How could I compare a SYSTEM of linear second-order partial differential equations with two different functions within them to the heat equation? Please also provide a reference that I can cite in my paper.
In particular, for a paper, I am looking to compare this equation
ie^(4omegat) (delPhi)/(delt)+(del^2Phi)/(dely^2) = 0
to the forward heat equation in one dimension,
(delu)/(delt)-(del^2u)/(delx^2) = 0 ,
and the backward heat equation in one dimension,
(delu)/(delt)+(del^2u)/(delx^2) = 0 ,
where omega is a constant and i is the familiar imaginary unit.
My problem is, anytime I multiply by i , it can look like either the forward or backward heat equation, and I can't just have it look like either one arbitrarily...
I tried rewriting Phi in terms of parts with real and imaginary coefficients:
\Phi(y,t) = N \text{exp}(\frac{i\epsilon}{4\omega}e^{-4\omega t})sin(\sqrt{\epsilon}y)
= N [cos(\frac{\epsilon}{4\omega}e^{-4\omega t})+isin(\frac{\epsilon}{4\omega}e^{-4\omega t})]sin(\sqrt{\epsilon}y)
= stackrel(Phi_{re})overbrace(Ncos(\frac{\epsilon}{4\omega}e^{-4\omega t})sin(\sqrt{\epsilon}y)) + istackrel(Phi_{im})overbrace(Nsin(\frac{\epsilon}{4\omega}e^{-4\omega t})sin(\sqrt{\epsilon}y))
where epsilon and N are constants too. I could then write this as:
= \Phi_{re} + i\Phi_{im}
However, when I plug it back into the PDE, I get a system of PDEs with mixed functions...
e^{4\omega t}\frac{\partial\Phi_{im}}{\partial t} -\frac{\partial^2\Phi_{re}}{\partial y^2} = 0
e^{4\omega t}\frac{\partial\Phi_{re}}{\partial t} + \frac{\partial^2\Phi_{im}}{\partial y^2} = 0
How can I still compare to the forward and/or backward heat equation? Please help soon, this is due by Friday April 28 for a 15-page paper. I am almost done, except for this.
Classifying these wasn't a problem (they are both parabolic). It's the comparison to the heat equation that's giving me trouble.
In particular, for a paper, I am looking to compare this equation
ie^(4omegat) (delPhi)/(delt)+(del^2Phi)/(dely^2) = 0
to the forward heat equation in one dimension,
(delu)/(delt)-(del^2u)/(delx^2) = 0 ,
and the backward heat equation in one dimension,
(delu)/(delt)+(del^2u)/(delx^2) = 0 ,
where
My problem is, anytime I multiply by
I tried rewriting
\Phi(y,t) = N \text{exp}(\frac{i\epsilon}{4\omega}e^{-4\omega t})sin(\sqrt{\epsilon}y)
= N [cos(\frac{\epsilon}{4\omega}e^{-4\omega t})+isin(\frac{\epsilon}{4\omega}e^{-4\omega t})]sin(\sqrt{\epsilon}y)
= stackrel(Phi_{re})overbrace(Ncos(\frac{\epsilon}{4\omega}e^{-4\omega t})sin(\sqrt{\epsilon}y)) + istackrel(Phi_{im})overbrace(Nsin(\frac{\epsilon}{4\omega}e^{-4\omega t})sin(\sqrt{\epsilon}y))
where
However, when I plug it back into the PDE, I get a system of PDEs with mixed functions...
e^{4\omega t}\frac{\partial\Phi_{im}}{\partial t} -\frac{\partial^2\Phi_{re}}{\partial y^2} = 0
e^{4\omega t}\frac{\partial\Phi_{re}}{\partial t} + \frac{\partial^2\Phi_{im}}{\partial y^2} = 0
How can I still compare to the forward and/or backward heat equation? Please help soon, this is due by Friday April 28 for a 15-page paper. I am almost done, except for this.
Classifying these wasn't a problem (they are both parabolic). It's the comparison to the heat equation that's giving me trouble.
1 Answer
Explanation:
