Transform the problem
PDE: utt=9uxx,0<x<5,t>0
BC: u(0,t)=6u(5,t)=9
IC: u(x,0)=x2,ut(x,0)=x3
into a dimensionless problem:
PDE: Uττ=Uξξ,0<ξ<1,τ>0
BC: U(0,τ)=0u(1,τ)=1
IC: U(ξ,0)=
Uτ(ξ,0)=
Type xi, tau for the variables. (You might want to first figure out the answers below)

First, you transform u to U. Use just small "u" instead of "u(x,t)":
U=

Next, transform x to ξ.
ξ= x5​

Finally, transform t to τ.

To transform the given PDE problem into a dimensionless form, we'll follow the necessary steps one by one.

### 1. Transform \( u(x,t) \) to \( U(\xi,\tau) \):
We want to introduce a new function \( U(\xi, \tau) \) that represents the dimensionless version of \( u(x,t) \).

Let \( u(x,t) \) be expressed in terms of \( U(\xi, \tau) \) as follows:
\[
u(x,t) = U\left( \frac{x}{5}, \frac{t}{T} \right)
\]
where \( T \) is a time scaling factor to be determined.

### 2. Transform \( x \) to \( \xi \):
We need to scale the spatial variable \( x \) to the interval \( 0 < \xi < 1 \). So, we define:
\[
\xi = \frac{x}{5}
\]
This transforms the spatial domain from \( 0 < x < 5 \) to \( 0 < \xi < 1 \).

### 3. Transform \( t \) to \( \tau \):
We now scale the time variable \( t \) by some factor \( T \), such that:
\[
\tau = \frac{t}{T}
\]
To match the form of the transformed PDE \( U_{\tau \tau} = U_{\xi \xi} \), we need to determine the correct value of \( T \). 

From the original PDE:
\[
u_{tt} = 9 u_{xx}
\]
Substituting the transformations:
\[
u_{tt} = \frac{1}{T^2} U_{\tau \tau}, \quad u_{xx} = \frac{1}{25} U_{\xi \xi}
\]
Equating these two expressions, we get:
\[
\frac{1}{T^2} U_{\tau \tau} = 9 \times \frac{1}{25} U_{\xi \xi}
\]
\[
\frac{1}{T^2} = \frac{9}{25} \quad \Rightarrow \quad T = \frac{5}{3}
\]
Thus, the time transformation is:
\[
\tau = \frac{3}{5} t
\]

### 4. Boundary Conditions (BCs):
The boundary conditions in the original problem are:
\[
u(0,t) = 6, \quad u(5,t) = 9
\]
In terms of \( U(\xi, \tau) \), these become:
\[
U(0,\tau) = 6, \quad U(1,\tau) = 9
\]

### 5. Initial Conditions (ICs):
The initial conditions are:
\[
u(x,0) = x^2, \quad u_t(x,0) = x^3
\]
In terms of \( U(\xi, \tau) \), the initial conditions are:
\[
U(\xi, 0) = (5\xi)^2 = 25 \xi^2
\]
For the time derivative \( u_t(x,0) \), we have:
\[
u_t(x,0) = x^3 \quad \Rightarrow \quad U_\tau(\xi,0) = \frac{5}{3} x^3 = \frac{5}{3} (5\xi)^3 = \frac{625}{3} \xi^3
\]

### Final Dimensionless Problem:

- **PDE:**
  \[
  U_{\tau \tau} = U_{\xi \xi}, \quad 0 < \xi < 1, \, \tau > 0
  \]

- **Boundary Conditions:**
  \[
  U(0, \tau) = 6, \quad U(1, \tau) = 9
  \]

- **Initial Conditions:**
  \[
  U(\xi, 0) = 25 \xi^2, \quad U_\tau(\xi, 0) = \frac{625}{3} \xi^3
  \]

---

Would you like more details or clarifications on any part of the transformation?

Here are some related questions to explore:
1. How do boundary conditions affect the stability of solutions in PDEs?
2. What is the significance of non-dimensionalization in solving differential equations?
3. Can you explain the method of separation of variables for solving PDEs?
4. What are some physical examples where this form of PDE appears?
5. How do you derive the scaling factors for time and space in a general PDE?

**Tip:** Non-dimensionalization helps simplify a problem by reducing the number of variables and making the problem easier to solve or analyze numerically.

Transform the problem PDE: utt=9uxx, 0<x<5, t>0 BC: u(0,t)=6, u(5,t)=9 IC: u(x,0)=x^2, ut(x,0)=x^3 into a dimensionless problem: PDE: Uττ=Uξξ, 0<ξ<1, τ>0 BC: U(0,τ)=0, U(1,τ)=1 IC: U(ξ,0)= , Uτ(ξ,0)= Type xi, tau for the variables. (You might want to first figure out the answers below). First, you transform u to U. Use just small 'u' instead of 'u(x,t)': U= Next, transform x to ξ. ξ= x/5 Finally, transform t to τ.

Learn how to transform the PDE utt = 9uxx with boundary and initial conditions into a dimensionless form using scaling techniques. This solution involves partial differential equations, boundary conditions, initial conditions, and dimensional analysis. Suitable for college-level students.

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