# Thread: very basic PDE question

1. ## very basic PDE question

I want to find the general solution to $\frac{\partial u}{\partial t} - 2 \ \frac{\partial u}{\partial x} = 2$

so I let $\alpha = ax + bt$ and $\beta = cx+dt$

then $\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \alpha} \frac{\partial \alpha}{\partial t} + \frac{\partial u}{\partial \beta} \frac{\partial \beta}{\partial t} = a \frac{\partial u}{\partial \alpha} + c \ \frac{\partial u}{\partial \beta}$

similarly, $\frac{\partial u}{\partial x} = b \frac{\partial u}{\partial \alpha} + d \ \frac{\partial u}{\partial \beta}$

so $(b-2a) \ \frac{\partial u}{\partial \alpha} + (d-2c) \ \frac{\partial u}{\partial \beta} =2$

Would be OK to now do any of the following:

Let a=1,b=0,c=1, and d=2 (then $\frac{\partial u}{\partial \alpha} = -1$)

Let a=0,b=1, c=1,and d=2 (then $\frac{\partial u}{\partial \alpha} = 2$ )

Let a=1, b=0, c=1/2, d=1 (then then $\frac{\partial u}{\partial \alpha} = -1$)

Let a=1, b=2, c= -1, d=0 (then then $\frac{\partial u}{\partial \beta} = 1$ )

etc. ?

Are there any values that would somehow lead me to the incorrect solution?

2. Originally Posted by Random Variable
I want to find the general solution to $\frac{\partial u}{\partial t} - 2 \ \frac{\partial u}{\partial x} = 2$

so I let $\alpha = ax + bt$ and $\beta = cx+dt$

then $\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \alpha} \frac{\partial \alpha}{\partial t} + \frac{\partial u}{\partial \beta} \frac{\partial \beta}{\partial t} = a \frac{\partial u}{\partial \alpha} + c \ \frac{\partial u}{\partial \beta}$

similarly, $\frac{\partial u}{\partial x} = b \frac{\partial u}{\partial \alpha} + d \ \frac{\partial u}{\partial \beta}$

so $(b-2a) \ \frac{\partial u}{\partial \alpha} + (d-2c) \ \frac{\partial u}{\partial \beta} =2$

Would be OK to now do any of the following:

Let a=1,b=0,c=1, and d=2 (then $\frac{\partial u}{\partial \alpha} = -1$)

Let a=0,b=1, c=1,and d=2 (then $\frac{\partial u}{\partial \alpha} = 2$ )

Let a=1, b=0, c=1/2, d=1 (then then $\frac{\partial u}{\partial \alpha} = -1$)

Let a=1, b=2, c= -1, d=0 (then then $\frac{\partial u}{\partial \beta} = 1$ )

etc. ?

Are there any values that would somehow lead me to the incorrect solution?
They will all lead to the same solution. The only choice would be if the Jacobian of the transformation vanishes. The method of characteristics is probably the quickest way

$\frac{dt}{1} = \frac{dx}{-2} = \frac{du}{2}$

and pick in pair

$1) \frac{dt}{1} = \frac{dx}{-2}\;\; \Rightarrow\;\; I_1 = x + 2t$

$2) \frac{dt}{1} = \frac{du}{2}\;\; \Rightarrow\;\; I_2 = u - 2t$

Solution $I_2 = f(I_1) \; \text{or}\; u = 2t + f(x+2t)$.

3. The solution could also be written as -x + f(x+2t), correct?

4. Originally Posted by Random Variable
The solution could also be written as -x + f(x+2t), correct?
Absolutely! As $f(\lambda)$ is arbitrary you could let $f(\lambda) = \tilde{f}(\lambda) - \lambda$ where mine becomes yours.