# Thread: Linear PDEq in two variables

1. ## Linear PDEq in two variables

Here's the equation:
$\frac{\partial v}{\partial t} - \frac{\partial v}{\partial z} = 0$

Boundary conditions: Not much. v has to be bounded for all values of t, z.

There's a simple way to approach this...I just can't think of it. And yes, I could look up solution by characteristic curves but I'm convinced there is an easier way.

As it happens, I like the solution v = constant, but I need to see if there are any others.

Thanks for the help!

-Dan

2. ## Re: Linear PDEq in two variables

This is the same as $\frac{\partial v}{\partial t}= \frac{\partial v}{\partial z}$.

One method: which is really based on the "characteristics" which are z+ t= constant and z- t= constant. Let x= z+ t and y= z- t. Then, adding the two equations, $x+ y= 2z$ so $z= \frac{x+ y}{2}$. Subtracting the two equations, x- y= 2t so $t= \frac{x- y}{2}$. $\frac{\partial v}{\partial t}= \frac{\partial v}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial v}{\partial y}\frac{\partial y}{\partial t}= \frac{1}{2}\frac{\partial v}{\partial x}- \frac{1}{2}\frac{\partial v}{\partial y}$.
$\frac{\partial v}{\partial z}= \frac{\partial v}{\partial x}\frac{\partial x}{\partial z}+ \frac{\partial v}{\partial y}\frac{\partial y}{\partial z}= \frac{1}{2}\frac{\partial v}{\partial x}+ \frac{1}{2}\frac{\partial v}{\partial y}$.

So the differential equation becomes $\frac{\partial v}{\partial y}= 0$. The integral, with respect to y is a costant. But, since taking the partial derivative with respect to y treats x as a constant, that "constant" might be any function of x. That is, v(x, y)= F(x) where F can be any differentiable function of x. Since x= t+ z, We have v(t, z)= F(t+ z). That is, for F any differentiable function of a single variable, v(x, t)= F(t+ z) satisfies the differential equation.

What that function, F, is depends upon those boundary conditions of which you say "Not much"!

3. ## Re: Linear PDEq in two variables

Thanks again, HoI. I don't know why I didn't think of it earlier, but a useful solution based on your form for the answer is a 2D Gaussian:
$v(t, z) = A e^ { -i(t + z)^2 / \sigma ^2 }$.

-Dan