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2.1 HW 1

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2.1.1 Problem 1.8a

Find all quadratic polynomial solutions of the 3D Laplace equation $\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}+\frac{\partial ^{2}u}{\partial z^{2}}=0$

Solution

A quadratic polynomial in variables $x,y,z$ is

$$\begin{equation} u=a_{1}+a_{2}x+a_{3}y+a_{4}z+a_{5}x^{2}+a_{6}y^{2}+a_{7}z^{2}+a_{8}xy+a_{9}xz+a_{10}yz \tag{1} \end{equation}$$ Hence $u_{x}=a_{2}+2a_{5}x+a_{8}y+a_{9}z$ which implies that $u_{xx}=2a_{5}$. Similarly $u_{y}=a_{3}+2a_{6}y+a_{8}x+a_{10}z$, therefore $u_{yy}=2a_{6}$. And finally $u_{z}=a_{4}+2a_{7}z+a_{9}x+a_{10}y$ and $u_{zz}=2a_{7}$. Substituting these results in the Laplace equation gives above result in $$\begin{align*} 2a_{5}+2a_{6}+2a_{7} & =0\\ a_{5}+a_{6}+a_{7} & =0 \end{align*}$$

Therefore $a_{5}=-\left ( a_{6}+a_{7}\right )$. Using this relation back in (1) gives

$$\begin{align*} u & =a_{1}+a_{2}x+a_{3}y+a_{4}z-\left ( a_{6}+a_{7}\right ) x^{2}+a_{6}y^{2}+a_{7}z^{2}+a_{8}xy+a_{9}xz+a_{10}yz\\ & =a_{1}+a_{2}x+a_{3}y+a_{4}z+a_{6}\left ( -x^{2}+y^{2}\right ) +a_{7}\left ( -x^{2}+z^{2}\right ) +a_{8}xy+a_{9}xz+a_{10}yz \end{align*}$$

Which can be written as

$$u\left ( x,y,z\right ) =A_{1}+A_{2}x+A_{3}y+A_{4}z+A_{5}\left ( y^{2}-x^{2}\right ) +A_{6}\left ( z^{2}-x^{2}\right ) +A_{7}xy+A_{8}xz+A_{9}yz$$

2.1.2 Problem 1.7

Find all real solutions to 2D Laplace equation $u_{xx}+u_{yy}=0$ of the form $u=\log \left ( p\left ( x,y\right ) \right )$ where $p\left ( x,y\right )$ is a quadratic polynomial.

Solution

A quadratic polynomial $p\left ( x,y\right )$ in variables $x,y$ is

$$p\left ( x,y\right ) =a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy$$ Therefore $$u\left ( x,y\right ) =\log \left ( a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy\right )$$ Hence $$u_{x}=\frac{a_{2}+2a_{4}x+a_{6}y}{p\left ( x,y\right ) }$$ and $$\begin{equation} u_{xx}=\frac{2a_{4}}{p\left ( x,y\right ) }-\frac{\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}}{p\left ( x,y\right ) ^{2}} \tag{1} \end{equation}$$ Similarly $$u_{y}=\frac{a_{3}+2a_{5}y+a_{6}x}{p\left ( x,y\right ) }$$ And $$\begin{equation} u_{yy}=\frac{2a_{5}}{p\left ( x,y\right ) }-\frac{\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}}{p\left ( x,y\right ) ^{2}} \tag{2} \end{equation}$$ Substituting (1,2) into $u_{xx}+u_{yy}=0$ gives $$\begin{align*} \left ( \frac{2a_{4}}{p\left ( x,y\right ) }-\frac{\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}}{p\left ( x,y\right ) ^{2}}\right ) +\left ( \frac{2a_{5}}{p\left ( x,y\right ) }-\frac{\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}}{p\left ( x,y\right ) ^{2}}\right ) & =0\\ 2a_{4}-\frac{\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}}{p\left ( x,y\right ) }+2a_{5}-\frac{\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}}{p\left ( x,y\right ) } & =0\\ 2a_{4}+2a_{5}-\frac{\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}+\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}}{p\left ( x,y\right ) } & =0 \end{align*}$$

Or

$$\left ( 2a_{4}+2a_{5}\right ) p\left ( x,y\right ) =\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}+\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}$$ But $p\left ( x,y\right ) =a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy$. Hence the above becomes $$\left ( 2a_{4}+2a_{5}\right ) \left ( a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy\right ) =\left ( a_{2}+2a_{4}x+a_{6}y\right ) ^{2}+\left ( a_{3}+2a_{5}y+a_{6}x\right ) ^{2}$$ Expanding and comparing coefficients gives

$$\begin{multline*} 2x^{2}a_{4}^{2}+2x^{2}a_{4}a_{5}+2a_{6}a_{4}xy+2a_{6}a_{5}xy+2a_{2}a_{4}x+2a_{2}xa_{5}+2y^{2}a_{4}a_{5}+2y^{2}a_{5}^{2}+2a_{3}a_{4}y+2a_{3}a_{5}y+2a_{1}a_{4}+2a_{1}a_{5}=\\ 4x^{2}a_{4}^{2}+x^{2}a_{6}^{2}+4a_{4}a_{6}xy+4a_{5}a_{6}xy+4xa_{2}a_{4}+2a_{3}a_{6}x+4y^{2}a_{5}^{2}+y^{2}a_{6}^{2}+2a_{2}a_{6}y+4a_{3}a_{5}y+a_{2}^{2}+a_{3}^{2} \end{multline*}$$ Simplifying $$\begin{multline*} 2a_{4}a_{5}x^{2}+2a_{2}a_{5}x+2a_{4}a_{5}y^{2}+2a_{3}a_{4}y+2a_{1}a_{4}+2a_{1}a_{5}=\\ 2x^{2}a_{4}^{2}+a_{6}^{2}x^{2}+2a_{4}a_{6}xy+2a_{5}a_{6}xy+2a_{2}a_{4}x+2a_{3}a_{6}x+2a_{5}^{2}y^{2}+a_{6}^{2}y^{2}+2a_{2}a_{6}y+2a_{3}a_{5}y+a_{2}^{2}+a_{3}^{2} \end{multline*}$$ Comparing coefficients of terms that contain no $x,y$ and coefficients of $x,y,xy,x^{2},y^{2}$ gives the following equations in order $$\begin{align*} 2a_{1}a_{4}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\\ 2a_{2}a_{5} & =2a_{2}a_{4}+2a_{3}a_{6}\\ 2a_{3}a_{4} & =2a_{2}a_{6}+2a_{3}a_{5}\\ 0 & =4a_{4}a_{6}\\ 2a_{4}a_{5} & =2a_{4}^{2}+a_{6}^{2}\\ 2a_{4}a_{5} & =2a_{5}^{2}+a_{6}^{2} \end{align*}$$

Equation $0=4a_{4}a_{6}$ above implies that $a_{4}=0$ or $a_{6}=0$ or both are zero. But if both are zero, there is no solution. On the other hand, if $a_{4}=0,$ then this also leads to no solution as all equations reduce to $0=0$. Therefore only choice left is $a_{6}=0$. Now the above equations become

$$\begin{align*} 2a_{1}a_{4}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\\ 2a_{2}a_{5} & =2a_{2}a_{4}\\ 2a_{3}a_{4} & =2a_{3}a_{5}\\ 0 & =0\\ 2a_{4}a_{5} & =2a_{4}^{2}\\ 2a_{4}a_{5} & =2a_{5}^{2} \end{align*}$$

Or

$$\begin{align*} 2a_{1}a_{4}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\\ a_{5} & =a_{4}\\ a_{4} & =a_{5}\\ 0 & =0\\ a_{5} & =a_{4}\\ a_{4} & =a_{5} \end{align*}$$

Hence

$$\begin{align} a_{4} & =a_{5}\tag{3}\\ a_{6} & =0\tag{4}\\ 2a_{1}a_{4}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\nonumber \end{align}$$

Since $a_{4}=a_{5}$ then

$$\begin{align} 2a_{1}a_{5}+2a_{1}a_{5} & =a_{2}^{2}+a_{3}^{2}\nonumber \\ a_{5} & =\frac{a_{2}^{2}+a_{3}^{2}}{2a_{1}} \tag{5} \end{align}$$

Using (3,4,5) in $p\left ( x,y\right ) =a_{1}+a_{2}x+a_{3}y+a_{4}x^{2}+a_{5}y^{2}+a_{6}xy$ gives

$$\begin{align*} p\left ( x,y\right ) & =a_{1}+a_{2}x+a_{3}y+a_{5}x^{2}+a_{5}y^{2}\\ & =a_{1}+a_{2}x+a_{3}y+a_{5}\left ( x^{2}+y^{2}\right ) \\ & =a_{1}+a_{2}x+a_{3}y+\frac{a_{2}^{2}+a_{3}^{2}}{2a_{1}}\left ( x^{2}+y^{2}\right ) \end{align*}$$

Only three arbitrary constants are needed. Let $a_{1}=a,a_{2}=b,a_{3}=c$ the above becomes

$$p\left ( x,y\right ) =a+bx+cy+\frac{b^{2}+c^{2}}{2a}\left ( x^{2}+y^{2}\right )$$ And the solution becomes $$u\left ( x,y\right ) =\log \left ( a+bx+cy+\frac{b^{2}+c^{2}}{2a}\left ( x^{2}+y^{2}\right ) \right )$$

2.1.3 Problem 1.13

Find all solutions $u=f\left ( r\right )$ of the 3D Laplace equation $u_{xx}+u_{yy}+u_{zz}=0$ that depends only on radial coordinates $r=\sqrt{x^{2}+y^{2}+z^{2}}$

Solution

The Laplacian in 3D in spherical coordinates is

$$\nabla ^{2}u\left ( r,\theta ,\phi \right ) =u_{rr}+\frac{2}{r}u_{r}+\frac{1}{r^{2}}\left ( \frac{\cos \theta }{\sin \theta }u_{\theta }+u_{\theta \theta }\right ) +\frac{1}{r^{2}\sin ^{2}\theta }u_{\phi \phi }$$ The above shows that the terms that depend only on $r$ makes the laplacian $$\nabla ^{2}u\left ( r\right ) =u_{rr}+\frac{2}{r}u_{r}$$ Hence the PDE $\nabla ^{2}u\left ( r\right ) =0$ becomes an ODE now since there is only one dependent variable giving $$u^{\prime \prime }\left ( r\right ) +\frac{2}{r}u^{\prime }\left ( r\right ) =0$$ Let $v=u^{\prime }\left ( r\right )$ and the above becomes $$v^{\prime }\left ( r\right ) +\frac{2}{r}v\left ( r\right ) =0$$ This is linear first order ODE. The integrating factor is $I=e^{\int \frac{2}{r}dr}=e^{2\ln r}=r^{2}$. Therefore the above becomes $\frac{d}{dr}\left ( vr^{2}\right ) =0$ or $vr^{2}=C_{1}$ or $v\left ( r\right ) =\frac{C_{1}}{r^{2}}$. Therefore $$\begin{align*} u^{\prime } & =\frac{C_{1}}{r^{2}}\\ du & =\frac{C_{1}}{r^{2}}dr \end{align*}$$

Integrating gives the solution

$$u=-\frac{C_{1}}{r}+C_{2}$$ The above is the required solution. Hence $$\fbox{$f\left ( r\right ) =-\frac{C_1}{r}+C_2$}$$ Where $C_{1},C_{2}$ are arbitrary constants.

2.1.4 Problem 1.20

The displacement $u\left ( t,x\right )$ of a forced violin string is modeled by the PDE $u_{tt}=4u_{xx}+F\left ( t,x\right )$. When the string is subjected to the external force $F\left ( t,x\right ) =\cos x$, the solution is $u\left ( t,x\right ) =\cos \left ( x-2t\right ) +\frac{1}{4}\cos x$, while when $F\left ( t,x\right ) =\sin x$, the solution is $u\left ( t,x\right ) =\sin \left ( x-2t\right ) +\frac{1}{4}\sin x$. Find a solution when the forcing function is (a) $\cos x-5\sin x$, (b) $\sin \left ( x-3\right )$

Solution

2.1.4.1 Part (a)

Since the PDE is linear, superposition can be used. When the input is $F\left ( t,x\right ) =\cos x-5\sin x$ then the solution is

$$\begin{align*} u\left ( t,x\right ) & =\left ( \cos \left ( x-2t\right ) +\frac{1}{4}\cos x\right ) -5\left ( \sin \left ( x-2t\right ) +\frac{1}{4}\sin x\right ) \\ & =\cos \left ( x-2t\right ) +\frac{1}{4}\cos x-5\sin \left ( x-2t\right ) -\frac{5}{4}\sin x \end{align*}$$

2.1.4.2 Part (b)

Since the PDE is linear, superposition can be used. When the input is $F\left ( t,x\right ) =\sin \left ( x-3\right )$ then the solution same as when the input is $\sin x$ but shifted by $3$. Hence

$$u\left ( t,x\right ) =\sin \left ( \left ( x-3\right ) -2t\right ) +\frac{1}{4}\sin \left ( x-3\right )$$

2.1.5 Problem 1.27b

Solve the following inhomogeneous linear ODE $5u^{\prime \prime }-4u^{\prime }+4u=e^{x}\cos x$

Solution

First the homogeneous solution $u_{h}$ is found, then a particular solution $u_{p}$ is found. The general solution will be the sum of both $u=u_{h}+u_{p}$. Since this is a constant coefficient ODE, the characteristic equation is $5\lambda ^{2}-4\lambda +4=0$. The roots are $\lambda _{1}=\frac{2}{5}+\frac{4}{5}i,\lambda _{1}=\frac{2}{5}-\frac{4}{5}i$, which implies the solution is

$$u_{h}\left ( x\right ) =e^{\frac{2}{5}x}\left ( c_{1}\cos \left ( \frac{4}{5}x\right ) +c_{2}\sin \left ( \frac{4}{5}x\right ) \right )$$ Using the method of undetermined coefficients, and since the forcing function is $e^{x}\cos x$, then let $$\begin{equation} u_{p}=Ae^{x}\left ( B\cos x+C\sin x\right ) \tag{1} \end{equation}$$ Hence $$\begin{align} u_{p}^{\prime } & =Ae^{x}\left ( B\cos x+C\sin x\right ) +Ae^{x}\left ( -B\sin x+C\cos x\right ) \tag{2}\\ u_{p}^{\prime \prime } & =Ae^{x}\left ( B\cos x+C\sin x\right ) +Ae^{x}\left ( -B\sin x+C\cos x\right ) +Ae^{x}\left ( -B\sin x+C\cos x\right ) +Ae^{x}\left ( -B\cos x-C\sin x\right ) \nonumber \\ & =Ae^{x}\left ( B\cos x+C\sin x-B\sin x+C\cos x-B\sin x+C\cos x-B\cos x-C\sin x\right ) \nonumber \\ & =Ae^{x}\left ( -B\sin x+C\cos x-B\sin x+C\cos x\right ) \nonumber \\ & =Ae^{x}\left ( -2B\sin x+2C\cos x\right ) \tag{3} \end{align}$$

Substituting (1,2,3) back into the original ODE gives

$$\begin{align*} 5Ae^{x}\left ( -2B\sin x+2C\cos x\right ) -4\left ( Ae^{x}\left ( B\cos x+C\sin x\right ) +Ae^{x}\left ( -B\sin x+C\cos x\right ) \right ) +4Ae^{x}\left ( B\cos x+C\sin x\right ) & =e^{x}\cos x\\ Ae^{x}\left ( -10B\sin x+10C\cos x\right ) -Ae^{x}\left ( 4B\cos x+4C\sin x\right ) -Ae^{x}\left ( -4B\sin x+4C\cos x\right ) +Ae^{x}\left ( 4B\cos x+4C\sin x\right ) & =e^{x}\cos x\\ Ae^{x}\left ( -10B\sin x+10C\cos x-4B\cos x-4C\sin x+4B\sin x-4C\cos x+4B\cos x+4C\sin x\right ) & =e^{x}\cos x \end{align*}$$

Hence

$$Ae^{x}\left ( 6C\cos x-6B\sin x\right ) =e^{x}\cos x$$ Comparing coefficients shows that $$\begin{align*} A & =1\\ B & =0\\ C & =\frac{1}{6} \end{align*}$$

Hence from (1)

$$u_{p}=e^{x}\frac{\sin x}{6}$$ Therefore the general solution is $$\begin{align*} u\left ( x\right ) & =u_{h}\left ( x\right ) +u_{p}\left ( x\right ) \\ & =e^{\frac{2}{5}x}\left ( c_{1}\cos \left ( \frac{4}{5}x\right ) +c_{2}\sin \left ( \frac{4}{5}x\right ) \right ) +e^{x}\frac{\sin x}{6} \end{align*}$$

2.1.6 Problem 2.1.6

Solve the PDE $\frac{\partial ^{2}u}{\partial x\partial y}=0$ for $u\left ( x,y\right )$

Solution

Integrating once w.r.t $x$ gives

$$\frac{\partial u}{\partial y}=F\left ( y\right )$$ Where $F\left ( y\right )$ acts as the constant of integration, but since this is a PDE, it becomes an arbitrary function of $y$ only. Integrating the above again w.r.t. $y$ gives $$u=\int F\left ( y\right ) dy+G\left ( x\right )$$ Where $G\left ( x\right )$ is an arbitrary function of $x$ only. If we let $\int F\left ( y\right ) dy=H\left ( y\right )$ where $H\left ( y\right )$ is the antiderivative for the indefinite integral which depends on $y$ only. Then the above can be written as $$\fbox{$u\left ( x,y\right ) =H\left ( y\right ) +G\left ( x\right ) $}$$ To verify, from the above $\frac{\partial u}{\partial y}=H^{\prime }\left ( y\right )$ and hence $$\begin{align*} \frac{\partial ^{2}u}{\partial x\partial y} & =\frac{d}{dx}\left ( H^{\prime }\left ( y\right ) \right ) \\ & =0 \end{align*}$$

2.1.7 Problem 2.2.2

Solve the following initial value problems and graph the solutions at $t=1,2,3$

a

$u_{t}-3u_{x}=0,u\left ( 0,x\right ) =e^{-x^{2}}$

b

$u_{t}+2u_{x}=0,u\left ( -1,x\right ) =\frac{x}{1+x^{2}}$

c

$u_{t}+u_{x}+\frac{1}{2}u=0,u\left ( 0,x\right ) =\arctan \left ( x\right )$

d

$u_{t}-4u_{x}+u=0,u\left ( 0,x\right ) =\frac{1}{1+x^{2}}$

Solution

2.1.7.1 Part a

Let $\xi$ be the characteristic variable defined such that $\xi =x-ct$. Where characteristic lines are given by $x=x_{0}+ct$. But $c=-3$ in this problem. Hence characteristic lines are

$$x=x_{0}-3t$$ Where $x_{0}$ means the same as $x\left ( 0\right )$, i.e. $x\left ( t\right )$ at time $t=0$. Since $c=-3$ then $$\xi =x+3t$$ Let $$u\left ( t,x\right ) \equiv v\left ( t,\xi \right )$$ $u_{t}-3u_{x}=0$ is now transformed to $v\left ( t,\xi \right )$ as follows $$\begin{align} \frac{\partial u}{\partial t} & =\frac{\partial v}{\partial t}\frac{\partial t}{\partial t}+\frac{\partial v}{\partial \xi }\frac{\partial \xi }{\partial t}\nonumber \\ & =\frac{\partial v}{\partial t}+3\frac{\partial v}{\partial \xi } \tag{1} \end{align}$$

And

$$\begin{align} \frac{\partial u}{\partial x} & =\frac{\partial v}{\partial t}\frac{\partial t}{\partial x}+\frac{\partial v}{\partial \xi }\frac{\partial \xi }{\partial x}\nonumber \\ & =0+\frac{\partial v}{\partial \xi }\nonumber \\ & =\frac{\partial v}{\partial \xi } \tag{2} \end{align}$$

Substituting (1,2) in $u_{t}-3u_{x}=0$ gives the transformed PDE as

$$\begin{align*} \frac{\partial v}{\partial t}+3\frac{\partial v}{\partial \xi }-3\frac{\partial v}{\partial \xi } & =0\\ \frac{\partial v}{\partial t} & =0 \end{align*}$$

Integrating w.r.t $\xi$ gives the solution in $v\left ( t,\xi \right )$ space as

$$v\left ( t,\xi \right ) =F\left ( \xi \right )$$ Where $F\left ( \xi \right )$ is an arbitrary continuous function of $\xi$. Transforming back to $u\left ( t,x\right )$ gives $$\begin{equation} u\left ( t,x\right ) =F\left ( x+3t\right ) \tag{3} \end{equation}$$ At $t=0$ the above becomes $$e^{-x_{0}^{2}}=F\left ( x_{0}\right )$$ This means that (3) becomes (since $x=x_{0}+ct$ or $x=x_{0}-3t$ or $x_{0}=x+3t$) $$u\left ( t,x\right ) =e^{-\left ( x+3t\right ) ^{2}}$$

2.1.7.2 Part b

$$\begin{align*} u_{t}+2u_{x} & =0\\ u\left ( -1,x\right ) & =\frac{x}{1+x^{2}} \end{align*}$$

Let $\xi$ be the characteristic variable defined such that $\xi =x-ct$. Where characteristic lines are given by $x=x_{0}+ct$. But $c=2$ in this problem. Hence characteristic lines are

$$x=x_{0}+2t$$ And $$\xi =x-2t$$ Let $u\left ( t,x\right ) \equiv v\left ( t,\xi \right )$. Then $u_{t}+2u_{x}=0$ is transformed to $v\left ( t,\xi \right )$ as was done in part (a) (will not be repeated) which results in $$\frac{\partial v}{\partial t}=0$$ Integrating w.r.t $\xi$ gives the solution $$v\left ( t,\xi \right ) =F\left ( \xi \right )$$ Where $F\left ( \xi \right )$ is an arbitrary continuous function of $\xi$. Transforming back to $u\left ( t,x\right )$ results in $$\begin{equation} u\left ( t,x\right ) =F\left ( x-2t\right ) \tag{3} \end{equation}$$ At $t=-1$ the above becomes $$\frac{x_{0}}{1+x_{0}^{2}}=F\left ( x_{0}+2\right )$$ Let $x_{0}+2=z$. Then $x_{0}=z-2$. And the above becomes $$\frac{z-2}{1+\left ( z-2\right ) ^{2}}=F\left ( z\right )$$ This means that (3) becomes $$\begin{align*} u\left ( t,x\right ) & =\frac{\left ( x-2t\right ) -2}{1+\left ( \left ( x-2t\right ) -2\right ) ^{2}}\\ & =\frac{x-2t-2}{1+\left ( x-2t-2\right ) ^{2}} \end{align*}$$

2.1.7.3 Part c

$$\begin{align} u_{t}+u_{x}+\frac{1}{2}u & =0\tag{1}\\ u\left ( 0,x\right ) & =\arctan \left ( x\right ) \nonumber \end{align}$$

Let $\xi$ be the characteristic variable defined such that $\xi =x-ct$. Where characteristic lines are given by $x=x_{0}+ct$. But $c=1$ in this problem. Hence characteristic lines are given by solution to

$$\begin{align*} \frac{dx}{dt} & =1\\ x\left ( t\right ) & =x_{0}+t \end{align*}$$

And

$$\begin{align*} \xi & =x-ct\\ & =x-t \end{align*}$$

Then $u_{t}+u_{x}$ are transformed to $v\left ( t,\xi \right )$ as was done in part (a) (will not be repeated) which results in

$$u_{t}+u_{x}=\frac{\partial v}{\partial t}$$ Substituting the above into (1) gives (where now $v$ is used in place of $u$). $$\frac{\partial v}{\partial t}+\frac{1}{2}v=0$$ This is now first order ODE since it only depends on $t$. Therefore $v^{\prime }+\frac{1}{2}v=0$. This is linear in $v$. Hence the solution is $\frac{d}{dt}\left ( ve^{\int \frac{1}{2}dt}\right ) =0$ or $ve^{\frac{1}{2}t}=F\left ( \xi \right )$ where $F$ is arbitrary function of $\xi$. Hence $$v\left ( t,\xi \right ) =e^{\frac{-1}{2}t}F\left ( \xi \right )$$ Converting back to $u\left ( t,x\right )$ gives $$\begin{equation} u\left ( t,x\right ) =e^{\frac{-t}{2}}F\left ( x-t\right ) \tag{2} \end{equation}$$ At $t=0$ the above becomes $$\arctan \left ( x_{0}\right ) =F\left ( x_{0}\right )$$ From the above then (2) can be written as $$u\left ( t,x\right ) =e^{\frac{-t}{2}}\arctan \left ( x-t\right )$$

2.1.7.4 Part d

$$\begin{align*} u_{t}-4u_{x}+u & =0\\ u\left ( 0,x\right ) & =\frac{1}{1+x^{2}} \end{align*}$$

Let $\xi$ be the characteristic variable defined such that $\xi =x-ct$. Where characteristic lines are given by $x=x_{0}+ct$. But $c=-4$ in this problem. Hence characteristic lines are

$$x=x_{0}-4t$$ And $$\xi =x+4t$$ Then $u_{t}-4u_{x}$ are transformed to $v\left ( t,\xi \right )$ as was done in part (a) (will not be repeated) which results in $$u_{t}-4u_{x}=\frac{\partial v}{\partial t}$$ Substituting the above into (1) gives (where now $v$ is used in place of $u$). $$\frac{\partial v}{\partial t}+v=0$$ This is now first order ODE since it only depends on $t$. Therefore $v^{\prime }+v=0$. This is linear in $v$. Hence the solution is $\frac{d}{dt}\left ( ve^{\int dt}\right ) =0$ or $ve^{t}=F\left ( \xi \right )$ where $F$ is arbitrary function of $\xi$. Hence $$v\left ( t,\xi \right ) =e^{-t}F\left ( \xi \right )$$ Converting to $u\left ( t,x\right )$ gives $$\begin{equation} u\left ( t,x\right ) =e^{-t}F\left ( x+4t\right ) \tag{2} \end{equation}$$ At $u\left ( 0,x\right ) =\frac{1}{1+x^{2}}$ the above becomes $$\frac{1}{1+x_{0}^{2}}=F\left ( x_{0}\right )$$ From the above then (2) can be written as $$u\left ( t,x\right ) =\frac{e^{-t}}{1+\left ( x+4t\right ) ^{2}}$$

2.1.8 Problem 2.2.3

Graph some of the characteristic lines for the following equation and write down the formula for the general solution

(b) $u_{t}+5u_{x}=0\,$, (d) $u_{t}-4u_{x}+u=0$

Solution

2.1.8.1 Part b

$$u_{t}+5u_{x}=0$$

Let $\xi$ be the characteristic variable defined such that $\xi =x-ct$. Where characteristic lines are given by $x=x_{0}+ct$. But $c=5$ in this problem. Hence characteristic lines are

$$\begin{equation} \fbox{$x\left ( t\right ) =x_0+5t$} \tag{1} \end{equation}$$ And $$\xi =x-5t$$ Then $u_{t}-5u_{x}=0$ is transformed to $v\left ( t,\xi \right )$ as was done in earlier (will not be repeated) which results in $$u_{t}-5u_{x}=\frac{\partial v}{\partial t}$$ Therefore $\frac{\partial v}{\partial t}=0$ which has the general solution $v\left ( t,\xi \right ) =F\left ( \xi \right )$ where $F$ is arbitrary function of $\xi$. Transforming back to $u\left ( t,x\right )$ gives $$\fbox{$u\left ( t,x\right ) =F\left ( x-5t\right ) $}$$ On the characteristic lines given by (1) the solution $u\left ( t,x\right )$ is constant. The slope of the characteristic lines is $5$ and intercept is $x_{0}$. The following is a plot of few lines using different values of $x_{0}$.

Figure 2.1:Showing some characteristic lines for part b

2.1.8.2 Part d

$$u_{t}-4u_{x}+u=0$$ Let $\xi$ be the characteristic variable defined such that $\xi =x-ct$. Where characteristic lines are given by $x=x_{0}+ct$. But $c=-4$ in this problem. Hence characteristic lines are $$\begin{equation} \fbox{$x\left ( t\right ) =x_0-4t$} \tag{1} \end{equation}$$ And $$\xi =x+4t$$ Then $u_{t}-4u_{x}$ is transformed to $v\left ( t,\xi \right )$ as was done in earlier (will not be repeated) which results in $$u_{t}-4u_{x}=\frac{\partial v}{\partial t}$$ Therefore the original PDE becomes $\frac{\partial v}{\partial t}+v=0$, where $u$ is replaced by $v$. This is linear first order ODE which has the solution $v\left ( t,\xi \right ) =e^{-t}F\left ( \xi \right )$ where $F$ is arbitrary function of $\xi$. Transforming back to $u\left ( t,x\right )$ gives the general solution as $$u\left ( t,x\right ) =e^{-t}F\left ( x+4t\right )$$ The following is a plot of few characteristic lines $x=x_{0}-4t$ using different values of $x_{0}$.

Figure 2.2:Showing some characteristic lines for part d

2.1.9 Problem 2.2.5

Solve $u_{t}+2u_{x}=\sin x,u\left ( 0,x\right ) =\sin x$

Solution

Let $\xi$ be the characteristic variable defined such that $\xi =x-ct$. Where characteristic lines are given by $x=x_{0}+ct$. But $c=2$ in this problem. Hence characteristic lines are

$$\begin{equation} \fbox{$x=x_0+2t$} \tag{1} \end{equation}$$ And $$\xi =x-2t$$ Then $u_{t}+2u_{x}$ is transformed to $v\left ( t,\xi \right )$ as was done in earlier (will not be repeated) which results in $$u_{t}+2u_{x}=\frac{\partial v}{\partial t}$$ Substituting this into the original PDE gives $$\frac{\partial v\left ( t,\xi \right ) }{\partial t}=\sin \left ( \xi +2t\right )$$ Integrating w.r.t $t$ gives $$\begin{align*} v\left ( t,\xi \right ) & =\int \sin \left ( \xi +2t\right ) dt+F\left ( \xi \right ) \\ & =-\frac{\cos \left ( \xi +2t\right ) }{2}+F\left ( \xi \right ) \end{align*}$$

Transforming back to $u\left ( t,x\right )$ gives

$$\begin{align} u\left ( t,x\right ) & =-\frac{\cos \left ( x-2t+2t\right ) }{2}+F\left ( x-2t\right ) \nonumber \\ & =\frac{-1}{2}\cos \left ( x\right ) +F\left ( x-2t\right ) \tag{1} \end{align}$$

When $t=0$, $u\left ( 0,x\right ) =\sin x$, therefore the above becomes

$$\begin{align*} \sin x_{0} & =F\left ( x_{0}\right ) -\frac{1}{2}\cos x_{0}\\ F\left ( x_{0}\right ) & =\sin x_{0}+\frac{1}{2}\cos x_{0} \end{align*}$$

Therefore the solution (1) becomes

$$\begin{align*} u\left ( t,x\right ) & =\left ( \sin \left ( x-2t\right ) +\frac{1}{2}\cos \left ( x-2t\right ) \right ) -\frac{1}{2}\cos x\\ & =\sin \left ( x-2t\right ) +\frac{1}{2}\cos \left ( x-2t\right ) -\frac{1}{2}\cos x \end{align*}$$

2.1.10 Problem 2.2.9

(a) Prove that if the initial data is bounded, $\left \vert f\left ( x\right ) \right \vert \leq M$ for all $x\in \mathbb{R}$, then the solution to the damped transport equation (2.14) $u_{t}+cu_{x}+au=0$ with $a>0$ satisfies $u\left ( t,x\right ) \rightarrow 0$ as $t\rightarrow \infty$. (b) Find a solution to (2.14) that is defined for all $\left ( t,x\right )$ but does not satisfy $u\left ( t,x\right ) \rightarrow 0$ as $t\rightarrow \infty$.

Solution

2.1.10.1 Part(a)

$u_{t}+cu_{x}+au=0$ is solved to show what is required. Let $\xi$ be the characteristic variable defined such that $\xi =x-ct$. Where characteristic lines are given by $x=x_{0}+ct$. Hence characteristic lines are

$$\begin{equation} x=x_{0}+ct \tag{1} \end{equation}$$ And $$\xi =x-ct$$ Then $u_{t}+cu_{x}$ is transformed to $v\left ( t,\xi \right )$ as was done in earlier (will not be repeated) which results in $$u_{t}+cu_{x}=\frac{\partial v}{\partial t}$$ Substituting this into the original PDE gives $$\frac{\partial v}{\partial t}+av=0$$ Where $u$ is replaced by $v$. This can be viewed as first order linear ODE since it depends on $t$ only. Its solution is $v\left ( t,\xi \right ) =e^{-at}F\left ( \xi \right ) \,$ where $F$ is arbitrary function of $\xi$. Transforming back to $u\left ( t,x\right )$ gives $$\begin{equation} u\left ( t,x\right ) =e^{-at}F\left ( x-ct\right ) \tag{1} \end{equation}$$ At $t=0$ initial data is $f\left ( x\right )$. Hence the above becomes at $t=0$ $$f\left ( x\right ) =F\left ( x\right )$$ Hence (1) now becomes $$\begin{equation} u\left ( t,x\right ) =e^{-at}f\left ( x-ct\right ) \tag{2} \end{equation}$$ But since $\left \vert f\left ( x\right ) \right \vert$ is bounded, and since $a>0$ then $e^{-at}\rightarrow 0$ as $t\rightarrow \infty$. Which implies the solution itself $u\left ( t,x\right )$ goes to zero as well. This is the reason why initial data needed to be bounded for this to happen.

2.1.10.2 Part(b)

Keeping $a>0$. If initial data have the form $f\left ( x\right ) e^{-bx}$ where $\left \vert b\right \vert >a$, then at $t=0$ the solution found in (1) becomes

$$f\left ( x_{0}\right ) e^{-bx_{0}}=F\left ( x_{0}\right )$$ Then the solution (2) now becomes, after replacing $x_{0}$ by $x-ct$ $$\begin{align*} u\left ( t,x\right ) & =e^{-at}e^{-b\left ( x-ct\right ) }f\left ( x-ct\right ) \\ & =e^{-at+bct}e^{-bx}f\left ( x-ct\right ) \\ & =e^{\left ( bc-a\right ) t}e^{-bx}f\left ( x-ct\right ) \end{align*}$$

The problem is asking to show that this does not go to zero for all $x\in \mathbb{R}$ as $t\rightarrow \infty$. Since $\left \vert b\right \vert >a$ then $bc-a$ is positive quantity ($c$ is assumed positive)¹ .

Therefore $e^{\left ( bc-a\right ) t}$ will blow up as $t\rightarrow \infty$. And therefore the whole solution will not go to zero. For any $x$, no matter how large $x$ is, a large enough $t$ can be found to make the product $e^{\left ( bc-a\right ) t}e^{-bx}$ blow up.

2.1.11 Key solution for HW 1

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