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PDE and heat equation 
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#1
Jan1513, 02:46 AM

P: 320

given the heat equation [tex]\frac{\partial u}{\partial x}=\frac{\partial^2 u}{\partial x^2}[/tex]
what does [itex]\frac{\partial^2 u}{\partial x^2}[/itex] represent on a practical, physical level? im confused because this is not timespace acceleration, but rather a temperaturespacial derivative. thanks all! 


#2
Jan1513, 03:08 AM

P: 428

you're missing a time derivative,
[tex]\partial_tu=\partial_{xx}u[/tex]. We have the second partial of temperature in the spatial direction. One way to see this is, consider a local maximum in temperature in a spatial distribution of temp. Then all points nearby are warmer, so as time proceeds, the temp should increase. In other words, [itex]\partial_tu>0[/itex]. if you can understand this, i think you'll have a decent understanding of the idea behind the heat equation. 


#3
Jan1513, 04:37 PM

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#4
Jan1513, 05:50 PM

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PDE and heat equation
The rate of heat flow along x is proportional to the temperature gradient along x, which is ##\partial_x u##.
##\partial_{xx} u## measures the rate of change of the temperature gradient along x, which is the rate at which heat "accumulates" at a point (because more heat is floiwing towards the point from one side than is flowing away from the other side). The rate at which the heat "accumulates" also measures the rate of change of temperature at the point, or ##\partial_t u##. So the differential equation for the system is ##\partial_t u = K \partial_{xx} u## for some constant ##K##, and if you measure length and time in suitable units, you can make ##K = 1##. 


#5
Jan1513, 05:51 PM

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#6
Jan1513, 05:56 PM

P: 428

I would associate it with the shape of the distribution. It is the second derivative, so what does this mean for a sine function? A quadratic equation? A straight line?
Are you okay on what u_xx means as far as the shape of u(t,x)? And the shape I correlate strongly with the idea that spatial convexity in the temperature leads to heat, that is, a changing temp in time. I'm not sure where exactly your question is, please let us know. 


#7
Jan2213, 09:52 PM

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As alephzero points out, this is one of the (rate of flow in) minus (rate of flow out) equals (rate of accumulation) balances. The rate that heat is flowing in the xdirection (the heat flux q) is given by [tex]q(x)=k\frac{\partial T}{\partial x}[/tex]
If you consider a small element of the material lying along the the xdirection between x and x + Δx, then the rate of heat flow into the element at x is [itex]k(\frac{\partial T}{\partial x})_x[/itex]; the rate of heat flow leaving the element at x + Δx is [itex]k(\frac{\partial T}{\partial x})_{x+\Delta x}[/itex]. The rate of heat entering minus the rate of mass leaving is [itex]k(\frac{\partial T}{\partial x})_x+k(\frac{\partial T}{\partial x})_{x+\Delta x}[/itex]. This is equal to the rate of heat accumulation within the element. An expression for the rate of heat accumulation within the element is related to the rate at which the temperature is rising: [tex]rate\ of\ accumulation=\Delta x \rho C_p\frac{\partial T}{\partial t}[/tex] where ρ is the material density and C_{p} is the heat capacity. If we set the rate of accumulation equal to the rate of heat flow in minus the rate of heat flow out, we obtain: [tex]\Delta x \rho C_p\frac{\partial T}{\partial t}=k(\frac{\partial T}{\partial x})_x+k(\frac{\partial T}{\partial x})_{x+\Delta x}[/tex] Taking the limit as Δx approaches zero gives: [tex]\frac{\partial T}{\partial t}=\kappa \frac{\partial^2 T}{\partial x^2}[/tex] where [itex]\kappa[/itex] is called the thermal diffusivity. 


#8
Jan2413, 11:23 PM

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