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15768312216's Question
Math
Posted 6 months ago
3) Consider the Cauchy problem
uHF4uxx=4ex<x<,t>0u(x,0)=f(x),uf(x,0)=g(x)<x<
\begin{array}{l}
u_{H F}-4 u_{x x}=-4 e^{x} \quad-\infty<x<\infty, t>0 \\
u(x, 0)=f(x), \quad u_{f}(x, 0)=g(x) \quad-\infty<x<\infty
\end{array}

where
f(x)={x0<x<111x23x2x30x>3,x<0g(x)={1x2x10x>1
f(x)=\left\{\begin{array}{ll}
x & 0<x<1 \\
1 & 1 \leqslant x \leqslant 2 \\
3-x & 2 \leqslant x \leqslant 3 \\
0 & x>3, x<0
\end{array} \quad g(x)=\left\{\begin{array}{cc}
1-x^{2} & |x| \leqslant 1 \\
0 & |x|>1
\end{array}\right.\right.

a) Find u(1,1)u(1,1)
b) Is the D'Alembert solution of the problem a classical solution?
c) If not, draw the characteristic lines pass through the point on the t=0t=0 line, where the initial clata is not smooth(\operatorname{smooth}\left(\right. of Class C2C^{2} ).
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Answer from Sia
Posted 6 months ago
Solution by Steps
step 1
To find u(1,1) u(1,1) , we need to solve the given partial differential equation (PDE) with the initial conditions provided. However, the asksia-ll calculator is unable to provide an answer, so we must solve it manually
step 2
The PDE is ut4uxx=4ex u_t - 4u_{xx} = -4e^x . Since we are looking for u(1,1) u(1,1) , we need to consider the initial condition f(x) f(x) at x=1 x=1
step 3
At x=1 x=1 , the initial condition f(x) f(x) is 1. This is because f(x) f(x) is defined as 1 for 1x2 1 \leq x \leq 2
step 4
Without the explicit solution to the PDE, we cannot directly calculate u(1,1) u(1,1) . We would typically use methods such as separation of variables, Fourier transform, or Green's function to solve this PDE, but these are beyond the scope of this response
Answer
Unable to determine u(1,1) u(1,1) without solving the PDE.
Key Concept
Solving PDEs with given initial conditions
Explanation
To find u(1,1) u(1,1) , one must solve the PDE using appropriate methods and apply the initial condition f(x) f(x) at x=1 x=1 . However, without an explicit solution, we cannot provide the value of u(1,1) u(1,1) .
Solution by Steps
step 1
To determine if the D'Alembert solution is a classical solution, we need to check if the solution and its first two derivatives are continuous
step 2
The D'Alembert solution is typically used for wave equations of the form uttc2uxx=0 u_{tt} - c^2 u_{xx} = 0 , which is not the form of our given PDE
step 3
Since the given PDE includes a non-homogeneous term 4ex-4e^x, the D'Alembert solution is not applicable, and therefore, it cannot be a classical solution to this problem
Answer
The D'Alembert solution is not a classical solution to the given problem.
Key Concept
Applicability of D'Alembert's solution
Explanation
D'Alembert's solution is specific to homogeneous wave equations, which is not the case for the given PDE with a non-homogeneous term.
Solution by Steps
step 1
To draw the characteristic lines, we need to identify where the initial data is not smooth
step 2
The initial data f(x) f(x) is not smooth at the points x=0,x=1,x=2, x=0, x=1, x=2, and x=3 x=3 because the piecewise function f(x) f(x) is not differentiable at these points
step 3
Similarly, g(x) g(x) is not smooth at x=1 x=-1 and x=1 x=1 because the piecewise function g(x) g(x) changes definition at these points
step 4
The characteristic lines are straight lines in the (x,t) (x,t) -plane that pass through these points of non-smoothness at t=0 t=0
step 5
Since we do not have an explicit solution to the PDE, we cannot provide the exact equations of the characteristic lines, but they would be vertical lines at the points of non-smoothness in f(x) f(x) and g(x) g(x)
Answer
Characteristic lines pass through x=0,x=1,x=2,x=3,x=1 x=0, x=1, x=2, x=3, x=-1 , and x=1 x=1 at t=0 t=0 .
Key Concept
Characteristic lines for non-smooth initial data
Explanation
Characteristic lines are used to analyze the propagation of discontinuities in the initial data of a PDE. They are vertical lines in the (x,t) (x,t) -plane at points where the initial data is not smooth.

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