MATH529: Mathematical methods for the physical sciences IIApril 28, 2024

Solve the following problems (5 course points each). Present a brief motivation of your method of solution. Problems 9 and 10 are optional; attempt them if you wis to improve your midterm examination score.

1. Solve the eigenvalue problem

   ${\displaystyle y^{\text{'}\text{'}}+3y^{\text{'}}+2y+\lambda y=0,y\left(0\right)=0,y\left(1\right)=0.}$

   Solution. Linear, second order, constant coefficient, homogeneous ODE with homogeneous boundary conditions (BCs), i.e., an eigenvalue problem. Try solutions of form $y\left(x\right)\sim e^{rx}$ to obtain characteristic equation

   ${\displaystyle r^2+3r+(\lambda +2)=0,}$

   with solutions

   ${\displaystyle r_{1,2}=\frac{1}{2}(-3\pm \sqrt{9-4(\lambda +2)})=-\frac{3}{2}\pm \frac{\sqrt{1-4\lambda }}{2}=-\alpha \pm \beta ,}$

   leading to

   ${\displaystyle y\left(x\right)=A{e}^{r_{1}x}+B{e}^{r_{2}x}.}$

   For $\beta =0$, obtain $y\left(x\right)=A{e}^{-\alpha x}+Bx{e}^{-\alpha x}$. Apply BCs

   ${\displaystyle \begin{array}{ll}x=0:& A=0\\ x=1:& A{e}^{-\alpha }+B{e}^{-\alpha }=0\end{array}\Rightarrow A=B=0,}$

   only a trivial solution.

   For $\beta \ne 0$, apply BCs

   ${\displaystyle \begin{array}{ll}x=0:& A+B=0\\ x=1:& A{e}^{r_1}+B{e}^{r_2}=0\end{array}.}$

   Non-trivial solutions (i.e., $A\ne 0$ or $B\ne 0$) obtained only if principal determinant of above is zero

   ${\displaystyle e^{r_2}-e^{r_1}=0\Rightarrow e^{-\alpha }(e^{\beta }-e^{-\beta })=0.}$

   Since $\alpha =-3/2$, $e^{\alpha }\ne 0$, hence

   ${\displaystyle e^{\beta }-e^{-\beta }=2\sinh \beta =0,}$

   with solutions only for $\Delta =1-4\lambda <0$ in which case $\beta =\frac{i}{2}\sqrt{4\lambda -1}$, and

   ${\displaystyle \sinh \beta =i\sin \frac{\sqrt{4\lambda -1}}{2}=0\Rightarrow \frac{\sqrt{4\lambda -1}}{2}=k\pi \Rightarrow {\lambda }_k=\frac{1}{4}[{\left(2k\pi \right)}^2+1],}$

   eigenvalues of the problem, with associated eigenfunctions

   ${\displaystyle y_k\left(x\right)=e^{-3x/2}\sin (k\pi x).}$

   Note: recall that eigenfunctions are determined up to a multiplicative constant.

2. Solve the eigenvalue problem

   ${\displaystyle y^{\text{'}\text{'}}+\lambda y=0,y\left(0\right)+y^{\text{'}}\left(0\right)=0,y\left(1\right)+3y^{\text{'}}\left(1\right)=0.}$

   Solution. Linear, second order, constant coefficient, homogeneous ODE with homogeneous boundary conditions (BCs), i.e., an eigenvalue problem. Try solutions of form $y\left(x\right)\sim e^{rx}$ to obtain characteristic equation $r^2+\lambda =0$, with solutions with solutions $r_{1,2}=\pm \sqrt{\lambda }$. When $\lambda =0$, $y\left(x\right)=A+Bx$ and BCs give only the trivial solution $y=0$. For $\lambda \ne 0$, $y\left(x\right)=A{e}^{\alpha x}+B{e}^{-\alpha x}$, $\alpha =\sqrt{\lambda }$, obtain $y^{\text{'}}\left(x\right)=\alpha (A{e}^{\alpha x}-B{e}^{-\alpha x})$, and BCs give

   ${\displaystyle \begin{array}{ll}x=0:& A+B+\alpha (A-B)=0\\ x=1:& A{e}^{\alpha }+B{e}^{-\alpha }+3\alpha (A{e}^{\alpha }-B{e}^{-\alpha })=0\end{array}\Rightarrow \{\begin{array}{lllll}(1+\alpha )A& +& (1-\alpha )B& =& 0\\ (1+3\alpha )e^{\alpha }A& +& (1-3\alpha )e^{-\alpha }B& =& 0\end{array}..}$

   Non-trivial solution obtained if

   ${\displaystyle (1-2\alpha -3{\alpha }^2)e^{-\alpha }-(1+2\alpha -3{\alpha }^2)e^{\alpha }=0\Rightarrow -2(1-3{\alpha }^2)\sinh \alpha -4\alpha \cosh \alpha =0\Rightarrow \tanh \alpha =-\frac{2\alpha }{1-3{\alpha }^2}.}$

   Sturm-Liouville theorem guarantees existence of a countably infinite number of eigenvalues, impossible for $\alpha \in ℝ$, thus implying $\alpha =i\beta$ ($\lambda <0$), $\beta \in ℝ$, in which case eigenvalues ${\beta }_k$ are solutions of

   ${\displaystyle \tan \beta =-\frac{2\beta }{1+3{\beta }^2},}$

   with associated ODE solution

   ${\displaystyle y_k\left(x\right)=A{e}^{i\beta x}+B{e}^{-i\beta x}=(A+B)\cos (\beta x)+i(A-B)\sin (\beta x).}$

   Use $x=0$ BC, $A+B=-\alpha (A-B)$ to obtain $A+B=-i\beta (A-B)$, and the eigenfunctions are

   ${\displaystyle y_k\left(x\right)={\beta }_k\cos ({\beta }_{k}x)-\sin ({\beta }_{k}x).}$

3. Find the Fourier series

   ${\displaystyle F\left(x\right)=a_0+\sum _{k=1}^{\infty }(a_k\cos (kx)+b_k\sin (kx))}$

   of $f:[0,\pi ]\to ℝ$, $f\left(x\right)=2x-3x^2$.

   Solution. The system $\{1,\cos x,\sin x,\cos 2x,\sin 2x,\dots \}$ is an orthogonal basis. Take scalar products

   ${\displaystyle \underset{0}{\overset{\pi }{\int }}F\left(x\right)\cdot 1dx=\pi a_0=\underset{0}{\overset{\pi }{\int }}(2x-3x^2)\cdot 1dx={\pi }^2-{\pi }^3={\pi }^2(1-\pi )\Rightarrow a_0=\pi (1-\pi )}$ ${\displaystyle \underset{0}{\overset{\pi }{\int }}F\left(x\right)\cdot \cos (kx)dx=a_k\underset{0}{\overset{\pi }{\int }}{\cos }^2(kx)dx=\frac{a_k\pi }{2}=\underset{0}{\overset{\pi }{\int }}(2x-3x^2)\cdot \cos (kx)dx\Rightarrow a_k=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}(2x-3x^2)\cdot \cos (kx)dx}$ ${\displaystyle b_k=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}(2x-3x^2)\cdot \sin (kx)dx.}$

   Form

   ${\displaystyle c_k=\frac{\pi }{2}(a_k+i{b}_k)=\underset{0}{\overset{\pi }{\int }}(2x-3x^2)e{}^{ikx}dx.}$

   Integrate by parts, $u=2x-3x^2$, $dv=e^{ikx}dx\Rightarrow v=e^{ikx}/(ik)$

   ${\displaystyle \underset{0}{\overset{\pi }{\int }}(2x-3x^2)e{}^{ikx}dx={\left[(2x-3x^2)\frac{e^{ikx}}{ik}\right]}_{x=0}^{x=\pi }-\frac{1}{ik}\underset{0}{\overset{\pi }{\int }}(2-6x)e{}^{ikx}dx=\frac{3{\pi }^2-2\pi }{ik}-\frac{1}{ik}\underset{0}{\overset{\pi }{\int }}(2-6x)e{}^{ikx}dx.}$

   Another integration by parts, $u=2-6x$, $dv=e^{ikx}dx\Rightarrow v=e^{ikx}/(ik)$ gives

   ${\displaystyle \underset{0}{\overset{\pi }{\int }}(2-6x)e{}^{ikx}dx={\left[(2-6x)\frac{e^{ikx}}{ik}\right]}_{x=0}^{x=\pi }+\frac{6}{ik}\underset{0}{\overset{\pi }{\int }}e{}^{ikx}dx=\frac{6\pi -2}{ik}-\frac{2}{ik}+\frac{12}{k^2}=\frac{6\pi -4}{ik}+\frac{12}{k^2}.}$

   Obtain

   ${\displaystyle c_k=\frac{3{\pi }^2-2\pi }{ik}-\frac{1}{ik}(\frac{6\pi -4}{ik}+\frac{12}{k^2})=\frac{3{\pi }^2-2\pi }{ik}-\frac{6\pi -4}{k^2}-\frac{12}{i{k}^3}=\frac{6\pi -4}{k^2}+i(\frac{12}{k^3}-\frac{3{\pi }^2-2\pi }{k})\Rightarrow }$ ${\displaystyle a_k=\frac{2(6\pi -4)}{\pi k^2},b_k=\frac{2}{\pi }(\frac{12}{k^3}-\frac{3{\pi }^2-2\pi }{k}).}$

4. Find $u(x,t)$, $u:[0,1]\times [0,\infty )\to ℝ$ by solving the problem

   ${\displaystyle u_t=u_{xx},u(0,t)=0,u(1,t)=0,u(x,0)=x(1-x).}$

   Solution. Linear, second order, homogeneous PDE with homogeneous BCs, inhomogeneous initial condition (IC). Separation of variables $u(x,t)=X\left(x\right)T\left(t\right)$ leads to

   ${\displaystyle \frac{T^{\text{'}}}{T}=\frac{X^{\text{'}\text{'}}}{X}=-{\left(n\pi \right)}^2,}$

   and solution given as superposition of eigenfunctions of the $x$-BVP

   ${\displaystyle u(x,t)=\sum _{n=1}^{\infty }{c}_n\sin (n\pi )e^{-{(n\pi )}^{2}t}.}$

   Initial condition gives

   ${\displaystyle c_n=\underset{0}{\overset{1}{\int }}x(1-x)\sin (n\pi x)dx.}$

   Integration by parts $u=x(1-x)$, $dv=\sin (n\pi x)dx\Rightarrow v=-\cos (n\pi x)/(n\pi )$

   ${\displaystyle c_n=-{\left[\frac{x(1-x)\cos (n\pi x)}{n\pi }\right]}_{x=0}^{x=1}+\frac{1}{n\pi }\underset{0}{\overset{1}{\int }}(1-2x)\cos (n\pi x)dx.}$

   Again, $u=(1-2x)$, $dv=\cos (n\pi x)\Rightarrow v=\sin (n\pi x)/(n\pi )$

   ${\displaystyle \underset{0}{\overset{1}{\int }}(1-2x)\cos (n\pi x)dx={\left[(1-2x)\frac{\sin (n\pi x)}{n\pi }\right]}_{x=0}^{x=1}+\frac{2}{n\pi }\underset{0}{\overset{1}{\int }}\sin (n\pi x)dx=-\frac{2}{{(n\pi )}^2}{\left[\cos (n\pi x)\right]}_{x=0}^{x=1}.}$

   Deduce

   ${\displaystyle c_{2k}=0,c_{2k+1}=\frac{4}{{((2k+1)\pi )}^3}.}$

5. For $z=x+iy$, $\mathrm{Re}\left(z\right)>0$ show that

   ${\displaystyle \mathrm{Ln}z=\frac{1}{2}{\log }_e(x^2+y^2)+i{\tan }^{-1}\frac{y}{x},}$

   and verify that $\mathrm{Ln}z$ thus defined is analytic in the right half-plane.

   Solution. From $f\left(z\right)=\ln z$, $z=r{e}^{i\theta }$, $r={(x^2+y^2)}^{1/2}$, $\theta ={\tan }^{-1}(y/x)$,

   ${\displaystyle f\left(z\right)=\ln r+i(\theta +2k\pi ).}$

   Choose principal branch $k=0$ and obtain above relation. Verify that $\mathrm{Ln}z=u+iv$ is analytic by Caucy-Riemann conditions

   ${\displaystyle u_x=\frac{x}{x^2+y^2},v_y=\frac{(1/x)}{1+{(y/x)}^2}=\frac{x}{x^2+y^2}=u_x?}$ ${\displaystyle u_y=\frac{y}{x^2+y^2},v_x=\frac{-(y/x^2)}{1+{(y/x)}^2}=-\frac{y}{x^2+y^2}=-u_y?}$

   for all points in the right half plane.

6. Show that the real and imaginary parts of $\mathrm{Ln}z$ defined above are harmonic.

   Solution. As above.

7. Determine the value of the integral

   ${\displaystyle I=\underset{1-i}{\overset{1+2i}{\int }}z{e}^{z^2}dz.}$

   Solution. Integrand has primitive $F\left(z\right)=e^{z^2}/2$ hence

   ${\displaystyle I=\frac{1}{2}{\left[e^{z^2}\right]}_{z=1-i}^{z=1+2i}=\frac{1}{2}(e^{1+4i-4}-e^{1-2i+1})=\frac{1}{2}(e^{4i-3}-e^{2-2i}).}$

8. Find the value of

   ${\displaystyle I=\underset{C}{\oint }\frac{dz}{z^2(z^2+1)},C:|z-i|=\frac{3}{2}.}$

   Solution. Integrand $f\left(z\right)$ has simple poles at $z_{1,2}=\pm i$, and a double pole at $z_3=0$. The pole $z_1=-i$ is outside the contour. Apply residue formula

   ${\displaystyle I=2\pi i[\mathrm{res}(f,i)+\mathrm{res}(f,0)].}$

   Compute

   ${\displaystyle \mathrm{res}(f,i)={\left[(z-i)\frac{1}{z^2(z^2+1)}\right]}_{z=i}={\left[\frac{1}{z^2(z+i)}\right]}_{z=i}=-\frac{1}{2i}=\frac{i}{2}.}$ ${\displaystyle \mathrm{res}(f,i)={\left[\frac{d}{dz}\left(z^2\frac{1}{z^2(z^2+1)}\right)\right]}_{z=0}=-{\left[\frac{2z}{{(z^2+1)}^2}\right]}_{z=0}=0.}$

   Deduce $I=\pi i$.

9. An elastic cylinder of radius $R=1$ is subjected to surface force $f(\theta ,t)=\cos \theta \sin (\omega t)$. Formulate the wave equation problem for radial displacements $u(\theta ,t)$ of the cylinder surface from its equilibrium position.

   Solution. Wave equation $u_{tt}=c^2\nabla \cdot (\nabla u)+f$ in polar coordinates $(r,\theta )$ gives

   ${\displaystyle u_{tt}=c^2\nabla \cdot [\frac{\partial u}{\partial r}{𝒆}_r+\frac{1}{r}\frac{\partial u}{\partial \theta }{𝒆}_{\theta }]=\frac{c^2}{r}[\frac{\partial }{\partial r}(\frac{\partial u}{\partial r}r)+\frac{\partial }{\partial \theta }\left(\frac{1}{r}\frac{\partial u}{\partial \theta }\right)]=\frac{c^2}{r^2}{u}_{\theta \theta }+f.}$

   On $r=R=1$ obtain, $u_{tt}=c^2{u}_{\theta \theta }+f$, with periodic BCs $u(0,t)=u(2\pi ,t)$, $u_{\theta }(0,t)=u_{\theta }(2\pi ,t)$ and initial conditions $u(\theta ,0)=0$, $u_t(\theta ,0)=0$.

10. Solve the above problem by the separation of variables $u(\theta ,t)=\Theta \left(\theta \right)T\left(t\right)$.

    Solution. The above is a second-order inhomogeneous PDE. Solve by expanding both $u(\theta ,t)$ and $f(\theta ,t)$ on the eigenfunctions of the homogeneous PDE $\{\frac{1}{2},\cos \theta ,\sin \theta ,\dots ,\cos (n\theta ),\sin (n\theta ),\dots \}$

    ${\displaystyle u(\theta ,t)=\frac{1}{2}{a}_0\left(t\right)+\sum _{n=1}^{\infty }[a_n\left(t\right)\cos (n\theta )+b_n\left(t\right)\sin (n\theta )],f(\theta ,t)=\sin (\omega t)\cos \theta \Rightarrow }$ ${\displaystyle u_{tt}=\frac{1}{2}{a}_0^{\text{'}\text{'}}\left(t\right)+\sum _{n=1}^{\infty }[a_n^{\text{'}\text{'}}\left(t\right)\cos (n\theta )+b_n^{\text{'}\text{'}}\left(t\right)\sin (n\theta )],}$ ${\displaystyle u_{\theta \theta }=-\sum _{n=1}^{\infty }{n}^2[a_n\left(t\right)\cos (n\theta )+b_n\left(t\right)\sin (n\theta )]}$

    Since the eigenfunctions are orthogonal obtain ODE system

    ${\displaystyle a_0^{\text{'}\text{'}}=0,a_1^{\text{'}\text{'}}=-c^2{a}_1+\sin (\omega t),}$ ${\displaystyle a_n^{\text{'}\text{'}}=-c^2{a}_n,b_n^{\text{'}\text{'}}=-c^2{b}_n,n>1.}$

    Applying initial conditions leads to only one none-zero term, $a_1\left(t\right)$.