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I have the following equation:

$$\frac{\partial U}{\partial t}=k\frac{\partial^2 U}{\partial x^2}-v_{0}\frac{\partial U}{\partial x}, x>0$$

with initial conditions:

$$U(0,t)=0$$

$$U(x,0)=f(x)$$

In the problem is requested to give an interpretation of each of the terms in the above equation, and noting that such systems can model, besides solving by Fourier Transform. The Fourier Transform solution is quite simple to do; however, I can not give a physical interpretation of the terms of the equation not to mention a system that can model it. So I wanted to ask your help to answer this question. Thank you very much for your help and attention.

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2 Answers 2

up vote 5 down vote accepted

The heat equation is an example of a convection-diffusion equation.

Your problem is one-dimensional in space (only $x$), which simplifies it a bit. The term on the left hand side is the time-rate of change of the internal energy $U$ (often a multiple of the temperature).

The second term is a diffusion term, as, in time, it diffuses or "smooths" peaks. The last term is a convective term, i.e. your medium is moving at constant velocity $v_0$ to the right.

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If you re-write the equation to take the form $$ \frac{\partial\psi}{\partial t}+v\frac{\partial\psi}{\partial x}=k\frac{\partial^2\psi}{\partial x^2} $$ Then we can note that the first term is the one-dimensional form of the material derivative: $$ \left(\frac{\partial}{\partial t}+\mathbf v\cdot\nabla\right)\equiv\frac{d}{dt}\equiv D_t $$

Using the material derivative, your equation becomes $$ \frac{d\psi}{dt}=k\frac{\partial^2\psi}{\partial x^2} $$ This equation describes the diffusion of a fluid along the flow in the flow's frame of reference (i.e., the Lagrangian description of continuum mechanics). The equation as you have it, then, describes the diffusion of a flow as it flows in the stationary frame (i.e., the Eulerian description)--which, as Bernhard says, is the convection-diffusion equation.

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