Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 3.4

1. Find dispersive wave solutions of the $n$-dimensional linear Klein-Gordon equation $u_{tt}-c^2\Delta u+m^2u=0$.
Solution

Suppose that the dispersive wave solution of the $n$-dimensional linear Klein-Gordon equaiton is of the form $u(x,t)=Ae^{i(k\cdot x-\omega t)}$ (cf. (55) and Remark at page 96 in the textbook), where $k=(k_1,\dots,k_n)$ and $k\cdot x=k_1x_1+\cdots+k_nx_n$. From the linear Klein-Gordon equation, we have

$\begin{aligned}u_{tt}-c^2\Delta u+m^2u&=Ae^{i(k\cdot x-\omega t)}(-i\omega)^2-c^2\cdot Ae^{i(k\cdot x-\omega t)}[(ik_1)^2+(ik_2)^2+\cdots+(ik_n)^2]+m^2\cdot Ae^{i(k\cdot x-\omega t)}\\&=Ae^{i(k\cdot x-\omega t)}\cdot(-\omega^2+c^2|k|^2+m^2)=0.\end{aligned}$

Thus, $\omega^2=m^2-c^2|k|^2$, i.e., $\omega(k)=\pm\sqrt{c^2|k|^2+m^2}$. Hence the dispersive wave solutions of the $n$-dimensional linear Klein-Gordon equaiton is given by

$u(x,t)=Ae^{i(k\cdot x-\omega(k)t)}\quad\text{and}\quad\omega(k)=\pm\sqrt{c^2|k|^2+m^2}.$



2. Show that each of the following linear equations has dispersive wave solutions $u=A\exp[i(kx-\omega t)]$:
   1. The flexible beam equation $u_{tt}+\gamma^2u_{xxxx}=0$;
   2. The linearized KdV equation $u_t+cu_x+u_{xxx}=0$;
   3. The Boussinesq equation $u_{tt}-c^2u_{xx}=\gamma^2u_{ttxx}$;
   4. The Schrödinger equation $u_t=i\Delta u$.
Solution

1. Plugging $u=A\exp[i(kx-\omega t)]$ into flexible beam equation, we get

   $u_{tt}+\gamma^2u_{xxxx}=Ae^{i(kx-\omega t)}(-i\omega)^2+\gamma^2\cdot Ae^{i(kx-\omega t)}(ik)^4=Ae^{i(kx-\omega t)}(-\omega^2+\gamma^2k^4)=0$,

   which gives $\omega=\pm\gamma k^2$. Thus, the dispersive wave solution of the flexible beam equation is given by

   $u(x,t)=Ae^{i(kx-\omega(k)t)}\quad\text{and}\quad\omega(k)=\pm\gamma k^2$.

2. Plugging $u=A\exp[i(kx-\omega t)]$ into linearized KdV equation, we get

   $\begin{aligned}u_t+cu_x+u_{xxx}&=Ae^{i(kx-\omega t)}(-i\omega)+c\cdot Ae^{i(kx-\omega t)}(ik)+Ae^{i(kx-\omega t)}(ik)^3\\&=-Aie^{i(kx-\omega t)}[\omega-ck+k^3]=0,\end{aligned}$

   which $\omega=ck-k^3$. Thus, the dispersive wave solution of the linearized KdV equation is given by

   $u(x,t)=Ae^{i(kx-\omega(k)t)}\quad\text{and}\quad\omega(k)=ck-k^3$.

3. Plugging $u=A\exp[i(kx-\omega t)]$ into Boussinesq equation, we get

   $\begin{aligned}u_{tt}-c^2u_{xx}&=Ae^{i(kx-\omega t)}(-i\omega)^2-c^2\cdot Ae^{i(kx-\omega t)}(ik)^2=Ae^{i(kx-\omega t)}(-\omega^2+c^2k^2)\\&=Ae^{i(kx-\omega t)}(\gamma^2\omega^2k^2)=\gamma^2\cdot Ae^{i(kx-\omega t)}(-i\omega)^2(ik)^2=\gamma^2u_{ttxx},\end{aligned}$

   which gives $\displaystyle\omega=\pm\frac{ck}{\sqrt{1+\gamma^2k^2}}$. Thus, the dispersive wave solution of the Boussinesq equation is given by

   $\displaystyle u(x,t)=Ae^{i(kx-\omega(k)t)}\quad\text{and}\quad\omega(k)=\pm\frac{ck}{\sqrt{1+\gamma^2k^2}}.$

4. Plugging $u=A\exp[i(k\cdot x-\omega t)]$ ($k=(k_1,\dots,k_n)$ and $k\cdot x=k_1x_1+\cdots+k_nx_n$) into Schrödinger equation, we get

   $u_t=Ae^{i(k\cdot x-\omega t)}(-i\omega)=Ae^{i(k\cdot x-\omega t)}\cdot(-i|k|^2)=i\cdot Ae^{i(k\cdot x-\omega t)}[(ik_1)^2+\cdots+(ik_n)^2]=i\Delta x$,

   which gives $\omega=|k|^2$. Thus, the dispersive wave solution of the Schrödinger equation is given by

   $u(x,t)=Ae^{i(k\cdot x-\omega(k)t)}\quad\text{and}\quad\omega(k)=|k|^2$.



3. Show that the heat equation $u_t=u_{xx}$ admits uniform wave solutions of the form (55) in which $\omega(k)$ is complex-valued and the wave is exponentially decaying in $t$. (Such uniform waves are called diffusive. Thus the heat equation is diffusive and not dispersive.)
Solution
Plugging $u=Ae^{i(kx-\omega t)}$ into the heat equation, we get
$u_t=Ae^{i(kx-\omega t)}(-i\omega)=Ae^{i(kx-\omega t)}(ik)^2=u_{xx}$,
which gives $\omega=-ik^2$. Hence the solution of heat equation becomes
$u(x,t)=Ae^{i(kx+ik^2t)}=Ae^{ikx}e^{-k^2t}$
Clearly, $u$ decays exponentially to zero in $t$.


4. Find two uniform wave solutions of (52) with $\lambda>0$, satisfying the initial condition $u(x,0)=3\cos2x$.
Solution

Suppose that $u(x,t)=U(kx-\omega t)$ (cf. (53)). Substituting into (52) yields

$(\omega^2-k^2)U''(kx-\omega t)+\lambda U(kx-\omega t)=0$.

Moreover, suppose that $\omega^2-k^2>0$. Then we have

$\displaystyle U(s)=A\cos\left(s\frac{\sqrt\lambda}{\sqrt{\omega^2-k^2}}\right)+B\sin\left(s\frac{\sqrt\lambda}{\sqrt{\omega^2-k^2}}\right)$.

which implies

$\displaystyle u(x,t)=A\cos\left((kx-\omega t)\frac{\sqrt\lambda}{\sqrt{\omega^2-k^2}}\right)+B\sin\left((kx-\omega t)\frac{\sqrt\lambda}{\sqrt{\omega^2-k^2}}\right)$.

To meet the initial condition, we assume that $A=3$ and $B=0$. Then the solution may simply be expressed as

$\displaystyle u(x,t)=3\cos\left((kx-\omega t)\frac{\sqrt\lambda}{\sqrt{\omega^2-k^2}}\right)$.

For $\omega=\pm k\sqrt{\lambda+4}/2$, we have

$\displaystyle u_\pm(x,t)=3\cos\left(2x\mp\frac{4t}{k^2\sqrt{\lambda+4}}\right)$.



5. Find a condition on $g$ and $h$ that is necessary for the existence of a uniform wave solution of (52) satisfying the initial condition $u(x,0)=g(x)$ and $u_t(x,0)=h(x)$.
Solution

The uniform wave solution of (52) is of the form $u(x,t)=U(kx-\omega t)$ (see (53)). Plugging it into the initial condition, we have

$g(x)=u(x,0)=U(kx),\quad h(x)=u_t(x,0)=-\omega U'(kx-\omega t)\Big|_{t=0}=-\omega U'(kx)$.

We can differentiate $g$ to find

$\displaystyle g'(x)=kU'(kx)=\frac k\omega\cdot\omega U'(kx)=-\frac k\omega h(x)$,

which means $\omega g'(x)+kh(x)\equiv0$ for $x\in\mathbb R$. For (52), it is known that $\omega(k)=\pm\sqrt{k^2+\lambda}$ (see (56)). Thus, the necessary condition on $g$ and $h$ is

$\pm\sqrt{k^2+\lambda}g'(x)+kh(x)\equiv0\quad\text{for}~x\in\mathbb R$.



6. Find the solution of the telegrapher's equation $u_{tt}-u_{xx}+u_t+m^2u=0$ satisfying the initial conditions $u(x,0)=g(x)$ and $u_t(x,0)=0$ where $g$ is an arbitrary $C^2$ function.
Solution

Let $v(x,t)=e^{t/2}u(x,t)$ for $(x,t)\in\mathbb R\times\bar{\mathbb R}_+$. Clearly, we have

$\begin{aligned} &v_t=\frac12e^{t/2}u+e^{t/2}u_t,\\&v_{tt}=\frac14e^{t/2}u+e^{t/2}u_t+e^{t/2}u_{tt}.\end{aligned}$

Then $v$ satisfies

$\displaystyle v_{tt}-v_{xx}+\left(m^2-\frac14\right)v=0$

with the initial conditions $v(x,0)=g(x)$ and $v_t(x,0)=g(x)/2$ for $x\in\mathbb R$.

• If $|m|>1/2$, then we denote $\tilde m=\sqrt{m^2-(1/4)}>0$ so $v_{tt}-v_{xx}+\tilde m^2v=0$, which is the linear Klein-Gordon equation. By Solution 3 of Section 3.1, we have

  $\begin{aligned}v(x,t)&=\frac1{2\pi}\frac\partial{\partial t}\left(t\int_0^{2\pi}\!\int_0^1\!\frac{r\cos(\tilde mtr\sin\theta)g(x+tr\cos\theta)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta\right)\\&\quad+\frac t{4\pi}\int_0^{2\pi}\!\int_0^1\!\frac{r\cos(\tilde mtr\sin\theta)g(x+tr\cos\theta)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta.\end{aligned}$

  Thus, the solution $u$ of telegrapher's equation is

  $\begin{aligned}u(x,t)&=\frac{e^{-t/2}}{2\pi}\frac\partial{\partial t}\left(t\int_0^{2\pi}\!\int_0^1\!\frac{r\cos(tr\sqrt{m^2-(1/4)}\sin\theta)g(x+tr\cos\theta)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta\right)\\&\quad+\frac{te^{-t/2}}{4\pi}\int_0^{2\pi}\!\int_0^1\!\frac{r\cos(tr\sqrt{m^2-(1/4)}\sin\theta)g(x+tr\cos\theta)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta.\end{aligned}$

• If $m=\pm1/2$, then $v$ satisfies the wave equation. Then by the d'Alembert's formula (6), we have

  $\displaystyle v(x,t)=\frac{g(x+t)+g(x-t)}2+\frac14\int_{x-t}^{x+t}\!g(\xi)\,\mathrm d\xi$.

  Thus, the solution $u$ of telegrapher's equation is

  $\displaystyle u(x,t)=\frac{e^{-t/2}(g(x+t)+g(x-t))}2+\frac{e^{-t/2}}4\int_{x-t}^{x+t}\!g(\xi)\,\mathrm d\xi.$

• If $|m|<1/2$, then we define

  $\displaystyle w(x,y,t)=\exp\left(y\sqrt{\frac14-m^2}\right)v(x,t)\quad\text{for}~(x,y,t)\in\mathbb R^2\times\bar{\mathbb R}_+$.

  Then it is easy to verify that

  $w_{tt}=\exp\left(y\sqrt{\frac14-m^2}\right)v_{tt}=\exp\left(y\sqrt{\frac14-m^2}\right)\left[v_{xx}-\left(m^2-\frac14\right)v\right)=w_{xx}+w_{yy}.$

  This means $w$ satisfies the two-dimensional wave equation with the initial condition

  $\displaystyle w(x,y,0)=\exp\left(y\sqrt{\frac14-m^2}\right)g(x)\quad w_t(x,y,0)=\frac12\exp\left(y\sqrt{\frac14-m^2}\right)g(x)\quad\text{for}~(x,y)\in\mathbb R^2$.

  Then by the formula (39) in Section 3.2, we obtain

  $\begin{aligned}w(x,y,t)&=\frac1{2\pi}\frac\partial{\partial t}\left(t\int_{\xi_1^2+\xi_2^2<1}\!\frac{g(x+t\xi_1)\exp[(y+t\xi_2)\sqrt{(1/4)-m^2}]}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)\\&\quad+\frac t{4\pi}\int_{\xi_1^2+\xi_2^2<1}\frac{g(x+t\xi_1)\exp[(y+t\xi_2)\sqrt{(1/4)-m^2}]}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2.\end{aligned}$

  Note that $u(x,t)=e^{-t/2}v(x,t)=e^{-t/2}w(x,0,t)$. Therefore, the solution $u$ of telegrapher's equation is

  $\begin{aligned}u(x,t)&=\frac{e^{-t/2}}{2\pi}\frac\partial{\partial t}\left(t\int_{\xi_1^2+\xi_2^2<1}\!\frac{g(x+t\xi_1)\exp[t\xi_2\sqrt{(1/4)-m^2}]}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)\\&\quad+\frac{te^{-t/2}}{4\pi}\int_{\xi_1^2+\xi_2^2<1}\!\frac{g(x+t\xi_1)\exp[t\xi_2\sqrt{(1/4)-m^2}]}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\\&=\frac{e^{-t/2}}{2\pi}\frac\partial{\partial t}\left(t\int_0^{2\pi}\!\int_0^1\!\frac{r\exp[tr\sin\theta\sqrt{(1/4)-m^2}]g(x+tr\cos\theta)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta\right)\\&\quad+\frac{te^{-t/2}}{4\pi}\int_0^{2\pi}\!\int_0^1\!\frac{r\exp[tr\sin\theta\sqrt{(1/4)-m^2}]g(x+tr\cos\theta)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta.\end{aligned}$