PDE - Solve heat equation with convection

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Homework Help Overview

The discussion revolves around solving the heat equation with convection, represented by the partial differential equation u_t - k u_xx + V u_x = 0, along with the initial condition u(x,0) = f(x). Participants are exploring the implications of a variable transformation y = x - Vt to simplify the problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to substitute the variable transformation into the equation and are discussing the relationships between the derivatives u_x, u_t, and u_y. There are questions about the correct forms of these derivatives under the new variable.

Discussion Status

Some participants have offered guidance on the variable transformation and its implications for the derivatives. There is an ongoing exploration of the correct formulation of the equation after the transformation, with various interpretations being considered.

Contextual Notes

There are indications of confusion regarding the definitions of the variables and the transformations being applied. Participants are also noting the complexity of the equations involved and the initial conditions that need to be satisfied.

Ratpigeon
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Homework Statement


Solve u_t -k u_xx +V u_x=0
With the initial condition, u(x,0)=f(x)

Use the transformation y=x-Vt

Homework Equations


The solution to the equation u_t - k u_xx=0 with the initial condition is
u(x,t)=1/Sqrt[4\pi kt] \int e^(-(x-y)^2 /4kt)f(y) dy

The Attempt at a Solution


I really just need help subbing in the change in variable.

I think it's something like
u_y= u_t dt/dy +u_x dx/dy with dx/dy=1/(dy/dx)=1, dt/dy=1/V
=-1/V u_t +u_x
But this doesn't put the equation into a useful form...

the other thing I thought of was
u_x=u_t dt/dx +u_y dy/dx =0+u_y
and u_t=u_x dx/dt+u_y dy/dt =-Vu_y
And then we have that u_t+V u_x=-V u_y +V u_y=0, so the DE is just k u_xx=0; which I'm guessing isn't right either - because then it's just straight integration; (with the constants as functions of y?)...
Anyway, I'm fairly sure that the change of variables will result in either u_y=u_t+V u_x or possibly some multiple.
 
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Hi Ratpigeon! :smile:

(try using the X2 button just above the Reply box :wink:)
Ratpigeon said:
u_x=u_t dt/dx +u_y dy/dx =0+u_y
and u_t=u_x dx/dt+u_y dy/dt =-Vu_y

no, you have two sets of variables: (x,t) and (y,t') with t' = t

your ut should have uy and ut' parts, not uy and ux parts :wink:
 
So is s=x+Vt
Then is it ux= uy +us, uxx=uyy+2uys+uss
ut=-Vuy+Vus and the total equaton is 2Vuy+uyy+2uys+uss?
 
Ratpigeon said:
So is s=x+Vt

what on Earth is "s" ? :confused:

you have two sets of variables: (x,t) and (y,t') with y = x - Vt and t' = t

your ut should have uy and ut' parts, not uy and ux parts :wink:
 
Okay so i define h(x-Vt,t')=u(x,t) and get
ht=ut-Vuy and hy=ux for y=x-Vt
So:
ht-khyy=ut-kuxx+Vux=0
And the initial conditon u(x,0)=f(x) becomes h(x-Vt,0)=f(x-Vt); and since t=0 this is just h(x,0)=f(x)
and then i have an integral of...

exp(-(x-Vt-y)^2/4kt) f(y)dy (with the square root scalar out the front) and that's the solution for h and hence also u?
 
sorry, this is too difficult to read :redface:

but anyway wouldn't it be easier to find ux and ut, since they're actually mentioned in the question?
 
Im about 90% sure i got it. thankyou so much for your help :) i just wasnt getting the variable change until you explained it... :S
 

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