求解偏微分方程设U=U(x,t),满足Ut=Uxx+U,U(x,0)=xe^2x,求U(x,t) 答案 u(x,t)=(4t+x)e^(5t+2x).唯一性我没验证.我只能输100字,没法细说.大概就是分离变量,得到一个带可变常数c的解,比如说u_c.那么u_(c+h)也是解,(u_(c+h)-u_c)/h也是解,相当于可以对c求导,以满足边界条件....
先解相应的齐次方程的定解问题:-|||-ut =uxx-|||-u(x,0)=x(1-x),0≤x≤1-|||-ux(0,t)=ux(1,t)=0,t=0-|||-作变量分离,u(x,t)=∑fn(x)gn(t)对每个分量代入原方程,整理得:-|||-n=l-|||-g,()_f(x)-|||-一(两边分别是关于x,的函数,且相等,只能为一常数)-|||-8n(t...
数学物理方程题目,ut=uxx+x U(x,0)=x(1-x),0<=x<=1 Ux(0,t)=u 数学物理方程题目,ut=uxx+xU(x,0)=x(1-x),0<=x<=1Ux(0,t)=ux(1,t)=0,t>=0急求,必有重谢!... 数学物理方程题目,ut=uxx+xU(x,0)=x(1-x),0<=x<=1Ux(0,t)=ux(1,t)=0,t>=0急求,必有重谢! 展...
两类可化为Ut=Uxx形式的非线性偏微分方程
(1.1) forms the cubic equation ut − uxxt + 4u2ux = 3uux uxx + u2uxxx , (1.3) which has been derived by Novikov in [55]. It has been proven that (1.3) possesses a bi- Hamiltonian structure with peaked solutions of explicit an infinite form u(t, x )se=qu±en√ceceo−f|...
3.用正弦变换或余弦变换解下列定解问题:ut=a2uxx:t0,x0,(1)u(t,0)=p(t),u(0,x)=0.[提示:用正弦变换]utt=a2△3u,t0, r0,(2)ur=0有界, r=√(x^2+y^2+z^2)u|t=0=0, u_t|t=0=(1+r^2)^(-2) .[提示:采用球坐标,令v=ru.]3) Δuin=f_fx0,y0 u_x|x=0=0 . w...
7.23求初值问题 , -∞+∞0 , t0,u(x,0)=0,-∞+∞0 的格林函数(其中,b为常数),并且写出初值问题ut =a2uxx +bu + f (x, t), -∞+∞0 , t0,u(x,0)=(x),∞x+∞x 解的表达式,其中,(x)适当光滑和 f(x,t)连续. 相关知识点: ...
设U=U(x,t),满足Ut=Uxx+U,U(x,0)=xe^2x,求U(x,t) 扫码下载作业帮搜索答疑一搜即得 答案解析 查看更多优质解析 解答一 举报 u(x,t)=(4t+x)e^(5t+2x).唯一性我没验证.我只能输100字,没法细说.大概就是分离变量,得到一个带可变常数c的解,比如说u_c.那么u_(c+h)也是解,(u_(c+h)-u_...
摘要: 应用几种变换将形如ut(x,t)=a(t)uxx(x,t)+b(t)ux(x,t)+c(t)u(x,t)的一类抛物型方程转化为方程Ut=Uxx的形式,从而更有利于解决一维抛物型方程反问题.关键词: 反问题 ;抛物型方程 ;变换 ;未知参数 ; DOI: 10.13450/j.cnki.jzknu.2017.05.003 年份: 2017 ...
Equation ut = uxx + u ln2u: Regional Blow-upThe main goal of this chapter is to show that there exists a remarkable semilinear heat equation with regional blow-up described by an associated first-order Hamilton—Jacobi eqaution. In our asymptotic analysis, we obtain a singularly perturbed ...