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$$\frac{\partial U}{\partial t}=\alpha \frac{\partial^2 U}{\partial x^2}+\beta U$$

I have been given this partial differential equation and am asked to find an application for it. I can see that the partial terms correspond to the heat diffusion equation but I am having trouble understanding what the last term on the right represents. I am additionally tasked with finding relevant values for $\alpha$ and $\beta$, but if I can understand that far-right term or find an overall application then I can find the relevant ranges for these values. Any tips or help would be greatly appreciated.

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    $\begingroup$ It can be fit to match the Schrödinger equation, with a constant potential. $\endgroup$
    – secavara
    May 23, 2021 at 16:19
  • $\begingroup$ Thanks for the tip, I'll look into it. Is there any application to heat diffusion or chemical diffusion? $\endgroup$
    – SpiderCrawler
    May 23, 2021 at 16:25

4 Answers 4

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The last term on the right corresponds to heat transfer through Newton's Law of cooling, typically a convective term. As reviewed in a recent answer, $\alpha$ is the thermal diffusivity, and $\beta$ corresponds to $-h/L\rho c$, where $h$ is a heat transfer coefficient (e.g., the convective coefficient), $L$ is a characteristic length, $\rho$ is density, and $c$ is the specific heat. $U$ is a temperature difference, often the temperature difference $T-T_\infty$ between the object (at $T$) and the environment (at $T_\infty$).

The solution is reviewed here (starting at Eq. 4; note that $\beta$ is defined differently than in your equation).

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  • $\begingroup$ So in this case, all U represent $T-T_{\inf}$ where T is in Kelvin? $\endgroup$
    – SpiderCrawler
    May 23, 2021 at 18:23
  • $\begingroup$ Yes, and $T$ could be in °C, °F, or °R as well, as we're talking about temperature differences only, not absolute temperatures. Note that this assumes that $T_\infty$ is constant because we need to have $\partial U/\partial t=\partial T/\partial t$ and $\partial^2 U/\partial x^2=\partial^2 T/\partial x^2$. $\endgroup$
    – Chemomechanics
    May 23, 2021 at 18:24
  • $\begingroup$ It wasn't recognised by OP ('liking' being a lottery here!) but yours was the best answer. +1 from me. The PDE describes a heated rod with convective losses (for example, as well as other situations) $\endgroup$
    – Gert
    May 23, 2021 at 18:56
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The $\alpha$ would typically represent thermal diffusivity (dispersion of heat along the rod), and, in typical applications, $\beta$ would be negative and associated with cooling rate per unit length of the rod (in proportion to the temperature difference with the surroundings). This might be the equation for transient operation of a cooling fin, for example. If $\beta$ were positive, it would represent run-away heating of the rod in proportion to the existing temperature, which would typically not be encountered in practice.

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Another way to study this problem is to note that if we define $V(x,t) = e^{-\beta t}U(x,t) \iff U(x,t) = e^{\beta t} V(x,t)$, then the differential equation reduces to the standard diffusion equation for $V$. Therefore, the solutions to your equation are solutions to the standard diffusion equation, multiplied by an exponential factor.

If $\beta>0$ (resp. $<0$) then the solutions will be exponentially growing (resp. decaying) with time. If you want to think of an application for such an equation, think about systems which exhibit diffusive effects (loosely, those which tend to spread out from high "concentration" to low) as well as exponential growth or decay in time.

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This equation is simply a 1D diffusion equation including damping! beta describes the spatial "spread", whereas "alpha" (< 0) is an exponential damping factor. You can solve this equation with Fourier transformation. Application is for instance heat conductivity in a "lossy" piece of metal, or spatiotemporal propagation of a chemical reactant that can be absorbed by a chemical reaction.

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  • $\begingroup$ What would be the possible magnitudes of $\beta$? I would think it would be less than one, but am not quite positive what the actual range would be. $\endgroup$
    – SpiderCrawler
    May 23, 2021 at 18:26

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