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Partial Differential Equations - Basics

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uxx+uyy=0u_{xx} + u_{yy} = 0

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u(x,y)=f(x+iy)+g(x−iy)u(x, y) = f(x + iy) + g(x - iy), where ff and gg are analytic functions.

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ut−kuxx=0u_{t} - k u_{xx} = 0

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u(x,t)=14πkt∫−∞∞e−(x−ξ)24ktϕ(ξ)dξu(x, t) = \frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^{\infty} e^{\frac{-(x-\xi)^2}{4kt}} \phi(\xi) d\xi, where ϕ\phi is the initial condition.

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∇2u+k2u=0\nabla^2 u + k^2 u = 0

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u(x)=∫Rneix⋅pf^(p)e−k∣p∣dpu(\mathbf{x}) = \int_{\mathbb{R}^n} e^{i\mathbf{x} \cdot \mathbf{p}} \hat{f}(\mathbf{p}) e^{-k|\mathbf{p}|} d\mathbf{p}, where f^(p)\hat{f}(\mathbf{p}) is the Fourier transform of initial conditions.

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uxx+2uxy+uyy=0u_{xx} + 2u_{xy} + u_{yy} = 0

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Solutions are obtained by transforming the variables to decouple the equation into simpler PDEs.

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ur+1ru=0u_{r} + \frac{1}{r}u = 0

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u(r)=Cru(r) = \frac{C}{r}, where CC is a constant.

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uzz−k2u=0u_{zz} - k^2 u = 0

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u(z)=Cekz+De−kzu(z) = Ce^{kz} + De^{-kz}, where CC and DD are constants.

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ut=kuxx−huu_{t} = ku_{xx} - hu

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The solution usually requires numerical methods like finite difference or finite element analysis, depending on the initial and boundary conditions.

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ut+uux=0u_{t} + u u_{x} = 0

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u(x,t)=f(x−ut)u(x, t) = f(x - ut), where ff is the initial condition, and the solution is valid only until shock formation.

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ut=uxx+uyy+uzzu_{t} = u_{xx} + u_{yy} + u_{zz}

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The solution involves the Green's function for the heat equation in three dimensions and will generally require numerical approximation methods.

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uxx+uyy+uzz+k2u=0u_{xx} + u_{yy} + u_{zz} + k^2 u = 0

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Solution methods include separation of variables, spherical harmonics for radial symmetry, or using Green's function for specific boundary conditions.

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urr+1rur+1r2uθθ=0u_{rr} + \frac{1}{r}u_{r} + \frac{1}{r^2}u_{\theta\theta} = 0

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u(r,θ)=A0+∑n=1∞rn(Ancos⁡(nθ)+Bnsin⁡(nθ))u(r, \theta) = A_0 + \sum_{n=1}^{\infty} r^n (A_n \cos(n\theta) + B_n \sin(n\theta)), where AnA_n and BnB_n are constants.

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ur+n−1ru=0u_{r} + \frac{n-1}{r}u = 0

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u(r)=Crn−1u(r) = \frac{C}{r^{n-1}}, where CC is a constant and nn the dimension.

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utt−c2uxx=0u_{tt} - c^2 u_{xx} = 0

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u(x,t)=f(x−ct)+g(x+ct)u(x, t) = f(x - ct) + g(x + ct), where ff and gg are twice differentiable functions.

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uxxx−6uux=0u_{xxx} - 6uu_x = 0

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u(x)=2k2csech2(k(x−ct)−x0)u(x) = \frac{2k^2}{c} \mathrm{sech}^2\left( k(x - ct) - x_0 \right), where kk, cc, and x0x_0 are constants.

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ux−uyy=0u_{x} - u_{yy} = 0

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Solution methods include separation of variables or the method of characteristics, applied according to specific boundary and initial conditions.

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∂u∂t+∇⋅(uv)=μ∇2u\frac{\partial u}{\partial t} + \nabla \cdot (u \mathbf{v}) = \mu \nabla^2 u

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Analytic solutions are rare and often the problem is approached with numerical methods, especially finite difference or finite volume methods.

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∇2u−μ∂u∂t=0\nabla^2 u - \mu \frac{\partial u}{\partial t} = 0

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The general solution will depend on boundary and initial conditions, typically requiring Green's functions or numerical methods.

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∂2u∂t2−c2∇2u+ku=0\frac{\partial^2 u}{\partial t^2} - c^2 \nabla^2 u + k u = 0

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The solution often uses techniques from both wave equations and harmonic oscillator problems, and may require numerical approximation in complex scenarios.

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utt−c2(urr+2rur)=0u_{tt} - c^2 (u_{rr} + \frac{2}{r} u_{r}) = 0

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u(r,t)=f(r−ct)+g(r+ct)ru(r, t) = \frac{f(r - ct) + g(r + ct)}{r}, where ff and gg are arbitrary functions, representing incoming and outgoing waves.

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∇2u+αut=0\nabla^2 u + \alpha u_t = 0

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Depending on boundary and initial conditions, methods like separation of variables or the Laplace transform can be used to approach the problem.

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∇⋅(σ∇u)+q=0\nabla \cdot (\sigma \nabla u) + q = 0

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A general solution might not exist in closed form and often the problem needs to be approached with numerical methods, such as finite elements or finite volumes.

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∂u∂t−α∇4u=0\frac{\partial u}{\partial t} - \alpha \nabla^4 u = 0

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Numerical approaches or perturbation techniques are often needed to find a solution, especially for complex boundary conditions.

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utt−Δu=0u_{tt} - \Delta u = 0

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u(x,t)=1(2π)n/2∫Rneix⋅pf^(p)e−ict∣p∣dpu(\mathbf{x}, t) = \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} e^{i\mathbf{x} \cdot \mathbf{p}} \hat{f}(\mathbf{p}) e^{-ict|\mathbf{p}|} d\mathbf{p}, where f^(p)\hat{f}(\mathbf{p}) is the Fourier transform of initial displacement.

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ut+cux=0u_{t} + c u_{x} = 0

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u(x,t)=f(x−ct)u(x, t) = f(x - ct), where ff is an arbitrary function.

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utt−uxx−uyy=0u_{tt} - u_{xx} - u_{yy} = 0

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u(x,y,t)=f(x−t,y)+g(x+t,y)u(x, y, t) = f(x - t, y) + g(x + t, y), where ff and gg are arbitrary functions.

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∇2u−1c2utt=0\nabla^2 u - \frac{1}{c^2} u_{tt} = 0

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The general solution depends on the boundary conditions and is often found using separation of variables or integral transform methods.

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ut−∇⋅(D∇u)=0u_{t} - \nabla \cdot (D \nabla u) = 0

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The general solution depends on the boundary conditions and DD, and is often tackled using numerical methods such as the finite element method.

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ut+∇⋅F=0u_{t} + \nabla \cdot \mathbf{F} = 0

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The solution is often given by the method of characteristics which involves solving ordinary differential equations.

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uxxxx=0u_{xxxx} = 0

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u(x)=Ax3+Bx2+Cx+Du(x) = Ax^3 + Bx^2 + Cx + D, where AA, BB, CC, and DD are constants.

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