For a *n*-order *E*-valued differential equation

F(x,y,{y}^{\prime},\dots ,{y}^{(n)})=0,\phantom{\rule{1em}{0ex}}x\in I,

we say that it has the Hyers-Ulam stability or it is stable in the sense of Hyers-Ulam sense if for a given \u03f5>0 and a *n* times strongly differentiable function f:I\to E satisfying \parallel F(x,f,{f}^{\prime},\dots ,{f}^{(n)})\parallel \le \u03f5 for all x\in I, then there exists an exact solution *h* of this equation such that \parallel f(x)-h(x)\parallel \le K(\u03f5) for all x\in I, where K(\u03f5) depends only on *ϵ*, and {lim}_{\u03f5\to 0}K(\u03f5)=0. More generally, if *ϵ* and K(\u03f5) are replaced by two control functions \phi ,\mathrm{\Phi}:I\to [0,+\mathrm{\infty}), respectively, we say that the above differential equation has the Hyers-Ulam-Rassias stability or it is stable in the sense of Hyers-Ulam-Rassias.

The following theorem is the main result of this paper.

**Theorem 2.1** *Let* \phi :I\to [0,\mathrm{\infty}) *be a function such that* \phi (x) *is integrable on the interval* [r,s] *for each* r,s\in I *with* r\le s. *Suppose that* f\in {C}^{2}[I,E] *satisfies the differential inequality*

\parallel p(x){f}^{\u2033}(x)+q(x){f}^{\prime}(x)+\lambda f(x)\parallel \le \phi (x)

(2)

*for all* x\in I. *Moreover*, *assume that the equation*

{\omega}^{\prime}(t)={\left(p(\omega (t))\right)}^{\frac{1}{2}}

(3)

*has a solution on* J=(c,d), *where* t\in J *corresponds to* \omega (t)\in I. *Then there exists* h\in {C}^{2}[I,E] *such that* *h* *satisfies Eq*. (1) *and*

\parallel f(x)-h(x)\parallel \le |{\int}_{{a}_{0}}^{{\omega}^{-1}(x)}{\int}_{{a}_{0}}^{{\omega}^{-1}(\rho )}\frac{\phi (\sigma )}{{\omega}^{\prime}({\omega}^{-1}(\sigma ))}\cdot \frac{1}{{\omega}^{\prime}({\omega}^{-1}(\rho ))}\phantom{\rule{0.2em}{0ex}}d\sigma \phantom{\rule{0.2em}{0ex}}d\rho |

(4)

*for all* x\in I, *where* {a}_{0}\in I *is an arbitrary fixed point*.

*Proof* Firstly, we take the solution \omega (t) of Eq. (3) and make an appropriate substitution for the variable *x*. Obviously, the condition {\omega}^{\prime}(t)>0, t\in J, implies that the inverse function t={\omega}^{-1}(x) exists on *I*.

Now, we define the map \eta :J\to E by \eta (t):=f(\omega (t)) for each t\in J. Obviously, \eta \in {C}^{2}[J,E]. Then we can obtain

{\eta}^{\prime}(t)={\omega}^{\prime}(t){f}^{\prime}(\omega (t)),\phantom{\rule{2em}{0ex}}{\eta}^{\u2033}(t)={\omega}^{\u2033}(t){f}^{\prime}(\omega (t))+{\omega}^{\prime}{(t)}^{2}{f}^{\u2033}(\omega (t)).

Moreover, since {p}^{\prime}(x)=2q(x), we get {p}^{\prime}(\omega (t))=2q(\omega (t)) for each t\in J. Thus, it follows from Eq. (3) that

2{\omega}^{\prime}(t){\omega}^{\u2033}(t)={p}^{\prime}(\omega (t)){\omega}^{\prime}(t)=2{\omega}^{\prime}(t)q(\omega (t)),

which implies that {\omega}^{\u2033}(t)=q(\omega (t)) for each t\in J.

Therefore, we have

\begin{array}{rcl}{\eta}^{\u2033}(t)+\lambda \eta (t)& =& {\omega}^{\prime}{(t)}^{2}{f}^{\u2033}(\omega (t))+{\omega}^{\u2033}(t){f}^{\prime}(\omega (t))+\lambda f(\omega (t))\\ =& p(\omega (t)){f}^{\u2033}(\omega (t))+q(\omega (t)){f}^{\prime}(\omega (t))+\lambda f(\omega (t))\end{array}

for all t\in J. Note that \omega (t)\in I for each t\in J. Based on the inequality (2) and the preceding equality, we can obtain

\parallel {\eta}^{\u2033}(t)+\lambda \eta (t)\parallel \le \phi (\omega (t))

(5)

for all t\in J.

We set \alpha (t)={\eta}^{\u2033}(t)+\lambda \eta (t), \beta (t)={\eta}^{\prime}(t)+i\sqrt{\lambda}\eta (t). Then we have

\begin{array}{rl}{\beta}^{\prime}(t)& ={\eta}^{\u2033}(t)+i\sqrt{\lambda}{\eta}^{\prime}(t)\\ =(\alpha (t)-\lambda \eta (t))+i\sqrt{\lambda}{\eta}^{\prime}(t)\\ =i\sqrt{\lambda}({\eta}^{\prime}(t)+i\sqrt{\lambda}\eta (t))+\alpha (t)\\ =i\sqrt{\lambda}\beta (t)+\alpha (t).\end{array}

It follows that {\beta}^{\prime}(t)-i\sqrt{\lambda}\beta (t)=\alpha (t). Multiplying both sides of the previous equality by {e}^{-i\sqrt{\lambda}t}, t\in J, we obtain

{({e}^{-i\sqrt{\lambda}t}\beta (t))}^{\prime}={e}^{-i\sqrt{\lambda}t}({\beta}^{\prime}(t)-i\sqrt{\lambda}\beta (t))={e}^{-i\sqrt{\lambda}t}\alpha (t).

By integrating both sides of the above equality from {c}_{0} (here {c}_{0} is an arbitrary fixed point in *J*) to *s* with respect to *τ*, {c}_{0},s\in J, it follows that

\begin{array}{rl}{\int}_{{c}_{0}}^{s}{e}^{-i\sqrt{\lambda}\tau}\alpha (\tau )\phantom{\rule{0.2em}{0ex}}d\tau & ={\int}_{{c}_{0}}^{s}{({e}^{-i\sqrt{\lambda}\tau}\beta (\tau ))}^{\prime}\phantom{\rule{0.2em}{0ex}}d\tau \\ ={e}^{-i\sqrt{\lambda}s}\beta (s)-{e}^{-i\sqrt{\lambda}{c}_{0}}\beta ({c}_{0}).\end{array}

Furthermore, we can infer that

\beta (s)={e}^{i\sqrt{\lambda}s}({e}^{-i\sqrt{\lambda}{c}_{0}}\beta ({c}_{0})+{\int}_{{c}_{0}}^{s}{e}^{-i\sqrt{\lambda}\tau}\alpha (\tau )\phantom{\rule{0.2em}{0ex}}d\tau )

for all s\in J.

In view of \beta (s)={\eta}^{\prime}(s)+i\sqrt{\lambda}\eta (s), for each s\in J, we have

{e}^{i\sqrt{\lambda}s}\beta (s)={e}^{i\sqrt{\lambda}s}({\eta}^{\prime}(s)+i\sqrt{\lambda}\eta (s))={({e}^{i\sqrt{\lambda}s}\eta (s))}^{\prime}.

Analogously, by integrating both sides of the last equality from {c}_{0} to *t* with respect to *s*, we conclude that

\begin{array}{rl}{e}^{i\sqrt{\lambda}t}\eta (t)-{e}^{i\sqrt{\lambda}{c}_{0}}\eta ({c}_{0})& ={\int}_{{c}_{0}}^{t}{({e}^{i\sqrt{\lambda}s}\eta (s))}^{\prime}\phantom{\rule{0.2em}{0ex}}ds\\ ={\int}_{{c}_{0}}^{t}{e}^{i\sqrt{\lambda}s}\beta (s)\phantom{\rule{0.2em}{0ex}}ds\\ ={\int}_{{c}_{0}}^{t}{e}^{2i\sqrt{\lambda}s}({e}^{-i\sqrt{\lambda}{c}_{0}}\beta ({c}_{0})+{\int}_{{c}_{0}}^{s}{e}^{-i\sqrt{\lambda}\tau}\alpha (\tau )\phantom{\rule{0.2em}{0ex}}d\tau )\phantom{\rule{0.2em}{0ex}}ds\\ =\frac{{e}^{i\sqrt{\lambda}(2t-{c}_{0})}-{e}^{i\sqrt{\lambda}{c}_{0}}}{2\sqrt{\lambda}i}\beta ({c}_{0})+{\int}_{{c}_{0}}^{t}{\int}_{{c}_{0}}^{s}{e}^{-i\sqrt{\lambda}(\tau -2s)}\alpha (\tau )\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds.\end{array}

(6)

Define the map g:J\to E by

g(t):={e}^{-i\sqrt{\lambda}(t-{c}_{0})}\eta ({c}_{0})+\frac{{e}^{i\sqrt{\lambda}(t-{c}_{0})}-{e}^{-i\sqrt{\lambda}(t-{c}_{0})}}{2\sqrt{\lambda}i}\beta ({c}_{0}),\phantom{\rule{1em}{0ex}}t\in J.

(7)

Clearly, g\in {C}^{2}[J,E]. Therefore, we can infer that

{g}^{\prime}(t)=-i\sqrt{\lambda}{e}^{-i\sqrt{\lambda}(t-{c}_{0})}\eta ({c}_{0})+\frac{i\sqrt{\lambda}{e}^{i\sqrt{\lambda}(t-{c}_{0})}+i\sqrt{\lambda}{e}^{-i\sqrt{\lambda}(t-{c}_{0})}}{2\sqrt{\lambda}i}\beta ({c}_{0}),

and thus

{g}^{\u2033}(t)=-\lambda {e}^{-i\sqrt{\lambda}(t-{c}_{0})}\eta ({c}_{0})+\frac{-\lambda {e}^{i\sqrt{\lambda}(t-{c}_{0})}+\lambda {e}^{-i\sqrt{\lambda}(t-{c}_{0})}}{2\sqrt{\lambda}i}\beta ({c}_{0}).

(8)

From the equalities (7) and (8), it is easy to see that {g}^{\u2033}(t)=-\lambda g(t) for each t\in J. Furthermore, we can infer from Eq. (6) that

\begin{array}{rl}{e}^{i\sqrt{\lambda}t}\eta (t)& ={e}^{i\sqrt{\lambda}{c}_{0}}\eta ({c}_{0})+\frac{{e}^{i\sqrt{\lambda}(2t-{c}_{0})}-{e}^{i\sqrt{\lambda}{c}_{0}}}{2\sqrt{\lambda}i}\beta ({c}_{0})+{\int}_{{c}_{0}}^{t}{\int}_{{c}_{0}}^{s}{e}^{-i\sqrt{\lambda}(\tau -2s)}\alpha (\tau )\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\\ ={e}^{i\sqrt{\lambda}t}g(t)+{\int}_{{c}_{0}}^{t}{\int}_{{c}_{0}}^{s}{e}^{-i\sqrt{\lambda}(\tau -2s)}\alpha (\tau )\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\end{array}

for each t\in J. Then, by Eq. (5), we can obtain

\begin{array}{rl}\parallel \eta (t)-g(t)\parallel & =\parallel {e}^{-i\sqrt{\lambda}t}{\int}_{{c}_{0}}^{t}{\int}_{{c}_{0}}^{s}{e}^{-i\sqrt{\lambda}(\tau -2s)}\alpha (\tau )\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\parallel \\ \le \left|{\int}_{{c}_{0}}^{t}{\int}_{{c}_{0}}^{s}\parallel \alpha (\tau )\parallel \phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\right|\\ \le \left|{\int}_{{c}_{0}}^{t}{\int}_{{c}_{0}}^{s}\phi (\omega (\tau ))\phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\right|.\end{array}

(9)

Finally, we define the map h:I\to E by

h(x)=g({\omega}^{-1}(x)),\phantom{\rule{1em}{0ex}}x\in I.

Therefore, we get

{h}^{\prime}(x)={g}^{\prime}({\omega}^{-1}(x)){({\omega}^{-1}(x))}^{\prime}.

(10)

Furthermore, we can obtain

{h}^{\u2033}(x)={({\omega}^{-1}(x))}^{\u2033}{g}^{\prime}({\omega}^{-1}(x))+{\left[{({\omega}^{-1}(x))}^{\prime}\right]}^{2}{g}^{\u2033}({\omega}^{-1}(x)).

(11)

According to the derivative of the inverse function, it follows that

{({\omega}^{-1}(x))}^{\prime}=\frac{1}{{\omega}^{\prime}(t)},\phantom{\rule{2em}{0ex}}{({\omega}^{-1}(x))}^{\u2033}=\frac{-{\omega}^{\u2033}(t)}{{[{\omega}^{\prime}(t)]}^{2}}.

(12)

Moreover, in view of {\omega}^{\prime}(t)={[p(\omega (t))]}^{\frac{1}{2}} together with \omega (t) being a solution of Eq. (3), we can infer from Eq. (12) that

{({\omega}^{-1}(x))}^{\prime}={(p(x))}^{-\frac{1}{2}},\phantom{\rule{2em}{0ex}}{({\omega}^{-1}(x))}^{\u2033}=-q(x){(p(x))}^{-\frac{3}{2}}.

(13)

Then it follows from Eqs. (8), (10), (11), and (13) that

\begin{array}{rl}{h}^{\u2033}(x)& =-q(x)p{(x)}^{-\frac{3}{2}}{g}^{\prime}({\omega}^{-1}(x))+p{(x)}^{-1}{g}^{\u2033}({\omega}^{-1}(x))\\ =p{(x)}^{-1}(-q(x)p{(x)}^{-\frac{1}{2}}{g}^{\prime}({\omega}^{-1}(x))+{g}^{\u2033}({\omega}^{-1}(x)))\\ =p{(x)}^{-1}(-q(x){h}^{\prime}(x)-\lambda g({\omega}^{-1}(x)))\\ =p{(x)}^{-1}(-q(x){h}^{\prime}(x)-\lambda h(x)).\end{array}

Hence, we get

p(x){h}^{\u2033}(x)+q(x){h}^{\prime}(x)+\lambda h(x)=0

for all x\in I. This means that h\in {C}^{2}(I,E) is a solution of Eq. (1).

Recall that \eta (t)=f(\omega (t)) for each t\in J, namely, f(x)=\eta ({\omega}^{-1}(x)) for each x\in I. By Eq. (9), we can obtain

\begin{array}{rl}\parallel f(x)-h(x)\parallel & =\parallel \eta ({\omega}^{-1}(x))-g({\omega}^{-1}(x))\parallel \\ \le |{\int}_{{a}_{0}}^{{\omega}^{-1}(x)}{\int}_{{a}_{0}}^{{\omega}^{-1}(\rho )}\frac{\phi (\sigma )}{{\omega}^{\prime}({\omega}^{-1}(\sigma ))}\cdot \frac{1}{{\omega}^{\prime}({\omega}^{-1}(\rho ))}\phantom{\rule{0.2em}{0ex}}d\sigma \phantom{\rule{0.2em}{0ex}}d\rho |,\end{array}

(14)

where {a}_{0}={\omega}^{-1}({c}_{0})\in I. □

**Remark 1** In Theorem 2.1, the arbitrariness of {a}_{0} does not mean that the right-hand side of the inequality (4) may be arbitrarily changed with respect to an appropriate function *h*, because one can see from Eq. (7) that the desired function *h* depends on the choice of {a}_{0}. Furthermore, we can conclude from Eq. (4) that for an arbitrary fixed {a}_{0}\in I, there exists an appropriate solution *h* of Eq. (1) such that *h* can be used to approximate the function *f* that satisfies the inequality (2), and the error can be estimated by the control function of the right-hand side of Eq. (4).

The following result associated with the Hyers-Ulam stability of Eq. (1) is a direct consequence of Theorem 2.1.

**Corollary 2.2** *Let* I=(a,b) *be a finite interval*, *i*.*e*., -\mathrm{\infty}<a<b<+\mathrm{\infty} *and let* \u03f5>0 *be a given number*. *Suppose that* f\in {C}^{2}[I,E] *satisfies the differential inequality*

\parallel p(x){f}^{\u2033}(x)+q(x){f}^{\prime}(x)+\lambda f(x)\parallel \le \u03f5

*for all* x\in I. *Moreover*, *assume that Eq*. (3) *has a solution* \omega (t) *on* J=(c,d), *where* t\in J *corresponds to* \omega (t)\in I. *Then there exists* h\in {C}^{2}[I,E] *such that* *h* *satisfies Eq*. (1) *and*

\parallel f(x)-h(x)\parallel \le \frac{{(d-c)}^{2}}{2}\u03f5

(15)

*for all* x\in I.

*Proof* According to Theorem 2.1, it suffices to verify that the inequality (15) holds for all x\in I. By Eqs. (5) and (9), we have

\begin{array}{rl}\parallel f(x)-h(x)\parallel & =\parallel \eta (t)-g(t)\parallel \\ \le \left|{\int}_{{c}_{0}}^{t}{\int}_{{c}_{0}}^{s}\parallel \alpha (\tau )\parallel \phantom{\rule{0.2em}{0ex}}d\tau \phantom{\rule{0.2em}{0ex}}ds\right|\\ \le \frac{{(t-{c}_{0})}^{2}}{2}\u03f5\\ <\frac{{(d-c)}^{2}}{2}\u03f5\end{array}

(16)

for all x\in I. □

**Remark 2** In Corollary 2.2, since {c}_{0}\in (c,d) is an arbitrary fixed point, the right-hand side of the equality (15) can be further improved. In fact, since t\in (c,d) is arbitrary, if we take the midpoint of the interval (c,d), *i.e.*, putting {c}_{0}=\frac{c+d}{2}, then we can obtain a better upper bound of the equality (15). Furthermore, the inequality (15) can be improved as

\parallel f(x)-h(x)\parallel \le \frac{{(d-c)}^{2}}{8}\u03f5.