Video: Finding the Laplacian of a Multivariable Exponential Function of Three Variables in the Spherical Coordinates

Find the Laplacian of the function π(π₯, π¦, π§) = π^(βπ₯Β² β π¦Β² β π§Β²) in terms of the spherical coordinates π, π, and π.

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Video Transcript

Find the Laplacian of the function π of π₯, π¦, π§ equals π to the negative π₯ squared minus π¦ squared minus π§ squared in terms of the spherical coordinates π, π, and π.

There are two approaches we can take for this problem. The first approach is to take the Laplacian in Cartesian coordinates and then convert to spherical coordinates. Letβs try this method first. Recall that the Laplacian of a function β squared π is equal to the divergence of the gradient of π, which in Cartesian coordinates is given by β two π by βπ₯ squared plus β two π by βπ¦ squared plus β two π by βπ§ squared. So to begin with, partial π by partial π₯ is partial by partial π₯ of π to the negative π₯ squared minus π¦ squared minus π§ squared. So we differentiate with respect to π₯ and treat π¦ and π§ as a constant, giving negative two π₯ times π to the negative π₯ squared minus π¦ squared minus π§ squared.

Therefore, β two π by βπ₯ squared is given by taking the partial derivative with respect to π₯ again of negative two π₯π to the negative π₯ squared minus π¦ squared minus π§ squared. Differentiating partially with respect to π₯ and using the product rule gives us negative two times π to the negative π₯ squared minus π¦ squared minus π§ squared plus four π₯ squared π to the negative π₯ squared minus π¦ squared minus π§ squared.

We can take out a common factor of two times the exponent to give two π to the negative π₯ squared minus π¦ squared minus π§ squared times two π₯ squared minus one. Now, we could repeat this process for β two π by βπ¦ squared and β two π by βπ§ squared. But we can save some time by recognizing the symmetry of the function. We can see that the function π has the same dependence on π¦ and π§ as it does on π₯. So differentiating partially with respect to π¦ and π§ will have the same result but with π¦ and π§ replacing π₯, respectively.

So our expressions for β two π by βπ¦ squared and β two π by βπ§ squared will be the same as the expression for β two π by βπ₯ squared, but with π¦ and π§ replacing π₯. Adding these all together and taking out the exponent as a common factor gives us the Laplacian of π equal to two π to the negative π₯ squared minus π¦ squared minus π§ squared times two π₯ squared plus two π¦ squared plus two π§ squared minus three. Letβs take out a common factor of negative one from the exponent and a common factor of two from the π₯ squared plus π¦ squared plus π§ squared inside the parentheses.

In spherical coordinates, the radius π squared is given by π₯ squared plus π¦ squared plus π§ squared. So these two terms here are, in fact, just equal to π squared. So this gives us our answer. The Laplacian of π, β squared π, is equal to two π to the negative π squared times two π squared minus three. An alternative method is to first convert to spherical coordinates and then take the Laplacian. As mentioned before, negative π₯ squared minus π¦ squared minus π§ squared is just negative π squared. So in spherical coordinates, the function π of π, π, π becomes simply π to the negative π squared.

This greatly simplifies the function π. But the difficulty with this method is that you need to either remember or be able to derive the Laplacian in spherical coordinates. In spherical coordinates, the Laplacian of π, β squared π, is given by one over π squared partial by partial π of π squared partial π by partial π plus one over π squared sin π partial by partial π of sin π partial π by partial π plus one over π squared sin squared π β two π by βπ squared. The good news is that in this case, this will greatly simplify taking the Laplacian of π since π depends only on π. So any partial terms with respect to either π or π will be zero.

This means that we can disregard both the second and third term, as they will both be equal to zero. First, we will need to find partial π by partial π, which is given by negative two π times π to the negative π squared. So then the Laplacian of π is given by one over π squared times partial by partial π of π squared times negative two ππ to the negative π squared. This simplifies to one over π squared times partial by partial π of negative two π cubed π to the negative π squared.

Differentiating with respect to π, we get one over π squared times negative six π squared π to the negative π squared plus four π to the fourth times π to the negative π squared. Multiplying out the parentheses and taking out a factor of π to the negative π squared, but not a factor of two this time, gives us π to the negative π squared times four π squared minus six, as before.