# Thread: ODE (ar laplace transform)

1. ## ODE (ar laplace transform)

where L[u(x,t)] = U(x,s) with respect to t
Given u(x,o)=x u(0.t)=o

It is an imhonogeous equation. It need to find out general soland particular sol.

if it is a honogeous equation, just very easy to do.Am I correct in finding general sol.?
what is the yp(particular sol) in order to find U(x,s)

Finally, How to do inverse laplace transform to find u(x,t).
Thank

2. Give us the partial differencial equation with the boundary and initial value problems. The way you have it just makes no sense.

3. Originally Posted by ThePerfectHacker
Give us the partial differencial equation with the boundary and initial value problems. The way you have it just makes no sense.
updated
bivp is u(x,0)=x u(0,t)=0

4. That does not help. Give me the original equation without taking the Laplace Transform. The one you have is after you taken the Laplace Transformation.

5. Originally Posted by ThePerfectHacker
That does not help. Give me the original equation without taking the Laplace Transform. The one you have is after you taken the Laplace Transformation.
O..sorry

The original equation is

Uxx= Ut (-inf < x < + inf , t>0)
u(x,0)=x u(0,t)=0

Use the foruier transform and laplace transform to solve this PDE.

Thank, The above one I have is after I taken the Laplace Transformation.

6. $\displaystyle \mathcal{L}\left\{ \frac{\partial^2 u}{\partial x^2} \right\} = \mathcal{L} \left\{ \frac{\partial^2 u}{\partial t} \right\}$

$\displaystyle \frac{d^2 U(x,s)}{dx^2} = sU(x,s) - u(x,0) = sU(x,s)-x$

$\displaystyle \frac{d^2 U(x,s)}{dx^2} - sU(x,s) = x$

$\displaystyle U(x,s) = Ae^{x\sqrt{s}}+Be^{-x\sqrt{s}} - \frac{x}{s}$

Now solve for $\displaystyle A,B$ with the conditions given.

7. Originally Posted by ThePerfectHacker
$\displaystyle \mathcal{L}\left\{ \frac{\partial^2 u}{\partial x^2} \right\} = \mathcal{L} \left\{ \frac{\partial^2 u}{\partial t} \right\}$

$\displaystyle \frac{d^2 U(x,s)}{dx^2} = sU(x,s) - u(x,0) = sU(x,s)-x$

$\displaystyle \frac{d^2 U(x,s)}{dx^2} - sU(x,s) = x$

$\displaystyle U(x,s) = Ae^{x\sqrt{s}}+Be^{-x\sqrt{s}} - \frac{x}{s}$

Now solve for $\displaystyle A,B$ with the conditions given.

Now I have solve A=-B
$\displaystyle A = \frac{L}{s} \frac{1}{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}$

$\displaystyle U(x,s) = \frac{L}{s} \frac{1}{e^{L\sqrt{s}}-e^{-L\sqrt{s}}} [{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}] - \frac{x}{s}$
Am I correct?
Then how to do inverse laplace transform. The U(x,s) is very complex.....

8. Originally Posted by cyw1984
Now I have solve A=-B
$\displaystyle A = \frac{L}{s} \frac{1}{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}$

$\displaystyle U(x,s) = \frac{L}{s} \frac{1}{e^{x\sqrt{s}}-e^{-x\sqrt{s}}} [{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}] - \frac{x}{s}$
Am I correct?
Then how to do inverse laplace transform. The U(x,s) is very complex.....
I just realized that perhaps we need another point. You provided u(0,t)=0. Is there supposed to be another one? Because you cannot solve for A and B yet because there is 1 equation and two unknowns, if you provide one more conditions then we have 2 equations and 2 unknowns and it should solve.

9. Originally Posted by ThePerfectHacker
I just realized that perhaps we need another point. You provided u(0,t)=0. Is there supposed to be another one? Because you cannot solve for A and B yet because there is 1 equation and two unknowns, if you provide one more conditions then we have 2 equations and 2 unknowns and it should solve.
now if u(0,t)=0 u(L,t) =0

$\displaystyle U(x,s) = \frac{L}{s} \frac{1}{e^{L\sqrt{s}}-e^{-L\sqrt{s}}} [{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}] - \frac{x}{s}$