• Support PF! Buy your school textbooks, materials and every day products Here!

Solving a PDE with elementary methods

  • Thread starter Emspak
  • Start date
  • #1
243
1

Homework Statement



OK, a PDE: $$a\frac{\partial u}{\partial t} + b \frac{\partial u}{\partial x} = u$$

we want the general solution.

2. The attempt at a solution

So, I'll set up a couple of equations thus:
r = m11x + m21t
s = m12x + m22t

(We have a nice matrix of m here if we want).

partially differentiating both:[itex] \frac{\partial r}{\partial x}= m_{11},\frac{\partial r}{\partial t}= m_{12}, \frac{\partial s}{\partial x}= m_{21}, \frac{\partial s}{\partial t}= m_{22} [/itex]

and plugging them back into the original PDE we get: $$ \frac{\partial u}{\partial t}= \frac{\partial u}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial u}{\partial s}\frac{\partial s}{\partial t}= m_{12} \frac{\partial u}{\partial r}+ m_{22}\frac{\partial u}{\partial s} $$ and $$ \frac{\partial u}{\partial x}= \frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}= m_{21} \frac{\partial u}{\partial s}+ m_{11}\frac{\partial u}{\partial r}=u$$

from there I substitute in to the original PDE and got: $$a\left(m_{12} \frac{\partial u}{\partial r}+ m_{22} \frac{\partial u}{\partial s} \right) + b\left(m_{11} \frac{\partial u}{\partial r}+ m_{21} \frac{\partial u}{\partial s} \right) \Rightarrow (am_{12}+bm_{11}) \frac{\partial u}{\partial r}+ (am_{22}+bm_{21}) \frac{\partial u}{\partial s} =u$$

So I want to get rid of one of the partials. The simplest way to do that is to set one of the coefficients equal to zero. So:
am12+bm11=0. That makes m12=-b and m11=a. So my r from above is r=ax-bt.

That leaves me with
[itex]\left(am_{22}+bm_{21}\right)\frac{\partial u}{\partial s}=u[/itex]

and I can set [itex]\left(am_{22}+bm_{21}\right)= 1[/itex] leaving [itex]m_{22} = \frac{1}{a}[/itex] and [itex]m_{21}=0[/itex]. From the equation for [itex]s[/itex]above that leaves us with [itex]s=\frac{1}{a}t[/itex] and we go back to the partials.

Since we set the coefficient of [itex]\frac{\partial u}{\partial s}[/itex] equal to 1 we are left with [itex]\frac{\partial u}{\partial s}=u[/itex] and that is a simple ODE that reduces to [itex]u=Ce^{-s}[/itex] and
Since $$u=f(r)e^{-s} = f(ax-bt)e^{-s}= f(ax-bt)e^{-\frac{t}{a}}$$ as our general solution.

So how did I do? We're still in the beginning of the class, and this was one of the problems that is supposed to explore "easy" methods; I know that the method of characteristics would be better.
 

Answers and Replies

  • #2
pasmith
Homework Helper
1,740
412

Homework Statement



OK, a PDE: $$a\frac{\partial u}{\partial t} + b \frac{\partial u}{\partial x} = u$$

we want the general solution.

2. The attempt at a solution

So, I'll set up a couple of equations thus:
r = m11x + m21t
s = m12x + m22t

(We have a nice matrix of m here if we want).

partially differentiating both:[itex] \frac{\partial r}{\partial x}= m_{11},\frac{\partial r}{\partial t}= m_{12}, \frac{\partial s}{\partial x}= m_{21}, \frac{\partial s}{\partial t}= m_{22} [/itex]

and plugging them back into the original PDE we get: $$ \frac{\partial u}{\partial t}= \frac{\partial u}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial u}{\partial s}\frac{\partial s}{\partial t}= m_{12} \frac{\partial u}{\partial r}+ m_{22}\frac{\partial u}{\partial s} $$ and $$ \frac{\partial u}{\partial x}= \frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}= m_{21} \frac{\partial u}{\partial s}+ m_{11}\frac{\partial u}{\partial r}=u$$

from there I substitute in to the original PDE and got: $$a\left(m_{12} \frac{\partial u}{\partial r}+ m_{22} \frac{\partial u}{\partial s} \right) + b\left(m_{11} \frac{\partial u}{\partial r}+ m_{21} \frac{\partial u}{\partial s} \right) \Rightarrow (am_{12}+bm_{11}) \frac{\partial u}{\partial r}+ (am_{22}+bm_{21}) \frac{\partial u}{\partial s} =u$$

So I want to get rid of one of the partials. The simplest way to do that is to set one of the coefficients equal to zero. So:
am12+bm11=0. That makes m12=-b and m11=a. So my r from above is r=ax-bt.

That leaves me with
[itex]\left(am_{22}+bm_{21}\right)\frac{\partial u}{\partial s}=u[/itex]

and I can set [itex]\left(am_{22}+bm_{21}\right)= 1[/itex] leaving [itex]m_{22} = \frac{1}{a}[/itex] and [itex]m_{21}=0[/itex]. From the equation for [itex]s[/itex]above that leaves us with [itex]s=\frac{1}{a}t[/itex] and we go back to the partials.

Since we set the coefficient of [itex]\frac{\partial u}{\partial s}[/itex] equal to 1 we are left with [itex]\frac{\partial u}{\partial s}=u[/itex] and that is a simple ODE that reduces to [itex]u=Ce^{-s}[/itex]
The solution of [itex]\frac{\partial u}{\partial s} = u[/itex] is [itex]u = C(r)e^s[/itex]. Aside from that your work is correct.
 
  • Like
Likes 1 person
  • #3
243
1
Thanks a lot, I wanted to be sure I wasn't doing something dumb. One other question: I noticed that if I put the m terms into a matrix I get: [itex]
\begin{pmatrix}
a & -b \\
0 & \frac{1}{a}\\
\end{pmatrix}
[/itex]

and the determinant is 1. I get the sense that when looking for solutions with this method that could be important. Is there a good way to use this fact to come up with (relatively) easy coefficients (the mmn values)?
 
  • #4
pasmith
Homework Helper
1,740
412
Thanks a lot, I wanted to be sure I wasn't doing something dumb. One other question: I noticed that if I put the m terms into a matrix I get: [itex]
\begin{pmatrix}
a & -b \\
0 & \frac{1}{a}\\
\end{pmatrix}
[/itex]

and the determinant is 1. I get the sense that when looking for solutions with this method that could be important.
The determinant must be non-zero, or the substitution will not be invertible.
 
  • #5
243
1
so I could put any numbers there in the matrix as long as they didn't come out to zero, right? (I realize it would be more complicated sometimes but I am just checking).
 

Related Threads on Solving a PDE with elementary methods

Replies
1
Views
4K
Replies
7
Views
2K
Replies
0
Views
6K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
892
Replies
2
Views
2K
  • Last Post
Replies
2
Views
605
Replies
2
Views
2K
Top