What Value of b Simplifies the PDE into an ODE?

[Benny]

Homework Statement

<br /> \frac{{\partial u}}{{\partial t}} = \frac{{\partial ^2 u}}{{\partial x^2 }} + 1,0 &lt; x &lt; \infty ,t &gt; 0<br />

Let \xi = \frac{x}{{\sqrt t }} and write u = t^b f\left( \xi \right). Determine the value of b required for f\left( \xi \right) to satisfy an ordinary differential equation involving itself and \xi only.

The Attempt at a Solution

I just set u = (t^b)f and substituted into the PDE (using the chain rule). I obtained

<br /> \frac{{d^2 f}}{{d\xi ^2 }} + \frac{\xi }{2}\frac{{df}}{{d\xi }} - bf\left( \xi \right) = - t^{1 - b} <br />

I thought about setting b = 0 so that I could use reduction of order but then there will always be a 't' term. Setting b = 1 leaves me with a constant on the RHS which I can't get rid of. I've checked my working and the ODE I've arrived at seems to be correct. I don't know how to go any further. Any help would be good thanks.

 

[HallsofIvy]

Use "reduction of order" to do what? You are not, at least in what you posted here, asked to solve the equationl. You are only asked to determine the value of b that will give you an ordinary differential equation in f and \xi. If you take b= 1, the right hand side becomes -1 and you have no "t" in the equation.

 

[Benny]

If there was no f term then I'd be left with some combination of f'' and f'. I can rewrite that in the form (I*f')' = -I*t where I is an appropriate integrating factor. But like I said, the problem with that is that I'm left with a t and I also have a derivative, f'.

If I take b = 1 I can get rid of the t on the RHS. But I'm also left with some derivatives of f and I'm not sure if that is an acceptable answer.

 

[HallsofIvy]

? The problem said, "Determine the value of b required for f(\xi) to satisfy an ordinary differential equation involving itself and \xi only". What makes you think a differential equation shouldn't involve derivatives?

 

[Benny]

I probably took the wrong approach. I thought that I was supposed to go for something like d/dx(xg(x)) = 0. It seems right now though, thanks.

 

[iamazad24]

I probably took the wrong approach. I thought that I was supposed to go for something like d/dx(xg(x)) = 0. It seems right now though, thanks.

Hello,

I need to know how did you solve for b? Can you please let me know the procedure?

Thanks

Azad