Solving a PDE with elementary methods

[Emspak]

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:\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}

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 into 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
\left(am_{22}+bm_{21}\right)\frac{\partial u}{\partial s}=u

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

Since we set the coefficient of \frac{\partial u}{\partial s} equal to 1 we are left with \frac{\partial u}{\partial s}=u and that is a simple ODE that reduces to u=Ce^{-s} 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.

 

[pasmith]

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:\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}

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 into 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
\left(am_{22}+bm_{21}\right)\frac{\partial u}{\partial s}=u

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

Since we set the coefficient of \frac{\partial u}{\partial s} equal to 1 we are left with \frac{\partial u}{\partial s}=u and that is a simple ODE that reduces to u=Ce^{-s}

The solution of \frac{\partial u}{\partial s} = u is u = C(r)e^s. Aside from that your work is correct.

 

[Emspak]

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: <br /> \begin{pmatrix}<br /> a &amp; -b \\<br /> 0 &amp; \frac{1}{a}\\<br /> \end{pmatrix}<br />

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 m_mn values)?

 

[pasmith]

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: <br /> \begin{pmatrix}<br /> a &amp; -b \\<br /> 0 &amp; \frac{1}{a}\\<br /> \end{pmatrix}<br />

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.

 

[Emspak]

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).