To show that v(x,y)=exsin(y) is a harmonic function and to find u(x,y) such that f(z)=u+iv is analytic, we follow these steps:
Step 1: Verify that v(x,y)=exsin(y) is a harmonic function
A function v(x,y) is harmonic if it satisfies the Laplace equation:
βx2β2vβ+βy2β2vβ=0
Let's compute the second partial derivatives of v(x,y)=exsin(y):
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Compute βxβvβ:
βxβvβ=βxββ(exsin(y))=exsin(y)
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Compute βx2β2vβ:
βx2β2vβ=βxββ(exsin(y))=exsin(y)
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Compute βyβvβ:
βyβvβ=βyββ(exsin(y))=excos(y)
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Compute βy2β2vβ:
βy2β2vβ=βyββ(excos(y))=βexsin(y)
Now, add these second derivatives:
βx2β2vβ+βy2β2vβ=exsin(y)βexsin(y)=0
Since the sum of the second partial derivatives is zero, v(x,y)=exsin(y) is indeed a harmonic function.
Step 2: Find u(x,y) such that f(z)=u+iv is analytic
For f(z)=u(x,y)+iv(x,y) to be analytic, the functions u(x,y) and v(x,y) must satisfy the Cauchy-Riemann equations:
βxβuβ=βyβvβandβyβuβ=ββxβvβ
Given v(x,y)=exsin(y), we have already computed:
βyβvβ=excos(y)andβxβvβ=exsin(y)
According to the Cauchy-Riemann equations, we need:
βxβuβ=excos(y)
βyβuβ=βexsin(y)
Now, integrate βxβuβ with respect to x to find u(x,y):
u(x,y)=β«excos(y)dx=excos(y)+Ο(y)
where Ο(y) is an arbitrary function of y.
Next, use βyβuβ=βexsin(y) to determine Ο(y):
βyβuβ=βexsin(y)+Οβ²(y)=βexsin(y)
This implies:
Οβ²(y)=0βΟ(y)=C
where C is a constant.
Thus, u(x,y)=excos(y)+C. We typically take C=0 for simplicity, giving:
u(x,y)=excos(y)
Conclusion
The analytic function f(z) corresponding to the given harmonic function v(x,y)=exsin(y) is:
f(z)=u(x,y)+iv(x,y)=excos(y)+iexsin(y)=ex(cos(y)+isin(y))
This simplifies to:
f(z)=ex+iy=ez
where z=x+iy.
Would you like more details on any step or further questions?
Related Questions:
- What are the Cauchy-Riemann equations, and why are they important?
- How can we determine if a complex function is harmonic?
- What are the physical interpretations of harmonic functions?
- Can we find the harmonic conjugate for other functions?
- How does the complex exponential function ez behave in the complex plane?
Tip:
When dealing with harmonic functions, remember that their real and imaginary parts are often interconnected through the Cauchy-Riemann equations. This relationship ensures that the overall function is analytic.