Thread: Seperation of variables View Single Post
 P: 838 This is the the first time I've encountered seperation with partial differential equations. There are no worked examples, so I need some help to work through this problem. The question seems to be somewhat hand holding, since it seems to be THE introduction. Q: Apply seperation of variables $u_t = u_x$ by substituting $u=A(x)B(t)$ and then dividing by AB. If one side depends only on $t$ and the other only on $x$, they must equal a constant $k$; what are $A$ and $B$? $$\frac{\partial u}{\partial t}-\frac{\partial u}{\partial x} = 0$$ $$u = A(x)B(t)$$ $$\frac{\partial}{\partial t} \left[ A(x)B(t) \right] - \frac{\partial}{\partial x} \left[ A(x)B(t) \right] = 0$$ $$A(x)B'(t)-A'(x)B(t)=0$$ $$\frac{A(x)B'(t)-A'(x)B(t)}{A(x)B(t)}$$ $$\frac{B'(t)}{B(t)}-\frac{A'(x)}{A(x)}=0$$ Now I was reading on various websites, that I can set each independent term equal to seperation constants to make two coupled (is this the proper word to use?) differential equations. I don't understand where this step comes from. but... $$\frac{B'(t)}{B(t)}=k$$ $$\frac{A'(x)}{A(x)}=k$$ Now solving for $A(x)$ and $B(t)$. I'm a little rusty here, so I don't know if this part is correct. Rewriting the two equations above in Leibniz notation $$\frac{dB(t)}{dt} \cdot \frac{1}{B(t)} = k$$ Seperating: $$\frac{dB(t)}{B(t)} = k dt$$ $$\int \frac{dB(t)}{B(t)} = \int k\,\,dt$$ $$\ln B(t) = kt +c$$ $$B(t) = e^{kt+c}$$ And subsequently: $$A(x) = e^{kx+c}$$ Does this make sense? :) Thanks in advance.