Method of characteristics for a system of pdes

Question

I can do parts a) and b) as follows

$\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1\end{pmatrix}\frac{\partial}{\partial{}x}\begin{pmatrix} u \\ v \\ w\end{pmatrix}+\begin{pmatrix} 1&1&0 \\ 1&2&1 \\ 0&1&1\end{pmatrix}\frac{\partial}{\partial y}\begin{pmatrix} u \\ v \\ w\end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

and hence

$\bigg| \begin{pmatrix} 1-\lambda&1&0 \\ 1&2-\lambda&1 \\ 0&1&1-\lambda\end{pmatrix} \bigg| \implies \lambda = 0,1,3$

for part c) we have that $u(0,y)=1, v(0,y)=0$ and $w(0,y)=y$ on $x=0$, so I parameterise the conditions so that

$u(0,\xi)=1, v(0,\xi)=0$ and $w(0,\xi)=\xi$ on $x=0$

therefore

on $\frac{dy}{dx}=0 \implies y= constant \implies y=\xi$

now we know that $u-v+w=constant$ and so

$\implies u-v+w=1-0+\xi \implies u-v+w=1+y$

similarly I get that

on $y=x+\xi \implies u=w=1-y+x$ and on $y=3x+\xi$ I have that $u+2v+w=1+y-3x$.

However this system is not consistent with the equation at the top of the question and therefore clearly not correct. Could someone please show me how to approach these equations. Thanks

Answer 1

The solution is done in two steps.

First, diagonalise the linear system. We denote by $\mathbf{U}=[u;v;w].$ The original system becomes $$\partial_x\mathbf{U}+M\partial_y\mathbf{U}=0,$$ where $$ M=\begin{pmatrix} 1 & 1 & 0\\ 1 & 2 & 1\\ 0 & 1 & 1 \end{pmatrix}. $$ We first diagonalise $M.$ We have (using Wolfram Alpha) $$ M=S\,J\,S^{-1}, $$ where $$ S = \begin{pmatrix} 1 & -1 & 1\\ -1 & 0 & 2\\ 1 & 1 & 1 \end{pmatrix}, $$ $$ S^{-1} = \begin{pmatrix} \frac{1}{3} & -\frac{1}{3} & \frac{1}{3}\\ -\frac{1}{2} & 0 & \frac{1}{2}\\ \frac{1}{6} & \frac{1}{3} & \frac{1}{6}, \end{pmatrix} $$ $$ J = \mathrm{diag}(0,1,3). $$ We denote by $\widetilde{\mathbf{U}}=S^{-1}\mathbf{U}.$ Then, the linear system becomes $$ \partial_x\widetilde{\mathbf{U}} + J\partial_y\widetilde{\mathbf{U}} = 0, $$ which consists of three independent scalar equations $$ \begin{aligned} \partial_x\widetilde{u} &= 0,\\ \partial_x\widetilde{v} + \partial_y\widetilde{v} &= 0,\\ \partial_x\widetilde{w} + 3\partial_y\widetilde{w} &= 0. \end{aligned} $$ The initial condition $\mathbf{U}(0)$ gives $\widetilde{\mathbf{U}}(0)=S^{-1}\mathbf{U}(0).$ In the case of (c), $$ \begin{aligned} &\left\{ \begin{aligned} &\partial_x\tilde{u}=0,\\ &\tilde{u}(0,y)=\frac{1}{3}(1+y), \end{aligned} \right. &\left\{ \begin{aligned} &\partial_x\tilde{v}+\partial_y\tilde{v}=0,\\ &\tilde{v}(0,y)=\frac{1}{2}(-1+y), \end{aligned} \right. &\left\{ \begin{aligned} &\partial_x\tilde{w}+3\partial_y\tilde{w}=0,\\ &\tilde{w}(0,y)=\frac{1}{6}(1+y). \end{aligned} \right. \end{aligned} $$

Second step, we solve them independently, $$ \begin{aligned} &\tilde{u}(x,y)\equiv\tilde{u}(0,y)=\frac{1}{3}(1+y),\\ &\tilde{v}(x,y)=\tilde{v}(x,x+c)\equiv\tilde{v}(0,c)=\frac{1}{2}(-1+c)=\frac{1}{2}(-1+y-x),\\ &\tilde{w}(x,y)=\tilde{w}(x,3x+c)\equiv\tilde{w}(0,c)=\frac{1}{6}(1+c)=\frac{1}{6}(1+y-3x). \end{aligned} $$ Finally, we use $\mathbf{U}=S\,\tilde{\mathbf{U}}$ to obtain $$ \begin{pmatrix} u\\v\\w \end{pmatrix} = \begin{pmatrix} 1 & -1 & 1\\ -1 & 0 & 2\\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} \tilde{u}\\\tilde{v}\\\tilde{w} \end{pmatrix} =\begin{pmatrix} 1 \\ -x\\ y \end{pmatrix}. $$