How Does T Satisfy the Heat Equation?

[mrcleanhands]

Homework Statement

Where T(x,t)=T_{0}+T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)

\omega = \frac{\Pi}{365} and \lambda is a positive constant.

Show that T satisfies T_{t}=kT_{xx} and determine \lambda in terms of \omega and k.

I'm not to sure what is meant by the latter part of "determine \lambda in terms of \omega and k."

Homework Equations

The Attempt at a Solution

So I think I first have to find the partial derivatives of the first order.
\frac{\partial T}{\partial x}=-\lambda T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)+T_{1}e^{-\lambda x}(-\lambda)\cos(\omega t-\lambda x)

\frac{\partial T}{\partial t}=T_{1}e^{-\lambda x}(\omega t)\cos(\omega t-\lambda x)I then work out the second order partial derivative with respect to x and here it gets kind of messy and where I get confused.
<br /> T_{xx}=-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)+T_{1}e^{-\lambda x}(-\lambda)\cos(\omega t-\lambda x)+(-\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)+-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)

<br /> T_{xx}=-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)-T_{1}e^{-\lambda x}\lambda\cos(\omega t-\lambda x)-\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)

T_{xx}=-2\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)-\lambda T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)(1-\lambda)

T_{xx}=\lambda T_{1}e^{-\lambda x}(-2\lambda\sin(\omega t-\lambda x)-\cos(\omega t-\lambda x)(1-\lambda)<br />


This looks nothing like the partial derivative of the first order with respect to t...

 

[phyzguy]

You're just not doing the partial derivatives correctly. For example, when you take the partial derivative with respect to t, the multiplier should be ω, not (ωt). Also, in Txx, each term should be multiplied by λ^2. You missed a factor of λ in the second term. Also, don't forget that (-λ)^2 is equal to λ^2, not -λ^2.

 

[mrcleanhands]

Ok I've improved it a little but still stuck...

T_{xx}=\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)+\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)+(\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)

T_{xx}=\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)+\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)+(\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)

T_{xx}=\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)+(\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)=2\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)


However, this still doesn't look right to me?

Because if T_{t}=kT_{xx}

then

\frac{\partial T}{\partial t}=T_{1}e^{-\lambda x}\omega\cos(\omega t-\lambda x)=kT_{xx}=k(2\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x) which isn't true

 

[phyzguy]

It is true for certain values of lambda. This basically gives you an equation which defines λ in terms of ω and k, which is what you were asked to provide.

 

[mrcleanhands]

Ok I see, so it goes like this:

Since\lambda
is a positive constant =kT_{xx}=k(2\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)
and \frac{\partial T}{\partial t}=\omega T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)

For \frac{\partial T}{\partial t}=kT_{xx} to be true k(2\lambda^{2})=\omega therefore \lambda=\sqrt{\frac{\omega}{2k}}Then T_{t}=kT_{xx}
only when \lambda=\sqrt{\frac{\omega}{2k}}

 

[phyzguy]

You got it!