2025年大学《数学与应用数学》专业题库- 数学在海洋科学研究中的应用

2025年大学《数学与应用数学》专业题库——数学在海洋科学研究中的应用考试时间：______分钟总分：______分姓名：______一、1.设$\vec{a}=(1,2,-1),\vec{b}=(3,-1,2)$，求$\vec{a}\cdot\vec{b}$及$\vec{a}\times\vec{b}$。2.求函数$f(x)=\ln(x^2+1)$在$x=1$处的泰勒展开式（前三项）。3.计算$\int_0^1\frac{x^2}{\sqrt{1+x^2}}\,dx$。二、1.设$z=z(x,y)$由方程$x^2+y^2+z^2=xy+z$确定，求$\frac{\partialz}{\partialx}$和$\frac{\partial^2z}{\partialx^2}$。2.求解微分方程$y'+y=\sinx$。三、1.讨论级数$\sum_{n=1}^{\infty}\frac{n\sinn}{2^n}$的敛散性。2.将函数$f(x)=x^2$在$[-\pi,\pi]$上展开成傅里叶级数。四、1.证明：设$\vec{F}(x,y,z)=(y^2z,z^2x,x^2y)$，则$\vec{F}$是保守场。五、1.设海洋表面温度$T(x,y,t)$满足热传导方程$\frac{\partialT}{\partialt}=\alpha\left(\frac{\partial^2T}{\partialx^2}+\frac{\partial^2T}{\partialy^2}\right)$，其中$\alpha$为热扩散系数。假设$T(x,y,t)$满足以下边界条件和初始条件：*边界条件：$T(0,y,t)=T(a,y,t)=T(x,0,t)=T(x,b,t)=0$。*初始条件：$T(x,y,0)=f(x,y)$。求解该问题的分离变量解，并简述其物理意义。试卷答案一、1.$\vec{a}\cdot\vec{b}=1\cdot3+2\cdot(-1)+(-1)\cdot2=3-2-2=-1$；$\vec{a}\times\vec{b}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&2&-1\\3&-1&2\end{vmatrix}=(2\cdot2-(-1)\cdot(-1))\vec{i}-(1\cdot2-(-1)\cdot3)\vec{j}+(1\cdot(-1)-2\cdot3)\vec{k}=(4-1)\vec{i}-(2+3)\vec{j}+(-1-6)\vec{k}=3\vec{i}-5\vec{j}-7\vec{k}$。2.$f'(x)=\frac{2x}{x^2+1}$，$f''(x)=\frac{2(x^2+1)-4x^2}{(x^2+1)^2}=\frac{2-2x^2}{(x^2+1)^2}$；$f(1)=\ln2$，$f'(1)=1$，$f''(1)=0$；泰勒展开式为$f(x)=f(1)+f'(1)(x-1)+\frac{f''(1)}{2!}(x-1)^2+\cdots=\ln2+(x-1)$。3.令$u=1+x^2$，则$du=2xdx$；$\int_0^1\frac{x^2}{\sqrt{1+x^2}}\,dx=\frac{1}{2}\int_1^2\frac{u-1}{\sqrt{u}}\,du=\frac{1}{2}\left(\int_1^2u^{1/2}\,du-\int_1^2u^{-1/2}\,du\right)=\frac{1}{2}\left(\left[\frac{2}{3}u^{3/2}\right]_1^2-\left[2u^{1/2}\right]_1^2\right)=\frac{1}{2}\left(\frac{2}{3}(2\sqrt{2}-1)-2(\sqrt{2}-1)\right)=\frac{1}{2}\left(\frac{4\sqrt{2}}{3}-\frac{2}{3}-2\sqrt{2}+2\right)=\frac{1}{2}\left(-\frac{2\sqrt{2}}{3}+\frac{4}{3}\right)=\frac{2-\sqrt{2}}{3}$。二、1.方程两边对$x$求偏导，得$2x+2z\frac{\partialz}{\partialx}=y+\frac{\partialz}{\partialx}$，解得$\frac{\partialz}{\partialx}=\frac{y-2x}{2z-1}$；再对$x$求偏导，得$2+2\left(\frac{\partialz}{\partialx}\right)^2+2z\frac{\partial^2z}{\partialx^2}=\frac{\partialz}{\partialx}$，代入$\frac{\partialz}{\partialx}=\frac{y-2x}{2z-1}$，整理得$\frac{\partial^2z}{\partialx^2}=\frac{1}{2z-1}\left(-2-\left(\frac{y-2x}{2z-1}\right)^2\right)-\frac{2(y-2x)}{(2z-1)^2}\frac{\partialz}{\partialx}=\frac{-2(2z-1)^2-(y-2x)^2-4(y-2x)(2z-1)(y-2x)}{(2z-1)^3}$。2.此为一阶线性微分方程，$P(x)=1$，$Q(x)=\sinx$；$y=e^{-\intP(x)dx}\left(\intQ(x)e^{\intP(x)dx}dx+C\right)=e^{-x}\left(\inte^x\sinxdx+C\right)$；令$u=e^x$，则$du=e^xdx$；$\inte^x\sinxdx=\int\sinxdu=-\cosxe^x+\int\cosxdu=-\cosxe^x+\sinxe^x-\int\sinxdu=-\cosxe^x+\sinxe^x+\cosxe^x-\int\sinxdu=\sinxe^x-\int\sinxdu=\sinxe^x+\cosxe^x-\int\cosxdu=\sinxe^x+\cosxe^x-\sinxe^x+\int\sinxdu=\cosxe^x-\int\sinxdu=\cosxe^x+\sinxe^x-\int\sinxdu=\cosxe^x+\sinxe^x+\int\sinxdu=\cosxe^x+\sinxe^x-\sinxe^x+\int\sinxdu=\cosxe^x$；故$y=e^{-x}(\cosxe^x+C)=\cosx+Ce^{-x}$。三、1.令$u_n=\frac{n\sinn}{2^n}$；考虑$\sum_{n=1}^{\infty}|u_n|=\sum_{n=1}^{\infty}\frac{n|\sinn|}{2^n}$；由于$|\sinn|\leq1$，所以$\sum_{n=1}^{\infty}\frac{n|\sinn|}{2^n}\leq\sum_{n=1}^{\infty}\frac{n}{2^n}$；$\sum_{n=1}^{\infty}\frac{n}{2^n}=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\cdots=\sum_{n=1}^{\infty}\frac{n}{2^n}=\frac{1/2}{(1-1/2)^2}=2$；由比较判别法知，$\sum_{n=1}^{\infty}|u_n|$收敛，故$\sum_{n=1}^{\infty}u_n$绝对收敛。2.$f(x)=x^2$在$[-\pi,\pi]$上为偶函数，故其傅里叶级数为余弦级数；$b_n=0$；$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2dx=\frac{2}{\pi}\int_0^{\pi}x^2dx=\frac{2}{\pi}\left[\frac{x^3}{3}\right]_0^{\pi}=\frac{2}{\pi}\cdot\frac{\pi^3}{3}=\frac{2\pi^2}{3}$；$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx=\frac{2}{\pi}\int_0^{\pi}x^2\cos(nx)dx$；令$u=x$，$dv=\cos(nx)dx$，则$du=dx$，$v=\frac{\sin(nx)}{n}$；$a_n=\frac{2}{\pi}\left(\left[x^2\frac{\sin(nx)}{n}\right]_0^{\pi}-\int_0^{\pi}2x\frac{\sin(nx)}{n}dx\right)=\frac{2}{\pi}\left(0-\frac{2}{n}\int_0^{\pi}x\sin(nx)dx\right)$；令$u=x$，$dv=\sin(nx)dx$，则$du=dx$，$v=-\frac{\cos(nx)}{n}$；$a_n=-\frac{4}{n\pi}\left(\left[-x\frac{\cos(nx)}{n}\right]_0^{\pi}+\int_0^{\pi}\frac{\cos(nx)}{n}dx\right)=-\frac{4}{n\pi}\left(-\frac{\pi\cos(n\pi)}{n}+0+\left[\frac{\sin(nx)}{n^2}\right]_0^{\pi}\right)=-\frac{4}{n\pi}\left(-\frac{\pi(-1)^n}{n}\right)=\frac{4(-1)^n}{n^2}$；故$f(x)=x^2=\frac{\pi^2}{3}+\sum_{n=1}^{\infty}\frac{4(-1)^n}{n^2}\cos(nx)$。四、1.证明$\vec{F}$为保守场，需证明存在标量函数$\phi(x,y,z)$，使得$\vec{F}=\nabla\phi$；即$\frac{\partial\phi}{\partialx}=y^2z$，$\frac{\partial\phi}{\partialy}=z^2x$，$\frac{\partial\phi}{\partialz}=x^2y$；$\frac{\partial\phi}{\partialx}=y^2z$两边对$y$求偏导，得$\frac{\partial^2\phi}{\partialy\partialx}=2yz$；$\frac{\partial\phi}{\partialy}=z^2x$两边对$x$求偏导，得$\frac{\partial^2\phi}{\partialx\partialy}=z^2$；由于$\frac{\partial^2\phi}{\partialy\partialx}=\frac{\partial^2\phi}{\partialx\partialy}$，故上述推导正确；令$\phi(x,y,z)=y^2zx+\varphi(y,z)$，则$\frac{\partial\phi}{\partialx}=y^2z+\frac{\partial\varphi}{\partialx}=y^2z$，故$\frac{\partial\varphi}{\partialx}=0$，即$\varphi(y,z)$与$x$无关；令$\varphi(y,z)=\psi(y)+h(z)$，则$\frac{\partial\phi}{\partialy}=2yzx+\psi'(y)+\frac{\partialh}{\partialy}=z^2x$，由于$\frac{\partialh}{\partialy}=0$，故$\psi'(y)=z^2x-2yzx=zx(z-2y)$，由于$z$和$x$为独立变量，故$\psi'(y)=-2yz\psi'(y)$，故$\psi'(y)=0$，即$\psi(y)$为常数；同理，$\frac{\partial\phi}{\partialz}=y^2x+\psi'(y)+\frac{\partialh}{\partialz}=x^2y$，故$\frac{\partialh}{\partialz}=x^2y-y^2x=xy(x-y)$，故$h(z)=\frac{1}{2}x^2yz^2+C$；故$\phi(x,y,z)=y^2zx+C=y^2zx+\frac{1}{2}x^2yz^2+C$；因此，存在标量函数$\phi(x,y,z)=y^2zx+\frac{1}{2}x^2yz^2+C$，使得$\vec{F}=\nabla\phi$，故$\vec{F}$为保守场。五、1.令$T(x,y,t)=X(x)Y(y)T(t)$，代入热传导方程，得$\frac{1}{T}\frac{dT}{dt}=\alpha\left(\frac{1}{X}\frac{d^2X}{dx^2}+\frac{1}{Y}\frac{d^2Y}{dy^2}\right)$；由于左侧仅与$t$有关，右侧仅与$x,y$有关，故等式两边必为常数，设为$-\lambda^2$；得三个方程：$\frac{dT}{dt}+\lambda^2T=0$，$\frac{1}{X}\frac{d^2X}{dx^2}=-\lambda^2$，$\frac{1}{Y}\frac{d^2Y}{dy^2}=\lambda^2$；解得$T(t)=Ae^{-\lambda^2t}$，$X(x)=B\cos(\lambdax)+C\sin(\lambdax)$，$Y(y)=D\cosh(\muy)+E\sinh(\muy)$，其中$\mu=\sqrt{\lambda^2+\frac{a^2}{b^2}}$；由边界条件$T(0,y,t)=T(a,y,t)=0$，得$X(0)=0$，$X(a)=0$，故$B=0$，$C\sin(\lambdaa)=0$，由于$\lambda\neq0$，故$\sin(\lambdaa)=0$，即$\lambda=\frac{n\pi}{2a}$，$n=1,2,3,\cdots$；由边界条件$T(x,0,t)=T(x,b,t)=0$，得$Y(0)=0$，$Y(b)=0$，故$D=0$，$E\sinh(\mub)=0$，由于$E\neq0$，故$\sinh(\mub)=0$，但由于$\mu>0$，故此方程无解，故需调整假设，令$Y(y)=D\sinh(\muy)$；代入$\frac{1}{Y}\frac{d^2Y}{dy^2}=\lambda^2$，得$\mu^2=\lambda^2+\frac{\alpha\pi^2n^2}{4a^2}$；故$X_n(x)=\sin\left(\frac{n\pi}{2a}x\right)$，$Y_n(y)=\sinh\left(\sqrt{\frac{\alpha\pi^2n^2}{4a^2}+\frac{\pi^2n^2}{4b^2}}y\right)$，$T_n(t)=A_ne^{-\left(\frac{\alpha\pi^2n^2}{4a^2}+\frac{\alpha\pi^2n^2}{4b^2}\right)t}$；通解为$T(x,y,t