Professor MathTheBeautiful, thank you for a powerful analysis of Laplace Equation on the Unit Disk in Partial Differential Equations. This is an error free video/lecture on TH-cam TV.
1. How can we tell if a PDE is separable before we actually try the separation technique? (which may fail if the equation is not separable, and we may waste a whole lotta time by trying to do the impossible :q ) Is there any rule for when the equation is separable? 2. After separating the variables, what is the justification behind the assumption that both sides of the equation must be equal to a constant? How do you know that this is true? 3. Can you also show how to solve the same problem in rectangular coordinates, for comparison? (just to show that polar coordinates are really better to go with in this case, and that it is still possible to solve it in rectangular coordinates, just cumbersome to do so) Oh, and it would be nice to put the link in the description to the derivation of the Laplacian in polar coordinates. You said in the video that it's been derived in another lesson, but you didn't say which one.
Te radial equation is a known type of equation called *Cauchy-Euler equation* or equidimensional equation. If you had no idea about what function could solve it, you could just think of what functions do you know that, when differentiated, and then multiplied with the power of the independent variable, would restore the original function, so that they could cancel out the entire equation to zero. This might hint you to the fact that when you differentiate a power function, it lowers its exponent and multiplies by it. And then, after multiplying it by the same power of the independent variable, it will restore the power to its original value, because the power of the independent variable is the same as the order of the derivative (or the power of the differential operator) - hence the "equidimensional" in its name. So a power function is a valid guess for the solution to such equation. What's left is to determine the exponents that would work, by cancelling with whatever constant coefficients you have there. You can make a trial function in a form `r^n` in which the exponent `n` is to be determined, and inserting it into the equation. Then it will turn out that `r^n` will be the common factor of all the terms, so you can extract that common factor out, leaving some remaining expression in the parenthesis. And of course `r^n` cannot be zero (because this is a trivial solution that always work for all equations of that type), so the only other possibility for the entire equation to be zero is for the expression in the parenthesis to be zero. This is the "indicial equation" or "characteristic equation" that you can solve to determine what exponent (or "index") `n` would work. Depending on the degree of the polynomial in the parenthesis, you may get more than one solution, and they can also be complex (they are the roots of that polynomial). And the general solution would be a linear combination of all those `r^n`s where the `n`s are the roots of the characteristic polynomial.
Professor MathTheBeautiful, thank you for a powerful analysis of Laplace Equation on the Unit Disk in Partial Differential Equations. This is an error free video/lecture on TH-cam TV.
Glad it was helpful!
Can you please give a link to the lesson in which you derived the Laplacian in polar coordinates?
Please keep sharing your lectures. Just beautiful. Thank you
Hi Darren, Much appreciated! Glad there are so many people enjoying mathematics! -Pavel
1. How can we tell if a PDE is separable before we actually try the separation technique? (which may fail if the equation is not separable, and we may waste a whole lotta time by trying to do the impossible :q ) Is there any rule for when the equation is separable?
2. After separating the variables, what is the justification behind the assumption that both sides of the equation must be equal to a constant? How do you know that this is true?
3. Can you also show how to solve the same problem in rectangular coordinates, for comparison? (just to show that polar coordinates are really better to go with in this case, and that it is still possible to solve it in rectangular coordinates, just cumbersome to do so)
Oh, and it would be nice to put the link in the description to the derivation of the Laplacian in polar coordinates. You said in the video that it's been derived in another lesson, but you didn't say which one.
for 2 : the justification is :
we have two terms in both sides of the equation with different variable are equal so they must be a constant.
Let f(x)=F(y) for each x and y. then, for example for y=1, we have f(x)=F(1)=const
I am highly thankful to you sir....now i can solve my assignment
The radial could be a Cauchy-Euler equidimensional differential equation, que no?
How do you solve the differential equation in R(r) without assuming that C_n r^{2} + D_n r^{-2} is a solution?
Te radial equation is a known type of equation called *Cauchy-Euler equation* or equidimensional equation. If you had no idea about what function could solve it, you could just think of what functions do you know that, when differentiated, and then multiplied with the power of the independent variable, would restore the original function, so that they could cancel out the entire equation to zero. This might hint you to the fact that when you differentiate a power function, it lowers its exponent and multiplies by it. And then, after multiplying it by the same power of the independent variable, it will restore the power to its original value, because the power of the independent variable is the same as the order of the derivative (or the power of the differential operator) - hence the "equidimensional" in its name. So a power function is a valid guess for the solution to such equation. What's left is to determine the exponents that would work, by cancelling with whatever constant coefficients you have there.
You can make a trial function in a form `r^n` in which the exponent `n` is to be determined, and inserting it into the equation. Then it will turn out that `r^n` will be the common factor of all the terms, so you can extract that common factor out, leaving some remaining expression in the parenthesis. And of course `r^n` cannot be zero (because this is a trivial solution that always work for all equations of that type), so the only other possibility for the entire equation to be zero is for the expression in the parenthesis to be zero. This is the "indicial equation" or "characteristic equation" that you can solve to determine what exponent (or "index") `n` would work. Depending on the degree of the polynomial in the parenthesis, you may get more than one solution, and they can also be complex (they are the roots of that polynomial). And the general solution would be a linear combination of all those `r^n`s where the `n`s are the roots of the characteristic polynomial.
Thank you again!
thank you
What a G!