Showing similarity solution satisfies its ODE

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In summary, the author is trying to solve for the velocity by plugging in the similarity solution into the original PDE. However, they are lost and need someone to help them.
  • #1

K29

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Homework Statement



Working with a fluids problem I have derived a pde in [itex]v(y,t)[/itex]. It does not seem to matter but I'll write the PDE I derived, in case:

[itex]\frac{\partial v}{\partial t}=\upsilon \frac{\partial ^2 v}{\partial y^2}[/itex]

Assuming I know that the similarity solution below will work in solving the pde:

[itex]v(y,t)=F(\xi)[/itex] where [itex]\xi = \frac{y}{\sqrt{t}}[/itex]

I need to simply show that [itex]F(\xi)[/itex] satisfies the ODE [itex]\frac{d^2 F}{d \xi^{2} }+\frac{\xi}{2\upsilon }\frac{dF}{d \xi}=0[/itex]

subject to boundary conditions [itex]F(0)=U[/itex]
[itex]F(\infty)=0[/itex]

([itex]\upsilon[/itex] and [itex]U[/itex] are constants related to the original PDE problem & its boundary conditions)


The Attempt at a Solution



I don't quite understand what I am supposed to do here. I tried simply solving the ODE, and I get an answer [itex]F=C\frac{2\upsilon}{x}e^{\frac{-x^2}{4\upsilon}}[/itex]

It was just a quick page of scribbling to see the form of the ODEs solution. It might be slightly wrong, but it does not seem to allow me to show the similarity solution satisfies the ODE.

Please help, with some guidance on what to do. I don't have much experience with similarity solutions, but I have read up on how they are actually derived from PDEs. The above question seems to be simpler than actually deriving it. But I'm a bit lost as to where to start.

Thanks.
 
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  • #2
I think you just need to plug u expressed via F into the original PDE and the boundary conditions for u and derive the ODE and the boundary conditions for F.
 
  • #3
K29 said:

Homework Statement



Working with a fluids problem I have derived a pde in [itex]v(y,t)[/itex]. It does not seem to matter but I'll write the PDE I derived, in case:

[itex]\frac{\partial v}{\partial t}=\upsilon \frac{\partial ^2 v}{\partial y^2}[/itex]

Assuming I know that the similarity solution below will work in solving the pde:

[itex]v(y,t)=F(\xi)[/itex] where [itex]\xi = \frac{y}{\sqrt{t}}[/itex]

I need to simply show that [itex]F(\xi)[/itex] satisfies the ODE [itex]\frac{d^2 F}{d \xi^{2} }+\frac{\xi}{2\upsilon }\frac{dF}{d \xi}=0[/itex]

subject to boundary conditions [itex]F(0)=U[/itex]
[itex]F(\infty)=0[/itex]

([itex]\upsilon[/itex] and [itex]U[/itex] are constants related to the original PDE problem & its boundary conditions)


The Attempt at a Solution



I don't quite understand what I am supposed to do here. I tried simply solving the ODE, and I get an answer [itex]F=C\frac{2\upsilon}{x}e^{\frac{-x^2}{4\upsilon}}[/itex]

It was just a quick page of scribbling to see the form of the ODEs solution. It might be slightly wrong, but it does not seem to allow me to show the similarity solution satisfies the ODE.

Please help, with some guidance on what to do. I don't have much experience with similarity solutions, but I have read up on how they are actually derived from PDEs. The above question seems to be simpler than actually deriving it. But I'm a bit lost as to where to start.

Thanks.

That coefficient on the right hand side of the differential equation is the kinematic viscosity, not the velocity.

When you apply this methodology, the partial derivative of v with respect to y is the ordinary derivative of F with respect to [itex]\xi[/itex] times the partial derivative of [itex]\xi[/itex] with respect to y. The partial derivative of v with respect to t is the ordinary derivative of F with respect to [itex]\xi[/itex] times the partial derivative of [itex]\xi[/itex] with respect to t. I'm sure you can figure out how to extend this further. The whole problem is worked out in detail in Transport Phenomena by Bird, Stewart, and Lightfoot.
 
  • #4
Thanks for the help
 

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