Method of characteristics for a system of pdes

Solution 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 $$ \begin{pmatrix} u\\v\\w \end{pmatrix} = \begin{pmatrix} 1 & -1 & 1\\ -1 & 0 & 2\\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} \tilde{u}\\\tilde{v}\\\tilde{w} \end{pmatrix} =\begin{pmatrix} 1 \\ -x\\ y \end{pmatrix}. $$