Finding if a solution exists

[vladimir69]

hi
this is my first attempt at using the latex commands ..
consider this Cauchy Problem:
[tex]
u_x\exp y +u_y\exp(x)=1
[/tex]
[tex]
u(t,tk^2)=k\exp(-t)
[/tex]
where k is a constant.
forgive me i have given up using latex as i am too slow, it would take me a week to type it up at this rate. anyway
i am trying to find the values of k where
a) this problem has a unique solution
b) no solutions

for part a) i know that it all boils down to showing where the vector (1,k^2) is never parallel to ( exp(t*k^2), exp(t) )
i eventually get to:
t can't be equal to 2ln(k)/(1-k^2), however the question asks for values of k and i haven't done that. but i don't know any way to find explicit values of k

for part b) i just said t=2ln(k)/(1-k^2)

is it reasonable to give the answer in the form i have? I'm not sure
hope you can help

thanks,
vladimir

 

[jvicens]

Hello Vladimir,

Welcome to using LaTeX commands! It can take some time to become familiar with them, but they are very useful for writing mathematical expressions and equations.

Let's take a closer look at the Cauchy Problem you have presented:

\begin{align*}
u_x\exp y +u_y\exp(x)&=1 \\
u(t,tk^2)&=k\exp(-t)
\end{align*}

To find the values of k where this problem has a unique solution, we need to consider the initial condition and the characteristic curves of the problem. The initial condition, u(t,tk^2)=k\exp(-t), tells us that the solution must pass through the point (t,tk^2) for all values of t. The characteristic curves, given by the equations \frac{dx}{dt}=\exp y and \frac{dy}{dt}=\exp(x), can be used to trace out the solution curves. We can see that for any value of k, the vector (1,k^2) is never parallel to ( exp(t*k^2), exp(t) ). This means that for all values of k, the problem has a unique solution.

For part b), you have correctly identified that the solution is not defined when t=2ln(k)/(1-k^2). This means that for these values of k, there is no solution to the Cauchy Problem. It is reasonable to give the answer in this form, as it clearly states the values of k where there is no solution.

I hope this helps and good luck with your future LaTeX endeavors!

 

[M98Ranger]

Hello Vladimir,

Thank you for sharing your attempt at using LaTeX commands. It takes some practice, but it will get easier with time.

Now, let's address the problem at hand. You are correct in saying that the existence of a solution to this Cauchy problem depends on the parallelism of the vectors (1, k^2) and (exp(t*k^2), exp(t)). In order for there to be a unique solution, these vectors must never be parallel, meaning that they must always have a nonzero cross product.

To find the values of k where this is the case, we can set the cross product equal to zero and solve for k. This gives us the equation 2k/(1-k^2) = 0. However, this equation has no solution since k cannot be both 0 and nonzero at the same time. Therefore, we can conclude that there are no values of k that will result in a unique solution to this Cauchy problem.

For part b), you have correctly found that the value of t that satisfies the parallelism condition is t = 2ln(k)/(1-k^2). However, since k cannot be both 0 and nonzero, this value of t cannot be achieved. Therefore, there are no solutions to this Cauchy problem.

In summary, it is reasonable to give your answers in the form you have presented. Keep in mind that in mathematics, we are often interested in general solutions rather than explicit values. So in this case, stating that there are no values of k that result in a unique solution is a valid answer.

I hope this helps clarify the problem for you. Keep practicing with LaTeX and you will get faster at it. Good luck!