I have perhaps a simple question but I have to be sure:
I have a differential equation in form of: $$\dfrac{\partial y(t,x)}{\partial t} + \dfrac{\partial }{\partial x}\left(v(t,x)y(t,x)\right)=S(t,x,y)$$
Now I want to change the dependent variable from $$y(t,x)$$ to $$k(t,x)$$ with: $$y(t,x)=\dfrac{k(t,x)}{a(t,x) b(t,x)}$$
How do I perform this change exactly? My proposal is: $$\dfrac{\partial k(t,x)}{\partial t}\dfrac{1}{a(t,x) b(t,x)}-\dfrac{\partial a(t,x)}{\partial t}\dfrac{k(t,x)}{a^{2}(t,x) b(t,x)}-\dfrac{\partial b(t,x)}{\partial t}\dfrac{ k(t,x)}{ a(t,x) b^{2}(t,x)}+ \dfrac{\partial k(t,x)}{\partial x}\dfrac{ v(t,x)}{a(t,x) b(t,x)}-\dfrac{\partial a(t,x)}{\partial x}\dfrac{v(t,x)k(t,x)}{a^{2}(t,x) b(t,x)}-\dfrac{\partial b(t,x)}{\partial x}\dfrac{v(t,x) k(t,x)}{ a(t,x) b^{2}(t,x)}+\dfrac{\partial v(t,x)}{\partial x}\dfrac{ k(t,x)}{ a(t,x) b(t,x)}=S(t,x,\dfrac{k(t,x)}{a(t,x) b(t,x)})$$
Multiplying both sides by $$a(t,x)b(t,x)$$ and using the material time derivative $$\dfrac{\mathrm{d} \phi(t,x)}{\mathrm{d} t}= \dfrac{\partial \phi(t,x)}{ \partial t }+v(t,x)\dfrac{\partial \phi}{\partial x}$$ it result's:
$$\dfrac{\partial k(t,x)}{\partial t} +\dfrac{\partial}{\partial x}(v(t,x)k(t,x))-\dfrac{k(t,x)}{a(t,x)}\dfrac{\mathrm{d} a(t,x)}{\mathrm{d} t}-\dfrac{k(t,x)}{b(t,x)}\dfrac{\mathrm{d} b(t,x)}{\mathrm{d} t}=a(t,x) b(t,x)S(t,x,\dfrac{k(t,x)}{a(t,x) b(t,x)})$$
Is this transformation done correctly? And do you have a good source?