Predicting the form of solution of PDE

[gikiian]

Predicting the functional form of solution of PDE

How do you conclude that the solution of the PDE

$u(x,y)\frac{∂u(x,y)}{∂x}+\upsilon(x,y)\frac{∂u(x,y)}{∂y}=-\frac{dp(x)}{dx}+\frac{1}{a}\frac{∂^{2}u(x,y)}{dy^{2}}$

is of the functional form

$u=f(x,y,\frac{dp(x)}{dx},a)$ ?


I know this seems obvious, but I think it might not be necessary. How can you predict this by using reason, mathematical manipulation, etc?


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Update:
I get the fact that $u$ will be a function of $x, y,$ and $a$, but what about d/dx p(x)? Won't it change to something like $p(x)+C$ leaving $u$ to be no longer a function of d/dx p(x)?

 

[hunt_mat]

It's sort of obvious really. It has derivative in both x and y, so they must be part of the solution. If a is constant then you will know that will paramatrise the solution and it will be dependent on the forcing term (the pressure gradient)

 

[gikiian]

It's sort of obvious really. It has derivative in both x and y, so they must be part of the solution. If a is constant then you will know that will paramatrise the solution and it will be dependent on the forcing term (the pressure gradient)

But won't the term $\frac{d}{dx}p(x)$ change to something like $p(x)+constant$ ?

 

[hunt_mat]

You're not integrating anything, dp/dx will be your input function.

 

[gikiian]

You're not integrating anything, dp/dx will be your input function.

You will have to integrate it in order to obtain a solution :)

 

[bigfooted]

Substitute the functional form into the PDE, effectively replacing u(x,y) and its gradients. If the functional form is depending on P(x,y) directly, then P(x,y) should also appear in the PDE.

 

[bigfooted]

I'm quite sure that this is connected to the Buckingham Pi theorem as well, which would give you a mathematical way to get all the independent variables in a formal way.

 

[hunt_mat]

The pi theorem is okay if you know the general functional form but not for general problems