Consider the heat equation $$\frac{\partial \theta}{\partial t}=\kappa \frac{\partial ^2\theta}{\partial x}$$ for an infinite rod. We use the transformation $q_1=\frac{x^2}{kt}$ and $q_2=\frac{\theta \sqrt{\kappa t}}{\theta_0}$. And conversely $\theta =q_2 \frac{\theta_0}{\sqrt{\kappa t}}$. Now I want to rewrite the differential equation to put it in terms of $q_1$ and $q_2$.

So first off I use the product rule to find: $$\frac{\partial \theta}{\partial t} = \frac{\theta_0}{\sqrt{\kappa t}} \frac{\partial q_2}{\partial t}-\frac{q_2}{2}\frac{\theta_0}{t\sqrt{\kappa t}}$$ Then I find: $$\frac{\partial \theta}{\partial x}=\frac{\theta_0}{\sqrt{\kappa t}}\frac{\partial q_2}{\partial x}= \frac{\theta_0}{\sqrt{\kappa t}}\frac{\partial q_2}{\partial q_1}\frac{\partial q_1}{\partial x}=\frac{2x\theta_0}{(\kappa t)^{\frac{3}{2}}}\frac{\partial q_2}{\partial q_1} $$ Similarly $$\frac{\partial ^2\theta}{\partial x^2}= \frac{4x^2 \theta_0}{(\kappa t)^{\frac{5}{2}}}\frac{\partial ^2q_2}{\partial q_1 ^2}$$ This turns the differential equation into $$\frac{\theta_0}{\sqrt{\kappa t}} \frac{\partial q_2}{\partial t}-\frac{q_2}{2}\frac{\theta_0}{t\sqrt{\kappa t}} = \kappa \frac{4x^2 \theta_0}{(\kappa t)^{\frac{5}{2}}}\frac{\partial ^2q_2}{\partial q_1 ^2}$$ This reduces to the final differential equation for $q_1$ and $q_2$. \begin{equation}\frac{\partial q_2}{\partial t} = \frac{q_2}{t}\left[4 \frac{\partial ^2q_2}{\partial q_1 ^2} +\frac{1}{2}\right]\end{equation}

I am not 100% sure that this is correct so far, so if someone could confirm or show me any mistakes that would be great.

Furthermore, we assume that $q_2=f(q_1)$ for some function $f$. I am asked to show that this leads to the following differential equation for $f$ $$4q_1 \frac{\partial ^2 f}{\partial q_1^2} +(q_1+2) \frac{\partial f}{\partial q_1} +\frac{f}{2} =0 $$

I assume that I must have made a mistake earlier on because I can't reduce my differential equation to this. If anyone could help me out that would we much appreciated. Also, I hope I provided enough information about the problem, if not, feel free to ask. Thanks in advance.

  • $\begingroup$ $\theta=\theta(x,t)$. But then you consider a coordinate transformation $(x,t)\mapsto (q_1,q_2)$, with $q_1(x,t)$ and $q_2(x,t)$, $\theta=\theta(q_1(x,t),q_2(x,t))$? Is this your setting? $\endgroup$ – Avitus Sep 5 '13 at 17:52
  • $\begingroup$ Originally the question was about removing the dependency of $\kappa$ in the differential equation, and then solving for $f$, and back substituting to find the solution to the original problem. So $\theta$ is just removed from the DE and then we put it back in later. I hope this answers your question! $\endgroup$ – Slugger Sep 5 '13 at 17:57
  • $\begingroup$ Ok, so you are trying to remove the parameter $k$ in the PDE using a coordinate transformation. Ok, now I got it, thanks! $\endgroup$ – Avitus Sep 5 '13 at 18:00

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