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The Buckingham-pi theorem says that a dimensional quantity of the form

$p = f(p_1, \cdots ,p_k,q_1 \cdots, q_n )$

(where the $p_i$'s dimensions form the fundamental set of units) can be rescaled as

$\tilde{p} = f(1,\cdots ,1,\tilde{q}_1, \cdots \tilde{q}_n)$

where $\tilde{p} , \tilde{q_i}$ are dimensionless.

The application of this to PDE's is throwing me off

For example in Navier-Stokes we have that a solution $u = u(x,t, \nu, p_0, L, u_0(x))$, the solution depends on position, time, viscosity, density, a length scale, and your initial velocity.

Clearly $L, u_0(x), p_0$ form a fundamental set of units for mass, length, and time.

This suggests that $u$ should be a function of 6-3 =3 variables , but upon nondimensionalizing the PDE we find $\tilde{u} = \tilde{u}(\tilde{x},\tilde{t}, \tilde{u_0}(x), Re)$, a function of four variables.

Somehow I am applying the Buckingham-pi theorem wrong, but I can't see how.

The only thing I can think of is that most things I've read factor in some sort of "typical velocity scale", but this should be dependent on $u_0(x)$. How can we identify any other velocity scale before we have solved a problem?

Is using an "infinite dimensional" variable like $u_0(x)$ throwing this off? If so why does u_o(x) being infinite dimensional cause a problem but not x or t?

The Buckingham-pi theorem says that a dimensional quantity of the form

$p = f(p_1, \cdots ,p_k,q_1 \cdots, q_n )$

(where the $p_i$'s dimensions form the fundamental set of units) can be rescaled as

$\tilde{p} = f(1,\cdots ,1,\tilde{q}_1, \cdots \tilde{q}_n)$

where $\tilde{p} , \tilde{q_i}$ are dimensionless.

The application of this to PDE's is throwing me off

For example in Navier-Stokes we have that a solution $u = u(x,t, \nu, p_0, L, u_0(x))$, the solution depends on position, time, viscosity, density, a length scale, and your initial velocity.

Clearly $L, u_0(x), p_0$ form a fundamental set of units for mass, length, and time.

This suggests that $u$ should be a function of 6-3 =3 variables , but upon nondimensionalizing the PDE we find $\tilde{u} = \tilde{u}(\tilde{x},\tilde{t}, \tilde{u_0}(x), Re)$, a function of four variables.

Somehow I am applying the Buckingham-pi theorem wrong, but I can't see how.

The only thing I can think of is that most things I've read factor in some sort of "typical velocity scale", but this should be dependent on $u_0(x)$. How can we identify any other velocity scale before we have solved a problem?

Is using an "infinite dimensional" variable like $u_0(x)$ throwing this off? If so why does $u_o(x)$ being infinite dimensional cause a problem but not $x$ or $t$?

The Buckingham-pi theorem says that a dimensional quantity of the form

$p = f(p_1, \cdots ,p_k,q_1 \cdots, q_n )$

(where the $p_i$'s dimensions form the fundamental set of units) can be rescaled as

$\tilde{p} = f(1,\cdots ,1,\tilde{q}_1, \cdots \tilde{q}_n)$

where $\tilde{p} , \tilde{q_i}$ are dimensionless.

The application of this to PDE's is throwing me off

For example in Navier-Stokes we have that a solution $u = u(x,t, \nu, p_0, L, u_0(x))$, the solution depends on position, time, viscosity, density, a length scale, and your initial velocity.

Clearly $L, u_0(x), p_0$ form a fundamental set of units for mass, length, and time.

This suggests that $u$ should be a function of 6-3 =3 variables , but upon nondimensionalizing the PDE we find $\tilde{u} = \tilde{u}(\tilde{x},\tilde{t}, \tilde{u_0}(x), Re)$, a function of four variables.

Somehow I am applying the Buckingham-pi theorem wrong, but I can't see how.

The only thing I can think of is that most things I've read factor in some sort of "typical velocity scale", but this should be dependent on $u_0(x)$. How can we identify any other velocity scale before we have solved a problem?

Is using an "infinite dimensional" variable like $u_0(x)$ throwing this off? If so why does u_o(x) being infinite dimensional cause a problem but not x or t?

The Buckingham-pi theorem says that a dimensional quantity of the form

$p = f(p_1, \cdots ,p_k,q_1 \cdots, q_n )$

(where the $p_i$'s dimensions form the fundamental set of units) can be rescaled as

$\tilde{p} = f(1,\cdots ,1,\tilde{q}_1, \cdots \tilde{q}_n)$

where $\tilde{p} , \tilde{q_i}$ are dimensionless.

The application of this to PDE's is throwing me off

For example in Navier-Stokes we have that a solution $u = u(x,t, \nu, p_0, L, u_0(x))$, the solution depends on position, time, viscosity, density, a length scale, and your initial velocity.

Clearly $L, u_0(x), p_0$ form a fundamental set of units for mass, length, and time.

This suggests that $u$ should be a function of 6-3 =3 variables , but upon nondimensionalizing the PDE we find $\tilde{u} = \tilde{u}(\tilde{x},\tilde{t}, \tilde{u_0}(x), Re)$, a function of four variables.

Somehow I am applying the Buckingham-pi theorem wrong, but I can't see how.

The only thing I can think of is that most things I've read factor in some sort of "typical velocity scale", but this should be dependent on $u_0(x)$. How can we identify any other velocity scale before we have solved a problem?

Is using an "infinite dimensional" variable like $u_0(x)$ throwing this off? If so why does u_o(x) being infinite dimensional cause a problem but not x or t?

The Buckingham-pi theorem says that given a dimensional quantity of the form

$p = f(p_1, \cdots ,p_k,q_1 \cdots, q_n )$

(where the $p_i$'s dimensions form the fundamental set of units) can be rescaled as

$\tilde{p} = f(1,\cdots ,1,\tilde{q}_1, \cdots \tilde{q}_n)$

where $\tilde{p} , \tilde{q_i}$ are dimensionless.

The application of this to PDE's is throwing me off

For example in Navier-Stokes we have that a solution $u = u(x,t, \nu, p_0, L, u_0(x))$, the solution depends on position, time, viscosity, density, a length scale, and your initial velocity.

Clearly $L, u_0(x), p_0$ form a fundamental set of units for mass, length, and time.

This suggests that $u$ should be a function of 6-3 =3 variables , but upon nondimensionalizing the PDE we find $\tilde{u} = \tilde{u}(\tilde{x},\tilde{t}, \tilde{u_0}(x), Re)$, a function of four variables.

Clearly I'm applying the Buckingham-pi theorem wrong, but I can't see how.

The only thing I can think of is that most things I've read factor in some sort of "typical velocity scale", but this should be dependent on $u_0(x)$. How can we identify any other velocity scale before we have solved a problem?

Is using an "infinite dimensional" variable like $u_0(x)$ throwing this off? If so why does u_o(x) being infinite dimensional cause a problem but not x or t?

The Buckingham-pi theorem says that a dimensional quantity of the form

$p = f(p_1, \cdots ,p_k,q_1 \cdots, q_n )$

(where the $p_i$'s dimensions form the fundamental set of units) can be rescaled as

$\tilde{p} = f(1,\cdots ,1,\tilde{q}_1, \cdots \tilde{q}_n)$

where $\tilde{p} , \tilde{q_i}$ are dimensionless.

The application of this to PDE's is throwing me off

For example in Navier-Stokes we have that a solution $u = u(x,t, \nu, p_0, L, u_0(x))$, the solution depends on position, time, viscosity, density, a length scale, and your initial velocity.

Clearly $L, u_0(x), p_0$ form a fundamental set of units for mass, length, and time.

This suggests that $u$ should be a function of 6-3 =3 variables , but upon nondimensionalizing the PDE we find $\tilde{u} = \tilde{u}(\tilde{x},\tilde{t}, \tilde{u_0}(x), Re)$, a function of four variables.

Somehow I am applying the Buckingham-pi theorem wrong, but I can't see how.

The only thing I can think of is that most things I've read factor in some sort of "typical velocity scale", but this should be dependent on $u_0(x)$. How can we identify any other velocity scale before we have solved a problem?

Is using an "infinite dimensional" variable like $u_0(x)$ throwing this off? If so why does u_o(x) being infinite dimensional cause a problem but not x or t?