The following mathematical analysis is based on the consideration of a nonlinear ODE derived from the model of a two-phase system consisting of a small, single gas-filled bubble immersed in an incompressible, highly viscous liquid under different outer pressure impacts. The bubble is assumed to be spherically symmetric and the origin of the coordinate system is placed in the center of the bubble. Combination of the corresponding continuity and momentum equation connected with the condition of incompressibility, i.e.,
\(\frac{\partial }{\partial t}(r^2v(r,t))=0\), and the kinematic boundary condition on the bubble surface, i.e.,
\(v(R,t)={\dot{R}}\), yields a reduced equation of motion for the bubble radius
\(R=R(t)\); see [
5] for more details. The derived dimensionless ODE is then given by
$$\begin{aligned}&\ddot{{\bar{R}}} + \frac{4 \eta ^*}{M^* {\bar{R}}^2+4 B^*} \dot{{\bar{R}}} + \frac{\tfrac{3}{2} M^* {\bar{R}}^2-A^*}{M^* {\bar{R}}^3+4 B^* {\bar{R}}} \dot{{\bar{R}}}^2+ \frac{2 \sigma ^*({\bar{R}}^2-1) + p_c {\bar{R}}^3-1}{M^* {\bar{R}}^4 + 4 B^* {\bar{R}}^2} = \frac{(p_c-{\bar{p}}_\infty ({\bar{t}})) {\bar{R}}}{M^* {\bar{R}}^2+4 B^*} \end{aligned}$$
(1)
with the non-dimensionalization
$$\begin{aligned} {\bar{R}}=\dfrac{R}{R_0},\quad {\bar{p}}=\dfrac{p}{p_0},\quad {\bar{t}}=\dfrac{t}{\tau }, \end{aligned}$$
(2)
where
R,
p and
t represent the bubble radius, the pressure and the time variable, respectively. The index
\(_0\) denotes the corresponding values at rest, that are
\(R_0=R(0)\) and
\(p_0=p(0)\), and
\(\tau \) stands for the excitation time
\(\tau :=R_0^2/\nu _\mathrm{l}\) with the liquid kinematic viscosity
\(\nu _\mathrm{l}\). Moreover, the non-dimensional coefficients are defined by
$$\begin{aligned} A^*&:= \dfrac{4\alpha +2\beta }{p_0\,\tau ^2} ,\quad B^* := \dfrac{\beta }{p_0\,\tau ^2} ,\quad M^* := \dfrac{\varrho _\mathrm{l}\,R_0^2}{p_0\,\tau ^2}, \nonumber \\ \sigma ^*&:= \dfrac{\sigma }{p_0\,R_0} ,\quad \eta ^* := \dfrac{\eta }{p_0\,\tau } ,\quad p_{g_0}^* := \dfrac{p_{g_0}}{p_0}. \end{aligned}$$
(3)
Here
\(\alpha \) and
\(\beta \) are non-Newtonian material coefficients arising from the constitutive relation of a second-order liquid [
6‐
9], namely
$$\begin{aligned} \varvec{\sigma } = -p {\varvec{I}} + \eta {\varvec{A}}_1 + \alpha {\varvec{A}}_1^2 + \beta {\varvec{A}}_2 , \end{aligned}$$
(4)
with the Rivlin–Ericksen tensors
\({\varvec{A}}_1\) and
\({\varvec{A}}_2\). The remaining quantities are the liquid density
\(\varrho _\mathrm{l}\), the surface tension
\(\sigma \), the dynamic viscosity
\(\eta \), and the gas pressure at rest
\(p_{g_0}\).
For steady solutions assume
$$\begin{aligned} {\bar{R}}=:k_o>0 , \quad \dot{{\bar{R}}}=0 ,\quad \ddot{{\bar{R}}}=0 \end{aligned}$$
(5)
and let
$$\begin{aligned} {\bar{p}}_\infty \left( {\bar{t}}\right) =: p_c . \end{aligned}$$
(6)
Then, (
1) becomes a third-degree polynomial
$$\begin{aligned} 2 \sigma ^*(k_o^2-1) + p_c k_o^3 - 1 = 0 \end{aligned}$$
(7)
with the (stationary) solution
\(k_o >0 \) for
\(p_c \in [0,\infty )\).
Equations (
1) and (
7) are rewritten as
$$\begin{aligned} \ddot{{\bar{R}}} + \frac{4 \eta ^*}{M^* {\bar{R}}^2+4 B^*} \dot{{\bar{R}}} + \frac{\tfrac{3}{2} M^* {\bar{R}}^2 -A^*}{M^* {\bar{R}}^3+4 B^* {\bar{R}}} \dot{{\bar{R}}}^2 + \frac{2 \sigma ^*({\bar{R}}^2-k_o^2) + p_c ({\bar{R}}^3-k_o^3)}{M^* {\bar{R}}^4+4 B^*{\bar{R}}^2} = \frac{(p_c-{\bar{p}}_\infty ({\bar{t}})) {\bar{R}}}{M^* {\bar{R}}^2+4 B^*}. \end{aligned}$$
(8)
To qualitatively analyze the ODE (
8), one has to apply appropriate substitutions. Using
\(x({\bar{t}}) := {\bar{R}}({\bar{t}}) - k_o\) [thus
\({\bar{R}}({\bar{t}})=x({\bar{t}})+k_o\)] with
\({\dot{x}}=\dot{{\bar{R}}}\) and
\(\ddot{x}=\ddot{{\bar{R}}}\) in (
1) yields
$$\begin{aligned}&\ddot{x}\, + \,\frac{4\eta ^*}{M^*(x+k_o)^2+4B^*}\,{\dot{x}}+ \frac{\tfrac{3}{2} M^*(x+k_o)^2 -A^*}{M^*(x+k_o)^3+4 B^*(x+k_o)}\,{\dot{x}}^2 \nonumber \\&\quad +\frac{2\sigma ^*(x+2 k_o)+p_c ((x+k_o)^2+(x+k_o)k_o + k_o^2)}{M^*(x+k_o)^4+4 B^*(x+k_o)^2}x = \frac{(p_c-{\bar{p}}_\infty ({\bar{t}}))(x+k_o)}{M^* (x+k_o)^2+4 B^*}. \end{aligned}$$
(9)
For the sake of simplification, one defines for
\(x > -k_o\)$$\begin{aligned} f(x)&:= \frac{4\eta ^*}{M^*(x+k_o)^2+4B^*} \; (>0) , \end{aligned}$$
(10)
$$\begin{aligned} p(x)&:= \frac{\tfrac{3}{2} M^*(x+k_o)^2 -A^*}{M^*(x+k_o)^3+4 B^*(x+k_o)}\, ,\\ \nonumber g(x)&:= \frac{2\sigma ^*(x+2 k_o)+p_c ((x+k_o)^2+(x+k_o)k_o + k_o^2)}{M^*(x+k_o)^4+4 B^*(x+k_o)^2} \end{aligned}$$
(11)
$$\begin{aligned}&\, = \frac{p_c (x+k_o)^2+(p_c k_o +2\sigma ^*)(x+2 k_o)}{M^*(x+k_o)^4+4 B^*(x+k_o)^2} \; (>0) ,\end{aligned}$$
(12)
$$\begin{aligned} h(x)&:= \frac{x+k_o}{M^* (x+k_o)^2+4 B^*} \; (>0) . \end{aligned}$$
(13)
Thus (
9) becomes
$$\begin{aligned} \ddot{x} + (f(x)+p(x){\dot{x}})\,{\dot{x}} + g(x) \cdot x = h(x)(p_c-{\bar{p}}_\infty ({\bar{t}})) . \end{aligned}$$
(14)
As preparation for the subsequent theorems, one considers some help functions. For an appropriate constant
\(c_0 > 0\), the expression
$$\begin{aligned} q(x) := \exp \left( c_0 + \int \limits _0^{x} p({\tilde{x}})\mathrm{d}{\tilde{x}}\right) \end{aligned}$$
(15)
is computed by using partial fraction decomposition and integration:
$$\begin{aligned} p(x)&= -\frac{A^*}{4 B^*} \frac{1}{(x+k_o)} + \left( \frac{3}{4} + \frac{A^*}{8B^*}\right) \frac{2 M^*(x+k_o)}{M^*(x+k_o)^2+4 B^*} ,\nonumber \\ c_0 + \int \limits _0^{x} p({\tilde{x}})\mathrm{d}{\tilde{x}}&= -\frac{A^*}{4 B^*} \ln {(x+k_o)} + \left( \frac{3}{4} + \frac{A^*}{8B^*}\right) \ln {\left( M^*(x+k_o)^2+4 B^*\right) } ,\nonumber \\ q(x)&= (x+k_o)^{-\tfrac{A^*}{4 B^*}} \cdot \left( M^*(x+k_o)^2+4 B^*\right) ^{\left( \tfrac{3}{4} + \tfrac{A^*}{8B^*}\right) } . \end{aligned}$$
(16)
The term (
16) has the following behavior:
$$\begin{aligned} q(x) = {\mathcal {O}}(x^{3/2}) \;\; (x \rightarrow \infty ),\quad q(x) = {\mathcal {O}} \left( (x+k_o)^{-\tfrac{A^*}{4 B^*}}\right) \;\; (x \rightarrow -k_o), \end{aligned}$$
(17)
where
\({\mathcal {O}}(.)\) is the Big-O notation.
Moreover,
$$\begin{aligned} q'(x) = \frac{\mathrm{d} q}{\mathrm{d} x} = q(x) p(x) . \end{aligned}$$
(18)
By defining
$$\begin{aligned} G_{}(x)&:= \int \limits _0^{x} q^{2}({\tilde{x}}) g({\tilde{x}}) \; {\tilde{x}} {\mathrm{d}}{\tilde{x}}, \end{aligned}$$
(19)
$$\begin{aligned} F_{}(x)&:= \int \limits _0^{x} q({\tilde{x}}) f({\tilde{x}}) \; \mathrm{d}{\tilde{x}} , \end{aligned}$$
(20)
one deduces
$$\begin{aligned} G_{}(x)&= \int \limits _0^{x}({\tilde{x}}+k_o)^{-\tfrac{A^*}{2 B^*}} (M^*({\tilde{x}}+k_o)^2+4 B^*)^{\left( \tfrac{3}{2} + \tfrac{A^*}{4 B^*}\right) } \\&\quad \cdot \frac{2 \sigma ^*({\tilde{x}}+2 k_o)+p_c (({\tilde{x}}+k_o)^2+({\tilde{x}}+k_o)k_o + k_o^2)}{M^*({\tilde{x}}+k_o)^4+4 B^*({\tilde{x}}+k_o)^2} \, {\tilde{x}} \, {\mathrm{d}}{\tilde{x}} \\&= \int \limits _0^{x}({\tilde{x}}+k_o)^{-2-\tfrac{A^*}{2 B^*}} (M^*({\tilde{x}}+k_o)^2+4 B^*)^{\left( \tfrac{1}{2} + \tfrac{A^*}{4 B^*} \right) } \\&\quad \cdot \left( 2 \sigma ^*({\tilde{x}}+2 k_o)+p_c (({\tilde{x}}+k_o)^2+({\tilde{x}}+k_o)k_o + k_o^2)\right) {\tilde{x}} {\mathrm{d}}{\tilde{x}}. \end{aligned}$$
Hence
$$\begin{aligned} G(x)>0 \quad (x \ne 0) , \quad G(x) \rightarrow \infty \quad \text{ for } \;\; x \rightarrow - k_o \;\; \text{ or } \;\; x \rightarrow \infty . \end{aligned}$$
For
F, it is easy to check that
$$\begin{aligned} F(x)g(x) x>0 \quad (x \not = 0). \end{aligned}$$
(21)
In what follows, the subsequent behavior will be useful:
$$\begin{aligned} h(x)q(x)&= \frac{x+k_o}{M^* (x+k_o)^2+4 B^*}(x+k_o)^{-\tfrac{A^*}{4 B^*}} \left( M^*(x+k_o)^2+4 B^* \right) ^{\left( \tfrac{3}{4} + \tfrac{A^*}{8 B^*} \right) } \\&= (x+k_o)^{1-\tfrac{A^*}{4 B^*}} \left( M^*(x+k_o)^2+4 B^*\right) ^{\left( -\tfrac{1}{4} + \tfrac{A^*}{8 B^*} \right) } \\ h(x)q(x)&= {\mathcal {O}}\left( (x+k_o)^{1 - \tfrac{A^*}{4 B^*}} \right) \quad (x \rightarrow -k_o) \\ h(x)q(x)&= {\mathcal {O}} \left( x^{1/2} \right) \quad (x \rightarrow \infty ) \end{aligned}$$
and
$$\begin{aligned} g(x) q(x)&= \frac{2 \sigma ^*(x + 2 k_o)+p_c \left( (x + k_o)^2+(x+k_o)k_o+k_o^2 \right) }{M^*(x+k_o)^4+4 B^*(x+k_o)^2} \nonumber \\&\quad \cdot (x+k_o)^{-\tfrac{A^*}{4 B^*}} \left( M^*(x+k_o)^2+4 B^* \right) ^{ \left( \tfrac{3}{4} + \tfrac{A^*}{8 B^*}\right) } \nonumber \\&= \frac{p_c (x+k_o)^2+(p_c k_o+2 \sigma ^*)(x+2k_o)}{\left( M^* (x+k_o)^2+4 B^* \right) ^{\left( \tfrac{1}{4}-\tfrac{A^*}{8 B^*}\right) } \left( x+k_o \right) ^{2+\tfrac{A^*}{4 B^*}}} . \end{aligned}$$
(22)