Solutions (level 4) In this video we are going to verify the solution
of partial differential equations also known as PDE’s. Be advised that a typical first
course in differential equations does not usually cover partial differential equations.
So if your class will not cover PDE’s you can skip this video and move along to the
next one. You are also free to check this video out. The examples in this video will serve
as a good calculus III refresher especially if you forgot how take partial derivatives
of functions with multiple variables, this skill will be required when we dive deeper into
this course. With that said let’s go over the first example.
Verify that the indicated function is a solution of the given partial differential equation.
The partial differential equation is the second partial derivative of u with respect to x
plus the second partial dertivative of u with respect to y equals 0, and the solution to
verify is u(x,y) = ln(x^2+y^2). Similar to Ordinary Differential equations, we first
want to find the partial derivatives before we attempt to verify the solution. Scanning
the PDE we see that we need to find the second partial derivative of the u with respect to
x and the second partial derivative of u with respect to y. let's first find the first partial
derivative of u with respect to x, recall that when taking the partial derivative of
a function with two or more variables we need to consider any variable for which we are
not taking the partial derivative as constants. Then we proceed to take the partial derivative
with respect to the specific variable, at the same time we need to consider the rest
of the variables as constants. To take the partial derivative of u with respect to x,
we need to consider the variable y as a constant, and then we go ahead and apply the chain rule
to find the partial derivative. Carrying out the partial derivative we obtain the following
expression. Next we need to take the partial derivative of this expression with respect
to x once more, to obtain the second partial derivative of u with respect to x. Doing that
we obtain the following expression.
Now let’s find the second partial derivative of u with respect to y, we first need to the
find the first derivative of u with respect to y, similar to the previous partial derivative
we need to consider x as a constant and proceed with the derivative step. Applying the chain
rule with obtain the following expression. Then we take the partial derivative of u with
respect to y once more, obtaining the following expression.
Having found the second partial derivative we proceed with the substitution step and
make sure that the left and right hand side of the equation are equal to each other. Carrying
out the substitutions we obtain the following expression. Next let’s go ahead and add
the two expressions on the left hand side of the equation, notice that we have to add
fractions, conveniently they both have a common denominator, so it’s just a matter of adding
the numerators. Collecting like terms we see that the numerator simplifies to 0, 0 divided
by the quantity x squared plus y squared, raised to the power of 2 is just equal 0.
Notice that the left hand side matches the right hand side of the equation; this means
that this function is indeed a solution to the partial differential equation. In these
examples we are not going to bother assigning an interval of definition. The main purpose
of these examples is to show how you can verify a solution to a partial differential equation.
Alright let’s try the next example.
The partial differential equation is a^2*u_xx = u_tt and the solution we need to verify
is u(x,y) = sin(x-at). Notice that this PDE is written using subscript notation as oppose
to Leibniz notation from the previous example. Also notice that we have the constant a squared,
keep in mind that this is just a constant. Similar to the
last example let's go ahead and find the partial derivatives. Let's first find the partial derivative
of u with respect to x, keeping the variable t constant and applying the chain rule we
obtain the following expression, next lets once again take the partial derivative of
u with respect to x keeping the variable t constant. Carrying out the partial derivative
we obtain the following expression. Now let’s find the partial derivative of u with respect
to t, here we need to keep the variable x constant and apply the chain rule, doing that
we obtain the following expression. Then we need to once again take the partial derivative
of u with respect to t, making sure we keep the variable x constant, carrying out the
partial derivative we obtain the following expression. Having found the second partial
derivative we proceed with the substitution step, replacing the second partial derivatives,
we obtain the following expression, and simplifying both sides of the equation we obtain the following.
Notice that both sides of the equation are identical this means that this function is
indeed a solution to the partial differential equation. Alright let’s go over the final example.
The differential equation is alpha^2*u_xx = u_t and the solution to verify is u(x,y) = e^(-alpha^2*t)*sin(x),
Alright notice that we have the constant alpha squared on both the PDE and the solution,
let’s keep this in mind when start verifying the solution to the PDE. Let’s first find
the partial derivative of u with respect to x, keeping the variable t constant and applying
the chain rule we obtain the following expression. Remember that we are keeping the variable
t constant this means that the entire exponential expression is also a constant; there is no
need to apply the product rule because we are treating this exponential term as a constant
and not a function. Taking the partial derivative we obtain the following expression. Next we
take the partial derivative of this function with respect to x once again, doing that we
obtain the following expression. Now we need to find the partial derivative of u with respect
t, keep in mind that we need to treat the variable x as a constant when taking the partial
derivative. Carrying out the partial derivative we obtain the following expression. Having
found all the partial derivatives we go ahead and substitute them into the PDE. We need
to make sure that the left hand side and right hand side of the equation match. Substituting
the partial derivatives we obtain the following expression. Finally, we simplify the left
hand side of the equation and obtain the following. Inspecting the PDE we see that both sides
are equivalent hence this function is indeed a solution to the PDE.
Alright in our next video we go over implicit solutions of ODE’s and show how to verify them.