How Can Variable Transformation Solve a Non-Homogeneous PDE?

[kgal]

Homework Statement

Find the solution of:

u_tt-u_xx = sin(∏x) for 0<x<1
u(x,0)=0 for 0<=x<=1
u_t(x,0)=0 for 0<=x<=1
u(0,t)=0
u(1,t)=0

Homework Equations

u_tt-u_xx = sin(∏x) for 0<x<1
u(x,0)=0 for 0≤x≤1
u_t(x,0)=0 for 0≤x≤1
u(0,t)=0
u(1,t)=0

The Attempt at a Solution

I was thinking that I would need to somehow make a change of variables, converting the PDE to a homogenous PDE and continuing from there. I don't really know how I can do this or solve it at all without some assistance!

 

[pasmith]

Homework Statement

Find the solution of:

u_tt-u_xx = sin(∏x) for 0<x<1
u(x,0)=0 for 0<=x<=1
u_t(x,0)=0 for 0<=x<=1
u(0,t)=0
u(1,t)=0

Homework Equations

u_tt-u_xx = sin(∏x) for 0<x<1
u(x,0)=0 for 0≤x≤1
u_t(x,0)=0 for 0≤x≤1
u(0,t)=0
u(1,t)=0

The Attempt at a Solution

I was thinking that I would need to somehow make a change of variables, converting the PDE to a homogenous PDE and continuing from there. I don't really know how I can do this or solve it at all without some assistance!

The strategy is the same as for linear ODEs with constant coefficients: Find a particular solution which satisfies the PDE but not necessarily the boundary conditions. Then add a complementary solution (a solution of $u_{tt} - u_{xx} = 0$) in order to satisfy the boundary conditions.

 

[kgal]

The strategy is the same as for linear ODEs with constant coefficients: Find a particular solution which satisfies the PDE but not necessarily the boundary conditions. Then add a complementary solution (a solution of $u_{tt} - u_{xx} = 0$) in order to satisfy the boundary conditions.

So basically something like Au_tt-B_xx= sin∏x and Cu_tt-Du_xx=0 and then sum the up into a solution u(x,t)?

 

[pasmith]

So basically something like Au_tt-B_xx= sin∏x and Cu_tt-Du_xx=0 and then sum the up into a solution u(x,t)?

No. You have $u(x,t) = u_p(x,t) + u_c(x,t)$ where $u_p$ is any solution of
$$u_{tt} - u_{xx} = \sin \pi x$$
and $u_c$ is the solution of
$$u_{tt} - u_{xx} = 0$$
subject to $u_c(x,t) = -u_p(x,t)$ on the boundary.

But here it is actually simplest to try a solution of the form $u(x,t) = f(t)\sin \pi x$, given that $(\sin(\pi x))'' = -\pi^2 \sin(\pi x)$. This reduces the problem to an ODE for f(t).

 

[kgal]

No. You have $u(x,t) = u_p(x,t) + u_c(x,t)$ where $u_p$ is any solution of
$$u_{tt} - u_{xx} = \sin \pi x$$
and $u_c$ is the solution of
$$u_{tt} - u_{xx} = 0$$
subject to $u_c(x,t) = -u_p(x,t)$ on the boundary.

But here it is actually simplest to try a solution of the form $u(x,t) = f(t)\sin \pi x$, given that $(\sin(\pi x))'' = -\pi^2 \sin(\pi x)$. This reduces the problem to an ODE for f(t).

So you're saying that I need to set u(x,t)= u_tt-u_xx = f(t)sin(∏x)?
How do I find the function f(t)?

What I did was this:
found u_xx, u_tt and plugged them into u_tt - u_xx = sin(∏x). What do I do with the boundary conditions and the initial conditions (how do they factor in)?

 

[pasmith]

So you're saying that I need to set u(x,t)= u_tt-u_xx = f(t)sin(∏x)?
How do I find the function f(t)?

What I did was this:
found u_xx, u_tt and plugged them into u_tt - u_xx = sin(∏x).

What do I do with the boundary conditions and the initial conditions (how do they factor in)?

The conditions $u(0,t) = f(t) \sin 0 = 0$ and $u(1,t) = f(t) \sin \pi = 0$ are satisfied automatically.

The remaining two conditions require that
$$u(x,0) = f(0) \sin\pi x = 0$$
for all $0 < x < 1$ and
$$u_t(x,0) = f'(0) \sin \pi x = 0$$
for all $0 < x < 1$. This gives you initial conditions for $f$ and $f'$.

 

[kgal]

Great! I got it!
How would I solve this problem using a technique like changing variables instead of separation of variables?