Basic partial differential problem

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The discussion centers on solving the partial differential equation ∂²u/∂x² = 0, where the solution involves integrating twice with respect to x. The first integration leads to ∂u/∂x = f(y), indicating that the partial derivative depends only on y. The second integration results in the general solution u(x, y) = f(y)x + g(y), where g(y) is another function of y. The key insight is that constants of integration in this context can be functions of other variables, reflecting the multidimensional nature of partial derivatives. Understanding this concept is crucial for correctly interpreting solutions to PDEs.
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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?
 
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miniradman said:

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".
 
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.
 
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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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