Basic partial differential problem

[miniradman]

Homework Statement

Find the solution of each of the following partial differential equation

$\frac{\partial^{2}u}{\partial x^{2}} = 0$

Homework Equations

assume the product form?
$u(x,y) = f(x)g(y)$ ? (not 100% sure)

The Attempt at a Solution

Hello, I'm only new to PDEs and I was wondering what kind of algorithm is used to solve a function such as this (with no boundary conditions).

I have the solution which states that
therefore:
$\frac{\partial u}{\partial x} = f(y)$

therefore:
$u(x,y) = xf(y)+g(y)$

but I cannot see the intuition behind this, does anyone know?

 

[pasmith]

Homework Statement

Find the solution of each of the following partial differential equation

$\frac{\partial^{2}u}{\partial x^{2}} = 0$

Homework Equations

assume the product form?
$u(x,y) = f(x)g(y)$ ? (not 100% sure)

The Attempt at a Solution

Hello, I'm only new to PDEs and I was wondering what kind of algorithm is used to solve a function such as this (with no boundary conditions).

I have the solution which states that
therefore:
$\frac{\partial u}{\partial x} = f(y)$

therefore:
$u(x,y) = xf(y)+g(y)$

but I cannot see the intuition behind this, does anyone know?

Where do constants of integration come from? From the fact that the derivative of a constant is zero.

So here, except that "derivative of a constant is zero" is replaced by "partial derivative with respect to $x$ of a function which doesn't vary with $x$ is zero".

 

[HallsofIvy]

You understand, I presume, that if the problem were an ordinary differential equation, $\frac{d^2}{dx^2}= 0$, you would solve it by integrating twice: since the second derivative is 0, the first derivative must be a constant: $\frac{du}{dx}= C$. And now, integrating again, $u(x)= Cx+ D$ where D is the constant of integration.

It's the same basic idea with partial derivatives except that the constant of integration may be a function of the other variables.

$\frac{\partial^{2}u}{\partial x^{2}} = 0$ can be read as
$$\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}\right)= 0$$

That is, that $\partial u/\partial x$ does not depend upon x. It does NOT, itself, say what variables the partial derivative does depend on. If we are given that the only variable are x and y, then $\partial u/\partial x$ must be a function of y only- and it could be any function of y: $\partial u/\partial u= f(y)$.

The point is that taking the partial derivative with respect to one variable, we treat the other variables as constants. So going the other way, taking the anti-derivative with respect to that variable, the ''constant of integration" may actually depend upon those other variables. If g(y) is a function of y only, $\frac{\partial g}{\partial x}= 0$.

Once we have
$$\frac{\partial u}{\partial x}= f(y)$$
integrating again (with respect to x, holding y constant) gives $$u(x, y)= f(y)x+ g(y)$$ where again, the "constant of integration" can be an arbitrary function of y.

If the independent variables were x, y, and z, the general solution to $\frac{\partial^2 u}{\partial x^2}= 0$ would be u(x, y, z)= f(y,z)x+ g(y, z) where, now, f and g can be any functions of y and z. You should be able to see by differentiating that these do satisfy the equation.

If you had $\frac{\partial^2 u}{\partial x\partial y}= 0$, again assuming that the only variables are x and y, integrating first, with respect to y would give $\partial u/\partial y= f(y)$ where f(y), the "constant of integration" can be any function of y. Integrating again, this time with respect to y, would give u= F(y)+ G(x) where F(y) is an anti-derivative of f(y) (since f(y) was arbitrary so is F(y)) and G(x) is the "constant of integration", an arbitrary function of x.

 

[miniradman]

Ahhhh, I see now. If I was doing this with in two dimensions, the integral of 0 is just constant c, however because I'm doing this in 3d (w.r.t 2 variables) my constant is now replaced with an arbitrary function. Who has it's function in terms the other variable. Right?