Sobolev spaces are essential in functional analysis and partial differential equations.
The space H1(−1,1) is a specific Sobolev space defined on the interval (−1,1) on the real line.
Definition of H1(−1,1)
The Sobolev space H1(−1,1) includes functions u that satisfy both the function itself and its first derivative being in L2(−1,1). Specifically, the space is defined as follows:
- A function u belongs to L2(−1,1) if:
∫1−1|u(x)|2dx<∞
- The first weak derivative u′ must also be in L2(−1,1), which means:
∫1−1|u′(x)|2dx<∞
Weak derivatives are used to accommodate functions that might not be differentiable in the classical sense everywhere on the interval, including functions that are continuous and differentiable almost everywhere but may have points of derivative discontinuity.
Norm in H1(−1,1)}
The norm in the Sobolev space H1(−1,1) is defined to incorporate both the function and its derivative, given by:
‖u‖H1=(∫1−1|u(x)|2+|u′(x)|2dx)1/2
Importance and Applications
Sobolev spaces such as H1(−1,1) are crucial in the study of partial differential equations (PDEs), as they frequently arise as the solution spaces for various PDEs. Understanding Sobolev spaces is foundational for advanced studies in fields such as mathematical physics, engineering, and applied mathematics.
Example Function for H1(−1,1)
Working with a concrete example clarifies above definitions.
Consider the function u(x)=x2 as an example for a function in the Sobolev space H1(−1,1). We will verify that both u and its first derivative u′ belong to L2(−1,1).
\Verification of u∈L2(−1,1)
First, we check if the function u(x)=x2 is square integrable over (−1,1):
∫1−1|u(x)|2dx=∫1−1x4dx
Calculate the integral:
∫1−1x4dx=[x55]1−1=15−(−15)=25
This result confirms that u(x)=x2 is in L2(−1,1) as the integral is finite (<∞).
Verification of u′∈L2(−1,1)
Next, compute the derivative of u(x) and check its square integrability:
u′(x)=2x
Check if u′(x)=2x is square integrable:
∫1−1|u′(x)|2dx=∫1−1(2x)2dx=4∫1−1x2dx
Computing the integral:
∫1−1x2dx=[x33]1−1=13−(−13)=23
Thus:
4∫1−1x2dx=4⋅23=83
Since this integral is also finite, u′(x)=2x is in L2(−1,1).
Thus, the function u(x)=x2 and its derivative u′(x)=2x are both in L2(−1,1), confirming that u(x)=x2 is a valid example of a function in the Sobolev space H1(−1,1).
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