My Math Forum Manipulating Differentials: Method of Characteristics

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December 14th, 2017, 12:58 PM   #11
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Quote:
 Originally Posted by zylo Edited
$\displaystyle \frac{dx}{x}=\frac{dy}{y}=\frac{du}{2xy}$

$\displaystyle \frac{dx}{x}=\frac{dy}{y} \rightarrow y=c_{1}x$
This gives the projection of the curve on the x,y plane. To get the rest of the curve, you need u=u(x,y(x)), which gives equation of curve in terms of parameter x: y=f(x), u=g(x).

$\displaystyle \frac{dx}{x}=\frac{du}{2xy}\rightarrow \frac{dx}{x}=\frac{du}{2c_{1}x^{2}} \rightarrow x^{2}= \frac{u}{c_{1}}+c_{2}$
$\displaystyle \rightarrow u=c_{1}x^{2}-c_{1}c_{2}$

If you subs $\displaystyle c_{1}=\frac{y}{x}$, you get

$\displaystyle u=xy-c_{2}\frac{y}{x}$
which looks like equation of surface (it sat's orig
PDE), but it is subject to $\displaystyle y=c_{1}x$.

The curve can also be specified by a parameter t and solving. x=x(t), y=y(t), u=u(t).

Then comes the challenge of specifying and incorporating initial conditions.

Is there a reference for barely readable studiot pictures? If I could read it I might be able to figure out what v8archie is saying in his last post, which is uninteligible to me.

 December 16th, 2017, 07:35 AM #12 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 In general, y=g(x) and u=h(x) are equation of a curve in terms of the parameter x. Previously, we had $\displaystyle \frac{dx}{x}=\frac{dy}{y}=\frac{du}{2xy}$ $\displaystyle \frac{dx}{x}=\frac{dy}{y} \rightarrow c_{1}=\frac{y}{x}$ $\displaystyle \frac{dx}{x}=\frac{du}{2xy}\rightarrow \frac{dx}{x}=\frac{du}{2c_{1}x^{2}} \rightarrow c_{2} =\frac{u}{c_{1}}-x^2$ At any point $\displaystyle P_{0}(x_{0},y_{0},u_{0})$ $\displaystyle c_{1}$ and $\displaystyle c_{2}$ can be evaluated, determining the characteristic curve through that point. Along any given curve cutting characteristic curves, a surface can be constructed by evaluating $\displaystyle c_{1}$ and $\displaystyle c_{2}$ at points along the curve. Along such a curve: $\displaystyle c_{2} =f( c_{1})=f(\frac{y}{x})$ $\displaystyle f(\frac{y}{x})=\frac{xu}{y}-x^{2}$ Now assume we are given a particular curve in space to determine a surface: $\displaystyle y=g(x)=x^{2}$ and $\displaystyle u=h(x)$=2. Then $\displaystyle f(\frac{x^{2}}{x})=\frac{x(2)}{x^{2}}-x^{2}$ $\displaystyle f(x)=\frac{2}{x}-x^{2}$. Then $\displaystyle f(\frac{y}{x})=\frac{2}{y/x}-(\frac{y}{x})^{2}=\frac{xu}{y}-x^{2}\rightarrow$ $\displaystyle u=xy+2-\frac{y^{3}}{x^{3}}$, which is equation of surface satisfying original OP equation and given Cauchy condition. Ref: https://www.math.ualberta.ca/~xinwei...cteristics.pdf Sec 2.1
 December 16th, 2017, 08:34 AM #13 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra One flaw with this method is that you don't usually define $f(x)$ by following that process, but rather $f\big(\gamma(x)\big)$ for some function $\gamma(x)$.
 December 19th, 2017, 12:24 PM #14 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle 2xyu_{x}+u_{y}=u$ $\displaystyle \frac{dx}{2xy}=\frac{dy}{1}=\frac{du}{u}$ Equation of curve generating surface from characteristics: y=0, u=x. Solving by simple integration: $\displaystyle x=c_{1}e^{y^{2}}$ $\displaystyle e^{y}=c_{2}u$, then $\displaystyle c_{1}=f(c_{2})\\$ $\displaystyle \frac{x}{e^{y^{2}}}=f(\frac{e^{y}}{u})$ and $\displaystyle y=0,u=x$ $\displaystyle \frac{x}{1}=f(\frac{1}{x}) \rightarrow f(x)=\frac{1}{x}$ $\displaystyle \frac{x}{e^{y^{2}}}=\frac{u}{e^{y}} \rightarrow u=x^{e^{y-y^{2}}}$ Came across a good example: start at time 7:10/10:03 The first part is bloated blah-blah and the application of BC confusing. So I used method of my previous post as an illustration.
 December 19th, 2017, 04:25 PM #15 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra Apart from the error.
 December 21st, 2017, 02:26 PM #16 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Another example from Ref suggested by skipjack http://1.droppdf.com/files/Epi4A/beg...rd-ed-2014.pdf pg 356 $\displaystyle xu_{x}+yu_{y}=\sec u$ $\displaystyle \frac{dx}{x}=\frac{dy}{y}=\frac{du}{\sec u} \rightarrow$ $\displaystyle y=c_{1}x$ $\displaystyle \ln x=\sin u+c$ $\displaystyle x=c_{2}e^{\sin u}$ $\displaystyle c_{1}=f(c_{2})$ $\displaystyle \frac{y}{x}=f(xe^{-\sin u})$ Equation of curve generating surface: $\displaystyle u=0, y=\sin \sqrt{x}\rightarrow$ $\displaystyle f(x)=\frac{\sin\sqrt{x}}{x}\rightarrow$ $\displaystyle f(xe^{-\sin u})=\frac{\sin\sqrt{xe^{-\sin u}}}{xe^{-\sin u}}\rightarrow$ $\displaystyle \frac{y}{x}=\frac{\sin\sqrt{xe^{-\sin u}}}{xe^{-\sin u}}\rightarrow$ $\displaystyle x=e^{\sin u}[\arcsin {ye^{-\sin u}}]^{2}$ Which is equation of surface in form x=F(y,u). Which is solution given in reference. If you don't like this method, feel free to use the one in the reference. Last edited by skipjack; December 23rd, 2017 at 08:59 AM.
 December 22nd, 2017, 12:39 PM #17 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Aside: If f(cosx) = x$\displaystyle ^{3}$, what is f(x)? Let t = cosx, then x = arccost and f(t) = (arccost)$\displaystyle ^{3}$ and f(x) = (arccosx)$\displaystyle ^{3}$ -------------------------------------- What if $\displaystyle \frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}$ can't be solved in pairs. Then if tangent curve is in parametric form, $\displaystyle \frac{dx}{dt}=a(x(t),y(t),u(t))$ $\displaystyle \frac{dy}{dt}=b(x(t),y(t),u(t))$ $\displaystyle \frac{du}{dt}=c(x(t),y(t),u(t))$ then $\displaystyle x=f(c_{1},c_{2},c_{3},t)$ $\displaystyle y=g(c_{1},c_{2},c_{3},t)$ $\displaystyle u=f(c_{1},c_{2},c_{3},t)$ Given a generating curve x = F(s), y = G(s)., and u = H(s), what is the characteristic curve through s? Assume t=0 at the intersection and solve $\displaystyle F(s)=f(c_{1},c_{2},c_{3},0)$ $\displaystyle G(s)=g(c_{1},c_{2},c_{3},0)$ $\displaystyle H(s)=f(c_{1},c_{2},c_{3},0)$ for $\displaystyle c_{1}, c_{2}$ and $\displaystyle c_{3}$ in terms of s. Example Ref Post #12: $\displaystyle dx/dt=x, x=c_{1}e^{t}$ $\displaystyle dy/dt=x, y=c_{2}e^{t}$ $\displaystyle du/dt= 2xy, u=c_{1}c_{2}e^{2t}+ c_{3}$ Generating curve: x=F(s)=s, y=G(s)=s$\displaystyle ^{2}$, u=H(s)=2. With t=0, $\displaystyle s=c_{1}, s^{2}=c_{2}, 2=c_{1}c_{2}+c_{3}$ or $\displaystyle c_{3}=2-s^{2}$ So equation of characteristic through generating curve at s is: $\displaystyle x=se^{t}, y=s^{2}e^{t}, u=(e^{2t}-1)s^{3}+2$ Eliminating s and t gives: $\displaystyle u=xy-(\frac{y}{x})^{3}+2$ Which is what we got in post #12 The insightful part of the example is the treatment of the constants $\displaystyle c_{1},c_{2}$ and $\displaystyle c_{3}$. Last edited by skipjack; December 22nd, 2017 at 02:48 PM.
 December 23rd, 2017, 06:22 AM #18 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Aside: A curve in space can be given either as x=x(t), y=y(t), u=u(t), previous post, or y=y(x), u=u(x), this post. I prefer method of previous post: Choose constants so that the characteristic (c) curve intersects generating (g) curve at a particular point. For the OP case again, $\displaystyle \frac{dx}{x}=\frac{dy}{y}=\frac{d6}{2xy}$ gives by simple integration the c curve: $\displaystyle y=c_{1}x$ $\displaystyle u=c_{1}c_{2}+c_{1}x^{2}$ Assume a g curve: $\displaystyle y=x^{2}$ $\displaystyle u=2$ They intersect at $\displaystyle x_{0}$ if $\displaystyle x_{0}^{2}=c_{1}x_{0}$ and $\displaystyle 2=c_{1}c_{2}+c_{1}x_{0}^{2}$ Solving for $\displaystyle c_{1}$ and $\displaystyle c_{2}$ gives $\displaystyle y=x_{0}x$ $\displaystyle u=(2-x_{0}^{2})+x_{0}x^{2}$ as the c curve intersecting g curve at arbitrary point $\displaystyle x_{0}$. Eliminating $\displaystyle x_{0}$ gives, again, $\displaystyle u=xy-(\frac{y}{x})^{3}+2$ Last edited by skipjack; December 23rd, 2017 at 08:54 AM.
 December 23rd, 2017, 06:43 AM #19 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra Zylo, your enthusiasm is laudable, but the purpose of this thread (and the site as a whole) is not for individuals to post a never-ending stream of their private study. You are just regurgitating what you have understood, and, in my opinion, not very clearly. If people want to research they can easily find many resources on the web created by people who have a much deeper understanding and are professional educators. Thanks from greg1313

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