# Another PDE and boundary conditions

#### Markov

##### Member
1) Solve

\begin{aligned} {{u}_{t}}&=K{{u}_{xx}},\text{ }0<x<L,\text{ }t>0, \\ {{u}_{x}}(0,t)&=0,\text{ }{{u}_{x}}(L,t)=0,\text{ for }t>0, \\ u(x,0)&=6+\sin \frac{3\pi x}{L} \end{aligned}

2) Transform the problem so that the boundary conditions get homogeneous:

\begin{aligned} {{u}_{t}}&=K{{u}_{xx}},\text{ }0<x<L,\text{ }t>0, \\ {{u}_{x}}(0,t)&=Ae^{-at},\text{ }{{u}_{x}}(L,t)=B,\text{ for }t>0, \\ u(x,0)&=0 \end{aligned}

Attempts:

1) No ideas for this one, I don't know how to proceed when the initial conditions have the first derivative.

2) I think I need to define a new function right? But how?

#### dwsmith

##### Well-known member
1) Solve

\begin{aligned} {{u}_{t}}&=K{{u}_{xx}},\text{ }0<x<L,\text{ }t>0, \\ {{u}_{x}}(0,t)&=0,\text{ }{{u}_{x}}(L,t)=0,\text{ for }t>0, \\ u(x,0)&=6+\sin \frac{3\pi x}{L} \end{aligned}

2) Transform the problem so that the boundary conditions get homogeneous:

\begin{aligned} {{u}_{t}}&=K{{u}_{xx}},\text{ }0<x<L,\text{ }t>0, \\ {{u}_{x}}(0,t)&=Ae^{-at},\text{ }{{u}_{x}}(L,t)=B,\text{ for }t>0, \\ u(x,0)&=0 \end{aligned}

Attempts:

1) No ideas for this one, I don't know how to proceed when the initial conditions have the first derivative.

2) I think I need to define a new function right? But how?
What book are you using? If it isn't helpful, you should get Elementary Partial Differential Equations by Berg and McGregor.

#### Markov

##### Member
I have no chances to get books, the only source I have left is the forum.

Think you could help me please?

#### dwsmith

##### Well-known member
I have no chances to get books, the only source I have left is the forum.

Think you could help me please?
The book I suggested is relatively cheap and it is a good book. I suggest you think about picking a book up whether it is that one or another one.

#### HallsofIvy

##### Well-known member
MHB Math Helper
For 1, because the boundary conditions are homogeneous, try a solution of the form $\sum_{n=0}^\infty A_n(t)cos(n\pi x/L)$.
Use cosine because the derivative of cosine is sine which is 0 at 0 and L.
For 2, yes, you need to change the function. And you want to make the boundary conditions 0 so find a function, f(x,t), that satisfies the boundary conditions and subtract f from u.

Last edited:

#### Markov

##### Member
Okay but on (1) why can we adopt that kind of solution?

As for second question, I have $v(x,t)=A{{e}^{-at}}-\left( A{{e}^{-at}}-B \right)\dfrac{x}{L}$ so I need $f(x,t)=v(x,t)+u(x,t),$ does this work? (I had a typo, the boundary conditions don't have the first derivative.)

Last edited: