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Transform the differential equation $\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 0 $ by introducing new variables $x = u+v$ and $y=u-v$. then solve it.

I which I could show some effort but I don't even know where to start and I can't find anything about this in my course literature (the subject is very basic multi-variable calculus and I found the question in an old exam).

Could anyone show me how to solve this?

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  • $\begingroup$ Typo? I think you want to have $\partial_x z + \partial_{\bf y}z = 0$. $\endgroup$
    – martini
    Jul 14 '14 at 8:15
  • $\begingroup$ @martini ah yes of course. thank you! $\endgroup$
    – John Smith
    Jul 14 '14 at 8:18
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So, let's start as given in the hint. We define a "transformed" version of our solution $z$ by introducing $g(u,v) = z(u+v, u-v)$. Then $$ z(x,y) = g\left(\frac 12(x+y), \frac 12 (x-y)\right) $$ This gives \begin{align*} \partial_x z &= \frac 12 \partial_u g + \frac 12 \partial_v g\\ \partial_y z &= \frac 12 \partial_u g - \frac 12 \partial_v g \end{align*} Hence $$ 0 = \partial_x z + \partial_y z = \partial_u g $$ So $g(u,v) = \phi(v)$ for some function $\phi$, that is $$ z(x,y) = \phi\left(\frac 12(x-y)\right). $$

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  • $\begingroup$ Is $\phi$ really a solution of the PDE? Note that: $\phi_x + \phi_y = \phi'(x+y)$. Am I wrong? $\endgroup$
    – Dmoreno
    Jul 14 '14 at 8:42
  • $\begingroup$ @Dmoreno Thanks, corrected. $\endgroup$
    – martini
    Jul 14 '14 at 8:43
  • $\begingroup$ You're welcome @martini! $\endgroup$
    – Dmoreno
    Jul 14 '14 at 8:43
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If you are purely interested in how to solve this kind of equations, you might find this useful.

There's no need of using this sort of 'out of the blue' change of variables if you are introduced to the method of characteristics for solving 1st order PDEs. In your case, the PDE is also linear and can be written as follows:

$$a(x,y) \, p + b(x,y)\, q = f(x,y,z), \quad p = \partial_x z, \quad q = \partial_y z,$$ and $a = 1, b = 1$ are known functions of your independent variables and $f = 0$ is a linear function on $z$. The method of characteristics then reads:

$$\frac{\mathrm{dx}}{a} = \frac{\mathrm{dy}}{b} = \frac{\mathrm{dz}}{f} \Rightarrow \frac{\mathrm{dx}}{1} = \frac{\mathrm{dy}}{1} = \frac{\mathrm{d}z}{0}.$$ The two first equalities tell us that:

$$ \mathrm{d}x - \mathrm{d}y = 0,$$ so $x-y=c$ (constant) somewhere. The fraction $\mathrm{d}z/0$ leads to setting $z=k$ (constant), again, somewhere. If you put $k$ as a function of $c$, then you solution becomes:

$$z = k(c) = k(x-y),$$

where $k(\cdot)$ is an arbitrary function of $x-y$, i.e., the characteristic curve of the equation.

Hope this helps!

Cheers.

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If you change your variables using a Lie group, you can find the most general solution of this pde. Use $$G(x,y,z)=(\lambda x,\lambda^\beta y,\lambda^\alpha z)\lambda_o=1$$ such that $x'=\lambda x$, $y'=\lambda^\beta y$, $z'=\lambda^\alpha z$, then $$ \frac{\partial z'}{\partial x'}+\frac{\partial z'}{\partial y'}=\frac{\lambda^\alpha \partial z}{\lambda \partial x}+\frac{\lambda^\alpha \partial z}{\lambda^\beta \partial y}=0$$ For invariance, $\lambda^{\alpha -1}=\lambda^{\alpha -\beta}$ or $\beta=1$ and there are no restrictions on $\alpha$. Thus we have found an invariant Lie group, one which preserves the structure of the smooth manifold and leaves the pde unchanged. Since $$ \lambda^\alpha z(x,y)=z(\lambda x,\lambda y) $$we can take partial derivatives of both sides, setting $\lambda=\lambda_o=1$, and get $$ \alpha z=x z_x + y z_y $$The characteristic equations $$ \frac{dz}{\alpha z}=\frac{dx}{x}=\frac{dy}{y} $$have two independent integrals, $$\frac{z}{y^\alpha}$$ and $$\frac{y}{x}$$ These are constants of integration which are group stabilizers for polynomials (which include DEQ's), and this is verified by noting that $$\frac{z'}{(y')^\alpha}=\frac{\lambda^\alpha z}{(\lambda y)^\alpha}=\frac{z}{y^\alpha}$$ and $$\frac{y'}{x'}=\frac{y}{x}$$ Sophus Lie's idea was that if an invariant infinite continuous group can be found (a Lie group), then the DEQ is in the kernel of the group with the stabilizers which form an embedded subspace. The pde should then "fall apart" into the group stabilizers and thus be expressed as an ODE. Here's how it works.

The most general solution of your pde is found by setting one independent stabilizer equal to an arbitrary function of the other one.
$$ \frac{z}{y^\alpha}=f\bigg(\frac{y}{x}\bigg) $$If we rename $\frac{y}{x}=\mu$ where $\mu$ is the argument of the function, we get $$ z=y^\alpha f(\mu) $$ $$ z_x=y^\alpha f_\mu\mu_x=y^\alpha f_\mu \bigg(\frac{-y}{x^2}\bigg)=-y^{\alpha -1}\mu^2 f_\mu $$ $$ z_y =\alpha y^{\alpha -1} f+y^\alpha f_\mu \mu_y =\alpha y^{\alpha -1}f +y^{\alpha -1}\mu f_\mu $$ Putting these back into the pde, the y-terms drop out and we have $$ -\mu^2 f_\mu +\alpha f+\mu f_\mu =0 $$which simplifies to $$ \frac{f_\mu}{f}=\frac{\alpha}{\mu^2 -\mu} $$or $$ \frac{d}{d\mu}ln f=\frac{\alpha}{\mu^2 -\mu} $$After integration we have $$ lnf=\alpha\bigg[ln\bigg(\frac{\mu-1}{\mu}\bigg)\bigg]+ln C=lnC\bigg(\frac{\mu-1}{\mu}\bigg)^\alpha $$Thus, $$ f=C\bigg(\frac{\mu-1}{\mu}\bigg)^\alpha=C\bigg(\frac{\frac{y}{x}-1}{\frac{y}{x}}\bigg)^\alpha=C\bigg(1-\frac{x}{y}\bigg)^\alpha $$Because $z=y^\alpha f(x,y)$, the most general solution is $$ z=C(y-x)^\alpha $$This solution is easily checked.

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