1},
u_{0}, & t=b_{1}, \end{cases} \qquad \tilde\varphi_{2}(t)= \begin{cases} u_{0}, & t=b_{2},
\varphi_{2}(t), & b_{2}<t<c. \end{cases} ]

Because (\displaystyle \lim_{t\to b_{1}^{-}}\varphi_{1}(t)=u_{0}) and (\displaystyle \lim_{t\to b_{2}^{+}}\varphi_{2}(t)=u_{0}), the extensions are continuous at the added endpoints.

2.2 Continuity of (\varphi) on ((a,c))

By construction (\varphi) equals (\tilde\varphi_{1}) on ((a,b_{1}]), equals the constant (u_{0}) on ([b_{1},b_{2}]), and equals (\tilde\varphi_{2}) on ([b_{2},c)).
All three pieces are continuous, and they match at the joining points: [ \tilde\varphi_{1}(b_{1})=u_{0}=\tilde\varphi_{2}(b_{2}). ] Hence (\varphi) is continuous on the whole interval ((a,c)).

2.3 Differentiability away from the junction points

If (t\in (a,b_{1})), then (\varphi(t)=\varphi_{1}(t)). Since (\varphi_{1}) solves the ODE,

[ \varphi’(t)=\varphi_{1}’(t)=\alpha(t)g\bigl(\varphi_{1}(t)\bigr) =\alpha(t)g\bigl(\varphi(t)\bigr). ]

If (t\in (b_{2},c)), the same argument with (\varphi_{2}) gives the identity.

If (t\in (b_{1},b_{2})), (\varphi(t)\equiv u_{0}) is constant, so (\varphi’(t)=0). Because (g(u_{0})=0),

[ \alpha(t)g\bigl(\varphi(t)\bigr)=\alpha(t)g(u_{0})=0=\varphi’(t). ]

Thus the ODE holds everywhere except possibly at (b_{1}) and (b_{2}).

2.4 Differentiability at (b_{1})

We compute the derivative from the definition.
Recall that (\varphi(b_{1})=u_{0}).

Left‑hand limit ((h<0))

[ \frac{\varphi(b_{1}+h)-\varphi(b_{1})}{h} =\frac{\varphi_{1}(b_{1}+h)-u_{0}}{h},\qquad h\in(-\varepsilon,0). ]

Because (\varphi_{1}) is differentiable on ((a,b_{1})), the Mean Value Theorem provides a point (\xi_{h}\in(b_{1}+h,b_{1})) such that

[ \frac{\varphi_{1}(b_{1}+h)-u_{0}}{h}= \varphi_{1}’(\xi_{h}). ]

Since (h\to0^{-}) forces (\xi_{h}\to b_{1}) and (\varphi_{1}’) has a finite limit at (b_{1}) (see below), we obtain

[ \lim_{h\to0^{-}}\frac{\varphi(b_{1}+h)-\varphi(b_{1})}{h} =\lim_{h\to0^{-}}\varphi_{1}’(\xi_{h}) =\lim_{t\to b_{1}^{-}}\varphi_{1}’(t). ]

Now (\varphi_{1}) satisfies the ODE on ((a,b_{1})); thus for every (t\in(a,b_{1})),

[ \varphi_{1}’(t)=\alpha(t)g\bigl(\varphi_{1}(t)\bigr). ]

Take the limit as (t\to b_{1}^{-}). By continuity of (\alpha) and (g) and because (\varphi_{1}(t)\to u_{0}),

[ \lim_{t\to b_{1}^{-}}\varphi_{1}’(t) =\alpha(b_{1})\,g(u_{0})=0. ]

Hence the left‑hand derivative at (b_{1}) exists and equals (0).

Right‑hand limit ((h>0))

Two cases.

If (b_{1}<b_{2}), then for every sufficiently small (h>0) we have (b_{1}+h\in[b_{1},b_{2}]), and (\varphi(b_{1}+h)=u_{0}). Consequently

[ \frac{\varphi(b_{1}+h)-\varphi(b_{1})}{h}=0, ] so the right‑hand derivative is (0).

If (b_{1}=b_{2}), then for (h>0) small we are in the interval ((b_{2},c)) and (\varphi(b_{1}+h)=\varphi_{2}(b_{2}+h)). By the same mean‑value argument as above,

[ \frac{\varphi(b_{1}+h)-\varphi(b_{1})}{h} =\varphi_{2}’(\eta_{h}),\qquad \eta_{h}\in(b_{2},b_{2}+h). ]

Letting (h\to0^{+}) gives (\eta_{h}\to b_{2}) and, using the ODE for (\varphi_{2}),

[ \lim_{h\to0^{+}}\varphi_{2}’(\eta_{h}) =\alpha(b_{2})\,g(u_{0})=0. ]

Thus the right‑hand derivative is also (0).

Since the left and right limits coincide, (\varphi) is differentiable at (b_{1}) with

[ \boxed{\;\varphi’(b_{1})=0=\alpha(b_{1})g\bigl(\varphi(b_{1})\bigr)\; }. ]

2.5 Differentiability at (b_{2})

The argument is completely symmetric.
If (b_{1}<b_{2}) the left side uses the constant part (u_{0}), giving derivative (0).
If (b_{1}=b_{2}) we apply the mean‑value theorem to (\varphi_{1}) on the left. In all cases we obtain

[ \boxed{\;\varphi’(b_{2})=0=\alpha(b_{2})g\bigl(\varphi(b_{2})\bigr)\; }. ]

2.6 Conclusion

We have shown that

(\varphi) is continuous on ((a,c));
(\varphi) is differentiable at every point of ((a,c));
the derivative satisfies the differential equation everywhere:

[ \varphi’(t)=\alpha(t)g\bigl(\varphi(t)\bigr),\qquad \forall\,t\in(a,c). ]

Hence (\varphi) is a solution of the ODE on the whole interval ((a,c)), i.e. the two partial solutions can indeed be concatenated through the equilibrium value (u_{0}).

3. Final answer

The piecewise‑defined function

[ \varphi(t)= \begin{cases} \varphi_{1}(t), & a<t<b_{1},\[2mm] u_{0}, & b_{1}\le t\le b_{2},\[2mm] \varphi_{2}(t), & b_{2}<t<c, \end{cases} ]

is a global solution of the differential equation

[ u’(t)=\alpha(t)g\bigl(u(t)\bigr) ]

on the interval ((a,c)). The crucial points (b_{1}) and (b_{2}) are differentiable with derivative (0) because the right‑hand side of the ODE vanishes at the equilibrium (u_{0}) ((g(u_{0})=0)).

4. Common mistakes to avoid

Mistake	Why it’s wrong	How to fix it
Assuming differentiability at (b_{1},b_{2}) without proof.	The function changes definition there; one must check the limit of the difference quotient.	Use the definition of derivative, or the Mean Value Theorem, as done above.
Using (\varphi_{1}’(b_{1})) or (\varphi_{2}’(b_{2})) directly.	The original solutions are not defined at the endpoints, so those derivatives are not given a priori.	Take limits of (\varphi_{1}’(t)) as (t\to b_{1}^{-}) (and similarly for (\varphi_{2})).
Ignoring the continuity of (\alpha) and (g).	Without continuity we cannot pass limits through (\alpha(t)g(\varphi(t))).	Explicitly invoke continuity when computing (\displaystyle\lim_{t\to b_{1}^{-}}\alpha(t)g(\varphi_{1}(t))).
**Confusing the “gap’’ ([

Original question: Concatenation of solutions on Mathematics Stack Exchange, licensed CC BY-SA.