Method of characteristics please guide me in right direction

In summary: U_s + U_sS_x + U_sS_y = -\frac{1}{x}$$which doesn't make sense... I think I may be lost still...In summary, the method of characteristics is a technique for solving first-order partial differential equations by reducing them to first-order ordinary differential equations along flow lines. In this problem, the original equation is transformed into an ODE by setting up a parameterization and using the chain rule to rewrite the partial derivatives in terms of the parameter. This allows for a simpler solution process, as the constant coefficients make it easy to find the flow lines and determine the value of the function at any point along them. By using this method, the solution for U(x,t) is found
  • #1
fahraynk
186
6

Homework Statement


$$
U_t+U_x+\frac{1}{x} = 0\\
U(x,0)=\phi(x)
$$

Homework Equations

The Attempt at a Solution


I learned somewhat of an algorithm for method of characteristics. It works for a different problem :
$$U_t + U_x - KU = 0 \\
U(x,0)= \phi(x) \\$$
Method :
$$ x_s = 1\\
t_s = 1\\
x=s+x_0\\
t=s+t_0=s+0\\
x-t=x_0\\
U_s -KU = 0\\
U=Ae^{Ks}\\
U=\phi(x-t)e^{kt}$$

For this problem... its not working -_-
The first part is the same, mainly :
$$
x_s=1\\
t_s = 1\\
x=s+x_0\\
t=s\\
x-t=x_0\\
$$
But the next part...
$$U_s + \frac{1}{x} = 0$$
I think x should be x0, so:
$$U_s + \frac{1}{x_0} = U_s + \frac{1}{x-s} $$or$$ U_t +\frac{1}{x-t} = 0$$
(since x=s+x0 or x=t+x0)
$$U= ln(x-t)$$

I'm not sure what to do here... can someone point me in the right direction please...
The books answer is $$-ln(t+1) + \phi(x-t)$$

I think the main transformation from $$U_x+U_t +KU = U_s + KU$$ comes from the fact that if U(x,t) and x(s), t(s), than $$U_s = U_x*x_s + U_t*t_s$$
with $$x_s = 1 \\t_s=1$$
So shouldn't this be the same for the next problem, with x = x0... ?
 
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  • #2
fahraynk said:
I think x should be x0, so:
Why do you think that? It is x that appears in the original differential equation. You need to write it as a function of s in order to have an ODE in s.
 
  • #3
Also note that the answer given by the book is false. It does not satisfy the given differential equation.
 
  • #4
Orodruin said:
Why do you think that? It is x that appears in the original differential equation. You need to write it as a function of s in order to have an ODE in s.
Thanks for pointing that out...
x-s=x0.
So I guess if I keep it x it would become X0+S = X0 + t

Us+1/(X0+S)=0, U=Aln|X0+S|
U = F(x-t) ln |X0+t|

So since book is wrong... is that right? (since you said books answer is wrong, which is a relief)
 
  • #5
fahraynk said:
So I guess if I keep it x it would become X0+S = X0 + t
Do not put it in terms of t - the entire point is to exchange x and t for s!
fahraynk said:
Us+1/(X0+S)=0, U=Aln|X0+S|
something strange happened in this integration ...
fahraynk said:
So since book is wrong... is that right? (since you said books answer is wrong, which is a relief)

No, see above. Note that you can easily check whether you are right or not by inserting your result in the original differential equation.
 
  • #6
fahraynk said:
U = F(x-t) ln |X0+t|
That does not appear to be a solution to the equation either.
I am not familiar with the method you are trying to use. I don't understand how it works, since it implies x-t is constant, yet x and t are independent.

The problem is not hard if you apply the usual process of solving the homogeneous equattion then finding a particular integral.

Edit: I see Orodruin replied 30 minutes ago, yet that only showed up for me after I replied.
 
  • #7
$$U_s + \frac{1}{X_0+s} = 0\\
U(s)=-ln|x_0+s|+C\\
U(x,t)=-ln|x_0+t|+C\\
U(x,0) = -ln|x_0|+C = \phi(x_0)\\$$
Yeah... I think I am missing something still... I guess my book has not explained this method well enough for me... I feel like I'm stabbing in the dark. Ill try again tomorrow.

Also thanks for help all.
 
  • #8
fahraynk said:
$$U_s + \frac{1}{X_0+s} = 0\\
U(s)=-ln|x_0+s|+C\\
U(x,t)=-ln|x_0+t|+C\\
U(x,0) = -ln|x_0|+C = \phi(x_0)\\$$
Yeah... I think I am missing something still... I guess my book has not explained this method well enough for me... I feel like I'm stabbing in the dark. Ill try again tomorrow.

Also thanks for help all.

You have reached the conclusion that ##U(x_0,0) = - \ln(x_0) + C = \phi(x_0)## - this allows you to determine the constant in terms of ##x_0##. Once you have that, you can eliminate ##x_0## and ##s## in favour of ##t## and ##x##. You are almost there.

Note that the constant ##C## is constant with respect to ##s##. I would not write ##U(x,t) = -\ln(x_0+t) + C##.

haruspex said:
I am not familiar with the method you are trying to use. I don't understand how it works, since it implies x-t is constant, yet x and t are independent.

The method of characteristics. You have a differential equation of the form
$$
\vec v(\vec x) \cdot \nabla u = f(u,\vec x).
$$
If you write down the change in ##u## along the flow lines of the vector field ##\vec v##, you obtain a first order ordinary differential equation along each flow line. In order to determine ##u##, you need a condition on ##u## on one point on each flow line. The price you pay for reducing it to an ordinary differential equation is that you have to find the flow lines. However, in the case of constant coefficients (as in this case), this is trivial.
 
  • #9
fahraynk said:

Homework Statement


$$
U_t+U_x+\frac{1}{x} = 0\\
U(x,0)=\phi(x)
$$

Homework Equations

The Attempt at a Solution


I learned somewhat of an algorithm for method of characteristics. It works for a different problem :
$$U_t + U_x - KU = 0 \\
U(x,0)= \phi(x) \\$$

Since you are having trouble with the non-homogeneous pde, eliminate the trouble by looking at the new function
$$V(x,t) = U(x,t) + \ln x .$$
What is the pde satisfied by ##V?## What is the boundary condition on ##V(x,0)?##
 
Last edited:
  • #10
$$
U_s+\frac{1}{s+x_0}=0\\
U(s)=-ln|s+x_0| + C(x,t)\\
U(x,0)=-ln|x_0|+C(x,t)=\phi(x)=\phi(t+x_0)\\
C=\phi(t+x_0)+ln|x_0|\\
U(x,t)=-ln|t+x_0|+\phi(t+x_0)+ln|x_0|$$
Checking:
$$U_t+U_x=-\frac{1}{x}\\
U_t=-\frac{1}{t+x_0}\\
U_x=0\\
-\frac{1}{t+x_0}=-\frac{1}{x}\\
x=t+x_0\\
-\frac{1}{x} =-\frac{1}{x}$$

I guess that's right?
By the way... this book does not explain how $$U_x+U_t=-\frac{1}{x}$$ becomes $$U_s=-\frac{1}{x}$$
Can anyone give me some insight into that transformation please? is it something like :
$$U_x=U_sS_x\\
U_y=U_sS_y$$ with S' of x,y = 1, since the coefficients are both constants of 1 in this case... and then that is why by setting $$\frac{dx}{dS}=1\\
\frac{dt}{dS} = 1\\ $$ it works out... but it doesn't because then it the new equation would be $$2U_s=-\frac{1}{x}$$
 
  • #11
fahraynk said:
$$
U_s+\frac{1}{s+x_0}=0\\
U(s)=-ln|s+x_0| + C(x,t)$$
No, you are working in terms of ##s## and ##x_0## now. Your integration constant should depend on ##x_0##, not ##x## and ##t##.

fahraynk said:
$$
U(x,t)=-ln|t+x_0|+\phi(t+x_0)+ln|x_0|$$

You cannot write ##U## in terms of ##x## and ##t## and still keep ##x_0##. First find ##U(s)## and then change variables to ##x## and ##t##.

fahraynk said:
By the way... this book does not explain how $$U_x+U_t=-\frac{1}{x}$$ becomes $$U_s=-\frac{1}{x}$$
Can anyone give me some insight into that transformation please?

This is just the chain rule:
$$
\frac{df}{ds} = \frac{dt}{ds} \frac{\partial f}{\partial t} + \frac{dx}{ds} \frac{\partial f}{\partial x}
$$
 
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  • #12
Orodruin said:
No, you are working in terms of ##s## and ##x_0## now. Your integration constant should depend on ##x_0##, not ##x## and ##t##.
You cannot write ##U## in terms of ##x## and ##t## and still keep ##x_0##. First find ##U(s)## and then change variables to ##x## and ##t##.
This is just the chain rule:
$$
\frac{df}{ds} = \frac{dt}{ds} \frac{\partial f}{\partial t} + \frac{dx}{ds} \frac{\partial f}{\partial x}
$$
Thanks... I get this now. I don't know why I was messing up that ODE it was so easy if I would have thought about it.
 

FAQ: Method of characteristics please guide me in right direction

What is the method of characteristics?

The method of characteristics is a mathematical technique used in the field of partial differential equations to solve initial value problems. It involves constructing characteristic curves and using them to determine the solution of the equation.

How does the method of characteristics work?

The method of characteristics works by utilizing the idea that the solution of a partial differential equation can be represented by a family of curves, known as characteristic curves. These curves are constructed using the initial conditions of the equation and are used to determine the solution at any given point.

What types of problems can be solved using the method of characteristics?

The method of characteristics is commonly used to solve initial value problems for first-order partial differential equations. It can also be applied to certain second-order equations, as well as boundary value problems.

What are the advantages of using the method of characteristics?

One of the main advantages of the method of characteristics is that it can handle nonlinear equations, making it a powerful tool in solving a wide range of problems. It also provides a visual representation of the solution through the characteristic curves.

What are the limitations of the method of characteristics?

The method of characteristics is not always applicable to all types of partial differential equations. It may also become computationally expensive for complex equations with multiple independent variables. Additionally, the method may produce singular solutions in some cases.

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