MATH529: Mathematical methods for the physical sciences IIMay 13, 2021

Solve the following problems (4 course points each). Present a brief motivation of your method of solution. Answers without explanation of solution procedure are not awarded credit.

1. Use the Laplace transform $F\left(s\right)=ℒ\left\{f\left(t\right)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$ to solve the problem

   ${\displaystyle a^2\frac{{\partial }^{2}u}{\partial x^2}=\frac{{\partial }^{2}u}{\partial t^2},x>0,t>0,}$ ${\displaystyle u(0,t)=0,{lim}_{x\to \infty }\frac{\partial u}{\partial x}(x,t)=0,t>0,}$ ${\displaystyle u(x,0)=0,v(x,0)=\frac{\partial u}{\partial t}(x,0)=-v_0,x>0,v\in {ℝ}_{+}.}$

   Solution. With notation $u_t=\partial u/\partial t$, $u_{tt}={\partial }^{2}u/\partial t^2$, use integration by parts to evaluate

   ${\displaystyle ℒ\left\{u^{\text{'}}(x,t)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}{u}_t(x,t)dt={[e^{-st}u(x,t)]}_{t=0}^{t\to \infty }+s\underset{0}{\overset{\infty }{\int }}{e}^{-st}u(x,t)dt=-u(x,0)+sU(x,s)}$ ${\displaystyle ℒ\left\{u^{\text{'}}(x,t)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}{u}_{tt}(x,t)dt={[e^{-st}{u}_t(x,t)]}_{t=0}^{t\to \infty }+s\underset{0}{\overset{\infty }{\int }}{e}^{-st}{u}_t(x,t)dt=-u_t(x,0)+s[sU(x,s)-u(x,0)].}$

   Apply $ℒ$ to PDE to obtain

   ${\displaystyle a^2\frac{d^{2}U(x,s)}{d{x}^2}=s^{2}U(x,s)-su(x,0)-u_t(x,0).}$

   Apply initial conditions to obtain

   ${\displaystyle a^2\frac{d^{2}U(x,s)}{d{x}^2}=s^{2}U(x,s)+v_0.}$

   Solve the homogeneous ODE

   ${\displaystyle a^2\frac{d^{2}U(x,s)}{d{x}^2}=s^{2}U(x,s),}$

   to obtain the general solution

   ${\displaystyle U(x,s)=c_1{e}^{-sx/a}+c_2{e}^{sx/a}.}$

   A particular solution is

   ${\displaystyle U(x,s)=c_1{e}^{-sx/a}+c_2{e}^{sx/a}-\frac{v_0}{s^2}.}$

   Take the Laplace transform of the boundary conditions:

   ${\displaystyle {lim}_{x\to \infty }\frac{dU(x,s)}{dx}=0\Rightarrow c_2=0}$ ${\displaystyle U(0,s)=0\Rightarrow c_1-\frac{v_0}{s^2}=0\Rightarrow c_1=\frac{v_0}{s^2},}$

   leading to solution

   ${\displaystyle U(x,s)=\frac{v_0}{s^2}(e^{-sx/a}-1).}$

   Take the inverse Laplace transforms

   ${\displaystyle u(t,x)={ℒ}^{-1}\left\{U(x,s)\right\}=v_0[{ℒ}^{-1}\{\frac{1}{s^2}{e}^{-sx/a}\}-{ℒ}^{-1}\left\{\frac{1}{s^2}\right\}].}$

   From

   ${\displaystyle ℒ\left\{t\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}tdt=-{[\frac{1}{s}{e}^{-st}t]}_{t=0}^{t\to \infty }+\frac{1}{s}\underset{0}{\overset{\infty }{\int }}{e}^{-st}dt=\frac{1}{s^2},}$

   deduce

   ${\displaystyle {ℒ}^{-1}\left\{\frac{1}{s^2}\right\}=t.}$

   Recall that exponential factors in a Laplace transform arise from a delay that can be represented through the unit step function $H\left(t\right)=1$ for $t>0$, 0 otherwise, as

   ${\displaystyle ℒ\left\{H(t-c)\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}H(t-c)dt=\underset{c}{\overset{\infty }{\int }}{e}^{-st}dt=-\frac{1}{s}{\left[e^{-st}\right]}_{t=c}^{t\to \infty }=\frac{e^{-cs}}{s},(c>0).}$

   To obtain an $s^2$ denominator, combine above with integration by parts

   ${\displaystyle ℒ\{tH(t-c)\}=\underset{c}{\overset{\infty }{\int }}t{e}^{-st}dt=-\frac{1}{s}{[t{e}^{-st}]}_{t=c}^{t\to \infty }+\frac{1}{s}\underset{c}{\overset{\infty }{\int }}{e}^{-st}dt=\frac{c{e}^{-cs}}{s}+\frac{e^{-cs}}{s^2}=cℒ\left\{H(t-c)\right\}+\frac{e^{-cs}}{s^2},}$

   to obtain

   ${\displaystyle \frac{e^{-cs}}{s^2}=ℒ\{tH(t-c)\}-cℒ\left\{H(t-c)\right\}=ℒ\{(t-c)H(t-c)\}\Rightarrow (t-c)H(t-c)={ℒ}^{-1}\left\{\frac{e^{-cs}}{s^2}\right\}.}$

   Gathering the above results

   ${\displaystyle u(t,x)=v_0[{ℒ}^{-1}\{\frac{1}{s^2}{e}^{-sx/a}\}-{ℒ}^{-1}\left\{\frac{1}{s^2}\right\}]=v_0[(t-\frac{x}{a})H(t-\frac{x}{a})-t].}$

2. Use the Fourier transform $F\left(\alpha \right)=ℱ\left\{f\left(x\right)\right\}={\int }_{-\infty }^{\infty }{e}^{i\alpha x}f\left(x\right)dx$ to solve the problem

   ${\displaystyle \frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0,0<x<\pi ,y>0,}$ ${\displaystyle u(0,y)=f\left(y\right),\frac{\partial u}{\partial x}(\pi ,y)=0,y>0,}$ ${\displaystyle \frac{\partial u}{\partial y}(x,0)=0,0<x<\pi .}$

   Solution. The problem is defined on the half-infinite strip $y>0$, and the boundary condition at $y=0$ is on the derivative of the function, $u_y(x,0)=0$. The appropriate transform is the cosine transform

   ${\displaystyle U(x,\beta )=ℱ\left\{u(x,y)\right\}=\underset{0}{\overset{\infty }{\int }}\cos (\beta y)u(x,y)dy.}$

   Integration by parts, assuming $u,\partial u/\partial y\to 0$, as $y\to \infty$ leads to

   ${\displaystyle \underset{0}{\overset{\infty }{\int }}\cos (\beta y)\frac{{\partial }^{2}u}{\partial y^2}(x,y)dy={\left[\cos \left(\beta y\right)\frac{\partial u}{\partial y}(x,y)\right]}_{y=0}^{y\to \infty }+\beta \underset{0}{\overset{\infty }{\int }}\sin (\beta y)\frac{\partial u}{\partial y}(x,y)dy=}$ ${\displaystyle -\frac{\partial u}{\partial y}(x,0)+\beta \underset{0}{\overset{\infty }{\int }}\sin (\beta y)\frac{\partial u}{\partial y}(x,y)dy=-\frac{\partial u}{\partial y}(x,0)+\beta ({\left[\sin (\beta y)u\right]}_{y=0}^{y\to \infty }-\beta \underset{0}{\overset{\infty }{\int }}\cos (\beta y)u(x,y)dy)=}$ ${\displaystyle -\frac{\partial u}{\partial y}(x,0)-{\beta }^{2}U(x,\beta ).}$

   Replacing into Laplace PDE and using boundary condition at $y=0$, gives

   ${\displaystyle \frac{d^{2}U}{d{x}^2}-{\beta }^{2}U=-\frac{\partial u}{\partial y}(x,0)=0,}$

   with solution

   ${\displaystyle U(x,\beta )=c_1\left(\beta \right)\cosh (\beta x)+c_2\left(\beta \right)\sinh (\beta x).}$

   Cosine transform of the boundary condition at $x=0$ leads to

   ${\displaystyle U(0,\beta )=F\left(\beta \right)=\underset{0}{\overset{\infty }{\int }}\cos (\beta y)f\left(y\right)dy=c_1\left(\beta \right),}$

   specifying that $c_1\left(\beta \right)=F\left(\beta \right)$, the cosine transform of the boundary value $f\left(y\right)$. Compute the $x$-derivative

   ${\displaystyle \frac{dU}{dx}(x,\beta )=\beta [c_1\left(\beta \right)\sinh (\beta x)+c_2\left(\beta \right)\cosh (\beta x)].}$

   The boundary condition at $x=\pi$ specifies

   ${\displaystyle \frac{dU}{dx}(\pi ,\beta )=0=\beta [c_1\left(\beta \right)\sinh (\beta \pi )+c_2\left(\beta \right)\cosh (\beta \pi )]\Rightarrow c_2\left(\beta \right)=-c_1\left(\beta \right)\frac{\sinh (\beta \pi )}{\cosh \left(\beta \pi \right)}=-F\left(\beta \right)\tanh (\beta \pi ).}$

   The solution is obtained by taking the inverse cosine transform of

   ${\displaystyle U(x,\beta )=F\left(\beta \right)[\cosh (\beta x)-\tanh (\beta \pi )\sinh (\beta x)]\Rightarrow }$ ${\displaystyle u(x,y)=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}F\left(\beta \right)[\cosh (\beta x)-\tanh (\beta \pi )\sinh (\beta x)]\cos (\beta y)d\beta =\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}F\left(\beta \right)\cosh \left[\beta (x-\pi )\right]\frac{\cos (\beta y)}{\cos (\beta \pi )}d\beta .}$

3. Find all solutions of the equation $z^8-2z^4+1=0.$ Write the roots in both Cartesian and polar form.

   Solution. Introduce $w=z^4$ to obtain $w^2-2w+1={(w-1)}^2=0$ with double root $w_{1,2}=1$. The equation $z^4-1=0\Rightarrow z^4=e^{2\pi i}$, has roots $z_k=e^{ik\pi /2}$ for $k=0,1,2,3$. The original equation $z^8-2z^4+1=0$ as double roots at each $z_k$.

4. Sketch the region defined by $-1⩽\mathrm{Im}(1/z)<1$. Is this region a domain?

   Solution. With $z=x+iy$,

   ${\displaystyle \mathrm{Im}\left(\frac{1}{x+iy}\right)=\mathrm{Im}\left(\frac{x-iy}{x^2+y^2}\right)=-\frac{y}{x^2+y^2}.}$

   The inequality $-y/(x^2+y^2)<1$ leads to $-y<x^2+y^2\Rightarrow x^2+{(y+\frac{1}{2})}^2-\frac{1}{4}>0$, the exterior of a circle of radius $1/2$ centered at $(0,-1/2)$, excluding the circle. The inequality $-1⩽-y/(x^2+y^2)$ leads to $x^2+y^2-y⩾0\Rightarrow x^2+{(y-\frac{1}{2})}^2-\frac{1}{4}⩾0$, the exterior of a circle of radius 1/2, centered at (0,1/2). The region is not a domain since it is not an open set.

5. Is $f\left(z\right)=x^2-x+y+i(y^2-5y-x)$ an analytic function? Is it differentiable along the curve $y=x+2$?

   Solution. With $z=x+iy$, $f\left(z\right)=u(x,y)+iv(x,y)$,

   ${\displaystyle u(x,y)=x^2-x+y,v=y^2-5y-x}$

   the first Cauchy-Riemann condition is not verified, i.e.,

   ${\displaystyle u_x=2x-1\ne v_y=2y-5,u_y=1=-v_x}$

   so $f$ is not analytic. On the curve $C:y=x+2$ the function restriction is a quadratic polynomial in $x$

   ${\displaystyle f_C\left(x\right)=x^2-x+x+2+i[{(x+2)}^2-5(x+2)-x]=x^2+2+i(x^2-2x-6),}$

   and is differentiable, $f_C^{\text{'}}\left(x\right)=2x+2i(x-1)$.

6. Evaluate the integral

   ${\displaystyle \underset{C}{\oint }\frac{2z}{z^2+3}dz}$

   for $C$ defined as:

   1. $\left|z\right|=1$;

      Solution. Assume $C$ traversed in positive (CCW) direction. The integrand has isolated pole singularities at $z_{1,2}=\pm i\sqrt{3}$ not within the unit circle, hence

      ${\displaystyle \underset{C}{\oint }\frac{2z}{z^2+3}dz=0}$

   2. $|z-2i|=1$.

      Solution. The curve $C$ is a circle of radius 1 centered at $(0,2)$. The singularity $z_1=i\sqrt{3}$ is within the CCW-traversed contour, the singularity $z_2=-i\sqrt{3}$ is outside. The integral is therefore

      ${\displaystyle \underset{C}{\oint }\frac{2z}{z^2+3}dz=\underset{C}{\oint }f\left(z\right)dz=2\pi i\mathrm{Res}\left[f(z=i\sqrt{3})\right].}$

      From

      ${\displaystyle \frac{2z}{z^2+3}=\frac{1}{z-i\sqrt{3}}+\frac{1}{z+i\sqrt{3}}}$

      observe $\mathrm{Res}\left[f(z=i\sqrt{3})\right]=1$.

7. Expand

   ${\displaystyle f\left(z\right)=\frac{1}{z{(1-z)}^2}}$

   in a Laurent series valid for:

   1. $0<\left|z\right|<1$;

      Solution. Differentiation term-by-term of the convergent series for $\left|z\right|<1$

      ${\displaystyle \frac{1}{1-z}=1+z+z^2+z^3+\cdots \Rightarrow \frac{1}{{(1-z)}^2}=1+2z+3z^2+4z^3+\cdots }$

      and multiplying by $1/z$ ($\left|z\right|>0\Rightarrow z\ne 0$) gives

      ${\displaystyle f\left(z\right)=\frac{1}{z}+2+3z+4z^2+5z^3+\cdots +(k+2)z^k+\cdots }$

   2. $\left|z\right|>1$.

      Solution. Substitute $w=1/z$, $\left|w\right|<1$, and use above series to obtain

      ${\displaystyle \frac{1}{z{(1-z)}^2}=\frac{w}{{(1-\frac{1}{w})}^2}=\frac{w^3}{{(w-1)}^2}=\frac{w^3}{{(1-w)}^2}=w^3+2w^4+3w^5+\cdots .}$

      Substitue back $z=1/w$ and deduce

      ${\displaystyle f\left(z\right)=\frac{1}{z^3}+\frac{2}{z^4}+\frac{3}{z^5}+\cdots +\frac{k-2}{z^k}+\cdots .}$