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I am trying to solve the below-mentioned PDE that represents a damped heat diffusion in one-dimensional space. I am using the separation of variables to solve it, however when I try to find the solution. It results in $0$(trivial solution), whereas I am trying to find all the non-trivial solutions for this. Can someone help me with this.

Here considering this partial differential equation as,

$\frac{\partial{x}}{\partial t}(\zeta,t)=\alpha\frac{\partial^2{x}}{\partial^2 \zeta}(\zeta,t)+\beta\frac{\partial{x}}{\partial \zeta}(\zeta,t),~x(\zeta,0)=x_0(\zeta),\\x(0,t)=0=x(1,t)$

where $\zeta\in[0,~1]\forall~t\geq0$ and $\alpha,\beta\in\mathbb{R}^+$.

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    $\begingroup$ Try an ansatz of the form $x(\xi, t) = f(\xi + \beta t, t)$, it removes the $\partial x/\partial \xi$ term and leaves you with the heat equation. $\endgroup$ – Mattos Jun 25 '16 at 14:15
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    $\begingroup$ Also, looking at the problem through separation of variables, solving the ODE in $\xi$ is difficult without knowing the values of $\alpha$ and $\beta$. Taking the separation constant to be $- \lambda$ and making the ansatz $\Xi (\xi) = e^{r \xi}$, we get the eigenvalues to be $$r = \frac{- \beta \pm \sqrt{\beta^{2} - 4 \alpha \lambda}}{2 \alpha}$$ $\endgroup$ – Mattos Jun 25 '16 at 14:31
  • $\begingroup$ Thanks @Mattos and ya that is the same way I solved it, and got to the relation of $\lambda=\frac{\beta^2}{4\alpha}$, but when I substituted in the solution (where the differential constants $a_1=-a_2$, my solution of PDE goes to zero. $\endgroup$ – Raaja Jun 25 '16 at 14:35
  • $\begingroup$ What are the differential constants $a_{1}$ and $a_{2}$? Actually, it doesn't matter, just use the hint I gave in the first comment. Hopefully you can apply the chain rule correctly to show it reduces to the general heat equation which is then easily solved using separation of variables or other methods. $\endgroup$ – Mattos Jun 25 '16 at 14:38
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    $\begingroup$ An 'ansatz' is a german word for an educated guess. I'll make a post. You'll need to do the boundary conditions yourself. $\endgroup$ – Mattos Jun 25 '16 at 22:38
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If we set

$$x(\xi, t) = f(\xi + \beta t, t) = f(z, t)$$

we find (by the chain rule)

\begin{align} x_{t} &= f_{z} \cdot z_{t} + f_{t} \\ &= \beta f_{z} + f_{t} \\ x_{\xi} &= f_{z} \cdot z_{\xi} \\ &= f_{z} \\ x_{\xi \xi} &= f_{zz} \cdot z_{\xi} \\ &= f_{zz} \end{align}

Substituting into our original PDE, we get

\begin{align} x_{t} &= \alpha x_{\xi \xi} + \beta x_{\xi} \\ \implies \beta f_{z} + f_{t} &= \alpha f_{zz} + \beta f_{z} \\ \implies f_{t} &= \alpha f_{zz} \end{align}

which is just the homogeneous heat equation, which is easily solved by the using the method of characteristics. You'll want to convert your boundary conditions to your new 'coordinates' $(z,t)$ at some point too.

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  • $\begingroup$ Thanks it clears everything :) $\endgroup$ – Raaja Jun 26 '16 at 7:06
  • $\begingroup$ now when I solve and shift the coordinates I end up with $x(\zeta,t)=e^{-\alpha\pi^2k^2t+ik(\zeta+\beta t)}$. Now can you also suggest me how to apply the boundary conditions over here. $\endgroup$ – Raaja Jun 26 '16 at 7:13
  • $\begingroup$ But in the transformed coordinate system we have boundary conditions depending on time. How can I solve that, I request your help with that. Thanks. $\endgroup$ – Raaja Jun 26 '16 at 7:50
  • $\begingroup$ @RaajaG I agree with you. The given transformation results in boundary conditions which depends on time. Also the new initial conditions, has the spatial coordinates which depends on time. In addition, I do not understand the suggestion given in the answer above at the end to use the method of characteristics to solve the resulting transformed PDE. method of characteristics is used to solve wave PDE's and not heat PDE's which is what we have here. $\endgroup$ – Nasser May 27 '17 at 2:27

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