Another attempt at solving a PDE with the method of characteristics

Question

I want to use the method of characteristics to obtain the solution to this PDE,

$$\frac{\partial F}{\partial t}=\left(z-t\right)\left(\beta z-\gamma\right)\frac{\partial F}{\partial z}$$

which I've seen is of the form

$$F\left(t,z\right)=F\left(\left(\frac{\beta\left(z-1\right)}{\beta z-\gamma}\right)e^{\left(\beta-\gamma\right)t}\right).$$

Attempt:

I was trying by identifying the PDE with one of the form

$$a\left(t,z\right)F_{z}+b\left(t,z\right)F_{t}=g\left(t,z,F\right),$$

deriving the solution $h\left(t,z\right)=c_{1}$ for

$$\frac{dz}{dt}=\frac{a\left(t,z\right)}{b\left(t,z\right)}$$

then the solution $j\left(t,z,F\right)=c_{2}$ for

$$\frac{dF}{dt}=\frac{g\left(t,z,F\right)}{b\left(t,z\right)}$$

and writting $j\left(t,z,F\right)=K\left(h\left(t,z\right)\right)$.

However, I got stuck when I stomped upon the following ODE to solve,

$$z'\left(t\right)=-\beta z^{2}\left(t\right)+\beta tz\left(t\right)+\gamma z\left(t\right)-\gamma t.$$

How can I solve the problem?

Answer 1

$$z'\left(t\right)=-\beta z^{2}\left(t\right)+(\beta t+\gamma)z\left(t\right)-\gamma t.$$ Let $\quad z(t)=\frac{y'(t)}{\beta y(t)} \quad\to\quad z'=\frac{y''}{\beta y}-\frac{y'^2}{\beta y^2} =-\beta \left(\frac{y'}{\beta y} \right)^{2}+(\beta t+\gamma)\frac{y'}{\beta y} -\gamma t$ $$y''-(\beta t+\gamma)y' +\gamma\beta ty=0$$ This is a second order linear ODE easy to solve for $y(t)$ . Then, compute $z(t)=\frac{y'(t)}{\beta y(t)} $.

Answer 2

Riccati's equation $$z'\left(t\right)=-\beta z^{2}\left(t\right)+\beta tz\left(t\right)+\gamma z\left(t\right)-\gamma t$$ $$z=-\beta z(z-t)+\gamma( z- t)$$ $$z=(\gamma-\beta z)( z- t)$$ Substitute $u=(\gamma-\beta z) \implies z=\frac {(\gamma -u)} {\beta}$ $$-\frac {u'}{\beta}=u(\frac {\gamma}{\beta}-\frac u {\beta}-t)$$ $$u'=u( {u-\gamma} + t {\beta})$$ Which is simply Bernouilli's equation ... $$\boxed{u'=u(t {\beta}-\gamma)+ u^2}$$ $$-(\frac 1 u)'= \frac {(t {\beta}-\gamma)} u+ 1 $$ $$\boxed{w'= w (\gamma-t\beta)-1 \text{, where } w=\frac 1 u=\frac 1 {\gamma-\beta z} }$$ And thats a first ODE easy to solve...

Answer 3

Just to clarify the function substitution in Jaquelins answer: the substitution is related to the Logarithmic derivative which is very famouse : $$\frac{\partial \{\log(y(t))\}}{\partial t} = \frac{y'(t)}{y(t)}$$ Where $y'(t)$ is the inner derivative and $\frac{\partial\log(y)}{\partial y} = \frac{1}y$ the outer derivative according to the Chain rule.