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
方程ut=uxx+f(t,x,u)O(h^4)局部超收敛半离散格式 有限差分法局部超收敛半离散格式摘要:Stys,T不详施炳云不详vip数理译丛
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...
(n2-1)m 因此,原问题的解 u(x,t)=2/π-2/π∑_(n=0)^∞(1+(-1)^n)/(n^2-1)e^(-n^2)⋅cos^2cosn nx (3)设u(x,t)=X(x)T(t)是非平凡特解,可得T满足方程 T'+a^2λT=0 ,X满足特 征值问题 (2.4) X'(0)-σX(0)=0 X'(l)+σX(l)=0 . 易知方程(2.4)的...
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_...
(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|...
The equations of typeut=uxxx+G(u,ux,uxx) which describe η‐pseudospherical surfaces are characterized. A new class of such equations is obtained. This class together with the Korteweg–de Vries (KdV) equation, the modified Korteweg–de Vries (MKdV) equation, and the linear equation completes...