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How to solve $ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$ via method of characteristics?

My attempt.

These are equations with which I begin:

$\dfrac{dx}{ds} = y+u $; $\dfrac{dy}{ds} = x+u $; $\dfrac{du}{ds} = x+y $.

However, I am stuck, because I can not solve the equations for $\dfrac{dx}{ds}$ and $\dfrac{dy}{ds}$ because we have a dependence on $u$.

Thanks for any help.

Thom
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  • Add them together and we get $\frac{d(x+y+u)}{ds}=2(x+y+u)$. Let $z = x+y+u$ and it turns to be $\frac{dz}{ds}=2z$. Solve $z$ and replace $y+u$ with $z-x$. – Zerox Aug 25 '20 at 14:10

2 Answers2

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Your differential equations are linear, with constant coefficients, for which there are well-established methods of solution. First write them in matrix form as: $$ \pmatrix{\frac{dx}{ds}\\ \frac{dy}{ds}\\ \frac{du}{ds}}=\pmatrix{0&1&1\\ 1&0&1\\ 1&1&0}\pmatrix{x\\y\\u}\ . $$ Next, find the eigenvalues of the matrix $\ \pmatrix{0&1&1\\1&0&1\\1&1&0}\ $, which are $\ -1,-1,2\ $, and a complete set of eigenvectors, which can be taken to be $$ \pmatrix{1\\-1\\0}, \pmatrix{0\\-1\\1}\ \text{ and }\ \pmatrix{1\\1\\1}\ . $$ These allow you to diagonalise the matrix thus $$ \pmatrix{1&0&1\\-1&-1&1\\0&1&1}^{-1}\pmatrix{0&1&1\\1&0&1\\1&1&0} \pmatrix{1&0&1\\-1&-1&1\\0&1&1}=\pmatrix{-1&0&0\\0&-1&0\\0&0&2}\ . $$ This, if you put $$ z= \pmatrix{1&0&1\\-1&-1&1\\0&1&1}^{-1}\pmatrix{x\\y\\u}\ , $$ you can write your differential equations in the form \begin{align} \frac{dz}{ds}=& \pmatrix{1&0&1\\-1&-1&1\\0&1&1}^{-1}\pmatrix{0&1&1\\1&0&1\\1&1&0} \pmatrix{1&0&1\\-1&-1&1\\0&1&1}z\\ =& \pmatrix{-1&0&0\\0&-1&0\\0&0&2}z\ , \end{align} which have solutions \begin{align} z_1&=Ae^{-s}\\ z_2&=Be^{-s}\\ z_3&=Ce^{2s}\ , \end{align} and so we get \begin{align} \pmatrix{x\\y\\u}&= \pmatrix{1&0&1\\-1&-1&1\\0&1&1}z\ \text{, or}\\ \\ x&=Ae^{-s}+Ce^{2s}\\ y&=Ce^{2s}-(A+B) e^{-s}\ ,\text{ and}\\ u&=B e^{-s}+Ce^{2s}\ . \end{align}

lonza leggiera
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  • Note : This method gives some solutions, but is far to give all the solutions. In the general solution of the PDE they are not only arbitrary constants, but there must be an arbitrary function (in so far there is no boundary condition specified in the wording of the question). – JJacquelin Aug 25 '20 at 18:01
  • The quantities $\ A, B, C\ $ are not universal constants, merely independent of $\ s\ $ along any given characteristic. In general, they will be functions of a point on the curve where the boundary conditions are specified, or of the parameter in a pair of parametric equations describing that curve. With that proviso, I contend that any solution obtainable from the method of characteristics can be obtained from the equations I have given. – lonza leggiera Aug 29 '20 at 14:43
  • If, for instance, you set \begin{align} h(t)&=\frac{u(t,-t)-t}{u(t,-t)+t}\ \ \text{ and}\ G(t)&=4\left(\frac{h^{-1}(t)(1+t)}{(1-t)}\right)^3\ \end{align} (assuming that $\ h\ $ is invertible), then it follows from those equations that $\ u\ $ must satisfy $$ G\left(\frac{u-x}{u-y}\right)=(u+x+y)(x-y)^2 $$ —that is, the second of the expressions you give for the "general solution" in your answer. – lonza leggiera Aug 29 '20 at 14:43
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$$ (y+u)\dfrac{\partial u}{\partial x} + (x+u)\dfrac{\partial u}{\partial y} = x+y$$ Charpit-Lagrange system of characteristic ODEs : $$ds=\frac{dx}{y+u}=\frac{dy}{x+u}=\frac{du}{x+y}=\frac{dx-dy}{(y+u)-(x+u)}=\frac{dx+dy+du}{(y+u)+(x+u)+(x+y)}$$ $$\frac{dx-dy}{y-x}=\frac{dx+dy+du}{2(x+y+u)}$$ $$-\ln|x-y|=\frac12\ln|x+y+u|+\text{constant}$$ A first characteristic equation is : $$(x+y+u)(x-y)^2=c_1$$ A second characteristic equation comes from $$\frac{dx}{y+u}=\frac{dy}{x+u}=\frac{du}{x+y}=\frac{dx-du}{(y+u)-(x+y)}=\frac{dy-du}{(x+u)-(x+y)}$$ $$\frac{dx-du}{u-x}=\frac{dy-du}{u-y}$$ $$\ln|u-x|=\ln|u-y|+\text{constant}$$ A second characteristic equation is : $$\frac{u-x}{u-y}=c_2$$ The general solution of the PDE expresed on the form of implicit equation $c_2=F(c_1)$ is : $$\boxed{\frac{u-x}{u-y}=F\big((x+y+u)(x-y)^2\big)}$$ F is an arbitrary function until no boundary condition is specified.

Note that the same general solution could be expressed on a number of equivalent forms, for example $u=-x-y+\frac{1}{(x-y)^2}G\left(\frac{u-x}{u-y}\right)$ where G is an arbitrary function.

Of course the PDE has an infinity many solutions. Among them the linear one : $u=\frac{1}{1-c}(x-cy)$ which coresponds to the above second characteristc equation. Or for example another solution $u=-x-y+\frac{c}{(x-y)^2}$ which corresponds to the above first characteristic equation.

Depending on the kind of boundary condition, the function F could be (or not) determined explicitely. Then putting it into the above general solution the equation could be (or not) solved explicitely for $u$.

JJacquelin
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