Solving Partial Differential Equations with Substitution

In summary, the conversation discusses a problem with solving an equation involving derivatives. The solution involves substituting a new variable and separating the variables, with the only obstacle being dividing by a non-zero quantity. The use of LaTeX is also suggested for easier and more accurate expression typing.
  • #1
Boltzman Oscillation
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Member warned that some effort must be shown

Homework Statement


Hello I am given the equation:

ut - 2uxx = 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

 
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  • #2
Boltzman Oscillation said:
I tried to but ...

3. The Attempt at a Solution

Show your attempt at a solution.
 
  • #3
expand and simplify
##(\exp(t)V(x,t))_t-2(\exp(t)V(x,t))_{xx}=\exp(t)V(x,t)##
 
  • #4
lurflurf said:
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.
 
  • #5
Sorry for the late reply. I have the following after expanding using the product rule:

(1) etV(x,t) + etV(x,t)t -2etV(x,t)xx - etV(x,t) = 0

I can simplify to:

(2) etV(x,t)t -2etV(x,t)xx = 0

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

Plugging into (2) I get:

(4) etX(x)T'(t) - 2etX''(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]
 
  • #6
Boltzman Oscillation said:
Sorry for the late reply. I have the following after expanding using the product rule:

(1) etV(x,t) + etV(x,t)t -2etV(x,t)xx - etV(x,t) = 0

I can simplify to:

(2) etV(x,t)t -2etV(x,t)xx = 0

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

Plugging into (2) I get:

(4) etX(x)T'(t) - 2etX''(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.
 
  • Like
Likes jim mcnamara
  • #7
Ray Vickson said:
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!
 

1. What is a partial derivative?

A partial derivative is a mathematical concept that represents the rate of change of a function with respect to one of its variables, while holding all other variables constant. It is used to analyze how a function changes in response to changes in its inputs.

2. Why do we need partial derivatives?

Partial derivatives are useful in many scientific fields, particularly in physics, engineering, and economics. They allow us to analyze how a system or process changes in response to changes in its inputs, which is crucial for understanding and predicting real-world phenomena.

3. How do you calculate a partial derivative?

To calculate a partial derivative, we first need to determine which variable we are taking the derivative with respect to. Then, we treat all other variables as constants and use the rules of differentiation to find the derivative of the function with respect to the chosen variable.

4. What are some real-world applications of partial derivatives?

Partial derivatives have many applications in fields such as physics, engineering, economics, and statistics. For example, they are used in thermodynamics to analyze how temperature and pressure affect a system, in economics to study how changes in price and demand impact a market, and in machine learning to optimize algorithms and models.

5. Can you give an example of a partial derivative?

One example of a partial derivative is the heat equation, which describes the flow of heat in a medium. By taking the partial derivative of this equation with respect to time, we can analyze how the temperature at a specific point changes over time, while holding the other variables (such as position and material properties) constant.

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