Reduction of PDE to an ODE by means of linear change of variables

In summary, the problem at hand is to derive the general solution of the given differential equation using an appropriate change of variables. The difficulty lies in determining what the appropriate change of variable is, and whether there is a general rule to follow. The author of the problem used a specific example with a linear change of variables in two dimensions to find the general solution, but it is unclear how they determined the specific values for the variables.
  • #1
jianxu
94
0

Homework Statement



So it's been a really long time since I've done any ode/linear algebra and would like some help with this problem.

Derive the general solution of the given equation by using an appropriate change of variables

2[tex]\delta[/tex]u/[tex]\delta[/tex]t + 3[tex]\delta[/tex]u/[tex]\delta[/tex]x = 0

The thing that I'm really confused about is how do we decide what an appropriate change of variable is? Is there a general rule that I should go by?

The Attempt at a Solution



none yet because I'm not sure how to find the appropriate change of variable

Thank you
 
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  • #2
(2*d/dt+3*d/dx)u=0. Suppose u is a function only of 3t-2x?
 
  • #3
Dick said:
(2*d/dt+3*d/dx)u=0. Suppose u is a function only of 3t-2x?
I understand what that means but the book seems to want a different way to approach this. In their example, they used [tex]\delta[/tex]u/[tex]\delta[/tex]t + [tex]\delta[/tex]u/[tex]\delta[/tex]x = 0.

Next, they made two linear change of variables equations i think...

[tex]\alpha [/tex] = ax+bt and [tex]\beta[/tex] = cx + dt
where a,b,c,d will be chosen appropriately. Then they used chain rule in 2 dimension giving:

[tex]\delta[/tex]u/[tex]\delta[/tex]x=[tex]\delta[/tex]u/[tex]\delta[/tex][tex]\alpha[/tex]*[tex]\delta[/tex][tex]\alpha[/tex]/[tex]\delta[/tex]x + [tex]\delta[/tex]u/[tex]\delta[/tex][tex]\beta[/tex]*[tex]\delta[/tex][tex]\beta[/tex]/[tex]\delta[/tex]t

and then they did the same for [tex]\delta[/tex]u/[tex]\delta[/tex]t

which gives them (a+b)[tex]\delta[/tex]u/[tex]\delta[/tex][tex]\alpha[/tex]+(c+d)[tex]\delta[/tex]u/[tex]\delta[/tex][tex]\beta[/tex] = 0

Then they assumed a =1, b=0, c=1, d=-1 which gives them:
[tex]\delta[/tex]u/[tex]\delta[/tex][tex]\alpha[/tex]=0

and they were able to find the general solution from there.

So I guess what I meant is how were they able to determine what [tex]\alpha[/tex] and [tex]\beta[/tex] are?
 

1. What is a PDE and an ODE?

A PDE (partial differential equation) is an equation that involves partial derivatives of a function with respect to multiple independent variables. An ODE (ordinary differential equation) is an equation that involves derivatives of a function with respect to a single independent variable.

2. What is the purpose of reducing a PDE to an ODE?

Reducing a PDE to an ODE simplifies the equation and makes it easier to solve. This is because ODEs only involve derivatives with respect to one variable, whereas PDEs involve derivatives with respect to multiple variables.

3. How is a PDE reduced to an ODE using linear change of variables?

Linear change of variables involves transforming the coordinate system of a PDE to a new coordinate system where the new variables are linear combinations of the old variables. This transformation simplifies the PDE and reduces it to an ODE.

4. What are the benefits of using linear change of variables to reduce a PDE?

Using linear change of variables to reduce a PDE can make the equation simpler and easier to understand. It can also lead to more efficient and accurate solutions. Additionally, it allows for the use of well-known techniques for solving ODEs, rather than having to use more complex methods for solving PDEs.

5. Are there any limitations to using linear change of variables to reduce a PDE?

While linear change of variables is a useful technique for reducing PDEs to ODEs, it may not be applicable to all PDEs. Some PDEs may require more advanced methods for reduction. Additionally, the transformation process can be complex and time-consuming, so it may not always be the most efficient method for solving PDEs.

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