If \(U = \mathop e\nolimits^{\frac{{\mathop x\nolimits^2 }}{{\mathop y\nolimits^2 }}} + \mathop e\nolimits^{\frac{{\mathop y\nolimits^2 }}{{\mathop x\nolimits^2 }}} \)then \(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}}\)is
0
Understanding the Problem: Partial Derivatives and Euler's Theorem
The problem asks for the value of the expression \(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}\) for the given function \(U = \mathop e\nolimits^{\frac{{\mathop x\nolimits^2 }}{{\mathop y\nolimits^2 }}} + \mathop e\nolimits^{\frac{{\mathop y\nolimits^2 }}{{\mathop x\nolimits^2 }}} \). This specific form of expression \(x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}\) is strongly related to Euler's theorem on homogeneous functions.
Checking for Homogeneity
First, let's determine if the function \(U(x, y)\) is a homogeneous function. A function \(f(x, y)\) is said to be homogeneous of degree \(n\) if for any non-zero constant \(t\), \(f(tx, ty) = t^n f(x, y)\).
Let's substitute \(tx\) for \(x\) and \(ty\) for \(y\) in the function \(U\):
\(U(tx, ty) = \mathop e\nolimits^{\frac{{\mathop {(tx)}\nolimits^2 }}{{\mathop {(ty)}\nolimits^2 }}} + \mathop e\nolimits^{\frac{{\mathop {(ty)}\nolimits^2 }}{{\mathop {(tx)}\nolimits^2 }}} \)
\(U(tx, ty) = \mathop e\nolimits^{\frac{{\mathop t\nolimits^2 \mathop x\nolimits^2 }}{{\mathop t\nolimits^2 \mathop y\nolimits^2 }}} + \mathop e\nolimits^{\frac{{\mathop t\nolimits^2 \mathop y\nolimits^2 }}{{\mathop t\nolimits^2 \mathop x\nolimits^2 }}} \)
Since \(t \neq 0\), we can cancel out \(\mathop t\nolimits^2\) from the exponents:
\(U(tx, ty) = \mathop e\nolimits^{\frac{{\mathop x\nolimits^2 }}{{\mathop y\nolimits^2 }}} + \mathop e\nolimits^{\frac{{\mathop y\nolimits^2 }}{{\mathop x\nolimits^2 }}} \)
We observe that \(U(tx, ty) = U(x, y)\). This can be written as \(U(tx, ty) = t^0 U(x, y)\).
Thus, the function \(U\) is a homogeneous function of degree \(n = 0\).
Applying Euler's Theorem
Euler's theorem for homogeneous functions states that if \(U(x, y)\) is a homogeneous function of degree \(n\), then the following relationship holds:
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = nU\)
In our case, we found that \(U\) is a homogeneous function of degree \(n = 0\).
Substituting \(n=0\) into Euler's theorem, we get:
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 0 \cdot U\)
\(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 0\)
Conclusion
Based on Euler's theorem, for the given homogeneous function \(U\) of degree zero, the expression \(x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}}\) evaluates to 0.
The final answer is \textbf{0}.
Important Questions from Differential Equations
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The solution of the differential equation when the source f(t) = θ(t) (the Heaviside step function) is
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