Solving the Constant PDE ∂u/∂x=∂u/∂y

AlonsoMcLaren · May 25, 2013

∂u/∂x=∂u/∂y, can we ensure that u is a constant not dependent on x and y?

Simon Bridge · May 25, 2013

It says that the gradient of u in the x direction is the same as in the y direction for all (x,y).
What sort of function has that property?

What do you mean by "ensure"?

I like Serena · May 25, 2013

AlonsoMcLaren said:

∂u/∂x=∂u/∂y, can we ensure that u is a constant not dependent on x and y?

The general solution is u(x,y)=f(x+y), where f is a differentiable function.
So u can be dependent on x and y.
For instance u=2(x+y) is a solution.

Solving the Constant PDE ∂u/∂x=∂u/∂y

Discussion

Undergrad Why ##a^0=1##?

Undergrad Finding the minimum distance between two curves

Undergrad Mixed approximation vs. full approximation for a power series expansion

Undergrad Best way to numerically solve this system of equations?

Undergrad Why is ##x=e^y## the inverse of ##y=\int_1^x \frac{1}{t} dt##?

Undergrad Correct Upper/Lower limits for continuation of solutions for ODE

Graduate About ellipticity and a proof that a system of PDEs is elliptic

Insights Revisiting the Velocity-Time Function

Insights Remote Operated Gate Control System

Insights AI Enriched Problem Solving

Insights Thinking Outside The Box Versus Knowing What’s In The Box

Insights Why Entangled Photon-Polarization Qubits Violate Bell’s Inequality

Insights Quantum Entanglement is a Kinematic Fact, not a Dynamical Effect

Solving the Constant PDE ∂u/∂x=∂u/∂y

Discussion

Similar threads

Undergrad Solving ∂u/∂x = ∂u/∂t with boundary condition u(x,1) = e^(2(x-1)²)

Verifying Solution: U(x, y) = cos(x) cosh(y) for PDE U_xx + U_yy = 0

Graduate How Do You Solve the PDE u_x + x u_y = u + x with Initial Condition u(x,0)=x^2?

Graduate Proving u''(x)+u(x)=Wf(x) for Green's function integral

Show that ##f(x,y)=u(x+cy)+v(x-cy)## is a solution of the given PDE

Solving the ODE $u_t+u^2 u_x=0$ with Initial Condition $u(x,0)=2+x$

Graduate Solving a PDE: Deriving a Solution for (y^2)u'' + 2yu' - 2u = 0

Graduate Solution to PDE U(x,t) = y^2e^{-3x} + h(x)

Graduate Solving PDEs: Deriving Wave Equation from u(x ± ct)

Graduate PDE 2Ux + 3Uy + U = 0 with change of variables V(x,y)=ln[U(x,y)]