## 7 Solutions to Some General Cases  
### Case \(3\): A Problem Taken from [42; 43] with \(p=m+1\) and \(q=1\)  
As a first example, we consider [42] (see also in [43])  
\[\frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\bigg{(}D_{0}u^{m} \frac{\partial u}{\partial x}\bigg{)}+bu^{m+1}-cu,\quad m>1 \tag{40}\]  
The double integration applied to (40) (assuming \(b=c=1\)) yields  
\[\begin{split}&\frac{d\delta^{2}}{dt}=D_{0}\frac{(n+1)(n+2)}{m+1}+ \delta^{2}\bigg{(}\frac{(n+1)(n+2)}{(n(m+1)+1)(n(m+1)+2)}-1\bigg{)}=D_{0}N-M_{ 3}\delta^{2}\\ & N=\frac{(n+1)(n+2)}{m+1},\ M_{D}=\frac{2m}{1+3m}\end{split} \tag{41}\]  
The solution with \(n=1/m\) is  
\[\delta_{3}(t)=\sqrt{\frac{D_{0}}{b}}\sqrt{\frac{N}{M_{3}}}\sqrt{1-e^{-M_{3}F \circ}} \tag{42}\]  
As in Section 4.2, we have a scaled front  
\[\delta_{3}(t)/\sqrt{\frac{D_{0}}{b}}\ =\left(\sqrt{\frac{N}{M_{3}}}\sqrt{1-e^{-M_{3}F \circ}}\right)_{t\to\infty}\to\sqrt{\frac{N}{M_{3}}} \tag{43}\]  
with a finite length for a long time (large values of \(Fo\)) or for moderate times with large values of \(M_{3}\).  
Hence, we obtain a situation similar to the case when \(b=0\), \(c\
eq 0\) (see Section 4.2). Then, the approximate profile is  
\[u_{a(3)}=\left(1-\frac{x}{\sqrt{\frac{D_{0}}{b}}\sqrt{1-e^{-M_{3} F\circ}}}\right)^{n}=\left(1-\frac{X_{3}}{\sqrt{\frac{N}{M_{3}}}\sqrt{1-e^{-M_{3} F\circ}}}\right)^{n}=\left(1-\frac{y_{3}}{\sqrt{\frac{N}{M_{3}}}}\right)^{n} \tag{44}\] \[X_{3}=\frac{x}{\sqrt{\frac{N}{M_{3}}}},\quad\eta_{3}=\frac{x}{ \sqrt{1-e^{-M_{3}F\circ}}} \tag{45}\]  
That can be generalized as  
\[f_{6}(u)=u^{m-1}\Big{(}b-cu^{2}\Big{)} \tag{55}\]  
which, to some extent, resembles the formulations the non-linearities in the Zeldovich equation [44; 45; 46].  
For \(m=1\), we have  
\[f_{61}(u)=b-cu^{m+2} \tag{56}\]  
That is, with only a sink working term,  
For \(m=2\), the growth function is  
\[f_{62}(u)=u\Big{(}b-cu^{2}\Big{)} \tag{57}\]  
With \(b=c=1\), we obtain \(f_{62}(u)=u\big{(}1-u^{2}\big{)}\), that is, the growth function of the modified Fisher equation, known also as the growth function of the Newell-Whitehead-Segel reaction-diffusion model [47]  
For \(m=3\), we obtain  
\[f_{63}(u)=u^{2}\Big{(}b-cu^{2}\Big{)} \tag{58}\]  
For \(m=4\) and for \(m=5\) we obtain similar expressions, respectively  
\[f_{64}(u)=u^{3}\Big{(}b-cu^{2}\Big{)} \tag{59}\]  
\[f_{65}(u)=u^{4}\Big{(}b-cu^{2}\Big{)} \tag{60}\]  
Now, let us consider the result of the double integration to the general formulation (54), when \(u\) is replaced by the Barenblatt profile \(u(x,t)=(1-x/\delta)^{n}\). In this case the function controlling the rate of the time-dependent term in the front evolution equation is  
\[M_{6}=\int\limits_{0}^{\delta}\int\limits_{x}^{\delta}\Big{(}bu^{m-1}-cu^{m+1 }\Big{)}dxdx \tag{61}\]  
\[M_{6}=b\bigg{(}\frac{1}{[(m-1)n+1][(m-1)n+2]}-\frac{1}{[(m+1)n+1][(m+1)n+2]} \bigg{)} \tag{62}\]  
assuming for simplicity that \(b=c\).  
With \(n=1/m\) and applying some algebras, we obtain  
\[M_{6}(n=1/m\ )=m^{2}\frac{10m-2}{(4m^{2}-1)(9m^{2}-1)} \tag{63}\]  
The condition \(M_{6}(n=1/m\ )>0\) requires (it is easy to check this) \(m>2/10\) or \(m>1/2\) or \(m>1/3\). All these conditions are satisfied by the general postulation that the model needs \(m>1\).  
With this last case, we close the discussion on more general formulations of the source-sink function, and we believe that we released some fruitful ideas.