Eigen function

[Dranzer]

modified Bessel function

I have an equation, which has lambda in it. It is given lambda satisfies the first zero of k*[lambda*m] how do I solve this to get the lambda value?

Thanks

 

[Deleted member 4993]

I have an equation, which has lambda in it. It is given lambda satisfies the first zero of k*[lambda*m] how do I solve this to get the lambda value?

Thanks

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[Dranzer]

the temperature distribution is given by T(r,t)= Ti-[2(q/k*r0)](r/r0){exp(-alpha*lambda^2*t)(K0[lambda*r]

r0=0.05 m, q=50 k=0.8 Ti=27 alpha=3x10^-6
lambda satisfies first zero of k0[lambda*r]

Find heat in r=.02 at anytime t.

So now I know I have integrate w.r.t r to get in terms of t and substitute the values, but how do I obtain lambda??

 

[Ishuda]

the temperature distribution is given by T(r,t)= Ti-[2(q/k*r0)](r/r0){exp(-alpha*lambda^2*t)(K0[lambda*r]

r0=0.05 m, q=50 k=0.8 Ti=27 alpha=3x10^-6
lambda satisfies first zero of k0[lambda*r]

Find heat in r=.02 at anytime t.

So now I know I have integrate w.r.t r to get in terms of t and substitute the values, but how do I obtain lambda??

Is K0 (and k0) supposed to be the modified Bessel Function of the second kind?

Also eigenvalues generally come from boundary conditions so are you sure that isn't the first zero of K₀(\(\displaystyle \lambda\) r₀) where K₀ is the modified Bessel function of the second kind or some such?

 

[Dranzer]

Is K0 (and k0) supposed to be the modified Bessel Function of the second kind?

Also eigenvalues generally come from boundary conditions so are you sure that isn't the first zero of K₀(\(\displaystyle \lambda\) r₀) where K₀ is the modified Bessel function of the second kind or some such?

Ah yes it is a modified Bessel function. How do I proceed?

 

[Dranzer]

Never mind, I figured it out. Thanks for the help:smile: