数学物理方程考试试题及解答

数学物理方程考试试题及解答数学物理方程试题（一）一、填空题（每小题5分，共20分）1.长为$\pi$的两端固定的弦的自由振动，如果初始位移为$x\sin2x$，初始速度为$\cos2x$。则其定解条件是$u(0,t)=0,u(\pi,t)=0,u(x,0)=x\sin2x,u_t(x,0)=\cos2x$。2.方程$\frac{\partial^2u}{\partialt^2}-3\frac{\partial^2u}{\partialx^2}=0$的通解为$u(x,t)=f(x+3t)+g(x-3t)$。3.已知边值问题$\begin{cases}X''(x)+\lambdaX(x)=0\\X'(0)=0,X'(\pi)=0\end{cases}$，则其固有函数$X_n(x)=\sqrt{\frac{2}{\pi}}\sinnx$。4.方程$xy''+xy'+(\alphax-n)y=0$的通解为$y=c_1x^n+c_2x^{-\alpha}$。二.单项选择题（每小题5分，共15分）1.拉普拉斯方程$\frac{\partial^2u}{\partialx^2}+\frac{\partial^2u}{\partialy^2}=0$的一个解是（D）$u(x,y)=\ln(x^2+y^2)$。2.一细杆中每点都在发散热量，其热流密度为$F(x,t)$，热传导系数为$k$，侧面绝热,体密度为$\rho$，比热为$c$，则热传导方程是$\frac{\partialu}{\partialt}=\frac{k}{\rhoc}\frac{\partial^2u}{\partialx^2}+\frac{F(x,t)}{\rhoc}$。3.理想传输线上电压问题$\begin{cases}\frac{\partial^2u}{\partialt^2}=a^2\frac{\partial^2u}{\partialx^2}\\u(x,0)=A\cos\omegax,u_t(x,0)=0\end{cases}$的解为（A）$u(x,t)=A\cos\omega(x+at)$。三.解下列问题1.求问题$\begin{cases}\frac{\partialu}{\partialx}+3\frac{\partialu}{\partialy}=3x\\u(x,0)=8e^{3x}\end{cases}$的解。解：由题意得$\frac{dx}{1}=\frac{dy}{3}=\frac{du}{3x}$，故有$u=3x^2+C_1(y)$，$\frac{dy}{dx}=3$，故$y=3x+C_2$，代入初值条件得$u=3x^2+8e^{-9}$。2.求问题$\begin{cases}\frac{\partial^2u}{\partialx\partialy}=6x^2y\\\frac{\partialu}{\partialx}(0,y)=0\\u(x,0)=1-\cosx\end{cases}$的解。解：对$\frac{\partial^2u}{\partialx\partialy}=6x^2y$两边积分得$\frac{\partialu}{\partialy}=3x^2y^2+C_1(x)$，再积分得$u=x^3y^2+C_1(x)y+C_2$，代入初值条件得$u=x^3y^2+\frac{1}{2}(1-\cosx)y$。3.求问题$\begin{cases}\frac{\partialu}{\partialt}=a\frac{\partial^2u}{\partialx^2}\\u(x,0)=\sin2x\end{cases}$的解。解：设$u=X(x)T(t)$，代入方程得$\frac{X''}{X}=\frac{T'}{aT}$，两边都等于常数$-\lambda$，则有$X''+\lambdaX=0$，$T'+a\lambdaT=0$，由初值条件得$X(x)=\sin\sqrt{2\lambda}x$，$T(t)=e^{-a\lambdat}$，故$u=\sum\limits_{n=1}^{\infty}A_n\sinn\pixe^{-a(n\pi)^2t}$。四.用适当的方法解下列问题1.解问题$\begin{cases}\frac{\partialu}{\partialt}=\frac{\partial^2u}{\partialx^2}\\u(x,0)=1-2x+3x^2\end{cases}$。解：设$u=\sum\limits_{n=0}^{\infty}T_n(t)X_n(x)$，代入方程得$\frac{T_n'}{T_n}=\frac{X_n''}{X_n}=-\lambda_n$，则有$X_n(x)=1,x,x^2,\cdots$，$T_n(t)=e^{-n^2t}$，故$u=1+\sum\limits_{n=1}^{\infty}(2-4n+3n^2)e^{-n^2t}x^n$。2.解问题$\begin{cases}\frac{\partial^2u}{\partialt^2}=a^2(\frac{\partial^2u}{\partialx^2}+\frac{\partial^2u}{\partialy^2}+\frac{\partial^2u}{\partialz^2})\\u(x,0)=\sinx+2x^2,u_t(x,0)=0\end{cases}$。解：设$u=\sum\limits_{n=0}^{\infty}T_n(t)X_n(x)Y_n(y)Z_n(z)$，代入方程得$\frac{T_n''}{a^2T_n}=\frac{X_n''}{X_n}+\frac{Y_n''}{Y_n}+\frac{Z_n''}{Z_n}=-\lambda_n$，则有$X_n(x)=\sinnx$，$Y_n(y)=\sinny$，$Z_n(z)=\sinnz$，$T_n(t)=\cosn\piat$，故$u=\sum\limits_{n=1}^{\infty}(2n^2-1)\sinnx\sinny\sinnz\cosn\piat$。1.方程$\frac{\partial^2u}{\partialx^2}-3\frac{\partial^2u}{\partialx\partialy}+2\frac{\partial^2u}{\partialy^2}=\sin(x^2+y^2)$的特征线为$\frac{dy}{dx}=2x-3y$。2.定解条件为$u(x,0)=0,\frac{\partialu}{\partialt}(x,0)=2x,u(0,t)=u(l,t)=0$。3.通解为$u(x,y)=x^2+y^2-2xy+2xy\ln(x)+f(y-2x)$。4.固有函数为$X_n(x)=\cos(\frac{n\pi}{2})\cos(n\pix)$。5.通解为$y=\frac{c_1}{x^8}+\frac{c_2}{x^5}+\frac{1}{x^5}\int\frac{25t-64}{t}e^{-\frac{t^2}{2}}dt$。6.$\intxJ_2(x)dx=\frac{1}{2}(x^2J_1(x)-xJ_2(x))+C$。二、单项选择题（每小题4分，共20分）1.解微分方程$u_{xxx}+u_{xyy}=0$，其中$u(x,0,y)=\sin(x+y),u_y(x,0,y)=\cos(x+y)$，正确的结论是（B）(A)通解为$u(x,t,y)=\sin(x+y+t)+\sin(x+y-t)$；(B)通解为$u(x,t,y)=\frac{1}{4}[\sin(x+y+t)+2\sin(x-y)+\sin(x+y-t)]$；(C)通解为$u(x,t,y)=\frac{1}{2}[\sin(x+y+t)+\sin(x+y-t)]$；(D)通解为$u(x,t,y)=\sin(x+y+t)\cos(x-y)$。2.解微分方程$u_t=u_{xx}-u$，其中$u(x,0)=\cos(2x)$，正确的结论是（D）(A)通解为$u(x,t)=e^{-t}\cos(2x)$；(B)通解为$u(x,t)=\cos(2x-t)$；(C)通解为$u(x,t)=e^{t}\cos(2x)$；(D)通解为$u(x,t)=\frac{1}{2}[\cos(2x-t)+e^{t}\cos(2x)]$；(E)通解为$u(x,t)=\cos(2x-t)+\frac{1}{2}e^{t}\cos(2x)$。3.解偏微分方程$\frac{\partialu}{\partialt}=a^2\frac{\partial^2u}{\partialx^2}+u$，其中$u(x,0)=\sin^2(x)$，正确的结论是（A）(A)通解为$u(x,t)=\frac{1}{2}(1+e^{2t})\sin^2(x)$；(B)通解为$u(x,t)=\sin^2(x-t)$；(C)通解为$u(x,t)=e^{t}\sin^2(x)$；(D)通解为$u(x,t)=\frac{1}{2}[\sin^2(x-t)+e^{t}\sin^2(x)]$；(E)通解为$u(x,t)=\sin^2(x-t)+\frac{1}{2}e^{t}\sin^2(x)$。4.解偏微分方程$\frac{\partialu}{\partialt}=\frac{\partial^2u}{\partialx^2}+\frac{\partial^2u}{\partialy^2}$，其中$u(x,0,y)=\sin(x+y)$，正确的结论是（B）(A)通解为$u(x,t,y)=\frac{1}{2}[\sin(x+y+t)+\sin(x+y-t)]$；(B)通解为$u(x,t,y)=\frac{1}{2}[\sin(x+y+t)+\sin(x+y-t)+2e^{-2t}\sin(x+y)]$；(C)通解为$u(x,t,y)=\sin(x+y+t)\cos(x+y)$；(D)通解为$u(x,t,y)=\sin(x+y+t)\sin(x+y)$；(E)通解为$u(x,t,y)=\frac{1}{2}[\sin(x+y+t)\cos(x-y)+\sin(x+y-t)\cos(x+y)]$。5.解偏微分方程$\frac{\partialu}{\partialt}=\frac{\partial^2u}{\partialx^2}+\frac{\partial^2u}{\partialy^2}$，其中$u(x,0,y)=\cos(x^2+y^2)$，正确的结论是（D）(A)通解为$u(x,t,y)=\cos(x^2+y^2-t)$；(B)通解为$u(x,t,y)=\cos(x^2+y^2+t)$；(C)通解为$u(x,t,y)=e^{t}\cos(x^2+y^2)$；(D)通解为$u(x,t,y)=\frac{1}{4\pit}e^{-\frac{x^2+y^2}{4t}}$；(E)通解为$u(x,t,y)=\frac{1}{\sqrt{4\pit}}e^{-\frac{x^2+y^2}{4t}}$。3.解：设方程解为$u(r,t)=A\cos\theta+Br^3\sin3\theta$，又$u(R,t)=AR\cos\theta+BR\sin3\theta=6R\cos\theta+2R\sin3\theta$，解得$A=6$，$B=\frac{2}{R^2}$，因此$u(r,t)=6r\cos\theta+\frac{2}{R^2}r^3\sin3\theta$。5.解：方程化为$(D_x-D_y)(D_x-2D_y+1)u=0$，通解是$u(x,y)=f(x+y)+g(2x+y)e^{-x}$。8.解：设$u(x,t)=X(x)T(t)$代入方程及边界，得到$X''+\lambdaX=0$，$X'(0)=X'(\pi)=0$，解得$\lambda=n^2$，$X_n(x)=\sinnx$。一簇解$u_n(x,t)=(C_n\cosant+D_n\sinant)\sinnx$，叠加解$u(x,t)=\sum\limit