MATH529: Mathematical methods for the physical sciences IIMarch 9, 2021

Solve the following problems (4 course points each). Present a brief motivation of your method of solution.

1. Find the eigenvalues and eigenfunctions of the boundary value problem

   ${\displaystyle x^2{y}^{\text{'}\text{'}}+x{y}^{\text{'}}+9\lambda y=0,y^{\text{'}}\left(1\right)=0,y\left(e\right)=0.}$

   Solution. This is a Cauchy-Euler ODE. Trial solution of the form $y=x^r$ leads to

   ${\displaystyle [r(r-1)+r+9\lambda ]x^r=0,}$

   and non-trivial solutions are obtained from roots of quadratic

   ${\displaystyle r^2+9\lambda =0\Rightarrow r_{1,2}=\pm 3i\sqrt{\lambda },}$

   leading to the general solution

   ${\displaystyle y\left(x\right)=c_1{x}^{3i\sqrt{\lambda }}+c_2{x}^{-3i\sqrt{\lambda }},y^{\text{'}}\left(x\right)=3i\sqrt{\lambda }(c_1{x}^{3i\sqrt{\lambda }-1}-c_2{x}^{-3i\sqrt{\lambda }-1})}$

   The boundary condition $y^{\text{'}}\left(1\right)=0$ implies $c_1=c_2$, and $y\left(e\right)=0$ leads to

   ${\displaystyle c_1(e^{3i\sqrt{\lambda }}+e^{-3i\sqrt{\lambda }})=2c_1\cos \left(3\sqrt{\lambda }\right)=0.}$

   Non-null solutions are obtained for $\cos \left(3\sqrt{\lambda }\right)=0$, leading to eigenvalues

   ${\displaystyle 3\sqrt{{\lambda }_k}=k\pi +\frac{\pi }{2},k=0,1,2,\dots }$

   with associated eigenfunctions

   ${\displaystyle y_k\left(x\right)=x^{3i\sqrt{{\lambda }_k}}+x^{-3i\sqrt{{\lambda }_k}}=x^{i(k\pi +\frac{\pi }{2})}+x^{-i(k\pi +\frac{\pi }{2})}.}$

   Since $x^a=e^{a\ln x}$, obtain

   ${\displaystyle y_k\left(x\right)=e^{i(k\pi +\frac{\pi }{2})\ln x}+e^{-i(k\pi +\frac{\pi }{2})\ln x}=2\cos [(k\pi +\frac{\pi }{2})\ln x].}$

   Since eigenfunctions are undetermined up to a multiplicative constant, the eigenvalue, eigenfunction pairs are

   ${\displaystyle {\lambda }_k=\frac{1}{9}{(k\pi +\frac{\pi }{2})}^2,y_k\left(x\right)=\cos [(k\pi +\frac{\pi }{2})\ln x],k\in \mathbb{N}.}$

2. Use separation of variables to find the solution of

   ${\displaystyle \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+2\frac{\partial u}{\partial x}+2\frac{\partial u}{\partial y}=0.}$

   Solution. Replacing $u(x,y)=X\left(x\right)Y\left(y\right)$ leads to

   ${\displaystyle \frac{X^{\text{'}\text{'}}}{X}+2\frac{X^{\text{'}}}{X}=-(\frac{Y^{\text{'}\text{'}}}{Y}+2\frac{Y^{\text{'}}}{Y})=\lambda ,}$

   and the constant-coefficient ODEs

   ${\displaystyle X^{\text{'}\text{'}}+2X^{\text{'}}-\lambda X=0,Y^{\text{'}\text{'}}+2Y^{\text{'}}+\lambda Y=0,}$

   with characteristic equations obtained from trial solutions $X\left(x\right)=e^{rx}$, $Y\left(y\right)=e^{sy}$

   ${\displaystyle r^2+2r-\lambda =0,s^2+2s+\lambda =0.}$

   The roots of the characteristic equation are

   ${\displaystyle r_{1,2}=-1\pm \sqrt{1+\lambda },s_{1,2}=-1\pm \sqrt{1-\lambda }.}$

   The general solution of the homogeneous PDE is therefore

   ${\displaystyle u(x,y)=e^{-x-y}[c_1{e}^{x\sqrt{1+\lambda }}+c_2^{}{e}^{-x\sqrt{1+\lambda }}][c_3{e}^{y\sqrt{1-\lambda }}+c_4{e}^{-y\sqrt{1-\lambda }}].}$

   Further analysis requires specification of boundary conditions.

   .

3. Solve the boundary-value problem

   ${\displaystyle \frac{{\partial }^{2}u}{\partial x^2}=\frac{\partial u}{\partial t},0<x<\pi ,t>0,}$ ${\displaystyle u(0,t)=0,u(\pi ,t)=0,t>0,}$ ${\displaystyle u(x,0)=\sin x,0<x<\pi .}$

   Solution. Separation of variables $u(x,t)=X\left(x\right)T\left(t\right)$ leads to

   ${\displaystyle \frac{X^{\text{'}\text{'}}}{X}=\frac{T^{\text{'}}}{T}=-k^2,}$

   where the constant $-k^2$ has been chosen negative to avoid non-physical ${lim}_{t\to \infty }T\left(t\right)=\infty$. Solution of $X^{\text{'}\text{'}}+k^{2}X=0$ gives

   ${\displaystyle X\left(x\right)=c_1\cos (kx)+c_2\sin (kx).}$

   Left boundary condition $u(0,t)=0$ implies $X\left(0\right)=0\Rightarrow c_1=0$. Right boundary condition $X\left(\pi \right)=0$ leads to

   ${\displaystyle c_2\sin \left(k\pi \right)=0,}$

   satisfied for eigenvalues $k\in {\mathbb{N}}_{+}$. The value $k=0$ is excluded since it implies $T\left(t\right)=c_3$, $X\left(x\right)=c_4+c_{5}x$, and homogeneous boundary conditions on $X\left(x\right)$ would imply $c_4=c_5=0\Rightarrow X\left(x\right)=0.$

   The solution is therefore the series

   ${\displaystyle u(x,t)=\sum _{k=1}^{\infty }{A}_k{e}^{-k^{2}t}\sin (kx).}$

   The initial condition

   ${\displaystyle u(x,0)=\sin \left(x\right)=\sum _{k=1}^{\infty }{A}_k\sin (kx),}$

   implies $A_1=1$, $A_k=0$ for $k>1$, hence the problem solution is

   ${\displaystyle u(x,t)=e^{-t}\sin \left(x\right).}$