PDE - Solve heat equation with convection

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SUMMARY

The discussion focuses on solving the heat equation with convection, represented by the equation u_t - k u_xx + V u_x = 0, with the initial condition u(x,0) = f(x). The transformation y = x - Vt is utilized to simplify the equation. Participants clarify the relationships between the variables and derive the solution using integral forms, ultimately arriving at the expression for h(x-Vt,0) = f(x-Vt) and the solution for u as an integral involving the initial condition f(y).

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Familiarity with the heat equation and its properties
  • Knowledge of variable transformations in differential equations
  • Proficiency in integral calculus and solving integrals
NEXT STEPS
  • Study the derivation of solutions for the heat equation with convection
  • Learn about variable transformations in PDEs, specifically y = x - Vt
  • Explore the method of characteristics for solving convection-dominated PDEs
  • Investigate the application of integral transforms in solving PDEs
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Mathematicians, physicists, and engineering students focusing on heat transfer and fluid dynamics, as well as anyone interested in solving partial differential equations with convection terms.

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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