- #1
Chuck88
- 37
- 0
The method of separation of variables is used to solve the problem of partial diffrential equation. For example, when the partial differential equation is:
[tex]
\frac{\partial u}{\partial t}-\alpha\frac{\partial^{2}u}{\partial x^{2}}=0
[/tex]
We could suppose that [tex]u(x,t)[/tex] is a solution concerning both x and t. It could also be represented as the product of two functions:
[tex]
u(x,t) = X(x)T(t)
[/tex]
My question is that the method of separation of variables is based on the assumption that our solution could be represented as the product of two functions, which are with respect to x and t repsectively. My question is that what if the solution u could not be represented as the product of two functions of x and t, like:
[tex]
u(x,t) = \frac{1}{xt+1}
[/tex]
We then could not use the method of separation of variables.
I know that the solution I provided above is NOT RIGHT. But is it possible that the solution is of the form of the solution I provided above? Which means that the solution could not be separated that easy?
[tex]
\frac{\partial u}{\partial t}-\alpha\frac{\partial^{2}u}{\partial x^{2}}=0
[/tex]
We could suppose that [tex]u(x,t)[/tex] is a solution concerning both x and t. It could also be represented as the product of two functions:
[tex]
u(x,t) = X(x)T(t)
[/tex]
My question is that the method of separation of variables is based on the assumption that our solution could be represented as the product of two functions, which are with respect to x and t repsectively. My question is that what if the solution u could not be represented as the product of two functions of x and t, like:
[tex]
u(x,t) = \frac{1}{xt+1}
[/tex]
We then could not use the method of separation of variables.
I know that the solution I provided above is NOT RIGHT. But is it possible that the solution is of the form of the solution I provided above? Which means that the solution could not be separated that easy?