Assume that all the given functions have continuous...

Assume that all the given functions have continuous second-order partial derivatives. If $u=f(x, y),$ where $x=e^{\prime} \cos t$ and $y=e^{x} \sin t,$ show that $$ \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=e^{-2} \cdot\left[\frac{\partial^{2} u}{\partial s^{2}}+\frac{\partial^{2} u}{\partial t^{2}}\right] $$

Assume that all the given functions have continuous second-order partial derivatives.
If $u=f(x, y),$ where $x=e^{\prime} \cos t$ and $y=e^{x} \sin t,$ show that
$$
\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=e^{-2} \cdot\left[\frac{\partial^{2} u}{\partial s^{2}}+\frac{\partial^{2} u}{\partial t^{2}}\right]
$$

Key Concepts

Chain Rule in Multivariable Calculus

The chain rule is a central principle used to compute the derivatives of composite functions. When a function depends on intermediate variables that are functions of other variables, the chain rule enables us to express the derivatives of the outer function in terms of the inner function derivatives. This concept is essential when dealing with coordinate transformations and ensuring that all partial derivative terms are appropriately accounted for.

Coordinate Transformation

Coordinate transformations involve re-expressing a function or a differential operator in terms of a new set of variables. By transforming coordinates, complex problems can sometimes be simplified or made more symmetric. This strategy is especially useful when the geometry or symmetry of the problem is better captured in a different coordinate system, and it naturally involves changing the variables in the differential operators.

Laplacian Operator Transformation

The Laplacian operator, which is the sum of the unmixed second-order partial derivatives, is a key differential operator in many fields including physics and engineering. Its transformation under a change of variables typically involves scaling factors and additional terms as determined by the mapping. This concept is vital in understanding how differential equations behave under coordinate changes.

Second-Order Partial Derivatives

Second-order partial derivatives provide detailed information about the curvature and concavity of functions along different directions. They form the basis of second-order differential operators like the Laplacian. In the context of coordinate transformations, these derivatives must be carefully re-expressed using the chain rule to ensure that the properties of the operator, especially its invariance or scaling behavior, are preserved.

Transcript

00:01 In this problem of chain rule we have given if u is equals to f of x y that u is the function of x and y where x is equal to e to the power s cost t and y is equal to e to the power s say this is e to the power s sine t now we have to show that say d to u divided with d x square plus d to u divided with dy square is equals to e to the power say minus 2s and here this is d2u divided with ds square plus d2 u divided with dt square so let's start from the right hand side so this is d2 u and ds square so this would be here so first see the term this is d2 u u divide with d s square so as we know that d u divide with d s is calculated as differentiation of u with respect to x and multiplied with differentiation of x with respect to u and here u is the function of x and this would be differentiation of u with respect to y also and then differentiation of y with respect to u and now we have to again differentiate it so with respect to x with respect to s actually and also you can say for differentiation of you with respect to s is equals to this would be t here so here differentiation with respect to t is differentiation of you with respect to x and then differentiation of x with respect to t plus differentiation of u with respect to y and then differentiation of y with respect to t now see the term so differentiation of x with respect to t so here this is equals to minus sign so this is minus sign t and this is multiplied with e to the power x now differentiation of this term with again so this would be d to x d t square is simply this a differentiation of this term is cost so this would be minus e to the power s and this would be cost t similarly differentiation of say this would be y say this would be differentiation of y with respect to t is given y say this would be e to the power s and this would be cost t so this is cost and now d to y d t square is basically so this would be e to the power s and this would be sine t so this would be so from here you can say this term is equal to this term.

03:14 Now from here we can say d2x divided with dt square is equal to d2y divide with dt square which is minus e to the power s.

03:26 And this would be actually here this is equated with this term actually...

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