Approach to 'double convolution'?

[lutzB]

Hello,

I am trying to examine the following differential equation:

diff(M(a), a) =
Lambda * ( 1 -
epsilon * int(M(b)*exp(-sigma*(a-b)), b = 0 .. a) ) -
mu * M(a)

I found out that this kind of equation can be solved for M(a) by applying Laplace-Transformation, solving for Laplace(M(a)) and then applying inverse Laplace-Transformation. This works quite fine e.g. in Maple.

Now, I would like to extend above equation by sort of adding a second convolution term, so that

diff(M(a), a) =
Lambda * ( 1 -
epsilon * int(M(b)*exp(-sigma*(a-b)), b = 0 .. a) ) +
gamma * int(M(b)*exp(-tau*(a-b)), b = 0 .. a) ) -
mu * M(a)

But apparently this is not solvable in the same way. At least, Maple won't find the inverse Laplace that I need for the final solution.

I would appreciate very much any hints regarding whether a differential equation like this should be solvable by this method, or if there are other methods I could try, or if it simply cannot be solved?

Many thanks and best wishes-
Lutz

 

[jvicens]

Hello Lutz,

Thank you for your question. It is possible that your extended differential equation may not be solvable using the same method as the original equation. This could be due to the added complexity of the second convolution term, which may require a different approach to solve.

One method you could try is using numerical methods to approximate the solution. This involves breaking the equation down into smaller parts and using numerical algorithms to approximate the solution at each step. This approach may be more computationally intensive, but it can be effective in solving complex equations.

Another approach you could try is using a series expansion method, such as the Taylor series method. This involves approximating the solution as a series of polynomials and using this to solve the equation. This method may be more suitable for equations with multiple terms and can provide a more accurate solution compared to numerical methods.

It is also possible that your extended equation may not have a closed-form solution and may require numerical or approximate methods to solve. In this case, it may be helpful to consult with a mathematician or utilize specialized software to find a solution.

I hope this helps and good luck with your research.

 

[tacman]

Hello Lutz,

Thank you for sharing your approach to solving the differential equation. It seems like you have made good progress in solving the equation by using the Laplace-Transformation method. However, as you mentioned, adding a second convolution term may make it more challenging to solve using this method.

One possible approach to solving this kind of equation could be to use the method of double convolution, also known as the product-to-sum method. This method involves transforming the product of two functions into a sum of two functions, which can then be solved using the Laplace-Transformation method. This method is commonly used in control systems engineering and signal processing.

To apply this method to your equation, you can rewrite it as:

diff(M(a), a) =
Lambda * ( 1 -
epsilon * int(M(b)*exp(-sigma*(a-b)), b = 0 .. a) ) +
gamma * int(M(b)*exp(-tau*(a-b)), b = 0 .. a) ) -
mu * M(a)

= Lambda * ( 1 -
epsilon * int(M(b)*exp(-sigma*(a-b)), b = 0 .. a) ) +
gamma * int(M(b)*exp(-tau*(a-b)), b = 0 .. a) ) -
mu * M(a)

= Lambda * ( 1 -
epsilon * int(M(b)*exp(-sigma*(a-b)), b = 0 .. a) ) +
gamma * (1 - exp(-tau*a)) * int(M(b)*exp(-(tau+sigma)*(a-b)), b = 0 .. a) ) -
mu * M(a)

By using the double convolution formula, you can rewrite the equation as:

diff(M(a), a) =
Lambda * ( 1 -
epsilon * int(M(b)*exp(-sigma*(a-b)), b = 0 .. a) ) +
gamma * (1 - exp(-tau*a)) * (M(a) * (tau+sigma) - int(M(b)*exp(-(tau+sigma)*(a-b)), b = 0 .. a) ) -
mu * M(a)

Now, you can solve this equation using the Laplace-Transformation method and then apply the inverse Laplace-Transformation to get the final solution for M(a). This method may require some additional algebraic manipulation, but it should be solvable using the same approach you used for the original equation.

I hope this