观察方程 \Delta u=0 可以发现, 方程具有旋转不变性, 即 v(y)=0=u(x), \text{其中}y=O(x), O: \mathbb{R}^n\rightarrow\mathbb{R}^n 是一个等距变换. 于是假设 u 具有对称不变性, 从而做出假设函数有解的形式 u(x)=v(r), r=\vert x\vert=(x_1^2+\cdots x_n^2)^{\frac{1}{2}}\\ 为了表示出 \Delta u, …

We now turn to studying Laplace’s equation ∆u = 0 and its inhomogeneous version, Poisson’s equation, ¡∆u = f: We say a function u satisfying Laplace’s equation is a harmonic function. Clearly, there are a lot of functions u which satisfy this equation. In particular, any constant function is harmonic.

ut ¡ux = 0: This equation models simple fluid motion, where u(x;t) is the height of the wave at time t and position x. Clearly, u(x;t) = 0 satisfies this equation, but so does u(x;t) = c for any constant c. In fact, there are a lot of different solutions to this equation. (We will discuss this equation more later in the course.)

2018年8月28日 · u(x, t) = √ 2 4t + 1e − x2 4t + 1 = 2 √8t + 2e − x2 4t + 1. the last step is totally not mandatory. This solution requires you to know the Green's function on the standard heat equation. ut = Duxx. which is.

u(x,0) = (x if 0 < x < 1, 2 ¡ x if 1 < x < 2, u(0,t) = ux(2,t) = 0. (4.21) In this problem, we have a mixture of both fixed and no flux boundary conditions. Again, assuming separable solutions u(x,t) = X(x)T(t), (4.22) gives rise to T0 = lT, X00 = lX, (4.23) where l is a constant. The boundary conditions in (4.21) becomes, accord-ingly X(0 ...

2024年4月10日 · We can see that Laplace’s equation would correspond to finding the equilibrium solution (i.e. time independent solution) if there were not sources. So, this is an equation that can arise from physical situations. How we solve Laplace’s equation will depend upon the geometry of the 2-D object we’re solving it on.

We will express u(x,t) as the sum of the steady-state solution and another solution w(x,t) that depends on time: u(x,t) = v(x)+w(x,t). As shown in the discussion in 10.6 (p. 614), w(x,t) also satisfies α2w xx = w t and has the boundary conditions w(0,t) = w(L,t) = 0.

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答案： u=\varphi_1(x+at)+\varphi_2(x-at) 。 （在熟悉不过了，且注意映射名用哪个都可以，不影响） （因为得出 u_{\xi \eta}=0 后，说明 u 的交叉项为0，所以先对 \eta 积分得到一个只与 \eta 有关的一个函数 u_{\xi}=f(\eta) ，再对 \xi 积分得到 u=f(\eta)+g(\xi) ，即积出来的必然是 ...

其实是解一个微分方程，令P = 4x^3y, Q = xf (x)，显然P,Q有连续的偏dao导数，又存在可微函数u，满足上面条件，那么Q对x的偏导 = P对y的偏导，得到一个微分方程，f (1) = 2，实际就是初始条件，就全做出来。 扩展资料： 前面的U（Xo）只是一个函数。 就相当于袭平时的f（x）=2x+1这个类型的。 后面一个倒置的“U”是属于的意思。 就相当于。 A=（1,2,3），B=（1,2,3,4,5,6）B就包含了A，A就属于B。 注意到，上式右端第一项是对应的齐次线性 方程式 （式2）的通解，第 …