First oder pde using laplce transform

1. Oct 25, 2009

babs

Hi I am having a lot of trouble trying to solve this equation. Any help is appreciated

x^2 \[partial]u/\[partial]x + 2 x \[partial]u/\[partial]t = g (t)

2. Oct 29, 2009

jasonRF

So you have
$$x^2 \frac{\partial u(x,t)}{\partial x} + 2 x \frac{\partial u(x,t)}{\partial t}=g(t)$$

I am guessing there are some initial and/or boundary conditions associated with this? Does g have any constraints?

The details depend upon the answers to those questions. However, the title of your post says it all. If you transform this equation with respect to t, you get an ordinary differential equation for $$U(x,s)$$, which is just the Laplace transform of $$u(x,t)$$. Solve this ODE, and then you just invert the transform and you are done!

Good luck.

Jason

3. Oct 30, 2009

javicg

I have exactly the same problem. These are all the conditions:

By using the Laplace transform, obtain as an integral the solution of the first order PDE $$x^2 \frac{\partial u}{\partial x} + 2x \frac{\partial u}{\partial t} = g(t),$$ subject to u(x, 0) = 0, u(1, t) = 0. The function g is continuous and $$|g(t)| \le Ke^{at}$$ for some constants K, a and $$t > 0$$ (Hint: In the Laplace inversion recall that $$x^b = e^{b \ln x}$$).

4. Oct 30, 2009

jasonRF

Is this a question for a class? Is it from a textbook?

It seems like an improbable coincidence that two people who just started posting here are interested in the same problem ...

5. Oct 30, 2009

javicg

I had it proposed by my uncle, who is a teacher btw. It could be easily from a textbook, but I don't know which one. Also, my first post was actually on Sep6-09.

Last edited: Oct 31, 2009
6. Oct 31, 2009

jasonRF

Fair enough. How far have you gotten?

7. Oct 31, 2009

javicg

I have been working on it and this is what I have now.

We take the Laplace transform in the t variable.

$$\mathcal{L}\left(2x \frac{\partial u}{\partial t} \right) = 2x \int_0^{\infty} \frac{\partial u}{\partial t} e^{-st} dt = 2x (-u(x,0) + \bar{U}(x,s)),$$​

where $$\bar{U}(x,s) =\int_0^{\infty} u(x,t) e^{-st} dt$$. Also,

$$\mathcal{L}\left(x^2 \frac{\partial u}{\partial x} \right) = x^2 \frac{\partial}{\partial x}\mathcal{L}(u) = x^2 \frac{\partial}{\partial x} \bar{U} (x,s),$$​

and $$x$$ is treated as a constant.

Hence we have,

$$-2x u(x,0) + 2xs \bar{U} (x,s) + x^2 \frac{\partial \bar{U}}{\partial x} (x,s) = \mathcal{L} (g(t)).$$​

So,

$$x^2 \frac{d}{dx} \bar{U}(x,s) + 2xs \bar{U} (x,s) = \mathcal{L} (g(t)),$$​

or dividing by $$x^2$$

$$\frac{d}{dx} \bar{U}(x,s) + \frac1x 2s \bar{U} (x,s) = \frac{\mathcal{L} (g(t))}{x^2}$$​

We solve this ODE by multiplying through by the integrating factor of $$x^{2s}$$.
This gives

$$x^{2s} \frac{d\bar{U}}{dx} + 2sx^{2s-1} \bar{U} = \mathcal{L} (g(t)) x^{2s-2}.$$​

Hence,

$$\frac{d}{dx} (x^2s\bar{U}) = \mathcal{L} (g(t)) x^{2s-2}.$$​

Therefore,

$$x^2s\bar{U} = \mathcal{L} (g(t)) \int x^{2s-2} = \mathcal{L} (g(t)) \left( \frac{x^{2s-1}}{2s-1} + C \right)$$​

for some constant of integration $$C$$. Solving for $$\bar{U} (x,s)$$ gives,

$$\bar{U} (x,s) = \frac{\mathcal{L} (g(t))}{x(2s-1)} + \frac{C}{x^{2s}}.$$​

Now we need to determine $$C$$. To do this we take the Laplace transform of the
initial condition to get,

$$\bar{U} (1,s) = \mathcal{L} (u(1,t)) = \mathcal{L} (0) = 0.$$​

We can now find our constant $$C$$. We have

$$\bar{U} (1,s) = \frac{\mathcal{L} (g(t))}{2s-1} + C = 0 \Rightarrow C = - \frac{\mathcal{L} (g(t))}{2s-1}$$​

Hence

$$\bar{U} (x,s) = \frac{\mathcal{L} (g(t))}{x(2s-1)} - \frac{\mathcal{L} (g(t))}{x^{2s}(2s-1)}.$$​

To find $$u(x,t)$$ we now take the inverse Laplace transform:

\begin{align*} u(x,t)&= \mathcal{L}^{-1} \left( \frac{\mathcal{L} (g(t))}{x(2s-1)} - \frac{\mathcal{L} (g(t))}{x^{2s}(2s-1)} \right)\\ &= \mathcal{L} (g(t)) \left( \mathcal{L}^{-1} \left( \frac1{x(2s-1)} \right) - \mathcal{L}^{-1} \left( \frac1{x^{2s}(2s-1)} \right) \right) \\ &= \mathcal{L} (g(t)) \left( \frac1{sx(2s-1)} - \frac{x^{-2s}}{s(2s-1)} \right) \end{align*}​