Solving Partial Differential Equations with Substitution

[Boltzman Oscillation]

Homework Statement

Hello I am given the equation:

u_t - 2u_xx = u
I was given other equations (boundary, eigenvalue equations) but i don't think I need that to solve this first part:

The book says to get rid of the zeroth order term by substituting u = exp(t)V(x,t). I tried to but I can't find a way to get rid of the zeroth order term! Any help would be appreciated.

Homework Equations

The Attempt at a Solution

 

[George Jones]

I tried to but ...

3. The Attempt at a Solution

Show your attempt at a solution.

 

[lurflurf]

expand and simplify
$(\exp(t)V(x,t))_t-2(\exp(t)V(x,t))_{xx}=\exp(t)V(x,t)$

 

[Ray Vickson]

expand and simplify
$(\exp(t)V(x,t))_t-2(\exp(t)V(x,t))_{xx}=\exp(t)V(x,t)$

So, now expand the derivatives using the product rule, etc.

 

[Boltzman Oscillation]

Sorry for the late reply. I have the following after expanding using the product rule:

(1) e^tV(x,t) + e^tV(x,t)_t -2e^tV(x,t)_xx - e^tV(x,t) = 0

I can simplify to:

(2) e^tV(x,t)_t -2e^tV(x,t)_xx = 0

(3) Now I will let V(x,t) = X(x)T(t)

Plugging into (2) I get:

(4) e^tX(x)T'(t) - 2e^tX''(x)T(t) = 0

I can now separate the variables but they will both end up with e^t. Is there a way to get rid of the e^t or am I doing something wrong?[/SUP][/SUP]

 

[Ray Vickson]

Sorry for the late reply. I have the following after expanding using the product rule:

(1) e^tV(x,t) + e^tV(x,t)_t -2e^tV(x,t)_xx - e^tV(x,t) = 0

I can simplify to:

(2) e^tV(x,t)_t -2e^tV(x,t)_xx = 0

(3) Now I will let V(x,t) = X(x)T(t)

Plugging into (2) I get:

(4) e^tX(x)T'(t) - 2e^tX''(x)T(t) = 0

I can now separate the variables but they will both end up with e^t. Is there a way to get rid of the e^t or am I doing something wrong?[/SUP][/SUP]

What is stopping you from dividing equation (4) by the nonzero quantity $e^t?$

BTW: you will save yourself hours of complex, frustrating and and error-prone typing, just by using LaTeX to typeset your expressions and equations. For example, (4) is a snap to write and looks better as well:
$$e^t X(x) T'(t) - 2 e^t X''(x) T(t)=0 \hspace{4ex}(4)$$
Just right-click on the image and ask for a display of math as tex commands.

 

[Boltzman Oscillation]

What is stopping you from dividing equation (4) by the nonzero quantity $e^t?$

BTW: you will save yourself hours of complex, frustrating and and error-prone typing, just by using LaTeX to typeset your expressions and equations. For example, (4) is a snap to write and looks better as well:
$$e^t X(x) T'(t) - 2 e^t X''(x) T(t)=0 \hspace{4ex}(4)$$
Just right-click on the image and ask for a display of math as tex commands.

I am new to mathematics via forums. Thank you for twlltelme about latex. As for the problem, I realized that I could divide the e^t (I hadn't slept for a while). Thank you for the answers everyone! On to the next challenge!